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✓ 12 results · all verified · 9 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 3 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Metric Spaces: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

The metrics $d_{1}$ , $d_{2}$ and $d_{\infty}$ on $R^{n}$ are metrics and are Lipschitz equivalent, with explicit constants

Example

Let $n \geq 1$ be a natural number and let $R^{n}$ carry the three metrics

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

of $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, where $R^{n}$ is the set of functions from the von Neumann natural $n$ to $R$ . All three are metrics (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); that is $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it and is quoted here rather than reproved. What this example adds is that the three are Lipschitz equivalent with explicit constants (Topologically, uniformly and Lipschitz equivalent metrics on a set): for all $x, y \in R^{n}$ ,

$d_{\infty} (x, y) \leq d_{2} (x, y) \leq \sqrt{n} d_{\infty} (x, y), d_{\infty} (x, y) \leq d_{1} (x, y) \leq n d_{\infty} (x, y), d_{2} (x, y) \leq d_{1} (x, y) \leq \sqrt{n} d_{2} (x, y) .$

Consequently the three are uniformly equivalent and topologically equivalent (Lipschitz equivalence implies uniform equivalence implies topological equivalence), so they determine the same open sets, the same convergent sequences and the same continuous maps on $R^{n}$ .

The constants are best possible: taking $x - y$ with a single nonzero coordinate gives equality in $d_{\infty} = d_{2} = d_{1}$ , and taking all $n$ coordinates equal in absolute value gives $d_{1} = n d_{\infty}$ and $d_{2} = \sqrt{n} d_{\infty}$ . Those two remarks are not needed for the equivalence and are not proved below.

Facts & Assumptions

Given: A natural $n \geq 1$ , elements $x, y \in R^{n}$ , the list $u_{k} : = x_{k} - y_{k}$ for $k < n$ , and the abbreviations $M : = d_{\infty} (x, y)$ , $S : = d_{1} (x, y) = \sum_{k < n} ∣ u_{k} ∣$ and $Q : = \sum_{k < n} u_{k}^{2}$ , so that $d_{2} (x, y) = \sqrt{Q}$ ; the canonical natural $n$ is here read inside $R$ as $n \cdot 1_{R}$ .

[L1]

$d_{1}$ , $d_{2}$ and $d_{\infty}$ are metrics on $R^{n}$ ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Finite sums and finite products, by recursion).

[L2]

Laws of finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): monotonicity, scaling, $\sum_{k < n} λ = n λ$ , a sum of nonnegative terms is nonnegative, and each single term is at most such a sum.

[L3]

The maximum of a nonempty finite set of reals exists, is one of its elements and bounds the set above (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L4]

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 $(\sqrt{c})^{2} = c$ ; hence $\sqrt{t^{2}} = t$ for $t \geq 0$ and $\sqrt{a b} = \sqrt{a} \sqrt{b}$ for $a, b \geq 0$ , both by uniqueness. Squaring is monotone on the nonnegatives, $s \leq t ⟺ s^{2} \leq t^{2}$ (Squaring is monotone on the nonnegatives), so the same holds for square roots.

[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]

Absolute value (Basic properties of the absolute value, Absolute value in an ordered field, Integer powers $a^{m}$ ): $∣ t ∣ \geq 0$ , $∣ t ∣^{2} = t^{2}$ , and $∣ t ∣ = t$ for $t \geq 0$ .

[L7]

Order arithmetic: multiplying an inequality by a nonnegative element preserves it and inequalities may be added, in the strict forms of Sign rules for products and monotonicity of multiplication and Order is preserved by adding a constant and by adding inequalities together with the case of equality settled by totality (Ordered field, Complete ordered field (least-upper-bound property)); and $n \cdot 1_{R} > 0$ for $n \geq 1$ (Canonical naturals are positive and strictly increasing).

[L8]

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

Verification

technique · direct

1.1

Since $n \geq 1$ the set ${∣ u_{k} ∣ : k < n}$ is nonempty and finite, so $M$ exists, equals $∣ u_{j} ∣$ for some $j < n$ , satisfies $M \geq 0$ , and bounds every $∣ u_{k} ∣$ above.

L3L6

1.2

The reals $n \cdot 1_{R}$ , $\sqrt{n}$ , $Q$ and $S$ are all nonnegative, and $n \cdot 1_{R} > 0$ , so $\sqrt{n} > 0$ ; also $M^{2} = u_{j}^{2}$ and $u_{k}^{2} = ∣ u_{k} ∣^{2}$ for every $k$ .

L2L4L6L7

2.1

First chain: $M^{2} = u_{j}^{2} \leq Q$ because a single nonnegative term is at most the sum, so $M = \sqrt{M^{2}} \leq \sqrt{Q} = d_{2} (x, y)$ ; and $Q = \sum_{k < n} u_{k}^{2} \leq \sum_{k < n} M^{2} = n M^{2}$ by monotonicity and scaling, so $d_{2} (x, y) = \sqrt{Q} \leq \sqrt{n M^{2}} = \sqrt{n} M$ .

step 1.1step 1.2L2L4L7

2.2

Second chain: $M = ∣ u_{j} ∣ \leq S$ because a single nonnegative term is at most the sum; and $S = \sum_{k < n} ∣ u_{k} ∣ \leq \sum_{k < n} M = n M$ by monotonicity and scaling.

step 1.1step 1.2L2L7

2.3

Third chain: for every $k < n$ , multiplying $∣ u_{k} ∣ \leq S$ by the nonnegative $∣ u_{k} ∣$ gives $u_{k}^{2} = ∣ u_{k} ∣^{2} \leq ∣ u_{k} ∣ S$ , so summing and scaling gives $Q \leq S \sum_{k < n} ∣ u_{k} ∣ = S^{2}$ and hence $d_{2} (x, y) = \sqrt{Q} \leq \sqrt{S^{2}} = S = d_{1} (x, y)$ ; and Cauchy-Schwarz applied to the lists $a_{k} : = ∣ u_{k} ∣$ and $b_{k} : = 1$ gives $S = ∣ \sum_{k < n} ∣ u_{k} ∣ \cdot 1 ∣ \leq \sqrt{\sum_{k < n} u_{k}^{2}} \sqrt{\sum_{k < n} 1} = \sqrt{Q} \sqrt{n} = \sqrt{n} d_{2} (x, y)$ .

step 1.1step 1.2L2L4L5L6L7

3.1

The three chains are exactly Lipschitz equivalences with positive constants: $1 \cdot d_{\infty} \leq d_{2} \leq \sqrt{n} d_{\infty}$ , $1 \cdot d_{\infty} \leq d_{1} \leq n d_{\infty}$ and $1 \cdot d_{2} \leq d_{1} \leq \sqrt{n} d_{2}$ , the constants $1$ , $\sqrt{n}$ and $n$ all being positive.

step 2.1step 2.2step 2.3L7L8

4.1

Hence any two of $d_{1}$ , $d_{2}$ , $d_{\infty}$ are Lipschitz equivalent, and therefore uniformly equivalent and topologically equivalent; all three induce the same topology on $R^{n}$ .

step 3.1L1L8∎

Remarks

The constants blow up with the dimension, and that is the whole point of the distinction. The comparison $d_{1} \leq n d_{\infty}$ is Lipschitz for each fixed $n$ and useless uniformly in $n$ , so no pair of constants serves all dimensions at once. Whether an analogue survives on spaces of infinite sequences is a question for a later page and is not addressed here.
Only $n \geq 1$ is treated, because $d_{\infty}$ is a maximum over the index set ${0, \dots, n - 1}$ and that set is empty when $n = 0$ ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it). For $n = 0$ the space $R^{0}$ is a single point and $d_{1}$ , $d_{2}$ are identically $0$ on it, while $d_{\infty}$ is not defined there at all, so there is nothing to compare.
Minkowski is not used here. The triangle inequalities were settled in $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it; what this page needs is only the comparison of the three values, and that runs on the finite-sum laws and Cauchy-Schwarz.

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

The discrete metric induces the discrete topology, in which every subset is clopen

Example

Let $X$ be any set and define $δ : X \times X \to R$ by

$δ (x, y) : = {\begin{cases} 0 & if x = y, \\ 1 & if x \neq y . \end{cases}$

Then:

$δ$ is a metric on $X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), the discrete metric.
For $x \in X$ and a real $r > 0$ : $B (x, r) = {\begin{cases} {x} & 0 < r \leq 1 \\ X & r > 1 \end{cases}, \overset{ˉ}{B} (x, r) = {\begin{cases} {x} & 0 < r < 1 \\ X & r \geq 1 \end{cases}, S (x, 1) = X ∖ {x} .$
Every subset of $X$ is open, hence every subset is closed, hence every subset is clopen (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). The metric topology of $δ$ is the discrete topology, the collection of all subsets of $X$ .

The example is the standard source of counterexamples about balls: here the closed ball of radius $1$ is the whole space while the closure of the open ball of radius $1$ is a single point, and the sphere of radius $1$ is everything except the centre while the boundary of the ball is empty.

Facts & Assumptions

Given: A set $X$ , the function $δ$ above, points $x, y, z \in X$ , a real $r > 0$ , and a subset $U \subseteq X$ .

[L1]

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

[L2]

Balls: $B (x, r) = {y : δ (x, y) < r}$ , $\overset{ˉ}{B} (x, r) = {y : δ (x, y) \leq r}$ and $S (x, r) = {y : δ (x, y) = r}$ (Open ball, closed ball and sphere in a metric space).

[L3]

Open and closed: $U$ is open when every point of it has a ball around it inside it; $F$ 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).

[L4]

Closed balls are closed (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed).

[L5]

Order: $0 < 1$ , so $0 \neq 1$ ; a sum of a positive and a nonnegative real is positive, and inequalities may be compared by trichotomy and transitivity (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Ordered field, Complete ordered field (least-upper-bound property)).

Verification

technique · direct

1.1

Separation and symmetry: $δ (x, y) = 0$ holds exactly when $x = y$ , since the other value $1$ is different from $0$ ; and the defining clauses are unchanged when $x$ and $y$ are exchanged, since " $x = y$ " is.

L1L5

1.2

Triangle inequality: if $x = z$ then $δ (x, z) = 0$ and the right side is a sum of two values in ${0, 1}$ , hence at least $0$ ; and if $x \neq z$ then $y$ cannot equal both $x$ and $z$ , so at least one of $δ (x, y), δ (y, z)$ equals $1$ while the other is $0$ or $1$ , whence $δ (x, y) + δ (y, z) \geq 1 = δ (x, z)$ .

