TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)

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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.

Depends on

Used by

Dependency tree · two levels

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Sources

J. Cichon, Some remarks about two definitions of continuity, Section 1, Definition 1 and following paragraph (standard reference, not scraped)
Continuous function (Wikipedia) (standard reference, not scraped)
Sequential continuity (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §18 (standard reference, not scraped)
Closure (topology) (Wikipedia) (standard reference, not scraped)