How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 18 results · all verified · 12 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 6 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Metric Spaces

1 · Prerequisites

2 · Summary

Objective. This page opens the topology track. It takes the one structure that a first course in analysis uses without naming, the distance between two points, isolates it into three axioms, and shows how much of the vocabulary of analysis is already determined by them: open and closed sets, interior, closure and boundary, convergence of sequences, and continuity of maps. Nothing here assumes anything about $R$ beyond the complete ordered field built on the earlier pages, and everything proved here is available verbatim in every metric space that later pages construct.

The axioms, and what is not among them. A metric on a set $X$ is a real-valued function of two points satisfying separation, symmetry and the triangle inequality (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric). Nonnegativity is deliberately not an axiom: it follows from the other three, and Nonnegativity of a metric is a consequence of the other axioms, not an axiom proves it, so a verification that some candidate function is a metric has three things to check and not four. The values are real numbers, never $+ \infty$ ; Which metric axiom list this library uses, the live naming fork between semimetric and pseudometric, and why extended metrics are not treated here records that decision, together with the live naming fork between pseudometric and semimetric. The extended real line is introduced later, but no extended-metric restatement is made on this page.

Three metric spaces are established here, and they are the ones later pages cite. That the absolute value makes $R$ a metric space, with the open balls exactly the bounded open intervals and the space unbounded, is The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded. That $R^{n}$ , defined ZFC-natively as the set of functions from the von Neumann natural $n$ to $R$ , carries the three metrics $d_{1}$ , $d_{2}$ and $d_{\infty}$ for every $n \geq 1$ is $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it. That restriction is not decoration: at $n = 0$ the metric $d_{\infty}$ would be a maximum over the empty index set. The lemma is proved from Minkowski at exponent $1$ and from Cauchy-Schwarz, with no rational power anywhere. That the bounded real-valued functions on a nonempty set carry the supremum metric is The supremum metric $d_{\infty} (f, g) = \sup_{x} ∣ f (x) - g (x) ∣$ is a metric on the bounded real-valued functions on a nonempty set. None of these had a home in the library before, and each is stated here rather than on the companion page because an examples page is a leaf and nothing may depend on it.

From the metric to the topology. The open sets are those in which every point has a ball around it inside the set (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed proves that this collection is closed under arbitrary unions and finite intersections, that balls are open and that closed balls are closed. Interior, closure, boundary, limit points and density follow (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space), and The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset identifies the closure three ways at once: as the points at distance zero from the set, as the set together with its limit points, and as the smallest closed superset. The distance to a fixed nonempty set is $1$ -Lipschitz ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz), a refinement of the reverse triangle inequality (The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space) that makes the first of those three descriptions the zero set of a well-behaved function.

Sequences, and where choice is spent. Convergence in a metric space is convergence to zero of the real sequence of distances (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ), so it inherits the conventions of the sequences page, including that $N$ contains $0$ and that the definition quantifies over rational $ε$ . Limits are unique (A sequence in a metric space has at most one limit), and more is true: distinct points are separated by disjoint balls, so every metric space is Hausdorff (Distinct points of a metric space have disjoint balls around them). The balls of radius $1 / n$ form a countable neighbourhood base at each point (The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable), which is what makes sequences powerful enough to detect the closure: a point is adherent to a set exactly when some sequence in the set converges to it, and a set is closed exactly when it is sequentially closed (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed). That theorem is the one place on this page where a choice principle is spent, and it spends only countable choice; the dependence is flagged at the step that spends it rather than suppressed, and Metric continuity characterisations, with countable choice for the sequential converse inherits it from there rather than adding to it.

Continuity, embeddings and comparison of metrics. The $ε$ - $δ$ definition (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form) agrees with four other conditions: preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f [\overline{A}] \subseteq \overline{f [A]}$ (Metric continuity characterisations, with countable choice for the sequential converse). Isometric embeddings are injective and identify their source with the subspace they land on, topology and all (Isometry, isometric embedding, and the subspace metric on a subset, An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image), which is what licenses treating a subset of a metric space as a space in its own right. Finally, two metrics on one set may be compared at three strengths, Lipschitz, uniform and topological (Topologically, uniformly and Lipschitz equivalent metrics on a set), ranked by Lipschitz equivalence implies uniform equivalence implies topological equivalence. The ranking is strict, but the witnesses live on the companion page, so the theorem claims only the two implications.

Two false statements close the page, both about reading too much into the names. The closure of an open ball need not be the closed ball of the same radius (FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ ); and boundedness is not determined by the topology, because every metric space carries a bounded metric with the same open sets ( $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology, FALSE: boundedness of a metric space is determined by its topology). The witnesses for both, and for the strictness of the equivalence hierarchy, are on the companion examples page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric

Definition

Throughout, $R$ is the complete ordered field (Complete ordered field (least-upper-bound property), Ordered field) constructed in this library (The real numbers) and carrying its order (Order on the reals).

Let $X$ be a set. A metric on $X$ is a function $d : X \times X \to R$ such that for all $x, y, z \in X$ :

(M1) Separation. $d (x, y) = 0$ if and only if $x = y$ .
(M2) Symmetry. $d (x, y) = d (y, x)$ .
(M3) Triangle inequality. $d (x, z) \leq d (x, y) + d (y, z)$ .

A metric space is a pair $(X, d)$ consisting of a set $X$ and a metric $d$ on it. The elements of $X$ are its points and $d (x, y)$ is the distance from $x$ to $y$ . When only one metric is in play we write $X$ for $(X, d)$ ; when several are, the metric is always named.

The values of a metric are real numbers. The codomain is $R$ , so $d (x, y)$ is an honest element of the complete ordered field and every inequality above is an inequality there. No infinite value is permitted; Which metric axiom list this library uses, the live naming fork between semimetric and pseudometric, and why extended metrics are not treated here records why extended metrics are not treated in this library.

Nonnegativity is deliberately absent from the axiom list. Many texts add a fourth axiom $d (x, y) \geq 0$ . It is redundant: (M1), (M2) and (M3) already force it, as Nonnegativity of a metric is a consequence of the other axioms, not an axiom proves. Nothing below assumes it before that lemma is available.

Pseudometric. A pseudometric on $X$ is a function $p : X \times X \to R$ satisfying (M2), (M3) and the weakening

(M1') Reflexivity. $p (x, x) = 0$ for every $x \in X$

of (M1). A pseudometric may therefore assign distance $0$ to two distinct points. Every metric is a pseudometric, and a pseudometric is a metric exactly when $p (x, y) = 0$ forces $x = y$ .

Ultrametric. An ultrametric on $X$ is a metric $d$ that in addition satisfies

(M3') Strong triangle inequality. $d (x, z) \leq \max {d (x, y), d (y, z)}$

for all $x, y, z \in X$ , where the maximum is that of a two-element subset of $R$ , which exists and is one of the two elements (Maximum and minimum of a set, Every nonempty finite set of reals has a maximum and a minimum). An ultrametric space is a pair $(X, d)$ with $d$ an ultrametric.

Remarks

(M3') is a genuine strengthening of (M3), not an independent axiom on top of it. A function satisfying (M1), (M2) and (M3') automatically satisfies (M3): by Nonnegativity of a metric is a consequence of the other axioms, not an axiom such a function is nonnegative, and for nonnegative reals $a, b$ one has $\max {a, b} \leq a + b$ , since the maximum is one of $a, b$ and the other summand is $\geq 0$ . So "a metric satisfying (M3')" and "a function satisfying (M1), (M2), (M3')" describe the same objects, and the definition above may be read either way.
Why the biconditional form of (M1). Splitting (M1) into " $d (x, x) = 0$ " and " $d (x, y) = 0 \Rightarrow x = y$ " gives the same notion; the split form is what makes the pseudometric weakening above a matter of deleting one clause. The naming fork between pseudometric and semimetric, which is live in the literature, is settled for this library in Which metric axiom list this library uses, the live naming fork between semimetric and pseudometric, and why extended metrics are not treated here.
The metric is part of the data. Two different metrics on the same set are two different metric spaces, even when they have the same open sets. That is why Topologically, uniformly and Lipschitz equivalent metrics on a set compares metrics at three separate strengths rather than one, and why a property can be invariant under one of them and not under another (FALSE: boundedness of a metric space is determined by its topology).

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Nonnegativity of a metric is a consequence of the other axioms, not an axiom

Statement

Let $X$ be a set and let $p : X \times X \to R$ satisfy the reflexivity axiom (M1') $p (x, x) = 0$ and the symmetry axiom (M2) $p (x, y) = p (y, x)$ of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric. Then:

If $p$ satisfies the triangle inequality (M3), then $p (x, y) \geq 0$ for all $x, y \in X$ .
If $p$ satisfies the strong triangle inequality (M3'), then $p (x, y) \geq 0$ for all $x, y \in X$ .

In particular every metric, every pseudometric and every ultrametric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) takes only nonnegative values. Nonnegativity is therefore a theorem about the axiom list this library uses, not a fourth axiom, and no statement on this page needs to assume it separately.

Facts & Assumptions

Given: A set $X$ , points $x, y \in X$ , and a function $p : X \times X \to R$ satisfying (M1') $p (a, a) = 0$ for every $a \in X$ and (M2) $p (a, b) = p (b, a)$ for all $a, b \in X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[A1]

(M3) The triangle inequality $p (a, c) \leq p (a, b) + p (b, c)$ holds for all $a, b, c \in X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[A2]

(M3') The strong triangle inequality $p (a, c) \leq \max {p (a, b), p (b, c)}$ holds for all $a, b, c \in X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

Trichotomy of the order of $R$ : for reals $a, b$ exactly one of $a < b$ , $a = b$ , $b < a$ holds, so $a \geq 0$ fails exactly when $a < 0$ (Order on the reals, Complete ordered field (least-upper-bound property), Ordered field).

[L2]

Adding two strict inequalities: if $a < b$ and $c < d$ then $a + c < b + d$ (Order is preserved by adding a constant and by adding inequalities).

[L3]

A two-element subset ${a, b}$ of $R$ has a maximum, and that maximum is $a$ or $b$ ; if $a = b$ it is $a$ (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Proof

technique · direct

1.1

Instantiate [A1] at $a = x$ , $b = y$ , $c = x$ : $p (x, x) \leq p (x, y) + p (y, x)$ .

1.2

Instantiate [A2] at $a = x$ , $b = y$ , $c = x$ : $p (x, x) \leq \max {p (x, y), p (y, x)}$ .

1.3

Suppose, towards ruling it out, that $p (x, y) < 0$ .

assume-hyp

2.1

By (M1') the left side of step 1.1 is $0$ and by (M2) the right side is $p (x, y) + p (x, y)$ , so $0 \leq p (x, y) + p (x, y)$ .

step 1.1given

2.2

By (M2) the two entries of the maximum in step 1.2 are the same real number, so that maximum equals $p (x, y)$ by [L3], and (M1') turns step 1.2 into $0 \leq p (x, y)$ , which is claim 2.

step 1.2givenL3

2.3

Adding the supposed inequality of step 1.3 to itself gives $p (x, y) + p (x, y) < 0 + 0 = 0$ .

step 1.3L2

3.1

Steps 2.1 and 2.3 assert $0 \leq p (x, y) + p (x, y)$ and $p (x, y) + p (x, y) < 0$ , which trichotomy forbids; so the supposition of step 1.3 is untenable and $p (x, y) \geq 0$ , which is claim 1.

step 2.1step 2.2step 2.3L1∎

Remarks

What each claim uses. Claim 1 is the familiar two-line argument $0 = p (x, x) \leq p (x, y) + p (y, x) = 2 p (x, y)$ followed by the observation that a negative real added to itself stays negative. Claim 2 does not need that second half at all: the strong triangle inequality delivers $0 \leq p (x, y)$ in one step, because the maximum of a real number with itself is that number.
Symmetry is used in both claims and cannot be dropped. Without (M2) the instantiation of step 1.1 only gives $0 \leq p (x, y) + p (y, x)$ , which leaves the possibility that one of the two values is negative and the other larger and positive. Dropping (M2) instead of weakening (M1) gives the notion usually called a quasimetric, which this library does not treat; for it the argument above is unavailable, so nonnegativity is not redundant there and is imposed as part of the definition (Which metric axiom list this library uses, the live naming fork between semimetric and pseudometric, and why extended metrics are not treated here).

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Open ball, closed ball and sphere in a metric space

Definition

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $x \in X$ and let $r \in R$ with $r > 0$ (Order on the reals). Define

$B (x, r) : = {y \in X : d (x, y) < r}, \overset{ˉ}{B} (x, r) : = {y \in X : d (x, y) \leq r}, S (x, r) : = {y \in X : d (x, y) = r} .$

$B (x, r)$ is the open ball, $\overset{ˉ}{B} (x, r)$ the closed ball and $S (x, r)$ the sphere of centre $x$ and radius $r$ . The radius is always a strictly positive real; a ball of radius $0$ or of negative radius is never written in this library.

Immediate consequences of the definitions. For every $x \in X$ and $r > 0$ :

$x \in B (x, r)$ , because $d (x, x) = 0 < r$ (axiom (M1) of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); in particular open and closed balls are nonempty.
$B (x, r) \subseteq \overset{ˉ}{B} (x, r)$ and $S (x, r) \subseteq \overset{ˉ}{B} (x, r)$ , and $\overset{ˉ}{B} (x, r)$ is the disjoint union of $B (x, r)$ and $S (x, r)$ , by trichotomy of the order of $R$ (Complete ordered field (least-upper-bound property), Ordered field): each $y$ satisfies exactly one of $d (x, y) < r$ , $d (x, y) = r$ , $d (x, y) > r$ .
If $0 < s \leq r$ then $B (x, s) \subseteq B (x, r)$ and $\overset{ˉ}{B} (x, s) \subseteq \overset{ˉ}{B} (x, r)$ , by transitivity of the order.
Nonnegativity of the metric (Nonnegativity of a metric is a consequence of the other axioms, not an axiom) is what forces the radius convention, and it forces it for the open ball only: if $r \leq 0$ then $B (x, r) = {y : d (x, y) < r}$ is empty, because $d (x, y) \geq 0 \geq r$ for every $y$ . The other two sets behave differently at $r = 0$ , and the convention $r > 0$ excludes them for uniformity rather than for emptiness: $\overset{ˉ}{B} (x, 0) = S (x, 0) = {x}$ , since $d (x, y) \leq 0$ together with $d (x, y) \geq 0$ gives $d (x, y) = 0$ and hence $y = x$ by (M1). For $r < 0$ all three sets are empty.

A sphere may be empty, and so the three sets are not on a par. For $r > 0$ the open and closed balls always contain $x$ , but nothing in the definition produces a point at distance exactly $r$ from $x$ . If a metric takes only the values $0$ and $1$ , as the discrete metric on the companion page does, then $S (x, 2) = \emptyset$ while $B (x, 2) = \overset{ˉ}{B} (x, 2)$ is the whole space. So nonemptiness of a sphere is never available by convention: where it is used, it is proved.

The ambient space is part of the notation. $B (x, r)$ depends on $(X, d)$ and not on $x$ and $r$ alone. When more than one space or more than one metric is in play we write $B_{X} (x, r)$ , or $B_{d} (x, r)$ , and likewise for $\overset{ˉ}{B}$ and $S$ . This matters as soon as subspaces appear (Isometry, isometric embedding, and the subspace metric on a subset): a ball of a subspace is the trace on it of a ball of the ambient space, and the two are different sets.

Remarks

The names "open ball" and "closed ball" are justified, not merely suggestive. That $B (x, r)$ is an open set and $\overset{ˉ}{B} (x, r)$ a closed set in the metric topology is proved in Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed; the words are used here only as names for the three sets displayed above.
The closed ball is not in general the closure of the open ball, and the sphere is not in general the boundary of either. Both failures are recorded on this page as FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ and witnessed on the companion page. The safe reading of the three names is the displayed one and nothing more.

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-08-02 (claude-opus-5)Open item page →

Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space

Definition

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $A, B \subseteq X$ .

Bounded subset. $A$ is bounded if $A = \emptyset$ or there are $x_{0} \in X$ and a real $r > 0$ with $A \subseteq B (x_{0}, r)$ (Open ball, closed ball and sphere in a metric space). The space $(X, d)$ is a bounded metric space if $X$ is a bounded subset of itself.