L1L5

1.3

Balls: $δ (x, x) = 0 < r$ always, so $x \in B (x, r)$ ; and for $y \neq x$ one has $δ (x, y) = 1$ , so $y \in B (x, r)$ exactly when $1 < r$ , $y \in \overset{ˉ}{B} (x, r)$ exactly when $1 \leq r$ , and $y \in S (x, 1)$ always. This is claim 2.

L2L5

2.1

Every subset is open: for $U \subseteq X$ and $x \in U$ the ball $B (x, 1)$ is ${x}$ by the computation of step 1.3, and ${x} \subseteq U$ .

L2L3L5

2.2

Claim 1 holds: $δ$ satisfies (M1) and (M2) by step 1.1 and (M3) by step 1.2, so it is a metric on $X$ .

step 1.1step 1.2L1

3.1

Claim 3 holds: every subset is open by step 2.1, so for any $F \subseteq X$ the complement $X ∖ F$ is open and $F$ is closed; thus every subset is clopen and the metric topology is the full power set of $X$ . In particular each singleton ${x}$ is closed, in agreement with the closed-ball computation $\overset{ˉ}{B} (x, r) = {x}$ for $0 < r < 1$ and the fact that closed balls are closed.

step 1.3step 2.1L3L4

4.1

Claims 1, 2 and 3 hold by steps 2.2, 1.3 and 3.1.

step 1.3step 2.2step 3.1∎

Remarks

Every map out of a discrete space is continuous, since every preimage is open (For a map of metric spaces the following agree: $ε$ - $δ$ continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f (\overline{A}) \subseteq \overline{f (A)}$ ); so the discrete metric carries no information about $X$ beyond its cardinality, and is the extreme case at one end of the range of metrics on a set.
Convergence is eventual constancy. $x_{k} \to x$ in $(X, δ)$ means $δ (x_{k}, x) < 1$ eventually, that is $x_{k} = x$ for all large $k$ (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).
Boundedness is immediate: $X \subseteq B (x, 2)$ for any $x$ , so every discrete metric space is bounded, with diameter $1$ as soon as $X$ has two points (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

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

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

Example

Let $S$ be a nonempty set, let $B (S)$ be the set of bounded functions $S \to R$ and let $d_{\infty} (f, g) = \sup {∣ f (s) - g (s) ∣ : s \in S}$ be the supremum metric; that this is a 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 and is quoted here rather than reproved. This example records two things about it.

The constants form an isometric copy of the real line. For $c \in R$ let $\hat{c} \in B (S)$ be the constant function with value $c$ . Then $c \mapsto \hat{c}$ is an isometric embedding of $(R, d_{R})$ into $(B (S), d_{\infty})$ (Isometry, isometric embedding, and the subspace metric on a subset, 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): $d_{\infty} (\hat{c}, \hat{c}^{'}) = ∣ c - c^{'} ∣ (c, c^{'} \in R) .$
The supremum need not be attained. Take $S = N$ , so that $B (N)$ is the set of bounded sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences), and put $f (k) : = 1 - \frac{1}{k + 1} (k \in N), g : = \hat{0} .$ Then $f, g \in B (N)$ , $d_{\infty} (f, g) = 1$ , and $∣ f (k) - g (k) ∣ < 1$ for every $k \in N$ . So $d_{\infty} (f, g)$ is a supremum that is not a maximum, and no single point of $S$ realises the distance.

The index shift in $1 / (k + 1)$ is forced: $N$ contains $0$ (The natural numbers $N$ (von Neumann)) and sequences here are indexed from $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so $1 / k$ would be undefined at $k = 0$ .

Facts & Assumptions

Given: A nonempty set $S$ ; reals $c, c^{'}$ ; the constant functions $\hat{c}, {\hat{c}}^{'} : S \to R$ ; and, for $S = N$ , the functions $f (k) = 1 - 1 / (k + 1)$ and $g = \hat{0}$ , together with $D (f, g) : = {∣ f (k) - g (k) ∣ : k \in N}$ .

[L1]

The supremum metric is a metric on $B (S)$ for nonempty $S$ , and $d_{\infty} (u, v)$ is the least upper bound of ${∣ u (s) - v (s) ∣ : s \in S}$ (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, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L2]

Least upper bounds: a nonempty subset of $R$ bounded above has a unique least upper bound (Complete ordered field (least-upper-bound property), Suprema and infima are unique); bounded subsets of $R$ are as in Lower bound, bounded below, bounded set.

[L3]

Epsilon characterisation of the supremum: for a nonempty $T \subseteq R$ bounded above and an upper bound $u$ of $T$ , one has $u = \sup T$ if and only if for every real $ε > 0$ there is $t \in T$ with $u - ε < t$ (Epsilon characterisation of the supremum).

[L4]

Reciprocal Archimedean property: for every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ (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); and $m \cdot 1_{R} > 0$ with $1 / m > 0$ for $m \geq 1$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L5]

Absolute value: $∣ t ∣ = t$ for $t \geq 0$ , $∣ t ∣ \geq 0$ (Basic properties of the absolute value, Absolute value in an ordered field); and the usual metric of $R$ is $d_{R} (c, c^{'}) = ∣ c - c^{'} ∣$ (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).

[L6]

Isometric embedding: a map preserving distances exactly (Isometry, isometric embedding, and the subspace metric on a subset); trichotomy and transitivity of the order of $R$ (Ordered field, Complete ordered field (least-upper-bound property)).

Verification

technique · direct

1.1

Each constant function $\hat{c}$ has range ${c}$ , a bounded subset of $R$ , so $\hat{c} \in B (S)$ ; and ${∣ \hat{c} (s) - {\hat{c}}^{'} (s) ∣ : s \in S} = {∣ c - c^{'} ∣}$ , a nonempty one-element set whose least upper bound is $∣ c - c^{'} ∣$ itself.

givenL1L2L5

2.1

For $S = N$ : each $k + 1$ is a natural $\geq 1$ , so $1 / (k + 1)$ is a positive real and $0 \leq f (k) = 1 - 1 / (k + 1) < 1$ ; hence the range of $f$ is bounded and $f \in B (N)$ , while $g = \hat{0}$ is bounded by step 1.1, and $D (f, g) = {1 - 1 / (k + 1) : k \in N}$ is nonempty with $1$ as an upper bound.

step 1.1L4L5

2.2

Claim 1: by step 1.1, $d_{\infty} (\hat{c}, {\hat{c}}^{'}) = ∣ c - c^{'} ∣ = d_{R} (c, c^{'})$ for all reals $c, c^{'}$ , which is exactly the statement that $c \mapsto \hat{c}$ is an isometric embedding of $(R, d_{R})$ into $(B (S), d_{\infty})$ .

step 1.1L1L5L6

3.1

$d_{\infty} (f, g) = 1$ : the number $1$ is an upper bound of $D (f, g)$ by step 2.1, and for an arbitrary real $ε > 0$ choose a natural $n \geq 1$ with $1 / n < ε$ and put $k : = n - 1$ , a natural since $n \geq 1$ , so that $1 - 1 / (k + 1) = 1 - 1 / n > 1 - ε$ ; by the epsilon characterisation $1$ is the least upper bound.

step 2.1L3L4L6

4.1

The supremum is not attained: every element of $D (f, g)$ has the form $1 - 1 / (k + 1)$ with $1 / (k + 1) > 0$ , hence is $< 1 = d_{\infty} (f, g)$ , so no $k \in N$ satisfies $∣ f (k) - g (k) ∣ = d_{\infty} (f, g)$ .

step 2.1step 3.1L4L6

5.1

Claims 1 and 2 are established, by step 2.2 and by steps 3.1 and 4.1 respectively.

step 2.2step 3.1step 4.1∎

Remarks

Why the constants matter. The isometric copy of $R$ inside $B (S)$ shows that $(B (S), d_{\infty})$ is unbounded whenever $S \neq \emptyset$ , since $R$ 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, An isometric embedding is injective and carries the metric topology of the source onto the subspace topology of its image).
With $S = N$ this is the space usually written $ℓ^{\infty}$ , the bounded real sequences with the supremum metric. Its completeness and its separability are questions for later pages and are not touched here.
Attainment is a genuinely different question from existence. The supremum exists because the set is nonempty and bounded above (Complete ordered field (least-upper-bound property)); whether it lies in the set is exactly the question of whether the sup is a maximum, and claim 2 answers it negatively for a specific pair.

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The $2$ -adic absolute value gives an ultrametric on $Q$ , in which every triangle is isosceles and every point of a ball is a centre

Example

Write $2 : = 1 + 1$ . Call an integer (The integers as equivalence classes of pairs of naturals) even if it is $2 t$ for some integer $t$ , and odd if it is $2 t + 1$ for some integer $t$ .

The $2$ -adic valuation. Every nonzero integer $m$ can be written in exactly one way as

$m = 2^{j} u with j \in N and u odd,$

and $v_{2} (m) : = j$ is the $2$ -adic valuation of $m$ (Integer powers $a^{m}$ ). For a nonzero rational $x$ (The rationals as equivalence classes of pairs of integers), written $x = a / b$ with $a$ and $b$ nonzero integers, the integer

$v_{2} (x) : = v_{2} (a) - v_{2} (b)$

does not depend on the chosen representation. The $2$ -adic absolute value is

$∣ x ∣_{2} : = 2^{- v_{2} (x)} (x \neq 0), ∣ 0 ∣_{2} : = 0,$

read inside $R$ through the embeddings $Z \to Q \to R$ (The integers embed in the rationals, The unique embedding of ℚ into an ordered field), and the $2$ -adic distance is

$d_{2} (x, y) : = ∣ x - y ∣_{2} (x, y \in Q) .$

Claims.

$d_{2}$ is an ultrametric on $Q$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric): it satisfies (M1), (M2) and the strong triangle inequality $d_{2} (x, z) \leq \max {d_{2} (x, y), d_{2} (y, z)}$ .
Every triangle is isosceles: if $d_{2} (x, y) \neq d_{2} (y, z)$ then $d_{2} (x, z) = \max {d_{2} (x, y), d_{2} (y, z)}$ .
Every point of a ball is a centre: if $y \in B (x, r)$ then $B (y, r) = B (x, r)$ (Open ball, closed ball and sphere in a metric space).

