TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $X$ be a first countable topological space (First countable space: a countable neighbourhood base at every point) and let $Y$ be a topological space. Then:

$seqcl (A) = \overline{A}$ for every $A \subseteq X$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure, A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set);
for $f : X \to Y$ and $p \in X$ , $f$ is continuous at $p$ (Continuity of a map of topological spaces at a point and globally) if and only if $f$ is sequentially continuous at $p$ .

Where ${A C}_{ω}$ is spent, and that it is not decoration. Both directions that this theorem adds to The sequential closure is contained in the closure, continuity implies sequential continuity, and sequential limits need not be unique build a sequence by picking one point from each of countably many nonempty sets $M_{k} \cap A$ , respectively $M_{k} ∖ f^{- 1} [V]$ , and the first countability hypothesis supplies no rule for the pick. The two applications of ${A C}_{ω}$ below are the only uses of any choice principle in the proof; the inclusions already proved in The sequential closure is contained in the closure, continuity implies sequential continuity, and sequential limits need not be unique use none at all.

Facts & Assumptions

Given: A first countable space $X$ , a topological space $Y$ , a subset $A \subseteq X$ , a point $p \in X$ , a function $f : X \to Y$ , and the Axiom of Countable Choice as an explicit hypothesis.

[A1]

Every point of $X$ has an at most countable neighbourhood base (First countable space: a countable neighbourhood base at every point).

[A2]

$x_{k} \to p$ means that for every neighbourhood $N$ of $p$ there is $K$ with $x_{k} \in N$ for all $k \geq K$ ; $seqcl (A)$ collects the points to which some sequence in $A$ converges; sequential continuity at $p$ says $x_{k} \to p$ implies $f (x_{k}) \to f (p)$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

[A3]

$f$ is continuous at $p$ when $f^{- 1} [V]$ is a neighbourhood of $p$ for every neighbourhood $V$ of $f (p)$ (Continuity of a map of topological spaces at a point and globally).

[L1]

$seqcl (A) \subseteq \overline{A}$ , and continuity at $p$ implies sequential continuity at $p$ (The sequential closure is contained in the closure, continuity implies sequential continuity, and sequential limits need not be unique, claims 1 and 2).

[L2]

$p \in \overline{A}$ if and only if every neighbourhood of $p$ meets $A$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, clause (b)).

[L3]

A finite intersection of neighbourhoods of $p$ is a neighbourhood of $p$ ; every superset of a neighbourhood of $p$ is a neighbourhood of $p$ ; every point lies in each of its neighbourhoods; and $X$ itself is a neighbourhood of $p$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[L4]

A nonempty at most countable set is the image of a surjection from $N$ (A nonempty set is at most countable iff it is a surjective image of $N$ ).

[L5]

Recursion: for any set $Z$ , any $z_{0} \in Z$ and any $F : Z \to Z$ there is a function $h : N \to Z$ with $h (0) = z_{0}$ and $h (σ (k)) = F (h (k))$ for every $k$ (The recursion theorem).

[L6]

${A C}_{ω}$ : for every family $(Z_{k})_{k \in N}$ of nonempty sets there is $c$ with $c (k) \in Z_{k}$ for every $k$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1

Fix an at most countable neighbourhood base $B_{p}$ at $p$ ; it is nonempty, since $X \in N (p)$ forces some member of $B_{p}$ to lie inside $X$ , so by [L4] there is a surjection $k \mapsto N_{k}$ from $N$ onto $B_{p}$ .

A1L3L4choose

2.1

Apply [L5] with $Z : = N \times N (p)$ , with $z_{0} : = (0, N_{0})$ and with $F (k, M) : = (σ (k), M \cap N_{σ (k)})$ , which lands in $Z$ because an intersection of two neighbourhoods of $p$ is a neighbourhood of $p$ ; the resulting $h$ has first coordinate $h (k) = (k, M_{k})$ by induction, so $M_{0} = N_{0}$ and $M_{σ (k)} = M_{k} \cap N_{σ (k)}$ . Hence every $M_{k}$ is a neighbourhood of $p$ , the family is decreasing, $M_{0} \supseteq M_{1} \supseteq \dots$ , and $M_{k} \subseteq N_{k}$ for every $k$ .

step 1.1L3L5construct

3.1

The family $(M_{k})_{k \in N}$ is again a neighbourhood base at $p$ : given $N \in N (p)$ there is a member of $B_{p}$ inside $N$ , and that member is $N_{k}$ for some $k$ by surjectivity, so $M_{k} \subseteq N_{k} \subseteq N$ .