Diameter, for nonempty bounded $A$ only. Suppose $A$ is nonempty and bounded, and put

$D (A) : = {d (a, b) : a, b \in A} \subseteq R .$

Then $D (A)$ is nonempty, since $A$ is, and it is bounded above: fixing $x_{0}$ and $r$ with $A \subseteq B (x_{0}, r)$ , every $a, b \in A$ satisfy $d (a, b) \leq d (a, x_{0}) + d (x_{0}, b) < r + r$ by the triangle inequality, symmetry (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and addition of inequalities (Order is preserved by adding a constant and by adding inequalities, Ordered field), so $r + r$ is an upper bound of $D (A)$ (Lower bound, bounded below, bounded set). Hence $D (A)$ has a least upper bound in $R$ by the least-upper-bound property (Complete ordered field (least-upper-bound property)), and that bound is unique (Suprema and infima are unique). Define

$diam (A) : = \sup D (A) .$

Distance from a point to a set, for nonempty $A$ only. Let $x \in X$ and let $A$ be nonempty, and put $E (x, A) : = {d (x, a) : a \in A}$ . Then $E (x, A)$ is nonempty and bounded below by $0$ , since a metric is nonnegative (Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Lower bound, bounded below, bounded set), so it has a greatest lower bound (Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)), unique by Suprema and infima are unique. Define

$d (x, A) : = \inf E (x, A) .$

Distance between two sets, for nonempty $A$ and $B$ only. Put $E (A, B) : = {d (a, b) : a \in A, b \in B}$ , again nonempty and bounded below by $0$ , and define

$d (A, B) : = \inf E (A, B) .$

Every one of the three scope restrictions is load bearing. In this library $\sup$ and $\inf$ denote real numbers and are written only after existence has been established; the extended real line is introduced on a later page and is not used for the suprema and infima taken here, and no convention $\sup \emptyset = - \infty$ is in force in this development (Conventions: $\sup \emptyset$ , unbounded sets, and the extended reals). Accordingly:

$diam (A)$ is defined exactly when $A$ is nonempty and bounded. It is not defined for $A = \emptyset$ , and it is not defined, not even as an infinite value, for an unbounded $A$ .
$d (x, A)$ is defined exactly when $A \neq \emptyset$ , and $d (A, B)$ exactly when both $A$ and $B$ are nonempty. No boundedness is needed for these two, because $0$ is always a lower bound.

Remarks

Diameter and the distance functions are nonnegative. For nonempty bounded $A$ and any $a \in A$ we have $0 = d (a, a) \in D (A)$ , so $diam (A) \geq 0$ ; and $d (x, A) \geq 0$ , $d (A, B) \geq 0$ because $0$ is a lower bound of the sets they are infima of (Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Greatest lower bound (infimum)).
$d (A, B)$ is not a metric on the nonempty subsets of $X$ . It is symmetric and vanishes on $A = B$ , but two distinct disjoint sets can be at distance $0$ , so the separation axiom (M1) of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric fails; the witness is on the companion page. The letter $d$ is reused for three different functions here, the metric, the point-to-set distance and the set-to-set distance, only because the arguments make the intended one unambiguous.
$d (x, A)$ is the special case $d ({x}, A)$ , since $E ({x}, A) = E (x, A)$ , and the two infima therefore agree by uniqueness (Suprema and infima are unique).

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $x, y, z \in X$ . Then

$∣ d (x, z) - d (y, z) ∣ \leq d (x, y),$

where $∣ \cdot ∣$ is the absolute value of $R$ (Absolute value in an ordered field).

Facts & Assumptions

Given: A metric space $(X, d)$ and points $x, y, z \in X$ ; write $t : = d (x, z) - d (y, z)$ .

[A1]

The triangle inequality (M3): $d (a, c) \leq d (a, b) + d (b, c)$ for all $a, b, c \in X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[A2]

Symmetry (M2): $d (a, b) = d (b, a)$ for all $a, b \in X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

For every real $u$ , the value $∣ u ∣$ equals $u$ or $- u$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L2]

Adding a constant to an inequality: if $a \leq b$ then $a + c \leq b + c$ . Order is preserved by adding a constant and by adding inequalities states the strict form $a < b \Rightarrow a + c < b + c$ ; the nonstrict form used here is that strict form together with the case $a = b$ , in which the two sides are equal, the order being total (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1

By [A1] at $(a, b, c) = (x, y, z)$ : $d (x, z) \leq d (x, y) + d (y, z)$ .

1.2

By [A1] at $(a, b, c) = (y, x, z)$ : $d (y, z) \leq d (y, x) + d (x, z)$ , and by [A2] $d (y, x) = d (x, y)$ , so $d (y, z) \leq d (x, y) + d (x, z)$ .

A1A2

2.1

Adding $- d (y, z)$ to both sides of step 1.1 gives $t = d (x, z) - d (y, z) \leq d (x, y)$ .

step 1.1L2

2.2

Adding $- d (x, z)$ to both sides of step 1.2 gives $d (y, z) - d (x, z) \leq d (x, y)$ , that is $- t \leq d (x, y)$ .

step 1.2L2

3.1

By [L1] the real number $∣ t ∣$ is either $t$ or $- t$ , and both of these are at most $d (x, y)$ by steps 2.1 and 2.2, so $∣ d (x, z) - d (y, z) ∣ = ∣ t ∣ \leq d (x, y)$ .

step 2.1step 2.2L1∎

Remarks

Read with $z$ fixed, this says the function $u \mapsto d (u, z)$ does not increase distances: its values at $x$ and at $y$ differ by at most $d (x, y)$ . That is the model for $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, which proves the same estimate with the point $z$ replaced by a nonempty set.
The inequality specialises, on $R$ with $d (u, v) = ∣ u - v ∣$ (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded), to the familiar $∣ ∣ x - z ∣ - ∣ y - z ∣ ∣ \leq ∣ x - y ∣$ .

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

$∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $A \subseteq X$ be nonempty and let $x, y \in X$ . Then

$∣ d (x, A) - d (y, A) ∣ \leq d (x, y),$

with $d (\cdot, A)$ the distance to a nonempty set (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). Thus the real-valued function $u \mapsto d (u, A)$ changes by at most $d (u, v)$ between $u$ and $v$ : it is $1$ -Lipschitz.

Facts & Assumptions

Given: A metric space $(X, d)$ , a nonempty $A \subseteq X$ , and points $x, y \in X$ ; write $E (u) : = {d (u, a) : a \in A}$ for $u \in X$ .

[A1]

The triangle inequality (M3) of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric: $d (u, a) \leq d (u, v) + d (v, a)$ for all $u, v \in X$ and $a \in A$ .

[L1]

For nonempty $A$ the real number $d (u, A) = \inf E (u)$ exists, because $E (u)$ is nonempty and bounded below by $0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Every nonempty set bounded below has an infimum).

[L2]

The infimum is a lower bound of its set and is the greatest such: $\inf S \leq s$ for every $s \in S$ , and $ℓ \leq \inf S$ for every lower bound $ℓ$ of $S$ (Greatest lower bound (infimum)).

[L3]

Adding a constant to an inequality: if $a \leq b$ then $a + c \leq b + c$ . Order is preserved by adding a constant and by adding inequalities states the strict form only; the nonstrict form used here is that form together with the case $a = b$ , settled by totality of the order (Ordered field, Complete ordered field (least-upper-bound property)).

[L4]

For every real $u$ , $∣ u ∣$ equals $u$ or $- u$ (Basic properties of the absolute value, Absolute value in an ordered field).

Proof

technique · direct

1.1

Both $d (x, A)$ and $d (y, A)$ are defined real numbers, since $A$ is nonempty.

givenL1

1.2

For every $a \in A$ : $d (x, a) \leq d (x, y) + d (y, a)$ .

1.3

For every $a \in A$ : $d (y, a) \leq d (y, x) + d (x, a)$ , and $d (y, x) = d (x, y)$ by symmetry (M2), so $d (y, a) \leq d (x, y) + d (x, a)$ .

2.1

For every $a \in A$ : $d (x, A) \leq d (x, a)$ , since $d (x, A)$ is a lower bound of $E (x)$ and $d (x, a) \in E (x)$ ; combining with step 1.2 gives $d (x, A) \leq d (x, y) + d (y, a)$ , hence $d (x, A) - d (x, y) \leq d (y, a)$ .

step 1.1step 1.2L2L3

2.2

For every $a \in A$ : $d (y, A) \leq d (y, a) \leq d (x, y) + d (x, a)$ by the same reasoning with the roles of $x$ and $y$ exchanged, hence $d (y, A) - d (x, y) \leq d (x, a)$ .

step 1.1step 1.3L2L3

3.1

The real number $d (x, A) - d (x, y)$ is therefore a lower bound of $E (y)$ , so it is at most the greatest lower bound: $d (x, A) - d (x, y) \leq d (y, A)$ , that is $d (x, A) - d (y, A) \leq d (x, y)$ .

step 2.1L2L3

3.2

Symmetrically $d (y, A) - d (x, y)$ is a lower bound of $E (x)$ , so $d (y, A) - d (x, A) \leq d (x, y)$ .

step 2.2L2L3

4.1

By [L4] the value $∣ d (x, A) - d (y, A) ∣$ is $d (x, A) - d (y, A)$ or its negative $d (y, A) - d (x, A)$ , and steps 3.1 and 3.2 bound both by $d (x, y)$ ; hence $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ .

step 3.1step 3.2L4∎

Remarks

Where the nonemptiness is used. Twice, and both times essentially: it is what makes $d (x, A)$ and $d (y, A)$ exist at all (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and it is what makes $E (x)$ and $E (y)$ nonempty so that "greatest lower bound" has content. For $A = \emptyset$ the statement has no meaning in this library, since $\inf \emptyset$ is undefined.
The point case is The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space: taking $A = {z}$ gives $E (u) = {d (u, z)}$ , whose infimum is $d (u, z)$ , and the conclusion becomes $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ .
The constant $1$ is best possible in general: on $R$ with $A = {0}$ the function is $u \mapsto ∣ u ∣$ , and $∣ ∣ x ∣ - ∣ y ∣ ∣ = ∣ x - y ∣$ whenever $x$ and $y$ have the same sign.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-08-02 (claude-opus-5)Open item page →

The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement

Definition

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

A subset $U \subseteq X$ is open in $(X, d)$ if for every $x \in U$ there is a real $r > 0$ with $B (x, r) \subseteq U$ (Open ball, closed ball and sphere in a metric space). A subset $F \subseteq X$ is closed in $(X, d)$ if its complement $X ∖ F$ is open.

The collection

$T_{d} : = {U \subseteq X : U is open in (X, d)}$

of all open subsets is the metric topology of $d$ on $X$ . A subset of $X$ that is both open and closed is called clopen.

Two sets are open for trivial reasons. $\emptyset$ is open, because the defining condition quantifies over no points; and $X$ is open, because $B (x, r) \subseteq X$ for every $x$ and every $r > 0$ . Consequently $X$ and $\emptyset$ are also closed, and both are clopen.

A neighbourhood of a point $x$ is any open set containing $x$ . The condition above therefore reads: $U$ is open exactly when every point of $U$ has a ball around it inside $U$ , and it is the balls alone that have to be tested.

The metric, not the set, determines $T_{d}$ . Two metrics on the same set may have different metric topologies, and two different metrics may have the same one; the systematic comparison is Topologically, uniformly and Lipschitz equivalent metrics on a set.

Remarks

What "topology" means here. $T_{d}$ is defined above as a collection of subsets of $X$ ; the abstract notion of a topological space, a collection of subsets closed under arbitrary unions and finite intersections taken as primitive data, is introduced on a later page and is not used here. What is proved here is that $T_{d}$ has exactly those closure properties (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed), which is what licenses the word.
Open and closed are not opposites. A set may be neither ( $[0, 1)$ inside $R$ , once the usual metric is available from The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded) or both ( $\emptyset$ and $X$ always, and in some spaces every subset at once, as the discrete metric on the companion page shows). "Not open" is never a synonym for "closed".
Closedness is complementation, and nothing else, at this stage. The description of closed sets by limits of sequences, and the description of the closure as an infimum of distances, are theorems proved later on this page (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed), not part of the definition.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded

Statement

Define $d_{R} : R \times R \to R$ by $d_{R} (x, y) : = ∣ x - y ∣$ (Absolute value in an ordered field). Then:

$d_{R}$ is a metric on $R$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); it is called the usual metric of $R$ .
For $x \in R$ and $r > 0$ the open ball is the bounded open interval (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Open ball, closed ball and sphere in a metric space) $B (x, r) = (x - r, x + r),$ and the closed ball is $\overset{ˉ}{B} (x, r) = [x - r, x + r]$ .
Consequently $U \subseteq R$ is open in the metric topology of $d_{R}$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) exactly when for every $x \in U$ there is $r > 0$ with $(x - r, x + r) \subseteq U$ . This topology is called the usual topology of $R$ .
$(R, d_{R})$ is not a bounded metric space (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): no ball contains $R$ , so $diam (R)$ is not defined.

Facts & Assumptions

Given: The complete ordered field $R$ (Complete ordered field (least-upper-bound property), Ordered field) with its absolute value (Absolute value in an ordered field), and the function $d_{R} (x, y) = ∣ x - y ∣$ ; points $x, y, z \in R$ and a real $r > 0$ .

[L1]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ if and only if $u = 0$ ; $∣ - u ∣ = ∣ u ∣$ ; and for $c > 0$ one has $∣ u ∣ < c$ if and only if $- c < u < c$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L2]

Triangle inequality in an ordered field: $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

[L3]

Intervals: $(a, b) = {t : a < t < b}$ and $[a, b] = {t : a \leq t \leq b}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

Archimedean property: for every $w \in R$ there is a natural $n \geq 1$ with $w < n \cdot 1_{R}$ (Every complete ordered field is Archimedean); and $n \cdot 1_{R} > 0$ for $n \geq 1$ (Canonical naturals are positive and strictly increasing).

[L5]

Adding a constant to an inequality, in strict and nonstrict form: the strict form is Order is preserved by adding a constant and by adding inequalities and the nonstrict form is that together with the case of equality, the order being total (Ordered field).

[L6]

Trichotomy: for reals $a, b$ exactly one of $a < b$ , $a = b$ , $b < a$ holds (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · direct

1.1

Separation (M1): $d_{R} (x, y) = ∣ x - y ∣ = 0$ holds if and only if $x - y = 0$ , that is if and only if $x = y$ .

1.2

Symmetry (M2): $d_{R} (y, x) = ∣ y - x ∣ = ∣ - (x - y) ∣ = ∣ x - y ∣ = d_{R} (x, y)$ .

1.3

Triangle inequality (M3): $d_{R} (x, z) = ∣ x - z ∣ = ∣ (x - y) + (y - z) ∣ \leq ∣ x - y ∣ + ∣ y - z ∣ = d_{R} (x, y) + d_{R} (y, z)$ .

1.4

For $y \in R$ and $r > 0$ : $y \in B (x, r)$ means $∣ x - y ∣ < r$ , which by [L1] holds if and only if $- r < x - y < r$ , and adding $y - r$ respectively $y + r$ to the two halves shows this is equivalent to $x - r < y < x + r$ .

L1L5

1.5

For $y \in R$ and $r > 0$ : $y \in \overset{ˉ}{B} (x, r)$ means $∣ x - y ∣ \leq r$ , which by the same equivalence read with $\leq$ in place of $<$ holds if and only if $x - r \leq y \leq x + r$ .

L1L5

1.6

Let $x_{0} \in R$ and $r > 0$ be arbitrary, and use [L4] to fix a natural $n \geq 1$ with $x_{0} + r < n \cdot 1_{R}$ ; write $w : = n \cdot 1_{R}$ .

L4choose

2.1

By steps 1.1, 1.2 and 1.3 the function $d_{R}$ satisfies (M1), (M2) and (M3), so it is a metric on $R$ , which is claim 1.

step 1.1step 1.2step 1.3

2.2

By step 1.4 and [L3] the set $B (x, r)$ has exactly the elements of $(x - r, x + r)$ , and by step 1.5 and [L3] the set $\overset{ˉ}{B} (x, r)$ has exactly the elements of $[x - r, x + r]$ ; this is claim 2.

step 1.4step 1.5L3

2.3

Since $r > 0$ we have $x_{0} < x_{0} + r < w$ , so $w - x_{0} > r > 0$ and hence $d_{R} (x_{0}, w) = ∣ x_{0} - w ∣ = ∣ - (w - x_{0}) ∣ = w - x_{0} > r$ ; therefore $w \notin B (x_{0}, r)$ .

step 1.6L1L5L6

3.1

Substituting claim 2 into the definition of open in the metric topology gives claim 3: $U$ is open exactly when every $x \in U$ admits $r > 0$ with $(x - r, x + r) = B (x, r) \subseteq U$ .

step 2.2

4.1

Since $x_{0}$ and $r$ were arbitrary, step 2.3 exhibits for every ball $B (x_{0}, r)$ a real not in it, so no ball contains $R$ ; hence $R$ is not a bounded subset of itself and $diam (R)$ is not defined, which is claim 4.

step 2.1step 2.3∎

Remarks

This is the metric every later ceiling rests on. Every real-line example on the companion page, and every subspace of $R$ used there, takes its metric from $d_{R}$ through the subspace construction of Isometry, isometric embedding, and the subspace metric on a subset.
Unboundedness needs no Archimedean input, and no completeness either. No ordered field is bounded under $d (u, v) = ∣ u - v ∣$ , and the reason is a single element rather than any cofinality property: given a centre $c$ and a radius $r > 0$ , the element $c + r + 1$ satisfies $d (c, c + r + 1) = ∣ - (r + 1) ∣ = r + 1 > r$ , because $1 > 0$ and $r + 1 > 0$ (Basic properties of the absolute value, The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities); so it lies outside $B (c, r)$ and no ball contains the field. Step 1.6 above chooses its witness with Every complete ordered field is Archimedean instead, which is a convenience and not a necessity: it delivers a witness that is a canonical natural, and claim 4 needs no such thing. Claim 4 therefore holds verbatim in every ordered field with this $d$ , Archimedean or not. Note also that a radius is an element of $R$ , so "a ball of infinite radius" is not something that can be written here.
The claim that $diam (R)$ is "not defined" is a claim about the conventions of this development (Conventions: $\sup \emptyset$ , unbounded sets, and the extended reals, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space): suprema here are real numbers and the extended real line, which is introduced on a later page, is not used for them, so an unbounded set has no diameter at all rather than a diameter $+ \infty$ .