Why $p = 2$ and not a general prime. The general $p$ -adic valuation needs primality and unique factorisation in $Z$ , which are developed on Primes, Euclid's Lemma and the Fundamental Theorem of Arithmetic (The $p$ -adic valuation $v_{p} (a)$ of a nonzero integer: the greatest $k \in N$ with $p^{k} ∣ a$ , Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ , The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ , The $p$ -adic valuation extends to the nonzero rationals by $v_{p} (a / b) : = v_{p} (a) - v_{p} (b) \in Z$ , independently of the representation; it satisfies $v_{p} (x y) = v_{p} (x) + v_{p} (y)$ , and $v_{p} (x + y) \geq \min {v_{p} (x), v_{p} (y)}$ whenever $x$ , $y$ and $x + y$ are nonzero) and are therefore available here; this item nevertheless develops the case $p = 2$ from parity alone, so that the ultrametric geometry below rests on nothing but the discreteness of $Z$ . At $p = 2$ everything reduces to parity, which is available: The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ partitions $N$ into the ranges of its two index maps, and that is what claim 2 of the verification turns into "even or odd, never both".

Claims 2 and 3 use nothing about $d_{2}$ beyond the strong triangle inequality, so they hold in every ultrametric space.

Facts & Assumptions

Given: The integers $Z$ and rationals $Q$ with their arithmetic; the successor $σ$ on $N$ ; the element $2 = 1 + 1$ ; the index maps $e, o$ of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ; and integers $m, a, c$ and rationals $x, y, z$ as introduced in the steps.

[L1]

Ring and field arithmetic: $Z$ is a commutative ring (The integers form a commutative ring) and $Q$ a field (The rationals form a field, The rationals as equivalence classes of pairs of integers); $Z$ is totally ordered and its order is compatible with addition and with multiplication by positives (The integers form a totally ordered ring); nonzero integers have nonzero product and cancel (The integers have no zero divisors; multiplicative cancellation).

[L2]

Induction (The principle of mathematical induction) and strong induction (Strong (complete) induction) on $N$ (The natural numbers $N$ (von Neumann)).

[L3]

The index maps: $e_{0} = 0$ , $e_{σ (j)} = σ (σ (e_{j}))$ , $o_{0} = σ (0)$ , $o_{σ (j)} = σ (σ (o_{j}))$ , and $N$ is the disjoint union of the ranges of $e$ and $o$ , each natural lying in exactly one range and being hit exactly once (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L4]

Addition on $N$ : $m + 0 = m$ , $m + σ (k) = σ (m + k)$ (Addition of natural numbers) and $σ (m) + n = σ (m + n)$ (Left successor law for addition); the order is total and transitive (Order on the natural numbers, $\leq$ is a linear order on $N$ ), and $n \neq 0$ gives $n \geq 1$ (Discreteness: $σ (n)$ is the immediate successor).

[L5]

The embedding $N \to Z$ is injective, preserves addition, multiplication and order, and its image is exactly the set of integers $\geq 0$ (The naturals embed in the integers); the embeddings $Z \to Q \to R$ are injective and order preserving (The integers embed in the rationals, The unique embedding of ℚ into an ordered field).

[L6]

Powers: $a^{0} = 1$ , $a^{n + 1} = a^{n} a$ , $a^{m + n} = a^{m} a^{n}$ and $a^{- m} = (a^{m})^{- 1}$ , valid for integer exponents when $a \neq 0$ , and $a \neq 0$ gives $a^{n} \neq 0$ (Integer powers $a^{m}$ , Laws of integer exponents); for $a > 1$ and $k \geq 1$ one has $a^{k} \geq a > 1$ , and $a > 0$ gives $a^{n} > 0$ (Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

[L7]

Order in $R$ : $0 < 1$ hence $0 < 1 < 2$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities); products and inverses of positives are positive and scaling preserves inequalities (Sign rules for products and monotonicity of multiplication, Inverses of positives are positive, and reciprocation reverses order, Field, Ordered field, Complete ordered field (least-upper-bound property)).

[L8]

Metric notions: the axioms (M1), (M2), (M3), the strong form (M3'), and the fact that a function satisfying (M1), (M2), (M3') is nonnegative and hence satisfies (M3), the maximum of two nonnegative reals being at most their sum (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Maximum and minimum of a set, Every nonempty finite set of reals has a maximum and a minimum); balls are as in Open ball, closed ball and sphere in a metric space.

Verification

technique · direct

1.1

For every $j \in N$ one has $e_{j} = j + j$ and $o_{j} = σ (j + j)$ , by induction on $j$ : at $j = 0$ , $e_{0} = 0 = 0 + 0$ and $o_{0} = σ (0) = σ (0 + 0)$ ; and if $e_{j} = j + j$ and $o_{j} = σ (j + j)$ , then $e_{σ (j)} = σ (σ (e_{j})) = σ (σ (j + j)) = σ (j) + σ (j)$ and $o_{σ (j)} = σ (σ (o_{j})) = σ (σ (σ (j + j))) = σ (σ (j) + σ (j))$ , using $σ (p) + q = σ (p + q)$ and $p + σ (q) = σ (p + q)$ .

L2L3L4

1.2

No integer $k$ satisfies $0 < k < 1$ : such a $k$ would be positive, hence the image of a natural $n \neq 0$ , so $n \geq 1$ and, the embedding being order preserving, $k \geq 1$ , contradicting $k < 1$ .

L1L4L5

1.3

A product of two odd integers is odd: $(2 a + 1) (2 c + 1) = 2 (2 a c + a + c) + 1$ by ring arithmetic.

2.1

Every natural number is $j + j$ for exactly one $j$ , or $σ (j + j)$ for exactly one $j$ , and never both: this is the disjoint-union statement for the ranges of $e$ and $o$ , rewritten through step 1.1.

step 1.1L3

2.2

No integer is both even and odd: if $2 s = 2 t + 1$ then the integer $k : = s - t$ satisfies $2 k = 1$ , and $k \leq 0$ would give $2 k \leq 0 < 1$ while $k \geq 1$ would give $2 k \geq 2 > 1$ , so $0 < k < 1$ , which step 1.2 forbids.

step 1.2L1L7

3.1

Every integer is even or odd: an integer $m \geq 0$ is the image of a natural $n$ , which by step 2.1 is $j + j$ or $σ (j + j)$ , so $m$ is $2 t$ or $2 t + 1$ with $t$ the image of $j$ , the embedding preserving addition and successors; and if $m < 0$ then $- m > 0$ is $2 t$ or $2 t + 1$ , whence $m = 2 (- t)$ or $m = 2 (- t - 1) + 1$ .

step 2.1L1L5

3.2

The representation is unique: if $2^{j} u = 2^{j^{'}} u^{'}$ with $u, u^{'}$ odd and, without loss of generality, $j \leq j^{'}$ , then $2^{j^{'}} = 2^{j} 2^{j^{'} - j}$ and cancelling the nonzero factor $2^{j}$ gives $u = 2^{j^{'} - j} u^{'}$ ; if $j^{'} > j$ then $j^{'} - j \geq 1$ and $u = 2 (2^{j^{'} - j - 1} u^{'})$ is even as well as odd, which step 2.2 forbids; so $j = j^{'}$ and then $u = u^{'}$ .

step 2.2L1L4L6

4.1

Every nonzero integer $m$ is $2^{j} u$ with $j \in N$ and $u$ odd: apply strong induction to the property $P (n)$ that every nonzero integer whose absolute value is the image of $n$ , that is every nonzero $m$ equal to the image of $n$ or to its negative, has such a representation. Given $n$ and $P (k)$ for all $k < n$ , note $n \neq 0$ since $m \neq 0$ ; by step 2.1 either $n = σ (j + j)$ , in which case $m$ is $\pm (2 t + 1)$ and hence odd by the computation of step 3.1, so $m = 2^{0} m$ works, or $n = j + j$ with $j \neq 0$ , in which case $m = 2 m^{'}$ for the nonzero integer $m^{'}$ that is the image of $j$ or its negative, and $j < n$ because $j \geq 1$ gives $n = j + j \geq j + 1 > j$ , so $P (j)$ supplies $m^{'} = 2^{i} u$ and $m = 2^{i + 1} u$ . Every nonzero integer is the image of some natural or its negative, so the conclusion holds for all of them.

step 2.1step 3.1L1L2L4L5L6

5.1

For nonzero integers $a, c$ with $a + c \neq 0$ : $v_{2} (a + c) \geq \min {v_{2} (a), v_{2} (c)}$ . Indeed write $a = 2^{s} u$ and $c = 2^{s^{'}} u^{'}$ with $u, u^{'}$ odd and, without loss of generality, $s \leq s^{'}$ ; then $a + c = 2^{s} (u + 2^{s^{'} - s} u^{'})$ , the bracket is a nonzero integer, so by step 4.1 it equals $2^{i} q$ with $q$ odd, whence $a + c = 2^{s + i} q$ and, by the uniqueness of step 3.2, $v_{2} (a + c) = s + i \geq s = \min {s, s^{'}}$ .

step 3.2step 4.1L1L6

5.2

The valuation of a nonzero rational is well defined: if $a / b = c / d$ with $a, b, c, d$ nonzero integers then $a d = c b$ , and writing $a = 2^{s} u$ , $b = 2^{t} w$ , $c = 2^{s^{'}} u^{'}$ , $d = 2^{t^{'}} w^{'}$ with all four of $u, w, u^{'}, w^{'}$ odd gives $a d = 2^{s + t^{'}} (u w^{'})$ and $c b = 2^{s^{'} + t} (u^{'} w)$ with $u w^{'}$ and $u^{'} w$ odd by step 1.3; the uniqueness of step 3.2 applied to the nonzero integer $a d$ forces $s + t^{'} = s^{'} + t$ , that is $s - t = s^{'} - t^{'}$ .

step 1.3step 3.2step 4.1L1L6

6.1

Basic properties of $∣ \cdot ∣_{2}$ : for $x \neq 0$ the value $2^{- v_{2} (x)}$ is a positive real, since $2 > 0$ and powers and inverses of positives are positive; so $∣ x ∣_{2} > 0$ for $x \neq 0$ and $∣ x ∣_{2} = 0$ exactly when $x = 0$ . Moreover $∣ - x ∣_{2} = ∣ x ∣_{2}$ , because $- x = (- a) / b$ and $- (2^{s} u) = 2^{s} (- u)$ with $- u = 2 (- t - 1) + 1$ odd when $u = 2 t + 1$ , so $v_{2} (- x) = v_{2} (x)$ .

