TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

How statement and proof provenance work

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

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

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

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

Statement

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

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

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

Facts & Assumptions

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

[A1]

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

[A2]

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

[L1]

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

[L2]

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

[L3]

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

[L4]

Countable choice: for a family $(A_n)_{n \in \mathbb{N}}$ of nonempty sets there is a function $n \mapsto a_n$ with $a_n \in A_n$ for every $n$ (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )).

[L5]

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

Proof

technique · direct

1.1

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

A1A2

1.2

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

A1L3L4choose

2.1

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

step 1.2A2L1L2L3

2.2

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

step 1.1L5

3.1

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

step 1.1step 1.2step 2.1

4.1

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

step 3.1L5

5.1

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

step 2.2step 3.1step 4.1∎

Remarks

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

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 98 results over 28 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Closure (topology) (Wikipedia) (standard reference, not scraped)
Sequentially closed set (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)
Axiom of countable choice (Wikipedia) (standard reference, not scraped)
Sequential space (Wikipedia) (standard reference, not scraped)

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