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

$R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it

Statement

Let $n \in N$ with $n \geq 1$ . A von Neumann natural is the set of its predecessors, $n = {0, 1, \dots, n - 1}$ (The natural numbers $N$ (von Neumann)), so it can be used directly as an index set. Define

$R^{n} : = {x : x is a function n \to R},$

and write $x_{k}$ for $x (k)$ , $k < n$ . Two elements of $R^{n}$ are equal exactly when they agree at every $k < n$ , functions being equal when they have the same values. For $x, y \in R^{n}$ put

$d_{1} (x, y) : = \sum_{k < n} ∣ x_{k} - y_{k} ∣, d_{2} (x, y) : = \sqrt{\sum_{k < n} (x_{k} - y_{k})^{2}}, d_{\infty} (x, y) : = \max {∣ x_{k} - y_{k} ∣ : k < n} .$

All three are well defined: the finite sums are those of Finite sums and finite products, by recursion; the sum of squares is nonnegative (Laws of finite sums and finite products, Squares of nonzero elements are positive) so it has a unique nonnegative square root (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ); and ${∣ x_{k} - y_{k} ∣ : k < n}$ is a nonempty finite subset of $R$ , because $n \geq 1$ , so it has a maximum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Then $d_{1}$ , $d_{2}$ and $d_{\infty}$ are metrics on $R^{n}$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

Why $n \geq 1$ . For $n = 0$ the set $R^{0}$ has exactly one element, the empty function, and $d_{1}$ and $d_{2}$ are the empty sum $0$ and its root; but $d_{\infty}$ would be the maximum of the empty set, which does not exist. The hypothesis $n \geq 1$ is therefore not decoration, and it is carried by every statement about $d_{\infty}$ in this library.

Facts & Assumptions

Given: A natural $n \geq 1$ ; elements $x, y, z \in R^{n}$ ; and the lists $a_{k} : = x_{k} - y_{k}$ , $b_{k} : = y_{k} - z_{k}$ for $k < n$ , so that $a_{k} + b_{k} = x_{k} - z_{k}$ . Write $A : = \sum_{k < n} a_{k}^{2}$ , $C : = \sum_{k < n} b_{k}^{2}$ and $B : = \sum_{k < n} a_{k} b_{k}$ .

[L1]

Laws of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): additivity, scaling, monotonicity; a sum of nonnegative terms is nonnegative, every single term is at most the sum, and a sum of nonnegative terms that vanishes has every term $0$ .

[L2]

Absolute value (Basic properties of the absolute value, Absolute value in an ordered field): $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ if and only if $u = 0$ ; $∣ - u ∣ = ∣ u ∣$ ; and $u \leq ∣ u ∣$ .

[L3]

Two-term triangle inequality: $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

[L4]

Minkowski's inequality at the rational exponent $p = 1$ (Minkowski's inequality for finite sums (rational exponent)): $\sum_{k < n} ∣ a_{k} + b_{k} ∣ \leq \sum_{k < n} ∣ a_{k} ∣ + \sum_{k < n} ∣ b_{k} ∣$ .

[L5]

Cauchy-Schwarz in root form (The Cauchy-Schwarz inequality for finite sums): $∣ \sum_{k < n} a_{k} b_{k} ∣ \leq \sqrt{\sum_{k < n} a_{k}^{2}} \sqrt{\sum_{k < n} b_{k}^{2}}$ .

[L6]

Square roots (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ): every $c \geq 0$ has a unique $\sqrt{c} \geq 0$ with $(\sqrt{c})^{2} = c$ ; in particular $\sqrt{c} = 0$ if and only if $c = 0$ .

[L7]

Squares (Squares of nonzero elements are positive, Integer powers $a^{m}$ ): $u^{2} \geq 0$ always, and $u^{2} = 0$ only for $u = 0$ ; and monotonicity of squaring on the nonnegatives, $s \leq t ⟺ s^{2} \leq t^{2}$ for $s, t \geq 0$ (Squaring is monotone on the nonnegatives).

[L8]

Maximum of a nonempty finite set of reals: it exists, it belongs to the set, and it is an upper bound of the set (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L9]

Order arithmetic in $R$ : inequalities may be added and a constant added to both sides, in the strict form of Order is preserved by adding a constant and by adding inequalities and, together with the case of equality settled by totality (Ordered field, Complete ordered field (least-upper-bound property)), in the nonstrict form used below.

Proof

technique · direct

1.1

Separation for $d_{1}$ : $d_{1} (x, y) = \sum_{k < n} ∣ a_{k} ∣$ is a sum of nonnegative terms, so it vanishes exactly when every $∣ a_{k} ∣$ vanishes, that is exactly when $x_{k} = y_{k}$ for all $k < n$ , that is exactly when $x = y$ .

L1L2

1.2

Separation for $d_{2}$ : $d_{2} (x, y) = \sqrt{A}$ vanishes exactly when $A = 0$ ; $A$ is a sum of nonnegative terms, so $A = 0$ exactly when $a_{k}^{2} = 0$ for every $k < n$ , which happens exactly when every $a_{k} = 0$ , that is exactly when $x = y$ .

L1L6L7

1.3

Separation for $d_{\infty}$ : the maximum $d_{\infty} (x, y)$ belongs to ${∣ a_{k} ∣ : k < n}$ and bounds it above, so it is $0$ exactly when every $∣ a_{k} ∣ = 0$ , that is exactly when $x = y$ .

L2L8

1.4

Symmetry for all three: $∣ y_{k} - x_{k} ∣ = ∣ - (x_{k} - y_{k}) ∣ = ∣ x_{k} - y_{k} ∣$ and $(y_{k} - x_{k})^{2} = (x_{k} - y_{k})^{2}$ for every $k < n$ , so the three defining expressions are unchanged when $x$ and $y$ are exchanged.

L2L7

1.5

Triangle inequality for $d_{1}$ : applying [L4] to the lists $(a_{k})$ and $(b_{k})$ gives $d_{1} (x, z) = \sum_{k < n} ∣ a_{k} + b_{k} ∣ \leq \sum_{k < n} ∣ a_{k} ∣ + \sum_{k < n} ∣ b_{k} ∣ = d_{1} (x, y) + d_{1} (y, z)$ .

1.6

Expanding with additivity and scaling: $\sum_{k < n} (a_{k} + b_{k})^{2} = \sum_{k < n} (a_{k}^{2} + 2 a_{k} b_{k} + b_{k}^{2}) = A + 2 B + C$ .

L1algebra

1.7

By [L5] and $B \leq ∣ B ∣$ : $B \leq \sqrt{A} \sqrt{C}$ , and $A = (\sqrt{A})^{2}$ , $C = (\sqrt{C})^{2}$ with $\sqrt{A}, \sqrt{C} \geq 0$ .

L2L5L6

1.8

Triangle inequality for $d_{\infty}$ : for each $k < n$ , $∣ a_{k} + b_{k} ∣ \leq ∣ a_{k} ∣ + ∣ b_{k} ∣ \leq d_{\infty} (x, y) + d_{\infty} (y, z)$ because the two maxima bound their sets; so $d_{\infty} (x, y) + d_{\infty} (y, z)$ is an upper bound of ${∣ a_{k} + b_{k} ∣ : k < n}$ , and the maximum $d_{\infty} (x, z)$ of that set is one of its elements, whence $d_{\infty} (x, z) \leq d_{\infty} (x, y) + d_{\infty} (y, z)$ .

L3L8L9

2.1

Combining steps 1.6 and 1.7: $\sum_{k < n} (a_{k} + b_{k})^{2} = A + 2 B + C \leq (\sqrt{A})^{2} + 2 \sqrt{A} \sqrt{C} + (\sqrt{C})^{2} = (\sqrt{A} + \sqrt{C})^{2}$ .

step 1.6step 1.7L9algebra

3.1

Both $d_{2} (x, z) = \sqrt{\sum_{k < n} (a_{k} + b_{k})^{2}}$ and $\sqrt{A} + \sqrt{C}$ are nonnegative, and by step 2.1 the square of the first is at most the square of the second, so monotonicity of squaring on the nonnegatives gives $d_{2} (x, z) \leq \sqrt{A} + \sqrt{C} = d_{2} (x, y) + d_{2} (y, z)$ .

step 2.1L6L7

4.1

Each of $d_{1}$ , $d_{2}$ , $d_{\infty}$ satisfies (M1) by steps 1.1, 1.2 and 1.3, satisfies (M2) by step 1.4, and satisfies (M3) by steps 1.5, 3.1 and 1.8 respectively; hence all three are metrics on $R^{n}$ .

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.8step 3.1∎

Remarks

$R^{n}$ is defined ZFC-natively here, as the set of functions from the von Neumann natural $n$ to $R$ , precisely so that its coordinates are indexed by $k < n$ and the finite-sum machinery of Finite sums and finite products, by recursion, Minkowski's inequality for finite sums (rational exponent) and The Cauchy-Schwarz inequality for finite sums, all of which sum over $k < n$ , applies without any reindexing.
No rational power appears anywhere above. The triangle inequality for $d_{2}$ is obtained from Cauchy-Schwarz and the existence of square roots, not from Minkowski at $p = 2$ , so this lemma does not depend on the theory of rational exponents. Minkowski is used only at $p = 1$ , where its statement is the termwise sum of the two-term triangle inequality.
The three metrics are Lipschitz equivalent, with explicit constants, and in particular have the same topology; that computation is on the companion page and is not needed here.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

The supremum metric $d_{\infty} (f, g) = \sup_{x} ∣ f (x) - g (x) ∣$ is a metric on the bounded real-valued functions on a nonempty set

Statement

Let $S$ be a nonempty set. Call a function $f : S \to R$ bounded when its range $f [S] = {f (s) : s \in S}$ is a bounded subset of $R$ (Lower bound, bounded below, bounded set), and write

$B (S) : = {f : f is a bounded function S \to R} .$

For $f, g \in B (S)$ put $D (f, g) : = {∣ f (s) - g (s) ∣ : s \in S}$ and

$d_{\infty} (f, g) : = \sup D (f, g) .$

This is well defined: $D (f, g)$ is nonempty because $S$ is, and it is bounded above (step 1.1 below), so its least upper bound exists (Complete ordered field (least-upper-bound property)) and is unique (Suprema and infima are unique).

Then $d_{\infty}$ is a metric on $B (S)$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), the supremum metric (also called the uniform metric).

The hypotheses ensure that the formula is a finite real-valued metric for every pair in the stated function space. Boundedness of $f$ and $g$ makes $D (f, g)$ bounded above, and nonemptiness of $S$ makes it nonempty. Some unbounded pairs can still have a finite supremum, but allowing all real-valued functions would not give a finite-valued metric: for example, on $S = R$ the functions $f (s) = s$ and $g (s) = 0$ make $D (f, g)$ unbounded above (Conventions: $\sup \emptyset$ , unbounded sets, and the extended reals).

Facts & Assumptions

Given: A nonempty set $S$ and bounded functions $f, g, h \in B (S)$ , with $ℓ_{f} \leq f (s) \leq u_{f}$ , $ℓ_{g} \leq g (s) \leq u_{g}$ and $ℓ_{h} \leq h (s) \leq u_{h}$ for all $s \in S$ ; a fixed $s_{0} \in S$ .

[L1]

Bounded subset of $R$ : $T$ is bounded when there are $ℓ, u \in R$ with $ℓ \leq t \leq u$ for every $t \in T$ (Lower bound, bounded below, bounded set).

[L2]

Least-upper-bound property: a nonempty subset of $R$ that is bounded above has a least upper bound, that is an upper bound below every upper bound; it is unique (Complete ordered field (least-upper-bound property), Suprema and infima are unique).

[L3]

Absolute value: $∣ u ∣ \geq 0$ ; $∣ u ∣ = 0$ if and only if $u = 0$ ; $∣ - u ∣ = ∣ u ∣$ ; and $∣ u ∣$ equals $u$ or $- u$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L4]

Two-term triangle inequality: $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality).

[L5]

A two-element subset of $R$ has a maximum, which is one of the two elements and bounds both (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L6]

Order arithmetic: inequalities may be added and a constant added to both sides, in the strict form of Order is preserved by adding a constant and by adding inequalities and, together with the case of equality settled by totality (Ordered field, Complete ordered field (least-upper-bound property)), in the nonstrict form used below; and by trichotomy $a \leq 0$ together with $a \geq 0$ gives $a = 0$ .

Proof

technique · direct

1.1

For every $s \in S$ the value $∣ f (s) - g (s) ∣$ is $f (s) - g (s)$ or $g (s) - f (s)$ , and $f (s) - g (s) \leq u_{f} - ℓ_{g}$ while $g (s) - f (s) \leq u_{g} - ℓ_{f}$ ; so $M : = \max {u_{f} - ℓ_{g}, u_{g} - ℓ_{f}}$ bounds $D (f, g)$ above, and since $s_{0} \in S$ makes $D (f, g)$ nonempty, $d_{\infty} (f, g) = \sup D (f, g)$ exists and is unique.

givenL1L2L3L5L6

1.2

Symmetry (M2): $∣ g (s) - f (s) ∣ = ∣ - (f (s) - g (s)) ∣ = ∣ f (s) - g (s) ∣$ for every $s \in S$ , so $D (g, f)$ and $D (f, g)$ are the same subset of $R$ and therefore have the same supremum.

L2L3

2.1

Separation (M1): $d_{\infty} (f, g)$ bounds $D (f, g)$ above, so $d_{\infty} (f, g) \geq ∣ f (s_{0}) - g (s_{0}) ∣ \geq 0$ ; if $d_{\infty} (f, g) = 0$ then $∣ f (s) - g (s) ∣ \leq 0$ and $∣ f (s) - g (s) ∣ \geq 0$ for every $s$ , hence $f (s) = g (s)$ for every $s$ and $f = g$ ; conversely if $f = g$ then $D (f, g) = {0}$ , whose least upper bound is $0$ .

step 1.1L2L3L6

2.2

For every $s \in S$ : $∣ f (s) - h (s) ∣ = ∣ (f (s) - g (s)) + (g (s) - h (s)) ∣ \leq ∣ f (s) - g (s) ∣ + ∣ g (s) - h (s) ∣ \leq d_{\infty} (f, g) + d_{\infty} (g, h)$ , the last inequality because each supremum bounds its own set above.

step 1.1L2L4L6

3.1

Triangle inequality (M3): step 2.2 says the real number $d_{\infty} (f, g) + d_{\infty} (g, h)$ is an upper bound of $D (f, h)$ , and $d_{\infty} (f, h)$ is the least upper bound of that set, so $d_{\infty} (f, h) \leq d_{\infty} (f, g) + d_{\infty} (g, h)$ .

step 2.2L2

4.1

The function $d_{\infty}$ therefore satisfies (M1) by step 2.1, (M2) by step 1.2 and (M3) by step 3.1, so it is a metric on $B (S)$ .

step 1.2step 2.1step 3.1∎

Remarks

Why the bounded functions and not all functions. For unbounded $f$ the set $D (f, g)$ need not be bounded above and $\sup D (f, g)$ then does not exist, so $d_{\infty}$ would not be a function into $R$ at all. Texts that write $d_{\infty} (f, g) = + \infty$ in that case are working in the extended real line, which is introduced on a later page. The suprema taken here are real numbers, and the extended real line is not used for them (Conventions: $\sup \emptyset$ , unbounded sets, and the extended reals, Which metric axiom list this library uses, the live naming fork between semimetric and pseudometric, and why extended metrics are not treated here).
The supremum need not be attained, so $d_{\infty} (f, g)$ is genuinely a supremum and not a maximum; the companion page carries a witness.
The name "uniform metric" points at later material. The quantified definition of uniform convergence of functions appears in Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions ↗. The later Agreement of the quantified real-valued definition with the later uniform-metric and uniform-topology formulations ↗ records its agreement with convergence in $d_{\infty}$ ; this lemma proves only that $d_{\infty}$ is a metric on the stated bounded-function space.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), with open and closed sets as in The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement and balls as in Open ball, closed ball and sphere in a metric space. Then:

Balls are open. $B (x, r)$ is open, for every $x \in X$ and every $r > 0$ .
Arbitrary unions. If $U$ is any collection of open subsets of $X$ , then $⋃ U$ is open.
Finite intersections. If $n \geq 1$ and $U_{0}, \dots, U_{n - 1}$ are open, then $U_{0} \cap \dots \cap U_{n - 1}$ is open.
Closed balls are closed. $\overset{ˉ}{B} (x, r)$ is closed, for every $x \in X$ and every $r > 0$ .

Together with the fact that $\emptyset$ and $X$ are open, recorded already in The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, claims 2 and 3 say that $T_{d}$ has exactly the closure properties that the word topology names.