step 5.2L1L6L7

7.1

Strong triangle inequality for $∣ \cdot ∣_{2}$ : $∣ x + y ∣_{2} \leq \max {∣ x ∣_{2}, ∣ y ∣_{2}}$ . If $x = 0$ , or $y = 0$ , or $x + y = 0$ , this is immediate from step 6.1. Otherwise write $x = a / b$ and $y = c / b$ over a common nonzero denominator $b$ , with $a, c$ nonzero integers, so that $x + y = (a + c) / b$ with $a + c \neq 0$ ; then $v_{2} (x) = v_{2} (a) - v_{2} (b)$ , $v_{2} (y) = v_{2} (c) - v_{2} (b)$ and $v_{2} (x + y) = v_{2} (a + c) - v_{2} (b)$ , so step 5.1 gives $v_{2} (x + y) \geq \min {v_{2} (x), v_{2} (y)}$ ; finally $t \mapsto 2^{t}$ is strictly increasing on $Z$ , since for $t < t^{'}$ one has $2^{t^{'}} = 2^{t} 2^{t^{'} - t}$ with $2^{t} > 0$ and $2^{t^{'} - t} > 1$ , so $2^{- v_{2} (x + y)} \leq 2^{- \min {v_{2} (x), v_{2} (y)}} = \max {2^{- v_{2} (x)}, 2^{- v_{2} (y)}}$ .

step 5.1step 5.2L1L5L6L7

8.1

Claim 1: $d_{2} (x, y) = ∣ x - y ∣_{2}$ vanishes exactly when $x = y$ by step 6.1, is symmetric because $∣ y - x ∣_{2} = ∣ - (x - y) ∣_{2} = ∣ x - y ∣_{2}$ , and satisfies $d_{2} (x, z) = ∣ (x - y) + (y - z) ∣_{2} \leq \max {∣ x - y ∣_{2}, ∣ y - z ∣_{2}} = \max {d_{2} (x, y), d_{2} (y, z)}$ by step 7.1; being nonnegative, it also satisfies the ordinary triangle inequality, the maximum of two nonnegative reals being at most their sum. So $d_{2}$ is an ultrametric on $Q$ .

step 6.1step 7.1L8

9.1

Claim 2: suppose $d_{2} (x, y) \neq d_{2} (y, z)$ and, without loss of generality, $d_{2} (x, y) < d_{2} (y, z)$ . Then $d_{2} (x, z) \leq \max {d_{2} (x, y), d_{2} (y, z)} = d_{2} (y, z)$ ; and $d_{2} (y, z) \leq \max {d_{2} (y, x), d_{2} (x, z)} = \max {d_{2} (x, y), d_{2} (x, z)}$ , where the maximum cannot be $d_{2} (x, y)$ , since that would give $d_{2} (y, z) \leq d_{2} (x, y) < d_{2} (y, z)$ ; so $d_{2} (y, z) \leq d_{2} (x, z)$ and the two are equal, that is $d_{2} (x, z) = \max {d_{2} (x, y), d_{2} (y, z)}$ .

step 8.1L7L8

9.2

Claim 3: let $y \in B (x, r)$ , so $d_{2} (x, y) < r$ . For $z \in B (y, r)$ the strong triangle inequality gives $d_{2} (x, z) \leq \max {d_{2} (x, y), d_{2} (y, z)} < r$ , so $z \in B (x, r)$ ; and for $z \in B (x, r)$ it gives $d_{2} (y, z) \leq \max {d_{2} (y, x), d_{2} (x, z)} < r$ , so $z \in B (y, r)$ . Hence $B (y, r) = B (x, r)$ .

step 8.1L8

10.1

Claims 1, 2 and 3 are established by steps 8.1, 9.1 and 9.2, so the $2$ -adic distance is an ultrametric on $Q$ in which every triangle is isosceles and every point of a ball is a centre.

step 8.1step 9.1step 9.2∎

Remarks

Where parity is spent, and where it is not. The partition of $N$ into the two ranges of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ is used exactly twice: in step 2.1, to know that a natural is even or odd, and through it in step 3.1 for the integers. Uniqueness of the valuation (step 3.2) uses only that no integer is both even and odd, which step 2.2 derives from the discreteness of $Z$ rather than from the partition.
What an ultrametric costs and what it buys. Claims 2 and 3 are formal consequences of (M3') and hold in every ultrametric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric); they are recorded here because they are the two facts that make ultrametric geometry look unlike the real line, where a triangle need not be isosceles and a ball has exactly one centre.
The general $p$ -adic absolute value, and what this item does instead. Its well-definedness needs the primality of $p$ in the form of Euclid's lemma, that $p$ dividing a product divides one of the factors. That is Euclid's lemma: if $p$ is prime and $p ∣ a b$ then $p ∣ a$ or $p ∣ b$ , and the resulting valuation on $Q^{\times}$ is The $p$ -adic valuation extends to the nonzero rationals by $v_{p} (a / b) : = v_{p} (a) - v_{p} (b) \in Z$ , independently of the representation; it satisfies $v_{p} (x y) = v_{p} (x) + v_{p} (y)$ , and $v_{p} (x + y) \geq \min {v_{p} (x), v_{p} (y)}$ whenever $x$ , $y$ and $x + y$ are nonzero, both on the primes page and both available here. This item deliberately does not use them: at $p = 2$ the statement doing the same work is that a product of odd integers is odd, proved in step 1.3 above by a one-line ring computation, so the whole development below is self-contained from parity.

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

The post-office metric $d (x, y) = ∥ x ∥ + ∥ y ∥$ for $x \neq y$ on $R^{n}$ , and its isolated points

Example

Let $n \geq 1$ and let $R^{n}$ carry the Euclidean metric $d_{2}$ of $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it. Write $0$ for the element of $R^{n}$ with all coordinates $0$ and

$∥ x ∥ : = d_{2} (x, 0) = \sqrt{\sum_{k < n} x_{k}^{2}} .$

Define the post-office metric (also called the SNCF metric) by

$ρ (x, y) : = {\begin{cases} 0 & x = y, \\ ∥ x ∥ + ∥ y ∥ & x \neq y . \end{cases}$

The name is the picture: to travel between two places you must first go to the central post office at $0$ . Then:

$ρ$ is a metric on $R^{n}$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).
Every $x \neq 0$ is an isolated point of $(R^{n}, ρ)$ (Interior, closure, boundary, limit point, isolated point and dense subset of a metric space), because $B_{ρ} (x, ∥ x ∥) = {x}$ .
$0$ is not an isolated point: every ball $B_{ρ} (0, r)$ contains points other than $0$ .

So $(R^{n}, ρ)$ has exactly one non-isolated point, which no metric equivalent to $d_{2}$ could achieve: in the Euclidean topology no point of $R^{n}$ is isolated.

Facts & Assumptions

Given: A natural $n \geq 1$ ; elements $x, y, z \in R^{n}$ ; a real $r > 0$ ; and the element $w \in R^{n}$ with $w_{0} : = r / 2$ and $w_{k} : = 0$ for $1 \leq k < n$ .

[L1]

$d_{2}$ is a metric on $R^{n}$ ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), so $∥ x ∥ = d_{2} (x, 0) \geq 0$ (Nonnegativity of a metric is a consequence of the other axioms, not an axiom) and $∥ x ∥ = 0$ exactly when $x = 0$ .

[L2]

Finite sums (Laws of finite sums and finite products, Finite sums and finite products, by recursion): splitting a sum at an index $m \leq n$ , and $\sum_{k < p} λ = p λ$ , so a sum all of whose terms are $0$ is $0$ ; and $N$ contains $0$ , so the index $0$ exists as soon as $n \geq 1$ (The natural numbers $N$ (von Neumann)).

[L3]

Square roots: $\sqrt{t^{2}} = t$ for $t \geq 0$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L4]

Halving: for a real $r > 0$ the element $r / 2 : = r (1 + 1)^{- 1}$ is positive with $r / 2 + r / 2 = r$ , hence $r / 2 < r$ ; this uses $0 < 1$ , positivity of inverses of positives, and positivity of products of positives (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, Sign rules for products and monotonicity of multiplication, Field, Ordered field).

[L5]

Order arithmetic: inequalities may be added, a nonnegative term may be dropped from the larger side, and by trichotomy $a \geq b$ rules out $a < b$ (Order is preserved by adding a constant and by adding inequalities, Ordered field, Complete ordered field (least-upper-bound property)).

[L6]

Balls, isolated points: $B_{ρ} (u, t) = {v : ρ (u, v) < t}$ , and $u$ is isolated in a set $A ∋ u$ when some ball meets $A$ only in $u$ (Open ball, closed ball and sphere in a metric space, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

Verification

technique · direct

1.1

Basic values: $ρ (u, v) \geq 0$ for all $u, v$ , being $0$ or a sum of two nonnegative numbers; $ρ$ is symmetric, since both defining clauses are; and $ρ (u, v) = 0$ exactly when $u = v$ , because for $u \neq v$ the value $∥ u ∥ + ∥ v ∥$ vanishes only if $∥ u ∥ = ∥ v ∥ = 0$ , that is only if $u = v = 0$ , contradicting $u \neq v$ .

L1L5

1.2

Triangle inequality: if $x = z$ then $ρ (x, z) = 0 \leq ρ (x, y) + ρ (y, z)$ ; if $x \neq z$ and $y = x$ , then $y \neq z$ and $ρ (x, y) + ρ (y, z) = 0 + ∥ x ∥ + ∥ z ∥ = ρ (x, z)$ ; if $x \neq z$ and $y = z$ , the same computation applies with the roles exchanged; and if $x \neq z$ with $y \neq x$ and $y \neq z$ , then $ρ (x, y) + ρ (y, z) = ∥ x ∥ + ∥ y ∥ + ∥ y ∥ + ∥ z ∥ \geq ∥ x ∥ + ∥ z ∥ = ρ (x, z)$ , since $∥ y ∥ \geq 0$ .

L1L5

1.3

The element $w$ satisfies $\sum_{k < n} w_{k}^{2} = w_{0}^{2} + 0 = (r / 2)^{2}$ , splitting the sum at index $1 \leq n$ and using that all remaining terms are $0$ ; hence $∥ w ∥ = \sqrt{(r / 2)^{2}} = r / 2 < r$ , and $w \neq 0$ because $w_{0} = r / 2 \neq 0$ .

