TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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 every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set

Statement

Let $(X, T)$ be a topological space, let $B$ be a basis for $T$ (Basis and subbasis for a topology, and the topology generated by a family of sets), let $A \subseteq X$ and let $x \in X$ . Closure, derived set and limit points are as in Interior, closure, boundary, exterior, derived set and isolated point in a topological space. Then:

The following four conditions are equivalent.
- (a) $x \in \overline{A}$ ;
- (b) $N \cap A \neq \emptyset$ for every neighbourhood $N$ of $x$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open);
- (c) $U \cap A \neq \emptyset$ for every open $U$ with $x \in U$ ;
- (d) $B \cap A \neq \emptyset$ for every $B \in B$ with $x \in B$ .
$\overline{A}$ is closed, contains $A$ , and is contained in every closed $F \subseteq X$ with $A \subseteq F$ ; so it is the smallest closed superset of $A$ , and $A$ is closed if and only if $A = \overline{A}$ .
$\overline{A} = A \cup A^{'}$ .

Claim 2 is recorded here for reference and is discharged in Interior, closure, boundary, exterior, derived set and isolated point in a topological space, where it is what makes the definition of $\overline{A}$ well posed; claims 1 and 3 are proved below. Claim 1 is the form in which the closure is used everywhere afterwards, and clause (d) is what makes a closure computable from a basis rather than from all open sets.

Facts & Assumptions

Given: A topological space $(X, T)$ , a basis $B$ for $T$ , a subset $A \subseteq X$ and a point $x \in X$ .

[A1]

$\overline{A}$ is the intersection of all closed supersets of $A$ ; it is closed, contains $A$ , and is contained in every closed superset of $A$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[A2]

$x \in A^{'}$ means that $N \cap (A ∖ {x}) \neq \emptyset$ for every neighbourhood $N$ of $x$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L1]

$N$ is a neighbourhood of $x$ when $x \in U \subseteq N$ for some open $U$ ; an open set containing $x$ is a neighbourhood of $x$ (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[L2]

$B$ is a basis for $T$ : for every open $U$ and every $x \in U$ there is $B \in B$ with $x \in B \subseteq U$ , and every member of $B$ is open (Basis and subbasis for a topology, and the topology generated by a family of sets).

[L3]

A set is closed exactly when its complement is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1A1L3

(a) implies (c): let $U$ be open with $x \in U$ and suppose $U \cap A = \emptyset$ ; then $X ∖ U$ is closed and contains $A$ , so $\overline{A} \subseteq X ∖ U$ by [A1], whence $x \notin \overline{A}$ , contradicting (a); therefore $U \cap A \neq \emptyset$ .

1.2A1L3

(c) implies (a): suppose $x \notin \overline{A}$ ; then $U : = X ∖ \overline{A}$ is open by [A1] and [L3], contains $x$ , and satisfies $U \cap A = \emptyset$ because $A \subseteq \overline{A}$ , so (c) fails.

1.3L1

(b) implies (c): an open $U$ containing $x$ is a neighbourhood of $x$ , so (b) applies to it.

1.4L1

(c) implies (b): let $N$ be a neighbourhood of $x$ and fix open $U$ with $x \in U \subseteq N$ ; then $\emptyset \neq U \cap A \subseteq N \cap A$ .

1.5L2

1.6L2

(d) implies (c): let $U$ be open with $x \in U$ and fix $B \in B$ with $x \in B \subseteq U$ ; then $\emptyset \neq B \cap A \subseteq U \cap A$ .

1.7A1L3

$A \subseteq \overline{A}$ and $\overline{A}$ is closed, and $\overline{A}$ is contained in every closed superset of $A$ , which is claim 2; in particular $A$ is closed exactly when $A = \overline{A}$ , since one inclusion always holds and the other says that $A$ is a closed superset of itself.

2.1step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6

By steps 1.1 to 1.6 the four conditions (a), (b), (c) and (d) are equivalent, which is claim 1: (a) and (c) are equivalent by steps 1.1 and 1.2, (b) and (c) by steps 1.3 and 1.4, and (c) and (d) by steps 1.5 and 1.6.

3.1step 2.1A1A2

$A \cup A^{'} \subseteq \overline{A}$ : points of $A$ lie in $\overline{A}$ by [A1], and if $x \in A^{'}$ then every neighbourhood $N$ of $x$ meets $A ∖ {x}$ and hence meets $A$ , so $x \in \overline{A}$ by condition (b).

3.2step 2.1A2

$\overline{A} \subseteq A \cup A^{'}$ : let $x \in \overline{A}$ and suppose $x \notin A$ ; then for every neighbourhood $N$ of $x$ condition (b) gives $N \cap A \neq \emptyset$ , and $A = A ∖ {x}$ because $x \notin A$ , so $N \cap (A ∖ {x}) \neq \emptyset$ and $x \in A^{'}$ .

4.1step 1.7step 2.1step 3.1step 3.2∎

Steps 3.1 and 3.2 give $\overline{A} = A \cup A^{'}$ , which is claim 3; with step 2.1 for claim 1 and step 1.7 for claim 2 the theorem is proved.

Remarks

Clause (d) is relative to a basis and clause (c) is not. Different bases for one topology give different families of test sets in (d), and the theorem says all of them detect the same closure. This is why a closure in a metric space may be computed with balls alone, and a closure in $R$ with bounded open intervals alone.
The empty set and the whole space. $\overline{\emptyset} = \emptyset$ , since $\emptyset$ is closed, and $\overline{X} = X$ ; claim 1 reads correctly in both cases, no neighbourhood meeting $\emptyset$ and every neighbourhood meeting $X$ .
Nothing here assumes that singletons are closed. In the indiscrete topology on a set with at least two points, $\overline{{x}} = X$ for every $x$ , since the only neighbourhood of any point is $X$ ; claim 3 then says $A^{'} = X ∖ {x}$ is contained in $\overline{{x}}$ , which it is.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A point lies in the closure of AA iff every basic neighbourhood of it meets AA; the closure is the smallest closed superset and equals AA together with its derived set

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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