Facts & Assumptions

Given: A metric space $(X, d)$ ; a point $x \in X$ and a real $r > 0$ ; a collection $U$ of open subsets of $X$ ; a natural $n \geq 1$ and open sets $U_{0}, \dots, U_{n - 1} \subseteq X$ .

[A1]

Open: $U$ is open when every $u \in U$ admits $t > 0$ with $B (u, t) \subseteq U$ ; closed means the complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[A2]

Balls: $B (u, t) = {w : d (u, w) < t}$ and $\overset{ˉ}{B} (u, t) = {w : d (u, w) \leq t}$ , and $B (u, t) \subseteq B (u, t^{'})$ whenever $0 < t \leq t^{'}$ (Open ball, closed ball and sphere in a metric space).

[L1]

Triangle inequality and symmetry of $d$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

Reverse triangle inequality: $∣ d (a, c) - d (b, c) ∣ \leq d (a, b)$ , so in particular $d (a, c) - d (b, c) \leq d (a, b)$ (The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space).

[L3]

A nonempty finite set of reals has a minimum, which belongs to the set and is a lower bound of it (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L4]

Order arithmetic: a constant may be added to both sides of an inequality and inequalities may be chained by transitivity, in the strict form of Order is preserved by adding a constant and by adding inequalities and, with the case of equality settled by totality, in the nonstrict form (Ordered field, Complete ordered field (least-upper-bound property)); and by trichotomy $a < b$ and $b \leq a$ cannot both hold.

Proof

technique · direct

1.1

Claim 1: let $y \in B (x, r)$ , so $d (x, y) < r$ , and put $s : = r - d (x, y) > 0$ ; for $z \in B (y, s)$ the triangle inequality gives $d (x, z) \leq d (x, y) + d (y, z) < d (x, y) + s = r$ , so $B (y, s) \subseteq B (x, r)$ , and since $y$ was arbitrary $B (x, r)$ is open.

A1A2L1L4

1.2

Claim 2: let $y \in ⋃ U$ , so $y \in U$ for some $U \in U$ ; as $U$ is open there is $t > 0$ with $B (y, t) \subseteq U \subseteq ⋃ U$ , and since $y$ was arbitrary the union is open.

1.3

Claim 3: let $y \in U_{0} \cap \dots \cap U_{n - 1}$ and for each $k < n$ pick $t_{k} > 0$ with $B (y, t_{k}) \subseteq U_{k}$ , which is possible because each $U_{k}$ is open and $y$ lies in it.

A1choose

1.4

Claim 4: let $y \in X ∖ \overset{ˉ}{B} (x, r)$ , so $d (x, y) > r$ , and put $s : = d (x, y) - r > 0$ ; for $z \in B (y, s)$ the reverse triangle inequality applied to the points $y, z, x$ gives $d (y, x) - d (z, x) \leq d (y, z) < s$ , hence $d (z, x) > d (y, x) - s = r$ , so $d (x, z) > r$ by symmetry and $z \notin \overset{ˉ}{B} (x, r)$ .

A2L1L2L4

2.1

Since $n \geq 1$ , the set ${t_{0}, \dots, t_{n - 1}}$ is a nonempty finite set of reals, so $t : = \min {t_{0}, \dots, t_{n - 1}}$ exists, equals some $t_{j}$ and is therefore $> 0$ , and satisfies $t \leq t_{k}$ for every $k < n$ .

step 1.3L3

2.2

Step 1.4 shows $B (y, s) \subseteq X ∖ \overset{ˉ}{B} (x, r)$ for the $y$ and $s$ chosen there, and $y$ was an arbitrary point of $X ∖ \overset{ˉ}{B} (x, r)$ ; hence $X ∖ \overset{ˉ}{B} (x, r)$ is open and $\overset{ˉ}{B} (x, r)$ is closed, which is claim 4.

step 1.4A1

3.1

By step 2.1, $B (y, t) \subseteq B (y, t_{k}) \subseteq U_{k}$ for every $k < n$ , so $B (y, t) \subseteq U_{0} \cap \dots \cap U_{n - 1}$ ; as $y$ was arbitrary that intersection is open, which is claim 3.

step 2.1A1A2

4.1

Claims 1, 2, 3 and 4 are established by steps 1.1, 1.2, 3.1 and 2.2 respectively.

step 1.1step 1.2step 2.2step 3.1∎

Remarks

Finiteness in claim 3 is essential and is exactly what step 2.1 uses. An infinite family of positive radii need have no positive lower bound, and the minimum of an infinite set of reals need not exist at all (Every nonempty finite set of reals has a maximum and a minimum is stated for finite sets for that reason). The intersection of the balls $B (x, 1 / n)$ over all $n \geq 1$ is a standard example of an intersection of open sets that need not be open.
The empty intersection is not covered and does not need to be. Claim 3 is stated for $n \geq 1$ ; the conventional value of an empty intersection is $X$ , which is open anyway (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).
Claim 4 is not the statement that $\overset{ˉ}{B} (x, r)$ is the closure of $B (x, r)$ , which is false in general (FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ ). All that is proved here is that the closed ball is a closed set.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Interior, closure, boundary, limit point, isolated point and dense subset of a metric space

Definition

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $A \subseteq X$ and let $x \in X$ . Balls are as in Open ball, closed ball and sphere in a metric space and open sets as in The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement; recall that a real $r$ written as a radius is always $> 0$ .

$x$ is an interior point of $A$ if $B (x, r) \subseteq A$ for some $r$ . The set of interior points is the interior $int (A)$ .
$x$ is an adherent point of $A$ if $B (x, r) \cap A \neq \emptyset$ for every $r$ . The set of adherent points is the closure $\overline{A}$ .
$x$ is a limit point (accumulation point) of $A$ if $B (x, r) \cap (A ∖ {x}) \neq \emptyset$ for every $r$ . The set of limit points is the derived set $A^{'}$ .
$x$ is an isolated point of $A$ if $x \in A$ and $B (x, r) \cap A = {x}$ for some $r$ .
The boundary of $A$ is $\partial A : = \overline{A} ∖ int (A)$ .
$A$ is dense in $X$ if $\overline{A} = X$ .

The interior is open, and it is the largest open subset of $A$ . If $x \in int (A)$ , fix $r$ with $B (x, r) \subseteq A$ ; the ball $B (x, r)$ is itself open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed), so every $y \in B (x, r)$ has some $s$ with $B (y, s) \subseteq B (x, r) \subseteq A$ , which puts $y$ in $int (A)$ . Hence $B (x, r) \subseteq int (A)$ and $int (A)$ is open. It is contained in $A$ , since $x \in B (x, r) \subseteq A$ for an interior point $x$ ; and if $V \subseteq A$ is open then every $v \in V$ has a ball inside $V \subseteq A$ , so $V \subseteq int (A)$ .

Two descriptions of the boundary agree. $x \in \partial A$ says that every ball around $x$ meets $A$ and that no ball around $x$ is contained in $A$ ; the second half says exactly that every ball around $x$ meets $X ∖ A$ . So

$\partial A = {x \in X : B (x, r) \cap A \neq \emptyset and B (x, r) \cap (X ∖ A) \neq \emptyset for every r},$

from which $\partial A = \partial (X ∖ A)$ is immediate.

Elementary containments, straight from the definitions. $A \subseteq \overline{A}$ , because $x \in A$ lies in every $B (x, r) \cap A$ ; $A^{'} \subseteq \overline{A}$ , because a ball meeting $A ∖ {x}$ meets $A$ ; and $int (A) \subseteq A \subseteq \overline{A}$ . A point of $A$ is either isolated in $A$ or a limit point of $A$ , and not both, according to whether some ball meets $A$ only in $x$ .

Remarks

The closure is defined here by adherent points and by nothing else. That it is closed, that it is the smallest closed set containing $A$ , that for nonempty $A$ it is ${x : d (x, A) = 0}$ , and that it consists of the limits of sequences from $A$ , are theorems (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed) and are proved from this definition.
Limit point of a set is not the same notion as subsequential limit of a sequence (Subsequential limit of a real sequence, and the subsequential limit set), which this library deliberately keeps under a different name: the constant sequence $x_{k} = 0$ has $0$ as a subsequential limit, while its set of values ${0}$ has no limit point at all.
Dense is relative to the ambient space, and the ambient space is part of the data: $A$ is dense in $X$ when $\overline{A} = X$ , with $\overline{A}$ computed in $(X, d)$ . The same $A$ inside a larger space is a different question.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $A \subseteq X$ , with closure, derived set and limit points as in Interior, closure, boundary, limit point, isolated point and dense subset of a metric space. Then:

If $A \neq \emptyset$ , then $\overline{A} = {x \in X : d (x, A) = 0}$ , where $d (x, A)$ is the distance from a point to a nonempty set (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).
$\overline{A} = A \cup A^{'}$ .
$\overline{A}$ is closed, contains $A$ , and is contained in every closed $F \subseteq X$ with $A \subseteq F$ . So $\overline{A}$ is the smallest closed superset of $A$ , and $A$ is closed if and only if $A = \overline{A}$ .

Claims 2 and 3 hold for every $A$ , the empty set included: $\overline{\emptyset}$ is empty because no ball meets $\emptyset$ , and $\emptyset$ is closed because $X$ is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement). Claim 1 carries the hypothesis $A \neq \emptyset$ because $d (x, A)$ is defined only for nonempty $A$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Facts & Assumptions

Given: A metric space $(X, d)$ , a subset $A \subseteq X$ , a point $x \in X$ , and a closed set $F \subseteq X$ with $A \subseteq F$ ; when $A \neq \emptyset$ , the set $E (x) : = {d (x, a) : a \in A}$ , whose infimum is $d (x, A)$ .

[A1]

Closure and derived set: $x \in \overline{A}$ means $B (x, r) \cap A \neq \emptyset$ for every $r > 0$ ; $x \in A^{'}$ means $B (x, r) \cap (A ∖ {x}) \neq \emptyset$ for every $r > 0$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[A2]

Open and closed: $U$ is open when every point of $U$ has a ball around it inside $U$ ; $F$ is closed when $X ∖ F$ is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L1]

For nonempty $A$ , the set $E (x)$ is nonempty and bounded below by $0$ , so $d (x, A) = \inf E (x)$ exists and is a lower bound of $E (x)$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)).

[L2]

Epsilon characterisation of the infimum: for a nonempty $S \subseteq R$ bounded below and a lower bound $ℓ$ of $S$ , one has $ℓ = \inf S$ if and only if for every $ε > 0$ there is $s \in S$ with $s < ℓ + ε$ (Epsilon characterisation of the infimum).

[L3]

Balls are open, so a point of a ball has a ball around it inside that ball (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Open ball, closed ball and sphere in a metric space).

[L4]

Membership in a ball: $a \in B (x, r)$ means $d (x, a) < r$ , and $x \in B (x, r)$ always (Open ball, closed ball and sphere in a metric space); trichotomy of the order of $R$ (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · direct

1.1

Suppose $A \neq \emptyset$ and $x \in \overline{A}$ , and let $ε > 0$ be arbitrary; then $B (x, ε) \cap A \neq \emptyset$ , so there is $a \in A$ with $d (x, a) < ε = 0 + ε$ , and $0$ is a lower bound of $E (x)$ , so $d (x, A) = 0$ by the epsilon characterisation.

A1L1L2L4

1.2

Conversely suppose $A \neq \emptyset$ and $d (x, A) = 0$ , and let $r > 0$ be arbitrary; the epsilon characterisation supplies $a \in A$ with $d (x, a) < 0 + r = r$ , that is $a \in B (x, r) \cap A$ , so $x \in \overline{A}$ .

A1L1L2L4

1.3

$A \subseteq \overline{A}$ and $A^{'} \subseteq \overline{A}$ : a point $a \in A$ lies in $B (a, r) \cap A$ for every $r$ , and a ball meeting $A ∖ {x}$ meets $A$ .

A1L4

1.4

If $x \in \overline{A}$ and $x \notin A$ , then for every $r$ the nonempty set $B (x, r) \cap A$ equals $B (x, r) \cap (A ∖ {x})$ , since $x$ is not a member of $A$ ; hence $x \in A^{'}$ .

1.5

$\overline{A}$ is closed: let $x \in X ∖ \overline{A}$ and fix $r$ with $B (x, r) \cap A = \emptyset$ ; for $y \in B (x, r)$ there is $s$ with $B (y, s) \subseteq B (x, r)$ , so $B (y, s) \cap A = \emptyset$ and $y \notin \overline{A}$ , whence $B (x, r) \subseteq X ∖ \overline{A}$ and $X ∖ \overline{A}$ is open.

A1A2L3

1.6

$\overline{A} \subseteq F$ for every closed $F \supseteq A$ : if $x \in \overline{A}$ had $x \notin F$ , then $X ∖ F$ open would give $r$ with $B (x, r) \subseteq X ∖ F \subseteq X ∖ A$ , so $B (x, r) \cap A = \emptyset$ , contradicting $x \in \overline{A}$ .

A1A2given

2.1

Claim 1 follows: by step 1.1 every adherent point of a nonempty $A$ satisfies $d (x, A) = 0$ , and by step 1.2 every $x$ with $d (x, A) = 0$ is adherent.

step 1.1step 1.2

2.2

Claim 2 follows: $A \cup A^{'} \subseteq \overline{A}$ by step 1.3, and $\overline{A} \subseteq A \cup A^{'}$ by step 1.4, since a point of $\overline{A}$ either lies in $A$ or, not lying in $A$ , lies in $A^{'}$ .

step 1.3step 1.4

2.3

Claim 3 follows: $\overline{A}$ is closed by step 1.5, contains $A$ by step 1.3, and sits inside every closed superset of $A$ by step 1.6; in particular if $A$ is closed then $\overline{A} \subseteq A \subseteq \overline{A}$ , so $A = \overline{A}$ , and conversely if $A = \overline{A}$ then $A$ is closed.

step 1.3step 1.5step 1.6

3.1

Claims 1, 2 and 3 are therefore all established.

step 2.1step 2.2step 2.3∎

Remarks

Claim 1 is where the infimum does the work. Reading it right to left, $d (x, A) = 0$ says that $A$ has points arbitrarily close to $x$ without saying that any of them is $x$ ; reading it left to right, adherence says the same thing in the language of balls. The equivalence is exactly the epsilon characterisation of the infimum (Epsilon characterisation of the infimum) with the lower bound $0$ .
The distance function is $1$ -Lipschitz ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz), so claim 1 exhibits $\overline{A}$ as the zero set of a function that does not increase distances. That is not used above and is recorded only as orientation.
Claim 3 is the form that transfers to general topology, where no metric is available and the closure is defined outright as the intersection of all closed supersets. Claim 1 is the specifically metric statement, and claim 2 sits between them.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$

Definition

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

A sequence in $X$ is a function $x : N \to X$ , written $(x_{k})$ with $x_{k} : = x (k)$ . As everywhere in this library, $N$ contains $0$ (The natural numbers $N$ (von Neumann)) and a sequence is indexed from $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences); an index range copied from a text that starts at $1$ must be shifted before it is used here.

Let $(x_{k})$ be a sequence in $X$ and $p \in X$ . The function $k \mapsto d (x_{k}, p)$ is a sequence of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), and it is nonnegative (Nonnegativity of a metric is a consequence of the other axioms, not an axiom), so $∣ d (x_{k}, p) ∣ = d (x_{k}, p)$ (Absolute value in an ordered field). Define

$x_{k} ⟶ p in (X, d) : ⟺ d (x_{k}, p) ⟶ 0 in R,$

the convergence on the right being that of Limits and Cauchy sequences of reals. Unwound, this says: for every rational $ε > 0$ there is $K \in N$ with $d (x_{k}, p) < ε$ for every $k \geq K$ . We then call $p$ a limit of $(x_{k})$ , and say $(x_{k})$ converges in $(X, d)$ if it has a limit.

Rational and real $ε$ agree here, as they do on the real line. Limits and Cauchy sequences of reals tests convergence against rational $ε$ only, and its own remark, restated for sequences in Sequences of reals: bounded, eventually, frequently, tails, subsequences, records that nothing is lost: below any real $η > 0$ lies a positive rational (The rationals embed densely in the reals), and the index belonging to that rational serves for $η$ . So a proof may establish convergence by producing an index for every real $ε > 0$ , and may use a convergence hypothesis at a real $ε$ by first passing to a rational below it. Both moves are used on this page and are always cited.

Subsequences and subsequential limits. A subsequence of $(x_{k})$ is the composite $x \circ n$ for a strictly increasing $n : N \to N$ , written $(x_{n_{j}})$ , exactly as for sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences); and $p$ is a subsequential limit of $(x_{k})$ in $(X, d)$ when some subsequence converges to $p$ , which is the metric-space form of Subsequential limit of a real sequence, and the subsequential limit set.