L2L3L4

2.1

Claim 1: $ρ$ satisfies (M1) and (M2) by step 1.1 and (M3) by step 1.2, so it is a metric on $R^{n}$ .

step 1.1step 1.2

3.1

Claim 2: let $x \neq 0$ , so $∥ x ∥ > 0$ by [L1] and trichotomy. For $y \neq x$ we get $ρ (x, y) = ∥ x ∥ + ∥ y ∥ \geq ∥ x ∥$ , so $y \notin B_{ρ} (x, ∥ x ∥)$ ; and $x \in B_{ρ} (x, ∥ x ∥)$ since $ρ (x, x) = 0 < ∥ x ∥$ . Hence $B_{ρ} (x, ∥ x ∥) = {x}$ and $x$ is isolated in $R^{n}$ .

step 1.1step 2.1L1L5L6

3.2

Claim 3: let $r > 0$ . For $y \neq 0$ one has $ρ (0, y) = ∥ 0 ∥ + ∥ y ∥ = ∥ y ∥$ , so the element $w$ of step 1.3 satisfies $ρ (0, w) = ∥ w ∥ < r$ and $w \neq 0$ ; thus $B_{ρ} (0, r)$ contains a point other than $0$ , for every $r > 0$ , and $0$ is not isolated.

step 1.3step 2.1L1L6

4.1

Claims 1, 2 and 3 hold by steps 2.1, 3.1 and 3.2.

step 2.1step 3.1step 3.2∎

Remarks

The topology is almost discrete. Every singleton ${x}$ with $x \neq 0$ is open by claim 2, so every subset of $R^{n} ∖ {0}$ is open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed); the only points a set has to be careful about are those near $0$ .
This is not equivalent to any of $d_{1}$ , $d_{2}$ , $d_{\infty}$ , not even topologically. Those three share a topology (The metrics $d_{1}$ , $d_{2}$ and $d_{\infty}$ on $R^{n}$ are metrics and are Lipschitz equivalent, with explicit constants) in which no point is isolated: for $x \in R^{n}$ and $r > 0$ the element $y$ with $y_{0} : = x_{0} + r / 2$ and $y_{k} : = x_{k}$ for $1 \leq k < n$ satisfies $d_{2} (x, y) = r / 2 < r$ , by the computation of step 1.3, and differs from $x$ . In $ρ$ , by claim 2, every point but $0$ is isolated.
The same construction works over any metric space with a distinguished point, replacing $∥ x ∥$ by the distance to that point; nothing above uses more about $∥ \cdot ∥$ than nonnegativity and vanishing exactly at $0$ .

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

$\min (∣ x - y ∣, 1)$ on $R$ has the usual topology and diameter at most $1$

Example

On $R$ , let $d_{R} (x, y) = ∣ x - y ∣$ be the usual metric (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) and put

$ρ (x, y) : = \min {∣ x - y ∣, 1} (x, y \in R) .$

Then:

$ρ$ is a metric on $R$ , uniformly equivalent and therefore topologically equivalent to $d_{R}$ (Topologically, uniformly and Lipschitz equivalent metrics on a set, Lipschitz equivalence implies uniform equivalence implies topological equivalence); so $(R, ρ)$ has exactly the usual topology of the real line.
$(R, ρ)$ is a bounded metric space and $diam (R) = 1$ in the metric $ρ$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), whereas $(R, d_{R})$ is not bounded at all and has no diameter.

This is the concrete instance of $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology on the real line, and it is the witness used for the failure of two plausible-sounding claims: that boundedness is topological, and that topologically equivalent metrics are Lipschitz equivalent.

Facts & Assumptions

Given: The real line with $d_{R} (x, y) = ∣ x - y ∣$ and the function $ρ = \min {d_{R}, 1}$ ; the set $D : = {ρ (x, y) : x, y \in R}$ .

[L1]

$d_{R}$ is a metric on $R$ and $(R, d_{R})$ is not bounded, so it has no diameter (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, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[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).

[L3]

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

[L4]

The minimum of a two-element set of reals exists and is one of them (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set); $∣ 1 - 0 ∣ = 1$ and $0 < 1$ (Basic properties of the absolute value, Absolute value in an ordered field, The multiplicative identity is positive).

[L5]

The diameter is the least upper bound of the set of distances, so it is $\geq$ every distance and $\leq$ every upper bound of them; and it is unique (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Suprema and infima are unique, Complete ordered field (least-upper-bound property), Ordered field).

Verification

technique · direct

1.1

By [L1] the function $d_{R}$ is a metric on $R$ and $R$ is not bounded in it.

1.2

$ρ (0, 1) = \min {∣ 0 - 1 ∣, 1} = \min {1, 1} = 1$ , so $1 \in D$ .

2.1

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

step 1.1L2

3.1

Claim 2: the diameter of $R$ in $ρ$ is an upper bound of $D$ and $1 \in D$ by step 1.2, so it is $\geq 1$ ; combined with $\leq 1$ from step 2.1 this gives $diam (R) = 1$ in the metric $ρ$ , while by step 1.1 the space has no diameter at all in $d_{R}$ .

step 1.1step 1.2step 2.1L5

3.2

Claim 1: $ρ$ is a metric uniformly equivalent to $d_{R}$ by step 2.1, hence topologically equivalent to it, so the metric topology of $ρ$ is the usual topology of $R$ .

step 2.1L3

4.1

Claims 1 and 2 hold by steps 3.2 and 3.1.

step 3.1step 3.2∎

Remarks

Nothing about $R$ is special here except unboundedness. The same computation on any unbounded metric space gives a bounded metric with the same topology ( $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology); the real line is chosen because it is the space every reader already has.
The diameter is exactly $1$ and not less, which is what step 1.2 contributes: for the construction $\min {d, 1}$ the bound of $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology is attained as soon as $d$ takes some value $\geq 1$ , because the new metric then takes the value $1$ itself.
The two metrics are not Lipschitz equivalent, so this pair also witnesses the strictness of the first implication of Lipschitz equivalence implies uniform equivalence implies topological equivalence; that computation is On $R$ the metrics $∣ x - y ∣$ and $\min (∣ x - y ∣, 1)$ are uniformly but not Lipschitz equivalent.

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

In ${0} \cup [1, 2]$ with the metric of $R$ , the closure of $B (0, 1) = {0}$ is ${0}$ while the closed ball is ${0, 1}$

Statement refuted

Refuted claim: in every metric space, the closure of the open ball $B (x, r)$ is the closed ball $\overset{ˉ}{B} (x, r)$ (FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ ).

The witness is the metric subspace

$X : = {0} \cup [1, 2] \subseteq R$

of the real line with its usual metric (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, Isometry, isometric embedding, and the subspace metric on a subset, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), with $x = 0$ and $r = 1$ . In it

$B_{X} (0, 1) = {0}, \overline{B_{X} (0, 1)} = {0}, {\overset{ˉ}{B}}_{X} (0, 1) = {0, 1},$

so the closure of the open ball is a proper subset of the closed ball of the same centre and radius. The inclusion $\overline{B (x, r)} \subseteq \overset{ˉ}{B} (x, r)$ that does hold in general is proved in FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ and is not repeated.

Facts & Assumptions

Given: The real line with $d_{R} (u, v) = ∣ u - v ∣$ , and $X : = {0} \cup [1, 2]$ with the subspace metric $d : = d_{R} ↾ (X \times X)$ .

[L1]

The subspace metric makes $X$ a metric space, and its balls are traces of the balls of $R$ (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).

[L2]

Balls: $B_{X} (a, s) = {y \in X : d (a, y) < s}$ and ${\overset{ˉ}{B}}_{X} (a, s) = {y \in X : d (a, y) \leq s}$ (Open ball, closed ball and sphere in a metric space).

[L3]

Open and closed sets of a metric space, and the closure as the set of adherent points (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, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space).

[L4]

A closed set equals its own closure, the closure being the smallest closed superset (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).

[L5]

Absolute value and order: $∣ t ∣ = t$ for $t \geq 0$ , $∣ - t ∣ = ∣ t ∣$ , and $0 < 1$ ; by trichotomy $t \geq 1$ excludes $t < 1$ (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)).

[L6]

Intervals: $[1, 2] = {t \in R : 1 \leq t \leq 2}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct

1.1

Every $y \in [1, 2]$ has $y \geq 1$ , so $d (0, y) = ∣ 0 - y ∣ = ∣ - y ∣ = y \geq 1$ ; and $d (0, 0) = 0 < 1$ .

L1L5L6

2.1

Hence $B_{X} (0, 1) = {y \in X : d (0, y) < 1} = {0}$ , since no $y \in [1, 2]$ satisfies $d (0, y) < 1$ ; and ${\overset{ˉ}{B}}_{X} (0, 1) = {y \in X : d (0, y) \leq 1} = {0, 1}$ , since among the points of $[1, 2]$ exactly $y = 1$ satisfies $y \leq 1$ .

step 1.1L2L5L6

2.2

The set $[1, 2] = X ∖ {0}$ is open in $X$ : for $y \in [1, 2]$ the ball $B_{X} (y, 1)$ cannot contain $0$ , because $d (0, y) \geq 1$ by step 1.1, so $B_{X} (y, 1) \subseteq [1, 2]$ . Therefore ${0}$ is closed in $X$ .

step 1.1L2L3

3.1

Since ${0}$ is closed it equals its own closure, so $\overline{B_{X} (0, 1)} = \overline{{0}} = {0}$ , whereas ${\overset{ˉ}{B}}_{X} (0, 1) = {0, 1}$ contains the point $1 \neq 0$ .

step 2.1step 2.2L4L5

4.1

The two sets are therefore different, and $(X, d)$ with $x = 0$ , $r = 1$ refutes the claim: the closure of an open ball can be a proper subset of the closed ball of the same centre and radius.

step 3.1∎

Remarks

Nothing pathological is used. $X$ is an unremarkable bounded subset of the real line and the metric is the one inherited from $R$ ; the only feature exploited is that $X$ has a gap, so that the point $1$ of the closed ball cannot be approached from inside $B_{X} (0, 1)$ .
A starker version lives in the discrete metric (The discrete metric induces the discrete topology, in which every subset is clopen) on a set with at least two points, where $\overline{B (p, 1)} = {p}$ while $\overset{ˉ}{B} (p, 1)$ is the entire space; the two witnesses refute the claim in the same way, at different scales.
The equality does hold in $R^{n}$ with any of $d_{1}$ , $d_{2}$ , $d_{\infty}$ , which is where the false intuition comes from. This library does not prove it, and neither $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it nor The metrics $d_{1}$ , $d_{2}$ and $d_{\infty}$ on $R^{n}$ are metrics and are Lipschitz equivalent, with explicit constants contains it; the usual argument runs along the segment from the centre to the point in question, and no such segment need exist in a general metric space.