step 1.1step 2.1A1

3.2

Let $p \in \overline{A}$ . Each $M_{k}$ is a neighbourhood of $p$ , so $M_{k} \cap A \neq \emptyset$ by [L2]; by ${A C}_{ω}$ applied to the family $(M_{k} \cap A)_{k \in N}$ there is a sequence $(x_{k})$ with $x_{k} \in M_{k} \cap A$ for every $k$ .

step 2.1L2L6

3.3

Assume $f$ is sequentially continuous at $p$ , let $V$ be a neighbourhood of $f (p)$ , and suppose no $M_{k}$ satisfied $M_{k} \subseteq f^{- 1} [V]$ . Then every set $M_{k} ∖ f^{- 1} [V]$ would be nonempty, so ${A C}_{ω}$ would supply a sequence $(y_{k})$ with $y_{k} \in M_{k} ∖ f^{- 1} [V]$ for every $k$ .

step 2.1assume-hypL6

4.1

The sequence of step 3.2 converges to $p$ : given $N \in N (p)$ , step 3.1 gives $k_{0}$ with $M_{k_{0}} \subseteq N$ , and for $k \geq k_{0}$ the nesting of step 2.1 gives $x_{k} \in M_{k} \subseteq M_{k_{0}} \subseteq N$ . Its terms lie in $A$ , so $p \in seqcl (A)$ .

step 2.1step 3.1step 3.2A2

4.2

The sequence of step 3.3 converges to $p$ for the same reason, while $f (y_{k}) \notin V$ for every $k$ , so $(f (y_{k}))$ is not eventually in the neighbourhood $V$ of $f (p)$ and does not converge to $f (p)$ ; that contradicts sequential continuity at $p$ . Hence some $M_{k_{1}}$ satisfies $M_{k_{1}} \subseteq f^{- 1} [V]$ , and $f^{- 1} [V]$ is then a neighbourhood of $p$ by [L3], since it contains the neighbourhood $M_{k_{1}}$ of $p$ .

step 2.1step 3.1step 3.3A2L3

5.1

Step 4.1 gives $\overline{A} \subseteq seqcl (A)$ , and [L1] gives the reverse inclusion, so claim 1 holds.

step 4.1L1

6.1

Step 4.2 shows that sequential continuity at $p$ implies continuity at $p$ , and [L1] gives the converse, so claim 2 holds.

step 4.2A3L1∎

Remarks

The hypothesis cannot be dropped. Under the standing Axiom of Countable Choice assumption, the cocountable topology on $R$ is not first countable, and both conclusions fail there: the sequential closure of $[0, 1]$ is $[0, 1]$ while its closure is $R$ , and the identity onto the usual topology is sequentially continuous without being continuous. Both are on the companion page, and the second is recorded on this page as a false statement.
Every metrizable space satisfies the hypothesis. The balls of radius $1 / n$ form an at most countable neighbourhood base at each point (The balls $B (x, 1 / n)$ , $n \geq 1$ , form a countable neighbourhood base at $x$ , so every metric space is first countable), so claim 1 specialises to the sequential characterisation of the closure in a metric space (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), which spends countable choice in exactly the same one of its two directions. Nothing here is new in the metric setting; what is new is that first countability alone is enough.
Why the base is made decreasing. Without the nesting of step 2.1 the chosen points $x_{k} \in N_{k} \cap A$ need not converge to $p$ : the sets $N_{k}$ may oscillate, and a point chosen from a large $N_{k}$ carries no information about membership in a small one. The running intersections repair this and cost only a recursion.
The real-analysis track states the same phenomenon for function limits, at the same cost in choice. $R$ with its usual topology is metrizable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, 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), hence first countable by the bullet above, and the Heine criterion Heine criterion: $\lim_{x \to c} f (x) = L$ iff $f (x_{k}) \to L$ for every sequence in $A ∖ {c}$ converging to $c$ is the function-limit form of this theorem there: sequences detect the $ε$ - $δ$ limit at a limit point of the domain, its sequence-to- $ε$ direction spends countable choice on a shrinking-neighbourhood selection exactly as the proof above does, and its other direction is choice-free (The sequence-to- $ε$ direction of the Heine criterion uses countable choice for $R$ , and where this library records that cost). Neither result cites the other, and neither is proved from the other: the criterion is proved from the order and the absolute value of $R$ directly. This is a sufficient first-countable instance of the broader sequential phenomenon, not a characterisation.

Depends on

Used by

Dependency tree · two levels

31 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

First-countable space (Wikipedia) (standard reference, not scraped)
Sequential space (Wikipedia) (standard reference, not scraped)
Axiom of countable choice (Wikipedia) (standard reference, not scraped)