Remarks

A limit is a point of $X$ , and uniqueness is a theorem. Nothing in the definition rules out two limits; that a sequence in a metric space has at most one is A sequence in a metric space has at most one limit, and its proof is where the separation axiom (M1) is spent. Reading the same definition with a pseudometric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) that is not a metric, that is one with $p (a, b) = 0$ for some $a \neq b$ , limits are genuinely not unique: the constant sequence at $a$ converges to $b$ as well.
Convergence is defined from the metric but determined by the topology. It can be restated as "every open set containing $p$ contains $x_{k}$ for all large $k$ ", which follows from The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable; so it is unchanged by passing to a topologically equivalent metric (Topologically, uniformly and Lipschitz equivalent metrics on a set). That restatement is not made part of the definition, because the metric form is what every proof on this page uses.
The relation between convergence and closure is A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed: a point lies in the closure of $A$ exactly when some sequence in $A$ converges to it.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

A sequence in a metric space has at most one limit

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $(x_{k})$ be a sequence in $X$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ). If $x_{k} \to p$ and $x_{k} \to q$ , then $p = q$ .

So a convergent sequence in a metric space has exactly one limit, and the notation $\lim_{k} x_{k}$ is unambiguous.

Facts & Assumptions

Given: A metric space $(X, d)$ , a sequence $(x_{k})$ in $X$ , and points $p, q \in X$ with $x_{k} \to p$ and $x_{k} \to q$ .

[A1]

Convergence: $x_{k} \to p$ means that for every rational $ε > 0$ there is $K \in N$ with $d (x_{k}, p) < ε$ for all $k \geq K$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Limits and Cauchy sequences of reals); and $d (x, y) \geq 0$ for all $x, y \in X$ , so in particular $d (x_{k}, p) \geq 0$ and its absolute value is itself (Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Absolute value in an ordered field, Basic properties of the absolute value).

[L1]

Density of the rationals: strictly between any two reals lies a rational, so below any real $η > 0$ there is a rational $ε$ with $0 < ε < η$ (The rationals embed densely in the reals).

[L2]

Halving. For a real $c > 0$ set $2 : = 1 + 1$ and $c / 2 : = c \cdot 2^{- 1}$ . Then $2 > 0$ , so $2 \neq 0$ and $2^{- 1} > 0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Inverses of positives are positive, and reciprocation reverses order, Ordered field); hence $c / 2 > 0$ (Sign rules for products and monotonicity of multiplication); and $c / 2 + c / 2 = c (2^{- 1} + 2^{- 1}) = c (2 \cdot 2^{- 1}) = c$ (Field).

[L3]

Separation (M1) and the triangle inequality (M3) of $d$ , together with symmetry (M2) (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L4]

Trichotomy of the order of $R$ , and transitivity: $a < b$ and $b \leq a$ cannot both hold (Complete ordered field (least-upper-bound property), Ordered field).

[L5]

Adding two inequalities: $a < b$ and $c < d$ give $a + c < b + d$ (Order is preserved by adding a constant and by adding inequalities).

Proof

technique · contradiction

1.1

Suppose, for contradiction, that $p \neq q$ .

assume-contra

2.1

By (M1) $d (p, q) \neq 0$ , and $d (p, q) \geq 0$ , so $c : = d (p, q) > 0$ by trichotomy; put $η : = c / 2$ , a positive real with $η + η = c$ .

step 1.1A1L2L3L4

3.1

Fix a rational $ε$ with $0 < ε < η$ , and use the convergence hypotheses at $ε$ to fix $K_{1}, K_{2} \in N$ with $d (x_{k}, p) < ε$ for $k \geq K_{1}$ and $d (x_{k}, q) < ε$ for $k \geq K_{2}$ .

step 2.1A1L1choose

4.1

Let $m$ be any natural with $m \geq K_{1}$ and $m \geq K_{2}$ , for instance $m : = K_{1} + K_{2}$ ; then $d (x_{m}, p) < ε$ and $d (x_{m}, q) < ε$ .

step 3.1choose

5.1

By symmetry and the triangle inequality, $c = d (p, q) \leq d (p, x_{m}) + d (x_{m}, q) = d (x_{m}, p) + d (x_{m}, q) < ε + ε < η + η = c$ .

step 4.1step 2.1L3L5

6.1

Step 5.1 asserts $c < c$ , which trichotomy forbids; the supposition of step 1.1 is therefore untenable and $p = q$ .

step 5.1L4discharge-contradiction∎

Remarks

Where each axiom is spent. Separation (M1) is what turns $p \neq q$ into $d (p, q) > 0$ , and it is the only axiom that distinguishes a metric from a pseudometric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric). In a pseudometric space with $d (p, q) = 0$ and $p \neq q$ , the constant sequence $x_{k} = p$ converges to both, so the lemma is false there and this is exactly the step that fails.
The same argument proves more, namely that a metric space is Hausdorff: the balls $B (p, c / 2)$ and $B (q, c / 2)$ are disjoint. That is recorded separately as Distinct points of a metric space have disjoint balls around them, and uniqueness of limits follows from it as well.
Instantiating at a rational is not cosmetic. Limits and Cauchy sequences of reals quantifies over rational $ε$ , so a convergence hypothesis may only be applied at a rational; step 3.1 passes from the real $η$ to a rational below it using The rationals embed densely in the reals, which is the sanctioned move.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $x \in X$ . For a natural $n \geq 1$ write $1 / n$ for the inverse of the canonical natural $n \cdot 1_{R}$ , a positive real, and put

$β_{n} : = B (x, 1 / n), B_{x} : = {β_{n} : n \in N, n \geq 1} .$

Then:

$B_{x}$ is at most countable (Finite, countably infinite, countable, uncountable).
Every $β_{n}$ is an open subset of $X$ containing $x$ .
For every open $U \subseteq X$ with $x \in U$ there is $n \geq 1$ with $β_{n} \subseteq U$ .

The two names used in the title are introduced by this statement, not cited from elsewhere. A family of open sets each containing $x$ , such that every open set containing $x$ contains a member of the family, is a neighbourhood base at $x$ ; a space in which every point has an at most countable neighbourhood base is first countable. Claims 1 to 3 say that $B_{x}$ is an at most countable neighbourhood base at $x$ , so every metric space is first countable.

Facts & Assumptions

Given: A metric space $(X, d)$ , a point $x \in X$ , and for each natural $n \geq 1$ the ball $β_{n} = B (x, 1 / n)$ .

[L1]

Canonical naturals: $n \cdot 1_{R} > 0$ for $n \geq 1$ (Canonical naturals are positive and strictly increasing), hence $n \cdot 1_{R}$ is invertible with $1 / n > 0$ (Inverses of positives are positive, and reciprocation reverses order); and $N$ contains $0$ , so $j \mapsto j + 1$ runs over exactly the naturals $\geq 1$ as $j$ runs over $N$ (The natural numbers $N$ (von Neumann)).

[L2]

Balls are open and every ball contains its centre (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Open ball, closed ball and sphere in a metric space); and $B (x, s) \subseteq B (x, t)$ whenever $0 < s \leq t$ (Open ball, closed ball and sphere in a metric space).

[L3]

Open sets: $U$ is open when every point of $U$ has a ball around it inside $U$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L4]

Reciprocal Archimedean property: for every real $r > 0$ there is a natural $n \geq 1$ with $1 / n < r$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L5]

Countability: a nonempty set admitting a surjection from $N$ is at most countable (A nonempty set is at most countable iff it is a surjective image of $N$ , Finite, countably infinite, countable, uncountable, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , Injection, surjection, bijection).

Proof

technique · direct

1.1

For every natural $n \geq 1$ the real $1 / n$ is defined and positive, so $β_{n}$ is a legitimate ball of positive radius.

1.2

Let $U$ be open with $x \in U$ , and fix a real $r > 0$ with $B (x, r) \subseteq U$ ; then fix a natural $n \geq 1$ with $1 / n < r$ .

L3L4choose

2.1

Each $β_{n}$ is open and contains $x$ , which is claim 2.

step 1.1L2

2.2

The map $s : N \to B_{x}$ given by $s (j) : = β_{j + 1}$ is well defined by step 1.1 and is surjective, because every member of $B_{x}$ is $β_{n}$ for some $n \geq 1$ and $n = j + 1$ for the natural $j$ with $j + 1 = n$ ; moreover $B_{x}$ is nonempty, containing $β_{1}$ .

step 1.1L1

2.3

By step 1.2 and monotonicity of balls in the radius, $β_{n} = B (x, 1 / n) \subseteq B (x, r) \subseteq U$ , which is claim 3.

step 1.2L2

3.1

By [L5] applied to the surjection of step 2.2, the nonempty set $B_{x}$ is at most countable, which is claim 1.

step 2.2L5

4.1

Claims 1, 2 and 3 hold by steps 3.1, 2.1 and 2.3, so $B_{x}$ is an at most countable neighbourhood base at $x$ and $(X, d)$ is first countable.

step 2.1step 2.3step 3.1∎

Remarks

The family can be finite, and that is not a defect. In a discrete metric space $1 / n \leq 1$ for every $n \geq 1$ , so every $β_{n}$ is the single point ${x}$ and $B_{x}$ is a one-element family. "At most countable" in this library includes finite (Finite, countably infinite, countable, uncountable), which is exactly why claim 1 is stated in that form and not as "countably infinite".
Only the reciprocal form of the Archimedean property is used, in step 1.2, and it is the form recorded as For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ precisely so that the inversion never has to be redone inside a proof.
This is what makes sequences sufficient in metric spaces. First countability is the hypothesis under which closure can be described by sequences rather than by nets, and it is used in exactly that way by A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $A \subseteq X$ , let $x \in X$ and let $F \subseteq X$ . Call $F$ sequentially closed when every sequence in $F$ that converges in $X$ has its limit in $F$ . Then:

$x \in \overline{A}$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space) if and only if there is a sequence $(a_{k})$ with $a_{k} \in A$ for every $k$ and $a_{k} \to x$ in $(X, d)$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).
$F$ is closed (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) if and only if $F$ is sequentially closed.

The Axiom of Countable Choice is used, once. The direction of claim 1 that manufactures a sequence out of adherence makes one choice per natural number, and that is exactly ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The converse direction, and the direction of claim 2 that goes from closed to sequentially closed, are choice free. This is flagged at the step that spends it.

Facts & Assumptions

Given: A metric space $(X, d)$ , a subset $A \subseteq X$ , a point $x \in X$ , and a subset $F \subseteq X$ ; for $n \in N$ write $A_{n} : = B (x, 1 / (n + 1)) \cap A$ .

[A1]

Closure: $x \in \overline{A}$ means $B (x, r) \cap A \neq \emptyset$ for every real $r > 0$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, Open ball, closed ball and sphere in a metric space).

[A2]

Convergence in $(X, d)$ : $a_{k} \to x$ means that for every rational $ε > 0$ there is $K$ with $d (a_{k}, x) < ε$ for all $k \geq K$ , and it is enough to produce such a $K$ for every REAL $ε > 0$ , since below any positive real lies a positive rational (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Limits and Cauchy sequences of reals, The rationals embed densely in the reals, Nonnegativity of a metric is a consequence of the other axioms, not an axiom); and $d (u, v) = d (v, u)$ for all $u, v \in X$ , which is the symmetry axiom (M2) (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

The balls $B (x, 1 / n)$ , $n \geq 1$ , are open, contain $x$ , and form a neighbourhood base at $x$ : every open $U ∋ x$ contains one of them (The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable).

[L2]

Balls are open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed), and $B (x, s) \subseteq B (x, t)$ for $0 < s \leq t$ (Open ball, closed ball and sphere in a metric space).

[L3]

Canonical naturals and reciprocals: for naturals $1 \leq m \leq p$ one has $0 < m \cdot 1_{R} \leq p \cdot 1_{R}$ and hence $0 < 1 / p \leq 1 / m$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order); and $N$ contains $0$ , so $n + 1 \geq 1$ for every $n \in N$ (The natural numbers $N$ (von Neumann)).

[L4]

Countable choice: for a family $(A_{n})_{n \in N}$ of nonempty sets there is a function $n \mapsto a_{n}$ with $a_{n} \in A_{n}$ for every $n$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L5]

The closure is the smallest closed superset, so $F$ is closed if and only if $F = \overline{F}$ (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

Proof

technique · direct

1.1

Suppose $(a_{k})$ is a sequence with $a_{k} \in A$ for every $k$ and $a_{k} \to x$ , and let $r > 0$ be an arbitrary real; then there is $K$ with $d (a_{k}, x) < r$ for all $k \geq K$ , so $d (x, a_{K}) = d (a_{K}, x) < r$ by the symmetry axiom (M2) of [A2] and hence $a_{K} \in B (x, r) \cap A$ , and since $r$ was arbitrary $x \in \overline{A}$ .

A1A2

1.2

Suppose $x \in \overline{A}$ ; then for every $n \in N$ the radius $1 / (n + 1)$ is a positive real and $A_{n} = B (x, 1 / (n + 1)) \cap A$ is nonempty, so countable choice supplies a sequence $(a_{n})$ with $a_{n} \in A_{n} \subseteq A$ for every $n$ .

A1L3L4choose

2.1

That sequence converges to $x$ : given a real $ε > 0$ , the ball $B (x, ε)$ is open and contains $x$ , so there is a natural $N \geq 1$ with $B (x, 1 / N) \subseteq B (x, ε)$ ; for every $n \geq N$ we have $n + 1 \geq N$ , hence $1 / (n + 1) \leq 1 / N$ and $a_{n} \in B (x, 1 / (n + 1)) \subseteq B (x, 1 / N) \subseteq B (x, ε)$ , that is $d (x, a_{n}) < ε$ .

step 1.2A2L1L2L3

2.2

If $F$ is closed and $(a_{k})$ is a sequence in $F$ converging to some $x \in X$ , then $x \in \overline{F}$ by step 1.1 applied with $A = F$ , and $\overline{F} = F$ because $F$ is closed; so $x \in F$ and $F$ is sequentially closed.

step 1.1L5

3.1

Claim 1 holds: step 1.1 gives the implication from a convergent sequence in $A$ to adherence, and steps 1.2 and 2.1 give the converse by producing such a sequence.

step 1.1step 1.2step 2.1

4.1

If $F$ is sequentially closed, let $x \in \overline{F}$ ; by claim 1 there is a sequence in $F$ converging to $x$ , so $x \in F$ , whence $\overline{F} \subseteq F$ ; the reverse inclusion always holds, so $F = \overline{F}$ and $F$ is closed.

step 3.1L5

5.1

Claim 2 holds by steps 2.2 and 4.1, and claim 1 by step 3.1.

step 2.2step 3.1step 4.1∎

Remarks

Where first countability enters. Step 2.1 is the only place, and it uses The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable to convert an arbitrary ball around $x$ into one of the countably many balls $B (x, 1 / N)$ . Nothing here should be read as saying that sequences describe the closure in a general topological space; the tool that always works there is the net, and that is a later page.
The indexing is from $0$ . The radii used are $1 / (n + 1)$ for $n \in N$ , not $1 / n$ , precisely because $N$ contains $0$ here (The natural numbers $N$ (von Neumann), Sequences of reals: bounded, eventually, frequently, tails, subsequences) and $1 / 0$ does not exist. A version of this proof copied from a text that indexes sequences from $1$ has to be reindexed, and this is the reindexing.
The use of choice is not concealed. It enters at exactly one place, namely step 1.2, which makes one selection per natural number, and that is what ${A C}_{ω}$ licenses (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Whether some proof in ZF alone reaches the same conclusion for arbitrary metric spaces is a question this library does not settle; what it does is record the assumption at the step that spends it.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Distinct points of a metric space have disjoint balls around them

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $p, q \in X$ with $p \neq q$ . Put $r : = d (p, q) / 2$ . Then $r > 0$ and

$B (p, r) \cap B (q, r) = \emptyset .$

Both sets are open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed) and contain $p$ respectively $q$ (Open ball, closed ball and sphere in a metric space), so every metric space is Hausdorff: distinct points are separated by disjoint open sets (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Facts & Assumptions

Given: A metric space $(X, d)$ and points $p, q \in X$ with $p \neq q$ ; write $c : = d (p, q)$ .

[L1]

Separation (M1), symmetry (M2) and the triangle inequality (M3) of a metric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); and $d \geq 0$ (Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L2]

Halving. For a real $c > 0$ put $2 : = 1 + 1$ and $c / 2 : = c \cdot 2^{- 1}$ . Then $2 > 0$ , so $2 \neq 0$ and $2^{- 1} > 0$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Inverses of positives are positive, and reciprocation reverses order, Ordered field); hence $c / 2 > 0$ (Sign rules for products and monotonicity of multiplication); and $c / 2 + c / 2 = c (2 \cdot 2^{- 1}) = c$ (Field).

[L3]

Adding two strict inequalities: $a < b$ and $a^{'} < b^{'}$ give $a + a^{'} < b + b^{'}$ (Order is preserved by adding a constant and by adding inequalities).

[L4]

Trichotomy of the order of $R$ : $a < a$ is impossible, and $a \neq 0$ together with $a \geq 0$ gives $a > 0$ (Complete ordered field (least-upper-bound property), Ordered field).