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

In the discrete metric the boundary of $B (p, 1)$ is empty while the sphere of radius $1$ is everything but $p$

Statement refuted

Refuted claim: in every metric space, the boundary of the open ball is the sphere of the same centre and radius,

$\partial B (x, r) = S (x, r)$

(Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, Open ball, closed ball and sphere in a metric space).

The witness is any set $X$ with at least two points, carrying the discrete metric $δ$ (The discrete metric induces the discrete topology, in which every subset is clopen), together with $x = p \in X$ and $r = 1$ . There

$B (p, 1) = {p}, \partial B (p, 1) = \emptyset, S (p, 1) = X ∖ {p} \neq \emptyset .$

Facts & Assumptions

Given: A set $X$ with at least two points, the discrete metric $δ$ on it, a point $p \in X$ and a point $q \in X$ with $q \neq p$ .

[L1]

The discrete metric: $δ$ is a metric, $B (p, 1) = {p}$ , $S (p, 1) = X ∖ {p}$ , and every subset of $X$ is both open and closed (The discrete metric induces the discrete topology, in which every subset is clopen, 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, 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).

[L2]

Interior, closure, boundary: $int (A)$ is the largest open subset of $A$ , $\overline{A}$ the smallest closed superset of $A$ , and $\partial A = \overline{A} ∖ int (A)$ (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).

Counterexample

technique · direct

1.1

In $(X, δ)$ the open ball is $B (p, 1) = {p}$ and the sphere is $S (p, 1) = X ∖ {p}$ , which is nonempty because $q \neq p$ lies in it.

givenL1

1.2

The set ${p}$ is open and closed in $(X, δ)$ , every subset of a discrete metric space being clopen.

2.1

Hence $int ({p}) = {p}$ , since ${p}$ is an open subset of itself and the interior is the largest one; and $\overline{{p}} = {p}$ , since ${p}$ is a closed superset of itself and the closure is the smallest one.

step 1.2L2

3.1

Therefore $\partial B (p, 1) = \overline{{p}} ∖ int ({p}) = {p} ∖ {p} = \emptyset$ , while $S (p, 1)$ contains $q$ and is not empty.

step 1.1step 2.1L2

4.1

The two sets differ, so the discrete metric on any set with at least two points refutes the claim.

step 3.1∎

Remarks

One inclusion does survive. In any metric space $\partial B (x, r) \subseteq S (x, r)$ : the ball $B (x, r)$ is open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed), so it is its own interior and $\partial B (x, r) = \overline{B (x, r)} ∖ B (x, r)$ , which sits inside $\overset{ˉ}{B} (x, r) ∖ B (x, r) = S (x, r)$ because $\overline{B (x, r)} \subseteq \overset{ˉ}{B} (x, r)$ (FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ , Open ball, closed ball and sphere in a metric space). What the witness above shows is that the inclusion can be strict, and as strict as possible: empty on the left, everything but the centre on the right.
It is the same defect as FALSE: in every metric space the closure of $B (x, r)$ is the closed ball of radius $r$ : the names open ball, closed ball and sphere are labels for three sets defined by three inequalities (Open ball, closed ball and sphere in a metric space), and none of the topological relations suggested by the words is automatic.
Every point of a discrete space is isolated, so no ball has any boundary at all; the post-office metric (The post-office metric $d (x, y) = ∥ x ∥ + ∥ y ∥$ for $x \neq y$ on $R^{n}$ , and its isolated points) shows the intermediate case, where all but one point is isolated.

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

$Z$ and ${n + 1 / n : n \geq 2}$ are disjoint closed subsets of $R$ at distance $0$ , so the set-to-set distance is not a metric

Statement refuted

Refuted claim: the set-to-set distance $d (A, B) = \inf {d (a, b) : a \in A, b \in B}$ of Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space is a metric on the nonempty subsets of a metric space; specifically, that it satisfies the separation axiom (M1) of Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, so that $d (A, B) = 0$ forces $A = B$ .

Work in $R$ with its usual metric (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) and write $m$ for the canonical copy of an integer $m$ inside $R$ (The integers as equivalence classes of pairs of naturals, The integers embed in the rationals, The unique embedding of ℚ into an ordered field). Put

$A : = {m : m \in Z}, B : = {n + \frac{1}{n} : n \in N, n \geq 2} .$

Then $A$ and $B$ are nonempty, disjoint, closed in $R$ , and

$d (A, B) = 0 while A \neq B .$

So (M1) fails for the set-to-set distance, and it fails on a pair of closed sets: closedness is not the missing hypothesis. Moreover $d (b, A) > 0$ for every individual $b \in B$ , so the infimum over pairs is not attained anywhere.

Facts & Assumptions

Given: The real line with $d_{R} (u, v) = ∣ u - v ∣$ ; the sets $A$ and $B$ above; and, for a subset $E \subseteq R$ , the property of being $c$ -separated for a real $c > 0$ , meaning $∣ e - e^{'} ∣ \geq c$ whenever $e, e^{'} \in E$ with $e \neq e^{'}$ .

[L1]

The embeddings $Z \to Q \to R$ are injective and order preserving (The integers embed in the rationals, The unique embedding of ℚ into an ordered field, The integers as equivalence classes of pairs of naturals), and $Z$ is totally ordered (The integers form a totally ordered ring); every integer $\geq 0$ is the image of a unique natural under $N \to Z$ , which is injective and order preserving (The naturals embed in the integers); and a natural $n \neq 0$ satisfies $n \geq 1$ (Discreteness: $σ (n)$ is the immediate successor, The natural numbers $N$ (von Neumann)).

[L2]

Canonical naturals in $R$ : $n \cdot 1_{R} > 0$ for $n \geq 1$ and $n \mapsto n \cdot 1_{R}$ is strictly increasing (Canonical naturals are positive and strictly increasing); reciprocals of positives are positive and reverse the order (Inverses of positives are positive, and reciprocation reverses order); and $r > 1$ gives $0 < 1 / r < 1$ (Reciprocals and order: $1 / r$ against $1$ ).

[L3]

[L4]

Absolute value: $∣ t ∣ \geq 0$ , $∣ t ∣ = t$ for $t \geq 0$ , $∣ - t ∣ = ∣ t ∣$ , and $∣ t ∣ < c$ is equivalent to $- c < t < c$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L5]

Infima: $d (x, E)$ and $d (E, F)$ exist for nonempty sets, being infima of nonempty sets 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, Greatest lower bound (infimum)); and $ℓ = \inf S$ for a lower bound $ℓ$ of a nonempty $S$ bounded below exactly when for every real $ε > 0$ some $s \in S$ has $s < ℓ + ε$ (Epsilon characterisation of the infimum).

[L6]

The closure of a nonempty $E$ is ${x : d (x, E) = 0}$ , and $E$ is closed exactly when $E = \overline{E}$ (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, 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, Open ball, closed ball and sphere in a metric space, Isometry, isometric embedding, and the subspace metric on a subset).

[L7]

Order and field arithmetic in $R$ : halving a positive real, adding inequalities, scaling by a positive, the minimum of a two-element set, and trichotomy (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, The multiplicative identity is positive, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Field, Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct

1.1

No canonical integer lies strictly between $0$ and $1$ : if $0 < k < 1$ in $R$ for an integer $k$ , then $0 < k < 1$ already in $Z$ , because the embedding is order preserving and injective and the order of $Z$ is total; then $k > 0$ makes $k$ the image of a natural $n \neq 0$ , so $n \geq 1$ and hence $k \geq 1$ , contradicting $k < 1$ .

L1L7

1.2

$B$ is $\frac{1}{2}$ -separated: for naturals $p > q \geq 2$ one has $p - q \geq 1$ and $0 < 1 / p < 1 / q \leq 1 / 2$ , so $(p + 1 / p) - (q + 1 / q) = (p - q) - (1 / q - 1 / p) \geq 1 - 1 / q \geq 1 - 1 / 2 = 1 / 2$ ; hence distinct elements of $B$ differ by at least $1 / 2$ in absolute value.

L2L4L7

1.3

A $c$ -separated set $E \subseteq R$ is closed: let $x \notin E$ . The ball $B (x, c / 2)$ contains at most one point of $E$ , since two distinct points of it would be within $c / 2 + c / 2 = c$ of each other, contradicting $c$ -separation with a strict inequality. If it contains none, then $B (x, c / 2) \cap E = \emptyset$ . If it contains exactly one point $e$ , then $x \neq e$ gives $t : = ∣ x - e ∣ > 0$ , and the radius $r : = \min {c / 2, t} > 0$ has $B (x, r) \cap E = \emptyset$ , because any point of $E$ in $B (x, r) \subseteq B (x, c / 2)$ would have to be $e$ , while $∣ x - e ∣ = t \geq r$ puts $e$ outside $B (x, r)$ . So the complement of $E$ is open and $E$ is closed.