[L5]

Membership in a ball: $z \in B (u, t)$ means $d (u, z) < t$ ; balls are open and contain their centres (Open ball, closed ball and sphere in a metric space, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

Proof

technique · direct

1.1

Since $p \neq q$ , axiom (M1) gives $c = d (p, q) \neq 0$ , and $c \geq 0$ , so $c > 0$ by trichotomy; hence $r : = c / 2$ is a positive real with $r + r = c$ .

givenL1L2L4

2.1

Suppose some $z \in X$ lay in both $B (p, r)$ and $B (q, r)$ , that is $d (p, z) < r$ and $d (q, z) < r$ ; then symmetry and the triangle inequality give $c = d (p, q) \leq d (p, z) + d (z, q) = d (p, z) + d (q, z) < r + r = c$ , so $c < c$ , which trichotomy forbids.

step 1.1L1L3L4L5

3.1

No such $z$ exists, so $B (p, r) \cap B (q, r) = \emptyset$ ; both sets are open and contain $p$ respectively $q$ , so distinct points of $(X, d)$ are separated by disjoint open sets.

step 1.1step 2.1L5∎

Remarks

This is a strengthening of uniqueness of limits. A sequence converging to two distinct points would eventually be inside both $B (p, r)$ and $B (q, r)$ , which the theorem forbids; that is a second route to A sequence in a metric space has at most one limit, and the two proofs use the same halving.
Separation (M1) is what the proof spends. A pseudometric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) with $d (p, q) = 0$ for distinct $p, q$ gives $r = 0$ at step 1.1 and there is no ball to speak of; such a space is not Hausdorff, and no argument repairs that.
The radius $d (p, q) / 2$ is not the only choice, and it is not optimal in every space: in an ultrametric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) the balls $B (p, c)$ and $B (q, c)$ are already disjoint, because a common point would force $d (p, q) \leq \max {d (p, z), d (z, q)} < c$ .

DefinitionDefinition: AI-adaptedProof: Not applicableverified 2026-09-10 (Codex)Open item page →

Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form

Definition

Let $(X, d_{X})$ and $(Y, d_{Y})$ be metric spaces (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), let $f : X \to Y$ be a function and let $a \in X$ .

$f$ is continuous at $a$ if for every real $ε > 0$ there is a real $δ > 0$ such that

$d_{X} (x, a) < δ ⟹ d_{Y} (f (x), f (a)) < ε for all x \in X .$

$f$ is continuous (globally, or on $X$ ) if it is continuous at every point of $X$ .

The same condition in balls. Since $d_{X} (x, a) < δ$ says $x \in B_{X} (a, δ)$ and $d_{Y} (f (x), f (a)) < ε$ says $f (x) \in B_{Y} (f (a), ε)$ (Open ball, closed ball and sphere in a metric space), continuity at $a$ reads: for every $ε > 0$ there is $δ > 0$ with

$f [B_{X} (a, δ)] \subseteq B_{Y} (f (a), ε) .$

Both forms are used below and are the same statement written twice.

Both metrics matter, and both are named. Continuity is a property of the triple $(d_{X}, d_{Y}, f)$ , not of $f$ alone. When several metrics on the same underlying sets are in play, as in Topologically, uniformly and Lipschitz equivalent metrics on a set, the metrics are always written out.

Quantifier order. The $δ$ is allowed to depend on $ε$ and on the point $a$ . Requiring one $δ$ to work at every point simultaneously is a strictly stronger condition, uniform continuity; it is defined on a later page of this library, and at this point in the reading order it is written out in full where needed (Topologically, uniformly and Lipschitz equivalent metrics on a set).

Remarks

Nothing is claimed here beyond the definition. Without any choice principle, continuity is equivalent to preimages of open sets being open, to preimages of closed sets being closed, and to $f [\overline{A}] \subseteq \overline{f [A]}$ ; each of these conditions implies sequential continuity. Assuming Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), sequential continuity is equivalent to them as well. These statements are proved in Metric continuity characterisations, with countable choice for the sequential converse.
Continuity at a point is a local condition: it depends only on the values of $f$ on any one ball around $a$ , since the condition may always be tested with a smaller $δ$ .
Every isometric embedding is continuous, with $δ = ε$ (Isometry, isometric embedding, and the subspace metric on a subset, An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image), and so is every map that does not increase distances, such as $u \mapsto d (u, A)$ for a fixed nonempty subset $A \subseteq X$ ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz).

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)Open item page →

Metric continuity characterisations, with countable choice for the sequential converse

Statement

Let $(X, d_{X})$ and $(Y, d_{Y})$ be metric spaces (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $f : X \to Y$ be a function, with images and preimages written $f [\cdot]$ and $f^{- 1} [\cdot]$ (Injection, surjection, bijection). Conditions (a), (b), (c), and (e) below are equivalent without choice, and each implies (d). Assuming Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), all five are equivalent. The authorized Axiom of Choice (The Axiom of Choice) suffices; only its countable instance is used for the converse from (d).

(a) $f$ is continuous at every point of $X$ in the $ε$ - $δ$ sense (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).
(b) $f^{- 1} [V]$ is open in $X$ for every open $V \subseteq Y$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).
(c) $f^{- 1} [G]$ is closed in $X$ for every closed $G \subseteq Y$ .
(d) $f$ is sequentially continuous: whenever $x_{k} \to x$ in $(X, d_{X})$ , also $f (x_{k}) \to f (x)$ in $(Y, d_{Y})$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).
(e) $f [\overline{A}] \subseteq \overline{f [A]}$ for every $A \subseteq X$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

Where choice is used. Only the implication (d) $\Rightarrow$ (e) uses a choice principle, and it uses it only through A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, whose forward direction spends the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The cycle (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (e) $\Rightarrow$ (a) and the implication (a) $\Rightarrow$ (d) are choice free.

Facts & Assumptions

Given: Metric spaces $(X, d_{X})$ , $(Y, d_{Y})$ and a function $f : X \to Y$ ; a point $a \in X$ , a real $ε > 0$ , subsets $A \subseteq X$ , $V \subseteq Y$ open and $G \subseteq Y$ closed, and a sequence $(x_{k})$ in $X$ . Assume Countable Choice for (d) implies (e), and hence for the five-way equivalence; the other stated implications require no choice.

[A1]

Continuity at $a$ : for every real $ε > 0$ there is $δ > 0$ with $f [B_{X} (a, δ)] \subseteq B_{Y} (f (a), ε)$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Open ball, closed ball and sphere in a metric space).

[A2]

Open and closed: $U$ is open when every point of $U$ has a ball around it inside $U$ ; $G$ is closed when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L1]

Preimages respect complements: $f^{- 1} [Y ∖ G] = X ∖ f^{- 1} [G]$ , since $f (x) \in Y ∖ G$ holds exactly when $f (x) \notin G$ (Injection, surjection, bijection).

[L2]

Closure: $\overline{A}$ consists of the points every ball around which meets $A$ ; it is closed, contains $A$ , and is contained in every closed superset of $A$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[L3]

Under Countable Choice, $x \in \overline{A}$ if and only if some sequence in $A$ converges to $x$ ; the direction producing the sequence uses countable choice, while the converse is choice free (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, The Axiom of Countable Choice ( ${A C}_{ω}$ )). AC supplies the needed indexed choices: choose from the family of nonempty sets and compose that choice function with the indexing map (The Axiom of Choice).

[L4]

Convergence: $x_{k} \to x$ means that for every rational $ε > 0$ there is $K$ with $d_{X} (x_{k}, x) < ε$ for $k \geq K$ , and producing such a $K$ for every REAL $ε > 0$ is equivalent, since below any positive real lies a positive rational (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Limits and Cauchy sequences of reals, The rationals embed densely in the reals).

[L5]

Balls are open and contain their centres (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Open ball, closed ball and sphere in a metric space); and trichotomy of the order of $R$ , so the negation of $t < ε$ is $t \geq ε$ (Complete ordered field (least-upper-bound property), Ordered field).

Proof

technique · direct

1.1

(a) implies (b): let $V \subseteq Y$ be open and $x \in f^{- 1} [V]$ ; since $f (x) \in V$ there is $ε > 0$ with $B_{Y} (f (x), ε) \subseteq V$ , and continuity at $x$ supplies $δ > 0$ with $f [B_{X} (x, δ)] \subseteq B_{Y} (f (x), ε) \subseteq V$ , that is $B_{X} (x, δ) \subseteq f^{- 1} [V]$ ; as $x$ was arbitrary, $f^{- 1} [V]$ is open.

A1A2

1.2

(b) implies (c): let $G \subseteq Y$ be closed; then $Y ∖ G$ is open, so $f^{- 1} [Y ∖ G]$ is open by (b), and that set is $X ∖ f^{- 1} [G]$ , so $f^{- 1} [G]$ is closed.

A2L1

1.3

(c) implies (e): let $A \subseteq X$ ; the set $\overline{f [A]}$ is closed in $Y$ , so $G_{0} : = f^{- 1} [\overline{f [A]}]$ is closed in $X$ by (c), and $A \subseteq G_{0}$ because $f [A] \subseteq \overline{f [A]}$ ; hence $\overline{A} \subseteq G_{0}$ by minimality of the closure, which says exactly $f [\overline{A}] \subseteq \overline{f [A]}$ .

1.4

(e) implies (a): fix $a \in X$ and a real $ε > 0$ , put $A_{ε} : = {x \in X : d_{Y} (f (x), f (a)) \geq ε}$ , and suppose no $δ > 0$ satisfies the continuity condition at $a$ for this $ε$ , that is every ball $B_{X} (a, δ)$ contains a point of $A_{ε}$ ; then $a \in \overline{A_{ε}}$ , so (e) gives $f (a) \in f [\overline{A_{ε}}] \subseteq \overline{f [A_{ε}]}$ , so the ball $B_{Y} (f (a), ε)$ meets $f [A_{ε}]$ and there is $x \in A_{ε}$ with $d_{Y} (f (x), f (a)) < ε$ , contradicting the definition of $A_{ε}$ ; hence some $δ > 0$ works, and since $a$ and $ε$ were arbitrary $f$ is continuous everywhere.

assume-hypA1L2L5

1.5

(a) implies (d): let $x_{k} \to x$ and let a real $ε > 0$ be given; continuity at $x$ supplies $δ > 0$ with $f [B_{X} (x, δ)] \subseteq B_{Y} (f (x), ε)$ , and convergence supplies $K$ with $d_{X} (x_{k}, x) < δ$ , that is $x_{k} \in B_{X} (x, δ)$ , for all $k \geq K$ ; then $d_{Y} (f (x_{k}), f (x)) < ε$ for all $k \geq K$ , so $f (x_{k}) \to f (x)$ .

A1L4L5

1.6

Assume Countable Choice. For (d) implies (e), let $A \subseteq X$ and $y \in f [\overline{A}]$ , say $y = f (x)$ with $x \in \overline{A}$ . Apply [L3] using Countable Choice (the countable instance of AC) to select a sequence $(a_{k})$ in $A$ converging to $x$ . This is the only use of choice in this proof. By (d), $f (a_{k}) \to f (x)$ ; since $f (a_{k}) \in f [A]$ , the choice-free converse of [L3] gives $f (x) \in \overline{f [A]}$ .

givenL3

2.1

Steps 1.1–1.4 give the choice-free equivalence of (a), (b), (c), and (e); step 1.5 shows each implies (d) without choice. Under Countable Choice, step 1.6 closes the converse and all five conditions are equivalent.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6∎

Remarks

(b) is the definition of continuity in general topology, and the theorem is what makes the metric $ε$ - $δ$ definition agree with it. Once (b) is available, continuity can be discussed without ever mentioning a metric, which is what the later topology pages do.
The sequential converse uses first countability and countable choice. The proof uses that metric spaces are first countable (The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable), which is what A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed rests on. Nothing above should be read as saying that sequential continuity always suffices.
Preimages, not images. Nothing here says that $f [U]$ is open for open $U$ ; that is openness of the map, a different condition, which continuity does not imply: a constant map is continuous and its image of any nonempty open set is a single point. The image condition that does hold is the closure inclusion (e), and even that is an inclusion and not an equality.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Isometry, isometric embedding, and the subspace metric on a subset

Definition

Let $(X, d_{X})$ and $(Y, d_{Y})$ be metric spaces (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

Isometric embedding and isometry. A function $f : X \to Y$ is an isometric embedding if

$d_{Y} (f (x), f (x^{'})) = d_{X} (x, x^{'}) for all x, x^{'} \in X,$

and an isometry if it is in addition bijective (Injection, surjection, bijection). Two metric spaces are isometric if some isometry between them exists.

Subspace metric. Let $A \subseteq X$ and let

$d_{A} : = d_{X} ↾ (A \times A)$

be the restriction of $d_{X}$ to pairs from $A$ . Then $d_{A}$ is a metric on $A$ : the three axioms (M1), (M2), (M3) of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric are conditions on triples of points, and each holds for points of $A$ because it holds for points of $X$ . The pair $(A, d_{A})$ is the metric subspace $A$ of $X$ , and the inclusion $A \to X$ is an isometric embedding by construction. The metric topology of $d_{A}$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) is the subspace topology of $A$ .

Balls of a subspace are traces of balls of the ambient space. For $a \in A$ and $r > 0$ ,

$B_{A} (a, r) = B_{X} (a, r) \cap A,$

directly from the definitions: a point $z$ lies in the left side exactly when $z \in A$ and $d_{A} (a, z) = d_{X} (a, z) < r$ (Open ball, closed ball and sphere in a metric space). This is why the ambient space is always written into the ball notation, and it is the source of every apparent paradox about balls in subspaces.

Remarks

An isometric embedding is automatically injective, and it identifies $X$ with the subspace $f [X]$ of $Y$ , topology and all; that is An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image. The word embedding is therefore justified rather than merely suggestive.
A bijective isometric embedding has an isometric inverse. If $f$ is an isometry then $f^{- 1}$ satisfies $d_{X} (f^{- 1} (y), f^{- 1} (y^{'})) = d_{Y} (y, y^{'})$ , because writing $y = f (x)$ and $y^{'} = f (x^{'})$ turns that into the defining identity of $f$ . So "isometric" is a symmetric relation between metric spaces, and it is transitive because a composite of isometries is one.
Isometry is much finer than having the same topology. Isometric spaces are homeomorphic, but $R$ with $d (x, y) = ∣ x - y ∣$ and $R$ with $\min {∣ x - y ∣, 1}$ have the same topology and are not isometric, the second being bounded and the first not ( $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded).

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image

Statement

Let $(X, d_{X})$ and $(Y, d_{Y})$ be metric spaces and let $f : X \to Y$ be an isometric embedding (Isometry, isometric embedding, and the subspace metric on a subset). Write $Z : = f [X] \subseteq Y$ with its subspace metric $d_{Z}$ . Then:

$f$ is injective (Injection, surjection, bijection).
$f$ , viewed as a map $X \to Z$ , is an isometry.
$f [B_{X} (x, r)] = B_{Z} (f (x), r)$ for every $x \in X$ and $r > 0$ (Open ball, closed ball and sphere in a metric space).
A subset $U \subseteq X$ is open in $(X, d_{X})$ if and only if $f [U]$ is open in $(Z, d_{Z})$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement). So $U \mapsto f [U]$ is a bijection from the metric topology of $X$ onto the subspace topology of $f [X]$ , and $f$ is a homeomorphism onto its image.

Facts & Assumptions

Given: Metric spaces $(X, d_{X})$ , $(Y, d_{Y})$ , an isometric embedding $f : X \to Y$ , the image $Z : = f [X]$ with the subspace metric $d_{Z} = d_{Y} ↾ (Z \times Z)$ , and the map $g : Z \to X$ inverse to $f : X \to Z$ once claim 2 is available.

[A1]

Isometric embedding: $d_{Y} (f (x), f (x^{'})) = d_{X} (x, x^{'})$ for all $x, x^{'} \in X$ ; the subspace metric on $Z$ is the restriction of $d_{Y}$ (Isometry, isometric embedding, and the subspace metric on a subset).

[A2]

Separation (M1): $d_{X} (x, x^{'}) = 0$ if and only if $x = x^{'}$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

Balls: $B_{X} (x, r) = {x^{'} : d_{X} (x, x^{'}) < r}$ , and likewise in $Z$ with $d_{Z}$ (Open ball, closed ball and sphere in a metric space).

[L2]

Continuity in the $ε$ - $δ$ form (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form), and the equivalence of continuity with "preimages of open sets are open" (Metric continuity characterisations, with countable choice for the sequential converse, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L3]

A bijection $h$ and its inverse satisfy $h [S] = (h^{- 1})^{- 1} [S]$ and $h^{- 1} [h [S]] = S$ for every subset $S$ of the domain (Injection, surjection, bijection).

Proof

technique · direct

1.1

Injectivity: if $f (x) = f (x^{'})$ then $d_{X} (x, x^{'}) = d_{Y} (f (x), f (x^{'})) = 0$ , hence $x = x^{'}$ by (M1); this is claim 1.