L4L6L7

2.1

$A$ is $1$ -separated: for integers $m \neq m^{'}$ the difference $k : = m - m^{'}$ is a nonzero integer, so by step 1.1 and trichotomy either $k \geq 1$ or $- k \geq 1$ , and in both cases $∣ m - m^{'} ∣ = ∣ k ∣ \geq 1$ .

step 1.1L1L4L7

2.2

$A$ and $B$ are disjoint and both nonempty: $0 \in A$ and $2 + 1 / 2 \in B$ ; and if $n + 1 / n = m$ for a natural $n \geq 2$ and an integer $m$ , then the integer $m - n$ satisfies $m - n = 1 / n$ with $0 < 1 / n < 1$ by [L2], contradicting step 1.1.

step 1.1L1L2L7

3.1

$A$ and $B$ are closed subsets of $R$ , by steps 2.1, 1.2 and 1.3 with $c = 1$ and $c = 1 / 2$ respectively.

step 1.2step 1.3step 2.1

3.2

$d (A, B) = 0$ : the set ${∣ a - b ∣ : a \in A, b \in B}$ is nonempty and bounded below by $0$ , and for each natural $n \geq 2$ it contains $∣ n - (n + 1 / n) ∣ = 1 / n$ ; given a real $ε > 0$ , [L3] supplies a natural $n \geq 1$ with $1 / n < ε$ , and then $n + 1 \geq 2$ with $1 / (n + 1) < 1 / n < ε = 0 + ε$ , so the infimum is $0$ by the epsilon characterisation.

step 2.2L2L3L4L5

4.1

Yet every single point of $B$ is at positive distance from $A$ : for $b \in B$ we have $b \notin A = \overline{A}$ by steps 2.2 and 3.1, so $d (b, A) \neq 0$ by the description of the closure, and $d (b, A) \geq 0$ , hence $d (b, A) > 0$ .

step 2.2step 3.1L5L6L7

5.1

So $A$ and $B$ are nonempty disjoint closed subsets of $R$ with $d (A, B) = 0$ and $A \neq B$ : the separation axiom (M1) fails for the set-to-set distance, and it fails even on closed sets and even though each individual point-to-set distance is strictly positive.

step 2.2step 3.2step 4.1∎

Remarks

What survives. The set-to-set distance is symmetric and vanishes on $A = A$ , and on singletons it reduces to the metric. It satisfies no useful triangle inequality either: in $R$ the sets $A = {0}$ , $B = {0, 10}$ , $C = {10}$ have $d (A, B) = d (B, C) = 0$ while $d (A, C) = 10$ , so $d (A, C) \leq d (A, B) + d (B, C)$ fails. The construction that does give a metric on a family of sets is the Hausdorff distance, which is taken up on a later page and not defined here.
Where the intuition breaks. For a nonempty $A$ , the function $x \mapsto d (x, A)$ vanishes exactly on $\overline{A}$ (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), so a point at distance $0$ from a closed set does lie in it. The set-to-set distance takes an infimum over a second variable as well, and an infimum of a family of positive numbers can be $0$ ; step 4.1 is exactly the record of that.
The two sets approach each other only along their tails. The point $n + 1 / n$ of $B$ sits at distance exactly $1 / n$ from the integer $n$ ; those distances are all positive, and by step 3.2 their infimum is $0$ . That is the whole mechanism: the distance is driven to $0$ by pairs with $n$ arbitrarily large, never by any single pair.

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

On $(0, \infty)$ the metrics $∣ x - y ∣$ and $∣ 1 / x - 1 / y ∣$ have the same topology and are not uniformly equivalent

Statement refuted

Refuted claim: topologically equivalent metrics are uniformly equivalent; equivalently, the implication "uniformly equivalent implies topologically equivalent" of Lipschitz equivalence implies uniform equivalence implies topological equivalence reverses.

Let $X : = (0, \infty) = {t \in R : t > 0}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) carry the metric $d (x, y) : = ∣ x - y ∣$ inherited from the real line (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, Isometry, isometric embedding, and the subspace metric on a subset), and put

$ρ (x, y) : = ∣ \frac{1}{x} - \frac{1}{y} ∣ (x, y \in X) .$

Then $ρ$ is a metric on $X$ , the two metrics are topologically equivalent, and they are not uniformly equivalent (Topologically, uniformly and Lipschitz equivalent metrics on a set). So the second implication of the hierarchy is strict.

Facts & Assumptions

Given: The set $X = (0, \infty)$ , the metrics $d$ and $ρ$ above, the map $ι : X \to X$ with $ι (x) : = x^{- 1}$ , a point $a \in X$ , a real $ε > 0$ , and a real $δ > 0$ .

[L1]

$d_{R} (u, v) = ∣ u - v ∣$ is a metric on $R$ and restricts to a metric on any subset (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, 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).

[L2]

Inverses: $u > 0$ gives $u^{- 1} > 0$ and $(u^{- 1})^{- 1} = u$ ; and $0 < u < v$ gives $0 < v^{- 1} < u^{- 1}$ (Inverses of positives are positive, and reciprocation reverses order, Field).

[L3]

Absolute value: $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ , $∣ u ∣ \geq 0$ , $∣ u ∣ = 0$ exactly when $u = 0$ , $∣ - u ∣ = ∣ u ∣$ , and $∣ u ∣ < c$ is equivalent to $- c < u < c$ for $c > 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L4]

Order arithmetic: scaling a strict inequality by a positive element (Sign rules for products and monotonicity of multiplication), adding a constant to an inequality (Order is preserved by adding a constant and by adding inequalities), transitivity and trichotomy (Ordered field, Complete ordered field (least-upper-bound property)); $0 < 1$ (The multiplicative identity is positive); halving a positive real; and the minimum of a two-element set of reals, which is one of the two (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L5]

Reciprocal Archimedean property: for every real $t > 0$ there is a natural $n \geq 1$ with $1 / n < t$ (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); and $n \cdot 1_{R} > 0$ for $n \geq 1$ (Canonical naturals are positive and strictly increasing, The natural numbers $N$ (von Neumann)).

[L6]

Open sets, balls, and the two notions of equivalence (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, Open ball, closed ball and sphere in a metric space, Topologically, uniformly and Lipschitz equivalent metrics on a set, Injection, surjection, bijection).

Counterexample

technique · direct

1.1

The map $ι$ is a well defined bijection of $X$ onto itself with $ι \circ ι = {i d}_{X}$ , since $x > 0$ gives $x^{- 1} > 0$ and $(x^{- 1})^{- 1} = x$ ; and for $x, y \in X$ one has $ρ (x, y) = d (ι x, ι y)$ together with the identity $ρ (x, y) = ∣ x - y ∣ (x y)^{- 1}$ , because $x^{- 1} - y^{- 1} = (y - x) (x y)^{- 1}$ and $∣ y - x ∣ = ∣ x - y ∣$ .

L2L3L4

2.1

$ρ$ is a metric on $X$ : it inherits symmetry and the triangle inequality from $d$ through $ι$ , and $ρ (x, y) = 0$ gives $d (ι x, ι y) = 0$ , hence $ι x = ι y$ and, $ι$ being injective, $x = y$ ; conversely $ρ (x, x) = 0$ .

step 1.1L1L2L3

2.2

Estimate A: put $δ_{A} : = \min {a / 2, ε a^{2} / 2} > 0$ . If $x \in X$ and $∣ x - a ∣ < δ_{A}$ then $x > a - δ_{A} \geq a - a / 2 = a / 2 > 0$ , so $x a > a^{2} / 2 > 0$ and $(x a)^{- 1} < 2 a^{- 2}$ ; hence $ρ (x, a) = ∣ x - a ∣ (x a)^{- 1} < δ_{A} (x a)^{- 1} < 2 δ_{A} a^{- 2} \leq ε$ when $x \neq a$ , and $ρ (a, a) = 0 < ε$ otherwise. So $B_{d} (a, δ_{A}) \subseteq B_{ρ} (a, ε)$ .

step 1.1L2L3L4

2.3

Uniform equivalence fails: suppose some $δ > 0$ satisfied $d (x, y) < δ \Rightarrow ρ (x, y) < 1$ for all $x, y \in X$ . Choose a natural $n \geq 1$ with $1 / n < δ$ and put $x : = 1 / n$ , $y : = 1 / (n + 1)$ , both in $X$ . Then $d (x, y) = ∣ 1 / n - 1 / (n + 1) ∣ < 1 / n < δ$ , because $0 < 1 / (n + 1) < 1 / n$ ; but $ρ (x, y) = ∣ n - (n + 1) ∣ = ∣ - 1 ∣ = 1$ , so $ρ (x, y) < 1$ fails. Hence no such $δ$ exists, and the pair $d, ρ$ is not uniformly equivalent.

step 1.1L2L3L4L5L6

3.1

Estimate B: apply estimate A at the point $ι (a) = a^{- 1} \in X$ to get $δ_{B} > 0$ with $∣ u - a^{- 1} ∣ < δ_{B} \Rightarrow ρ (u, a^{- 1}) < ε$ for $u \in X$ ; substituting $u : = ι (x) = x^{- 1}$ , which runs over $X$ as $x$ does, and using $ι \circ ι = {i d}_{X}$ , this reads $ρ (x, a) < δ_{B} \Rightarrow d (x, a) < ε$ , that is $B_{ρ} (a, δ_{B}) \subseteq B_{d} (a, ε)$ .

step 1.1step 2.2L2L3

4.1

Topological equivalence: if $U$ is $ρ$ -open and $a \in U$ , take $ε > 0$ with $B_{ρ} (a, ε) \subseteq U$ and then $δ_{A}$ from step 2.2, so $B_{d} (a, δ_{A}) \subseteq U$ and $U$ is $d$ -open; if $U$ is $d$ -open and $a \in U$ , take $ε > 0$ with $B_{d} (a, ε) \subseteq U$ and then $δ_{B}$ from step 3.1, so $B_{ρ} (a, δ_{B}) \subseteq U$ and $U$ is $ρ$ -open. Hence the two metric topologies coincide.

step 2.2step 3.1L6

5.1

So $d$ and $ρ$ are topologically equivalent metrics on $X$ that are not uniformly equivalent, which refutes the claim and shows that the implication from uniform to topological equivalence in Lipschitz equivalence implies uniform equivalence implies topological equivalence does not reverse.

step 2.1step 2.3step 4.1∎

Remarks

Why the failure is at the origin end. The pairs $1 / n$ and $1 / (n + 1)$ get arbitrarily close in $d$ while their images $n$ and $n + 1$ under $ι$ stay a fixed distance apart. Uniform equivalence would have to control this with a single $δ$ , and no single $δ$ can, because $ι$ stretches by the unbounded factor $(x y)^{- 1}$ near $0$ .
The two metrics are isometric copies of each other, via $ι$ : the map $ι : (X, ρ) \to (X, d)$ satisfies $d (ι x, ι y) = ρ (x, y)$ by step 1.1, so the two spaces are isometric (Isometry, isometric embedding, and the subspace metric on a subset). Being isometric as spaces says nothing about the identity map being uniformly bicontinuous, and that is exactly the distinction this example draws.
Completeness is the usual casualty. Uniform equivalence preserves Cauchy sequences and topological equivalence does not; here the sequence $1 / n$ is Cauchy for $d$ and not for $ρ$ , since $ρ (1 / n, 1 / m) = ∣ n - m ∣$ . Cauchy sequences in a metric space are defined on a later page and are not available here, so that comparison is orientation only and belongs to the completeness page.

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

On $R$ the metrics $∣ x - y ∣$ and $\min (∣ x - y ∣, 1)$ are uniformly but not Lipschitz equivalent

Statement refuted

Refuted claim: uniformly equivalent metrics are Lipschitz equivalent; equivalently, the implication "Lipschitz equivalent implies uniformly equivalent" of Lipschitz equivalence implies uniform equivalence implies topological equivalence reverses.

On $R$ take the usual metric $d (x, y) : = ∣ x - y ∣$ (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) and

$ρ (x, y) : = \min {∣ x - y ∣, 1},$

the minimum being that of a two-element set of reals (Maximum and minimum of a set).

These are uniformly equivalent ( $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology) and are not Lipschitz equivalent (Topologically, uniformly and Lipschitz equivalent metrics on a set), because a Lipschitz bound $α d \leq ρ$ with $α > 0$ would force $d$ to be bounded by $1 / α$ , and $d$ is unbounded on the real line.

Facts & Assumptions

Given: The real line with $d (x, y) = ∣ x - y ∣$ and $ρ = \min {d, 1}$ .

[L1]

$d$ is a metric on $R$ (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, satisfies $\min {d, 1} \leq 1$ everywhere, 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, Topologically, uniformly and Lipschitz equivalent metrics on a set, Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[L3]

Lipschitz equivalence of $d$ and $ρ$ means there are reals $α, β > 0$ with $α d (x, y) \leq ρ (x, y) \leq β d (x, y)$ for all $x, y$ (Topologically, uniformly and Lipschitz equivalent metrics on a set).

[L4]

Archimedean property: for every real $w$ 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]

Inverses and scaling: $α > 0$ gives $α^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order), and multiplying an inequality by a positive preserves it (Sign rules for products and monotonicity of multiplication); $0 < 1$ (The multiplicative identity is positive); trichotomy and transitivity (Ordered field, Complete ordered field (least-upper-bound property)); $∣ t ∣ = t$ for $t \geq 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

Counterexample

technique · direct

1.1

$ρ$ is a metric on $R$ and is uniformly equivalent to $d$ .

L1L2

1.2

$ρ (x, y) \leq 1$ for all $x, y \in R$ .

2.1

Suppose $d$ and $ρ$ were Lipschitz equivalent, with constants $α, β > 0$ as in [L3]. Then for all $x, y$ we would have $α ∣ x - y ∣ \leq ρ (x, y) \leq 1$ .

step 1.2L3L5

3.1

Apply the Archimedean property to $w : = α^{- 1}$ , which is a positive real: there is a natural $n \geq 1$ with $α^{- 1} < n \cdot 1_{R}$ , and multiplying by $α > 0$ gives $1 < α (n \cdot 1_{R})$ . Taking $x : = n \cdot 1_{R}$ and $y : = 0$ , so that $∣ x - y ∣ = n \cdot 1_{R}$ since $n \cdot 1_{R} > 0$ , step 2.1 gives $α (n \cdot 1_{R}) \leq 1$ , contradicting $1 < α (n \cdot 1_{R})$ by trichotomy.

step 2.1L4L5

4.1

No such constants exist, so $d$ and $ρ$ are uniformly equivalent metrics on $R$ that are not Lipschitz equivalent; the implication from Lipschitz to uniform equivalence in Lipschitz equivalence implies uniform equivalence implies topological equivalence therefore does not reverse.

step 1.1step 3.1∎

Remarks

Only the lower Lipschitz bound fails. The upper one holds with $β = 1$ , since $ρ \leq d$ always; it is the requirement that $ρ$ dominate a positive multiple of $d$ that a bounded metric can never meet against an unbounded one.
The same pair also shows boundedness is not topological (FALSE: boundedness of a metric space is determined by its topology, $R$ carries both an unbounded and a bounded metric inducing the same topology), which is no accident: Lipschitz equivalence is exactly the level of the hierarchy that preserves boundedness, and this is the pair that sits just below it.
Every metric space furnishes such a pair unless it is already bounded, by $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology; the real line is chosen only because its unboundedness is already recorded (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).

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

$R$ carries both an unbounded and a bounded metric inducing the same topology

Statement refuted

Refuted claim: boundedness of a metric space is determined by its topology (FALSE: boundedness of a metric space is determined by its topology).

The witness is the real line carrying two metrics at once:

$d (x, y) = ∣ x - y ∣ and ρ (x, y) = \min {∣ x - y ∣, 1} .$

They induce the same topology, $(R, d)$ is unbounded and has no diameter, and $(R, ρ)$ is bounded with diameter exactly $1$ ( $\min (∣ x - y ∣, 1)$ on $R$ has the usual topology and diameter at most $1$ , 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, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

Facts & Assumptions

Given: The real line with the two metrics $d$ and $ρ$ above.

[L1]

$d$ is a metric on $R$ and $(R, d)$ is not bounded, so it has no diameter (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, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[L2]

$ρ$ is a metric on $R$ , uniformly equivalent to $d$ , and $(R, ρ)$ is bounded with $diam (R) = 1$ ( $\min (∣ x - y ∣, 1)$ on $R$ has the usual topology and diameter at most $1$ , $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology).

[L3]

Uniform equivalence implies topological equivalence, that is equality of the metric topologies (Lipschitz equivalence implies uniform equivalence implies topological equivalence, Topologically, uniformly and Lipschitz equivalent metrics on a 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).

Counterexample

technique · direct

1.1

By [L1] the space $(R, d)$ is a metric space that is not bounded.

1.2

By [L2] the space $(R, ρ)$ is a metric space that is bounded, with diameter $1$ , and $ρ$ is uniformly equivalent to $d$ .

2.1

By [L3] the two metrics are topologically equivalent, so $T_{d} = T_{ρ}$ : the two metric spaces have exactly the same open sets, the same closed sets, the same convergent sequences and the same continuous maps.

step 1.2L3

3.1

So one and the same set with one and the same topology carries a bounded metric and an unbounded metric; boundedness is therefore not determined by the topology, and the claim is refuted.

step 1.1step 2.1∎

Remarks

Which level of the hierarchy does preserve boundedness. Lipschitz equivalence does, since $ρ \leq β d$ turns a $d$ -ball into a $ρ$ -ball of radius $β$ times as large (Topologically, uniformly and Lipschitz equivalent metrics on a set). The present pair is uniformly but not Lipschitz equivalent (On $R$ the metrics $∣ x - y ∣$ and $\min (∣ x - y ∣, 1)$ are uniformly but not Lipschitz equivalent), which is exactly the room in which boundedness is lost.
Nor is the diameter a topological invariant, even up to a constant: on $R$ the same construction with $\min {d, c}$ in place of $\min {d, 1}$ , for any real $c > 0$ , gives diameter $c$ while leaving the topology alone.
The general statement behind this witness is that every metric space is boundedly remetrisable ( $\min (d, 1)$ and $d / (1 + d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology), so the failure is not a peculiarity of the real line but the normal state of affairs.

Sources

Standard references

Recommended treatments; not extraction sources.

Metric Spaces: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The metrics d1d1​, d2d2​ and d∞d∞​ on RnRn are metrics and are Lipschitz equivalent, with explicit constants

Example

Facts & Assumptions

Verification

Remarks

The discrete metric induces the discrete topology, in which every subset is clopen

Example

Facts & Assumptions

Verification

Remarks

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

Example

Facts & Assumptions

Verification

Remarks

The 22-adic absolute value gives an ultrametric on QQ, in which every triangle is isosceles and every point of a ball is a centre

Example

Facts & Assumptions

Verification

Remarks

The post-office metric d(x,y)=∥x∥+∥y∥d(x,y)=∥x∥+∥y∥ for x≠yx=y on RnRn, and its isolated points

Example

Facts & Assumptions

Verification

Remarks

min⁡(∣x−y∣,1)min(∣x−y∣,1) on RR has the usual topology and diameter at most 11

Example

Facts & Assumptions

Verification

Remarks

In {0}∪[1,2]{0}∪[1,2] with the metric of RR, the closure of B(0,1)={0}B(0,1)={0} is {0}{0} while the closed ball is {0,1}{0,1}

Statement refuted

Facts & Assumptions

Counterexample

Remarks

In the discrete metric the boundary of B(p,1)B(p,1) is empty while the sphere of radius 11 is everything but pp

Statement refuted

Facts & Assumptions

Counterexample

Remarks

ZZ and {n+1/n:n≥2}{n+1/n:n≥2} are disjoint closed subsets of RR at distance 00, so the set-to-set distance is not a metric

Statement refuted

Facts & Assumptions

Counterexample

Remarks

On (0,∞)(0,∞) the metrics ∣x−y∣∣x−y∣ and ∣1/x−1/y∣∣1/x−1/y∣ have the same topology and are not uniformly equivalent

Statement refuted

Facts & Assumptions

Counterexample

Remarks

On RR the metrics ∣x−y∣∣x−y∣ and min⁡(∣x−y∣,1)min(∣x−y∣,1) are uniformly but not Lipschitz equivalent

Statement refuted

Facts & Assumptions

Counterexample

Remarks

RR carries both an unbounded and a bounded metric inducing the same topology

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Sources

Standard references

The metrics $d_{1}$ , $d_{2}$ and $d_{\infty}$ on $R^{n}$ are metrics and are Lipschitz equivalent, with explicit constants

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

The $2$ -adic absolute value gives an ultrametric on $Q$ , in which every triangle is isosceles and every point of a ball is a centre

The post-office metric $d (x, y) = ∥ x ∥ + ∥ y ∥$ for $x \neq y$ on $R^{n}$ , and its isolated points

$\min (∣ x - y ∣, 1)$ on $R$ has the usual topology and diameter at most $1$

In ${0} \cup [1, 2]$ with the metric of $R$ , the closure of $B (0, 1) = {0}$ is ${0}$ while the closed ball is ${0, 1}$

In the discrete metric the boundary of $B (p, 1)$ is empty while the sphere of radius $1$ is everything but $p$

$Z$ and ${n + 1 / n : n \geq 2}$ are disjoint closed subsets of $R$ at distance $0$ , so the set-to-set distance is not a metric

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 but not Lipschitz equivalent

$R$ carries both an unbounded and a bounded metric inducing the same topology