A1A2

2.1

As a map $X \to Z$ the function $f$ is surjective, $Z$ being its image by definition, and it is injective by step 1.1, so it is a bijection $X \to Z$ ; and $d_{Z} (f (x), f (x^{'})) = d_{Y} (f (x), f (x^{'})) = d_{X} (x, x^{'})$ , since $d_{Z}$ is the restriction of $d_{Y}$ , so it is an isometry, which is claim 2.

step 1.1A1

3.1

Both $f : X \to Z$ and its inverse $g : Z \to X$ are continuous, with $δ : = ε$ serving at every point in both directions, because $d_{Z} (f (x), f (x^{'})) = d_{X} (x, x^{'})$ and, writing $z = f (x)$ , $z^{'} = f (x^{'})$ , also $d_{X} (g (z), g (z^{'})) = d_{Z} (z, z^{'})$ .

step 2.1A1L2

3.2

Claim 3: $f [B_{X} (x, r)] = {f (x^{'}) : d_{X} (x, x^{'}) < r} = {f (x^{'}) : d_{Z} (f (x), f (x^{'})) < r}$ , and as $f$ is onto $Z$ the latter set is ${z \in Z : d_{Z} (f (x), z) < r} = B_{Z} (f (x), r)$ .

step 2.1A1L1

4.1

By [L2] applied to the continuous maps of step 3.1, the preimage under $f : X \to Z$ of every open subset of $Z$ is open in $X$ , and the preimage under $g$ of every open subset of $X$ is open in $Z$ .

step 3.1L2

5.1

Claim 4: for $U \subseteq X$ we have $f [U] = g^{- 1} [U]$ , so if $U$ is open in $X$ then $f [U]$ is open in $Z$ by step 4.1; conversely $U = f^{- 1} [f [U]]$ , so if $f [U]$ is open in $Z$ then $U$ is open in $X$ by step 4.1. Hence $U \mapsto f [U]$ maps the topology of $X$ into that of $Z$ , is injective because $f$ is, and is onto because any open $W \subseteq Z$ equals $f [f^{- 1} [W]]$ with $f^{- 1} [W]$ open.

step 2.1step 4.1L3

6.1

Claims 1, 2, 3 and 4 are established by steps 1.1, 2.1, 3.2 and 5.1, so an isometric embedding identifies $X$ with the metric subspace $f [X]$ of $Y$ , as a metric space and hence as a topological one.

step 1.1step 3.2step 5.1∎

Remarks

Only the image carries the right topology. Claim 4 compares the topology of $X$ with the SUBSPACE topology of $f [X]$ , not with the topology of $Y$ . The image of an open set is in general not open in $Y$ : the inclusion of $[0, 1]$ into $R$ is an isometric embedding, and $[0, 1]$ is open in itself but not in $R$ (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded).
This is what licenses treating a subset of a metric space as a space in its own right, and it is used on the companion page whenever a subset of $R$ is called a metric space.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27Open item page →

Topologically, uniformly and Lipschitz equivalent metrics on a set

Definition

Let $X$ be a set and let $d$ and $d^{'}$ both be metrics on $X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric). Note that the underlying set is the same; nothing below compares metrics on different sets.

$d$ and $d^{'}$ are topologically equivalent if they have the same metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement): $T_{d} = T_{d^{'}} .$
$d$ and $d^{'}$ are uniformly equivalent if for every real $ε > 0$ there are reals $δ > 0$ and $δ^{'} > 0$ such that, for all $x, y \in X$ , $d (x, y) < δ ⟹ d^{'} (x, y) < ε and d^{'} (x, y) < δ^{'} ⟹ d (x, y) < ε .$
$d$ and $d^{'}$ are Lipschitz equivalent if there are reals $α, β > 0$ with $α d (x, y) \leq d^{'} (x, y) \leq β d (x, y) for all x, y \in X .$

What the middle condition says in words. It is the statement that both identity maps $i d : (X, d) \to (X, d^{'})$ and $i d : (X, d^{'}) \to (X, d)$ are uniformly continuous: the same $δ$ works at every pair of points, not merely at each point separately as in Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form. Uniform continuity has no definition of its own at this point in the reading order, so the condition is written out in full above; a later page defines it, and until then this write-out is what earlier pages quote.

Each of the three is an equivalence relation on the metrics on $X$ . Reflexivity is immediate ( $δ = ε$ , and $α = β = 1$ ); symmetry is built into the statements, the uniform one being symmetric by construction and the Lipschitz one because $α d \leq d^{'} \leq β d$ gives $β^{- 1} d^{'} \leq d \leq α^{- 1} d^{'}$ ; and transitivity follows by composing the $δ$ s and multiplying the constants.

Remarks

The three are ranked, and the ranking is proved, not assumed: Lipschitz equivalence implies uniform equivalence implies topological equivalence (Lipschitz equivalence implies uniform equivalence implies topological equivalence). Neither implication reverses, and the witnesses live on the companion page.
Naming forks in the literature. Many texts say strongly equivalent for what is called Lipschitz equivalent here, and many say simply equivalent for what is called topologically equivalent here. This library always writes the qualifier, so that no statement depends on which convention a reader brings. A few texts define topological equivalence by "the identity is a homeomorphism", which is the same condition (Metric continuity characterisations, with countable choice for the sequential converse).
Topological equivalence preserves exactly the topological notions: open, closed, closure, interior, boundary, convergence of sequences, continuity of maps into and out of the space. It does not preserve boundedness, diameters or the Lipschitz constants, and FALSE: boundedness of a metric space is determined by its topology records the first of those failures.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

Lipschitz equivalence implies uniform equivalence implies topological equivalence

Statement

Let $d$ and $d^{'}$ be metrics on the same set $X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), with the three equivalences as in Topologically, uniformly and Lipschitz equivalent metrics on a set. Then:

If $d$ and $d^{'}$ are Lipschitz equivalent, they are uniformly equivalent.
If $d$ and $d^{'}$ are uniformly equivalent, they are topologically equivalent.

Strictness is not claimed here. The theorem asserts the two implications and nothing more; that neither reverses is witnessed by explicit pairs of metrics on the companion page, and those witnesses are not prerequisites of this theorem. See the first remark below.

Facts & Assumptions

Given: A set $X$ and two metrics $d, d^{'}$ on it; a real $ε > 0$ .

[A1]

Lipschitz equivalence: there are reals $α, β > 0$ with $α d (x, y) \leq d^{'} (x, y) \leq β d (x, y)$ for all $x, y \in X$ (Topologically, uniformly and Lipschitz equivalent metrics on a set).

[A2]

Uniform equivalence: for every real $ε > 0$ there are $δ, δ^{'} > 0$ such that $d (x, y) < δ$ implies $d^{'} (x, y) < ε$ and $d^{'} (x, y) < δ^{'}$ implies $d (x, y) < ε$ , for all $x, y \in X$ (Topologically, uniformly and Lipschitz equivalent metrics on a set).

[L1]

Inverses and products of positives: $γ > 0$ gives $γ^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order), and a product of positives is positive; multiplying an inequality by a positive preserves it, in the strict form of Sign rules for products and monotonicity of multiplication and, with the case of equality settled by totality, in the nonstrict form (Ordered field, Complete ordered field (least-upper-bound property)).

[L2]

Transitivity of the order, and addition of a constant to an inequality (Order is preserved by adding a constant and by adding inequalities, Ordered field).

[L3]

Continuity of a map at a point in the $ε$ - $δ$ form (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form); a map continuous at every point has open preimages of open sets (Metric continuity characterisations, with countable choice for the sequential converse).

[L4]

Open sets and the metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement); membership in a ball (Open ball, closed ball and sphere in a metric space).

Proof

technique · direct

1.1

Claim 1: assume [A1] and let $ε > 0$ . Put $δ : = ε β^{- 1}$ and $δ^{'} : = α ε$ , both positive since $α, β, ε$ are. If $d (x, y) < δ$ then $d^{'} (x, y) \leq β d (x, y) < β δ = ε$ ; and if $d^{'} (x, y) < δ^{'}$ then $α d (x, y) \leq d^{'} (x, y) < α ε$ , so $d (x, y) < ε$ after multiplying by $α^{- 1} > 0$ . Hence $d$ and $d^{'}$ are uniformly equivalent.

A1L1L2

1.2

Assume [A2]. Then the identity map $i d : (X, d) \to (X, d^{'})$ is continuous at every point $a \in X$ : given $ε > 0$ , the $δ$ of [A2] satisfies $d (x, a) < δ \Rightarrow d^{'} (x, a) < ε$ , which is the $ε$ - $δ$ condition at $a$ ; symmetrically $i d : (X, d^{'}) \to (X, d)$ is continuous at every point, using $δ^{'}$ .

A2L3

2.1

By [L3] applied to the two continuous identity maps of step 1.2: the preimage under $i d : (X, d) \to (X, d^{'})$ of a $d^{'}$ -open set $V$ is $V$ itself and is $d$ -open, so every $d^{'}$ -open set is $d$ -open; and symmetrically every $d$ -open set is $d^{'}$ -open. Hence $T_{d} = T_{d^{'}}$ , which is claim 2.

step 1.2L3L4

3.1

Claims 1 and 2 are established by steps 1.1 and 2.1, so Lipschitz equivalence implies uniform equivalence and uniform equivalence implies topological equivalence.

step 1.1step 2.1∎

Remarks

Neither implication reverses, and the witnesses are on the companion page. On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ have the same topology and are not uniformly equivalent (On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ have the same topology and are not uniformly equivalent ↗); on $R$ the metrics $∣ x - y ∣$ and $\min {∣ x - y ∣, 1}$ are uniformly equivalent and not Lipschitz equivalent (On $R$ the metrics $∣ x - y ∣$ and $\min (∣ x - y ∣, 1)$ are uniformly but not Lipschitz equivalent ↗). Those two items are read here as orientation only: this theorem does not depend on them, and its statement claims nothing about strictness.
Uniform equivalence is strictly more than "both identities are continuous". Continuity of both identity maps is exactly topological equivalence, by the argument of step 2.1 read in reverse; uniform equivalence additionally demands that one $δ$ serve at every point at once, and that is the whole difference between the two conditions.
What each level preserves. Lipschitz equivalence preserves boundedness and changes diameters by at most a constant factor; uniform equivalence preserves Cauchy sequences and uniform continuity, notions taken up on a later page and not defined here; topological equivalence preserves the open sets and everything defined from them, and nothing else.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

$\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and define, for $x, y \in X$ ,

$d^{'} (x, y) : = \min {d (x, y), 1}, d^{''} (x, y) : = \frac{d (x, y)}{1 + d (x, y)} .$

Both are well defined: $d (x, y) \geq 0$ (Nonnegativity of a metric is a consequence of the other axioms, not an axiom), so $1 + d (x, y) > 0$ and is invertible, and the minimum of a two-element set of reals exists (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set). Then:

$d^{'}$ and $d^{''}$ are metrics on $X$ .
$d^{'} (x, y) \leq 1$ and $d^{''} (x, y) < 1$ for all $x, y$ ; hence $(X, d^{'})$ and $(X, d^{''})$ are bounded metric spaces (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and if $X \neq \emptyset$ then $diam (X) \leq 1$ for both.
$d^{'}$ and $d^{''}$ are each uniformly equivalent to $d$ , hence topologically equivalent to it (Topologically, uniformly and Lipschitz equivalent metrics on a set, Lipschitz equivalence implies uniform equivalence implies topological equivalence).

Consequently every metric space carries a bounded metric with exactly the same topology, so boundedness cannot be read off the topology alone.

Facts & Assumptions

Given: A metric space $(X, d)$ , points $x, y, z \in X$ , a real $ε > 0$ , and the two functions $φ_{1} (t) : = \min {t, 1}$ and $φ_{2} (t) : = t (1 + t)^{- 1}$ , defined for reals $t \geq 0$ , so that $d^{'} = φ_{1} \circ d$ and $d^{''} = φ_{2} \circ d$ .

[L1]

A metric is nonnegative and satisfies (M1), (M2), (M3) (Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

The minimum of a two-element set of reals exists, is one of the two elements, and is a lower bound of both (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L3]

$0 < 1$ (The multiplicative identity is positive); a sum of positives is positive and inequalities may be added, in the strict form of Order is preserved by adding a constant and by adding inequalities and, with the case of equality settled by totality, in the nonstrict form (Ordered field, Complete ordered field (least-upper-bound property)).

[L4]

Inverses and order: $u > 0$ gives $u^{- 1} > 0$ , and $0 < u < v$ gives $0 < v^{- 1} < u^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order); Inverses of positives are positive, and reciprocation reverses order states only those strict forms, so the nonstrict version used below, that $0 < u \leq v$ gives $0 < v^{- 1} \leq u^{- 1}$ , is that statement together with the case $u = v$ , in which the two inverses are equal, the order being total (Ordered field, Complete ordered field (least-upper-bound property)). Multiplying an inequality by a positive preserves it, in the strict form of Sign rules for products and monotonicity of multiplication and, with the same equality case, in the nonstrict form; and $u u^{- 1} = 1$ (Field).

[L5]

Bounded subset and diameter: $A$ is bounded when it lies in some ball, and for nonempty bounded $A$ the diameter is the least upper bound of the distances, so any upper bound of those distances bounds the diameter (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Open ball, closed ball and sphere in a metric space, Suprema and infima are unique).

[L6]

Uniform and topological equivalence, and the implication between them (Topologically, uniformly and Lipschitz equivalent metrics on a set, Lipschitz equivalence implies uniform equivalence implies topological equivalence).

Proof

technique · direct

1.1

Properties of $φ_{1}$ on $t \geq 0$ : it is one of $t$ and $1$ , so $0 \leq φ_{1} (t) \leq t$ and $φ_{1} (t) \leq 1$ ; $φ_{1} (t) = 0$ exactly when $t = 0$ , since $1 \neq 0$ ; it is nondecreasing, because for $0 \leq s \leq t$ the value $φ_{1} (s)$ is $s \leq t$ or $1$ , and in both cases it is a lower bound of ${t, 1}$ , hence at most $φ_{1} (t)$ ; and it is subadditive, since for $s, t \geq 0$ either one of $s, t$ is $\geq 1$ , and then $φ_{1} (s) + φ_{1} (t) \geq 1 \geq φ_{1} (s + t)$ , or both are $< 1$ , and then $φ_{1} (s) + φ_{1} (t) = s + t \geq φ_{1} (s + t)$ .

L2L3

1.2

Properties of $φ_{2}$ on $t \geq 0$ : here $1 + t \geq 1 > 0$ , so $(1 + t)^{- 1} > 0$ and $(1 + t)^{- 1} \leq 1$ , whence $0 \leq φ_{2} (t) \leq t$ ; also $t < 1 + t$ gives $φ_{2} (t) < (1 + t) (1 + t)^{- 1} = 1$ ; $φ_{2} (t) = 0$ exactly when $t = 0$ ; $φ_{2}$ is strictly increasing, because $φ_{2} (t) = 1 - (1 + t)^{- 1}$ and $0 \leq s < t$ gives $0 < 1 + s < 1 + t$ , hence $(1 + t)^{- 1} < (1 + s)^{- 1}$ ; and it is subadditive, since for $s, t \geq 0$ one has $0 < 1 + s \leq 1 + s + t$ and $0 < 1 + t \leq 1 + s + t$ , so $φ_{2} (s + t) = s (1 + s + t)^{- 1} + t (1 + s + t)^{- 1} \leq s (1 + s)^{- 1} + t (1 + t)^{- 1} = φ_{2} (s) + φ_{2} (t)$ .

L3L4

1.3

Both $d^{'}$ and $d^{''}$ are symmetric, being $φ_{i}$ applied to the symmetric function $d$ , and both vanish exactly on the diagonal, since $φ_{i} (t) = 0$ exactly when $t = 0$ and $d (x, y) = 0$ exactly when $x = y$ .

2.1

$d^{'}$ is a metric: (M1) and (M2) are step 1.3, and (M3) follows because $d (x, z) \leq d (x, y) + d (y, z)$ with $φ_{1}$ nondecreasing and subadditive on nonnegatives gives $d^{'} (x, z) = φ_{1} (d (x, z)) \leq φ_{1} (d (x, y) + d (y, z)) \leq φ_{1} (d (x, y)) + φ_{1} (d (y, z)) = d^{'} (x, y) + d^{'} (y, z)$ .

step 1.1step 1.3L1

2.2

$d^{''}$ is a metric: identically, using that $φ_{2}$ is increasing and subadditive on nonnegatives, $d^{''} (x, z) \leq φ_{2} (d (x, y) + d (y, z)) \leq d^{''} (x, y) + d^{''} (y, z)$ .

step 1.2step 1.3L1

2.3

Boundedness: $d^{'} (x, y) \leq 1 < 2$ and $d^{''} (x, y) < 1 < 2$ for all $x, y$ , so if $X \neq \emptyset$ then fixing any $x_{0} \in X$ gives $X \subseteq B_{d^{'}} (x_{0}, 2)$ and $X \subseteq B_{d^{''}} (x_{0}, 2)$ , while $X = \emptyset$ is bounded outright; and $1$ is an upper bound of all the distances, so $diam (X) \leq 1$ in both metrics when $X \neq \emptyset$ . This is claim 2.

step 1.1step 1.2L3L5

3.1

$d^{'}$ is uniformly equivalent to $d$ : given $ε > 0$ , take $δ : = ε$ , so that $d (x, y) < δ$ gives $d^{'} (x, y) \leq d (x, y) < ε$ ; and take $δ^{'} : = \min {ε, 1} > 0$ , so that $d^{'} (x, y) < δ^{'} \leq 1$ forces $φ_{1} (d (x, y)) \neq 1$ , hence $d^{'} (x, y) = d (x, y)$ by [L2], hence $d (x, y) < δ^{'} \leq ε$ .

step 1.1step 2.1L2L3

3.2

$d^{''}$ is uniformly equivalent to $d$ : given $ε > 0$ , take $δ : = ε$ , so that $d (x, y) < δ$ gives $d^{''} (x, y) \leq d (x, y) < ε$ ; and take $δ^{'} : = φ_{2} (ε) > 0$ , so that $d^{''} (x, y) < δ^{'}$ forces $d (x, y) < ε$ , since $d (x, y) \geq ε$ would give $φ_{2} (d (x, y)) \geq φ_{2} (ε) = δ^{'}$ by monotonicity.

step 1.2step 2.2L4

4.1

Uniform equivalence implies topological equivalence, so $d^{'}$ and $d^{''}$ have exactly the metric topology of $d$ ; this completes claim 3.

step 3.1step 3.2L6

5.1

Claims 1, 2 and 3 hold by steps 2.1 and 2.2, step 2.3, and steps 3.1, 3.2 and 4.1; hence every metric space carries a bounded metric inducing the same topology.

step 2.1step 2.2step 2.3step 4.1∎

Remarks

Two constructions rather than one, on purpose. $\min {d, 1}$ is the shorter argument and is the one used by the counterexamples on the companion page; $d / (1 + d)$ is strictly less than $1$ everywhere and is strictly increasing in $d$ , which makes it the better behaved of the two when the value of the metric is to be compared, and it is the form that generalises to countable products.
Neither is Lipschitz equivalent to $d$ when $d$ is unbounded. A Lipschitz bound $α d \leq d^{'}$ with $α > 0$ would force $d \leq α^{- 1}$ everywhere, which fails as soon as $d$ takes arbitrarily large values; the real line is the witness (On $R$ the metrics $∣ x - y ∣$ and $\min (∣ x - y ∣, 1)$ are uniformly but not Lipschitz equivalent ↗).
Boundedness is therefore not a topological property, which is recorded as FALSE: boundedness of a metric space is determined by its topology with the real line as witness.
The bound $diam (X) \leq 1$ need not be an equality, and the two constructions differ on when it is. For a one-point space both new metrics are identically $0$ . For $d^{'} = \min {d, 1}$ the bound is attained as soon as $d$ takes some value $\geq 1$ , since then $d^{'}$ takes the value $1$ itself, and the companion page computes one such case. For $d^{''} = d / (1 + d)$ the value $1$ is never taken at all, by claim 2, so on a space where $d$ is bounded the diameter in $d^{''}$ is strictly below $1$ : for instance on a two-point space with $d = 1$ the new distance is $1 / 2$ .

RemarkRemark: AI-adaptedProof: Not applicableverified 2026-08-02 (claude-opus-5)Open item page →

Which metric axiom list this library uses, the live naming fork between semimetric and pseudometric, and why extended metrics are not treated here

The axiom list. Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric asks a metric $d : X \times X \to R$ for exactly three things: (M1) $d (x, y) = 0$ if and only if $x = y$ ; (M2) $d (x, y) = d (y, x)$ ; (M3) $d (x, z) \leq d (x, y) + d (y, z)$ . Many texts add a fourth, $d (x, y) \geq 0$ , or build it into the codomain by writing $d : X \times X \to [0, \infty)$ . That fourth condition is redundant: it follows from the other three, and Nonnegativity of a metric is a consequence of the other axioms, not an axiom proves it. The list is kept minimal here so that every verification of "is this a metric" has three things to check and not four, and so that no proof can quietly assume nonnegativity before it has been established.

Splitting (M1). Some texts state (M1) as two conditions, $d (x, x) = 0$ for all $x$ together with the implication $d (x, y) = 0 \Rightarrow x = y$ . That is the same notion, and the split form is convenient because deleting the second half is exactly the weakening that produces a pseudometric.

The naming fork, which is live and is why this library says pseudometric. Two different weakenings of the axiom list circulate under overlapping names.

Dropping the implication $d (x, y) = 0 \Rightarrow x = y$ , keeping symmetry and the triangle inequality, gives what most current sources, and this library, call a pseudometric (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric). This is the notion a seminorm induces.
Dropping the triangle inequality, keeping (M1) and (M2), gives what most current sources call a semimetric.

The fork is that a substantial part of the literature, especially in functional analysis and in older texts, uses semimetric for the first of these, that is as a synonym for pseudometric. There is no way to use the word semimetric here without inheriting the ambiguity, so this library does not use it at all: the first weakening is always called a pseudometric, and the second, which nothing here needs, is never named. Dropping symmetry instead gives a quasimetric, also not treated here; note that Nonnegativity of a metric is a consequence of the other axioms, not an axiom uses symmetry, so a quasimetric is not automatically nonnegative and the fourth axiom is not redundant for it.

Ultrametrics. The strong triangle inequality $d (x, z) \leq \max {d (x, y), d (y, z)}$ implies (M3) in the presence of (M1) and (M2), by Nonnegativity of a metric is a consequence of the other axioms, not an axiom and the fact that the maximum of two nonnegative reals is at most their sum. So an ultrametric is a metric, and the definition may be read either as "a metric that also satisfies (M3')" or as "a function satisfying (M1), (M2) and (M3')". The two readings pick out the same objects.

Why extended metrics are not treated here. An extended metric is allowed to take the value $+ \infty$ , so that its codomain is $[0, \infty]$ rather than $[0, \infty)$ ; the axioms are read with the usual arithmetic of $+ \infty$ . The construction is useful, for instance when one wants to glue metric spaces without connecting them, and it is standard in metric geometry. It is not treated here, for one reason: its values would have to live in the extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , whereas the axioms of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric are stated over the complete ordered field $R$ (Complete ordered field (least-upper-bound property)) and are never read anywhere else. Why they are kept there is set out in Conventions: $\sup \emptyset$ , unbounded sets, and the extended reals: $\overline{R}$ is not a field, the expressions $(+ \infty) + (- \infty)$ and $0 \cdot (+ \infty)$ have no definition compatible with the field axioms, and writing an infinite value silently moves the discussion into a different structure, after which every algebraic step needs its own justification. Every value of every metric in this library is therefore an element of $R$ .

Two consequences of that decision are visible on this page and are not oversights. First, an unbounded set has no diameter at all here, rather than a diameter $+ \infty$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). Second, the supremum metric is defined on the bounded real-valued functions only (The supremum metric $d_{\infty} (f, g) = \sup_{x} ∣ f (x) - g (x) ∣$ is a metric on the bounded real-valued functions on a nonempty set), where texts working in $\overline{R}$ define it on all of them.

Adding extended metrics honestly would mean restating Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric over a totally ordered set with a greatest element, carrying its own partial arithmetic, and re-proving over it everything this page proves over $R$ . No such restatement is made anywhere in this library, and until one is, every metric here takes real values.

5 · Examples, counterexamples and false statements

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)Open item page →

FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$

Statement

False claim: for every metric space $(X, d)$ , every $x \in X$ and every real $r > 0$ ,

$\overline{B (x, r)} = \overset{ˉ}{B} (x, r),$

that is, the closure of the open ball (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space) is the closed ball of the same centre and radius (Open ball, closed ball and sphere in a metric space).

One inclusion is a theorem and the other is false. The names open ball and closed ball do not by themselves license the equality, and the intuition behind it comes from $R^{n}$ with a Euclidean metric, where it happens to be true; it fails already in a subspace of the real line with a gap, and the witness used below is ${0} \cup [1, 2]$ .

Facts & Assumptions

Given: The real line with its usual metric $d_{R} (u, v) = ∣ u - v ∣$ (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded); the subset $X : = {0} \cup [1, 2] \subseteq R$ with the subspace metric $d : = d_{R} ↾ (X \times X)$ (Isometry, isometric embedding, and the subspace metric on a subset, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length); an arbitrary metric space $(X_{0}, d_{0})$ with $x \in X_{0}$ and a real $r > 0$ .

[L1]

The closed ball $\overset{ˉ}{B} (x, r)$ is a closed set, and it contains $B (x, r)$ (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Open ball, closed ball and sphere in a metric space).

[L2]

The closure of a set is the smallest closed superset of it, and a closed set equals its own closure (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset).

[L3]

The subspace metric makes $X$ a metric space and its balls are traces: $B_{X} (a, s) = B_{R} (a, s) \cap X$ (Isometry, isometric embedding, and the subspace metric on a subset, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Open ball, closed ball and sphere in a metric space).

[L4]

Absolute value and order: $∣ u ∣ = u$ when $u \geq 0$ , $∣ - u ∣ = ∣ u ∣$ , and $0 < 1$ ; and by trichotomy $u \geq 1$ rules out $u < 1$ (Absolute value in an ordered field, Basic properties of the absolute value, The multiplicative identity is positive, Ordered field, Complete ordered field (least-upper-bound property)).

[L5]

Open sets of a metric space: $U$ is open when every point of $U$ has a ball around it inside $U$ ; a set is closed when its complement is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Refutation

technique · direct

1.1

The inclusion that does hold, in every metric space: $\overset{ˉ}{B} (x, r)$ is closed and contains $B (x, r)$ , so the smallest closed superset of $B (x, r)$ satisfies $\overline{B (x, r)} \subseteq \overset{ˉ}{B} (x, r)$ .

L1L2

1.2

In the witness $X = {0} \cup [1, 2]$ , every $y \in [1, 2]$ satisfies $y \geq 1$ , hence $d (0, y) = ∣ 0 - y ∣ = ∣ - y ∣ = y \geq 1$ ; and $d (0, 0) = 0 < 1$ .

givenL3L4

2.1

Therefore $B_{X} (0, 1) = {y \in X : d (0, y) < 1} = {0}$ and ${\overset{ˉ}{B}}_{X} (0, 1) = {y \in X : d (0, y) \leq 1} = {0, 1}$ , since $1 \in [1, 2] \subseteq X$ has $d (0, 1) = 1$ while every other $y \in [1, 2]$ has $d (0, y) = y > 1$ or $y = 1$ .

step 1.2L3L4

2.2

The set $[1, 2] = X ∖ {0}$ is open in $X$ : for $y \in [1, 2]$ the ball $B_{X} (y, 1)$ omits $0$ , because $d (0, y) \geq 1$ by step 1.2, so $B_{X} (y, 1) \subseteq X ∖ {0} = [1, 2]$ . Hence ${0}$ is closed in $X$ .

step 1.2L3L5

3.1

Since ${0}$ is closed it equals its own closure, so $\overline{B_{X} (0, 1)} = \overline{{0}} = {0}$ by step 2.1, while ${\overset{ˉ}{B}}_{X} (0, 1) = {0, 1}$ ; and ${0} \neq {0, 1}$ because $1 \neq 0$ .

step 2.1step 2.2L2L4

4.1

The witness $(X, d)$ with $x = 0$ and $r = 1$ therefore refutes the claim; all that survives in general is the inclusion of step 1.1, and it can be strict.

step 1.1step 3.1∎

Remarks

Where the intuition comes from and why it does not transfer. In $R^{n}$ with the Euclidean metric ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it) the segment from the centre to a point of the closed ball lies in the space, and running along it approaches that point from inside the open ball; that is the usual route to the equality there, and this library does not prove it. A metric space need not contain any such segment: in the witness above, nothing of $X$ lies strictly between $0$ and $1$ , so the point $1$ of the closed ball is not approached from inside $B (0, 1) = {0}$ at all.
The failure is not exotic. A discrete metric on a set with at least two points produces the same phenomenon in a starker form, with $\overline{B (p, 1)} = {p}$ and $\overset{ˉ}{B} (p, 1)$ the whole space; the companion page carries both witnesses.
The sphere is not the boundary of the ball either, and that failure is recorded separately on the companion page.

False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26Open item page →

FALSE: boundedness of a metric space is determined by its topology

Statement

False claim: boundedness is a topological property of a metric space; that is, if $d$ and $d^{'}$ are topologically equivalent metrics on a set $X$ (Topologically, uniformly and Lipschitz equivalent metrics on a set) and $(X, d)$ is a bounded metric space (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), then $(X, d^{'})$ is bounded as well.

Equivalently, the false claim says that the metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) determines whether the space is bounded. It does not: every metric space carries a bounded metric with exactly the same topology, so as soon as one unbounded metric space exists the claim collapses.

Facts & Assumptions

Given: The real line $R$ with its usual metric $d_{R} (u, v) = ∣ u - v ∣$ , and the metric $ρ (u, v) : = \min {d_{R} (u, v), 1}$ (Maximum and minimum of a set).

[L1]

$d_{R}$ is a metric on $R$ and $(R, d_{R})$ is not a bounded metric space: no ball contains $R$ (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

For any metric $d$ the function $\min {d, 1}$ is a metric, is bounded with diameter at most $1$ on a nonempty space, and is uniformly equivalent to $d$ ( $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L3]

Uniform equivalence implies topological equivalence (Lipschitz equivalence implies uniform equivalence implies topological equivalence, Topologically, uniformly and Lipschitz equivalent metrics on a set).

Refutation

technique · direct

1.1

By [L1] the metric $d_{R}$ makes $R$ a metric space that is not bounded.

1.2

By [L2] the function $ρ = \min {d_{R}, 1}$ is a metric on $R$ , the space $(R, ρ)$ is bounded with $diam (R) \leq 1$ , and $ρ$ is uniformly equivalent to $d_{R}$ .

2.1

By [L3] the two metrics are therefore topologically equivalent: $T_{ρ} = T_{d_{R}}$ .

step 1.2L3

3.1

So $d_{R}$ and $ρ$ are topologically equivalent metrics on the same set, $(R, ρ)$ is bounded and $(R, d_{R})$ is not; the claim fails, and boundedness is a property of the metric and not of the topology.

step 1.1step 2.1∎

Remarks

What is true instead. Boundedness is preserved by Lipschitz equivalence, since $d^{'} \leq β d$ turns a ball for $d$ into a ball for $d^{'}$ (Topologically, uniformly and Lipschitz equivalent metrics on a set). It is the two weaker equivalences that lose it, and the witness above sits precisely in the gap between Lipschitz equivalence and uniform equivalence (Lipschitz equivalence implies uniform equivalence implies topological equivalence).
The diameter is even less topological than boundedness. Rescaling a metric by a positive constant is a Lipschitz equivalence and multiplies every diameter by that constant, so no numerical value of the diameter is determined even by the Lipschitz class.
This is why "bounded" is never used as a topological adjective in this library. The bounded subsets of $(X, d)$ are defined from $d$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and every statement about them names the metric.

Metric Spaces

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Metric space: d(x,y)=0d(x,y)=0 iff x=yx=y, symmetry, and the triangle inequality; pseudometric and ultrametric

Definition

Remarks

Nonnegativity of a metric is a consequence of the other axioms, not an axiom

Statement

Facts & Assumptions

Proof

Remarks

Open ball, closed ball and sphere in a metric space

Definition

Remarks

Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space

Definition

Remarks

The reverse triangle inequality ∣d(x,z)−d(y,z)∣≤d(x,y)∣d(x,z)−d(y,z)∣≤d(x,y) in any metric space

Statement

Facts & Assumptions

Proof

Remarks

∣d(x,A)−d(y,A)∣≤d(x,y)∣d(x,A)−d(y,A)∣≤d(x,y), so the distance to a fixed nonempty set is 11-Lipschitz

Statement

Facts & Assumptions

Proof

Remarks

The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement

Definition

Remarks

The absolute value makes RR a metric space: d(x,y)=∣x−y∣d(x,y)=∣x−y∣ is a metric, its open balls are the intervals (x−r,x+r)(x−r,x+r), and it is unbounded

Statement

Facts & Assumptions

Proof

Remarks

RnRn as the set of functions n→Rn→R, and d1d1​, d2d2​, d∞d∞​ are metrics on it

Statement

Facts & Assumptions

Proof

Remarks

The supremum metric d∞(f,g)=sup⁡x∣f(x)−g(x)∣d∞​(f,g)=supx​∣f(x)−g(x)∣ is a metric on the bounded real-valued functions on a nonempty set

Statement

Facts & Assumptions

Proof

Remarks

Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed

Statement

Facts & Assumptions

Proof

Remarks

Interior, closure, boundary, limit point, isolated point and dense subset of a metric space

Definition

Remarks

The closure of a nonempty AA is {x:d(x,A)=0}{x:d(x,A)=0}, equals AA together with its limit points, and is the smallest closed superset

Statement

Facts & Assumptions

Proof

Remarks

Convergence of a sequence in a metric space: xk→xxk​→x iff d(xk,x)→0d(xk​,x)→0 in RR

Definition

Remarks

A sequence in a metric space has at most one limit

Statement

Facts & Assumptions

Proof

Remarks

The balls B(x,1/n)B(x,1/n), n≥1n≥1, form a countable neighbourhood base at xx, so every metric space is first countable

Statement

Facts & Assumptions

Proof

Remarks

A point lies in the closure of AA iff some sequence in AA converges to it, and a set is closed iff it is sequentially closed

Statement

Facts & Assumptions

Proof

Remarks

Distinct points of a metric space have disjoint balls around them

Statement

Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric

The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space

$∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz

The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded

$R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it

The supremum metric $d_{\infty} (f, g) = \sup_{x} ∣ f (x) - g (x) ∣$ is a metric on the bounded real-valued functions on a nonempty set

The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset

Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$

The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable

A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed

Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form

$\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology

FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$