LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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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.

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A space is regular if and only if every point has a neighbourhood base of closed neighbourhoods, if and only if $x \in U$ open gives an open $V$ with $x \in V \subseteq \overline{V} \subseteq U$

Statement

Let $(X, T)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), with closures as in Interior, closure, boundary, exterior, derived set and isolated point in a topological space and neighbourhoods as in Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, so that a neighbourhood need not be open. The following three conditions are equivalent.

(a) $X$ is regular (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly).
(b) For every $x \in X$ and every open $U$ with $x \in U$ there is an open $V$ with $x \in V \subseteq \overline{V} \subseteq U .$
(c) Every point of $X$ has a neighbourhood base consisting of closed neighbourhoods: for every $x \in X$ and every neighbourhood $N$ of $x$ there is a closed neighbourhood $K$ of $x$ with $K \subseteq N$ .

Facts & Assumptions

Given: A topological space $(X, T)$ , a point $x \in X$ , an open set $U$ with $x \in U$ , a neighbourhood $N$ of $x$ , and a closed set $C$ with $x \notin C$ .

[A1]

$X$ is regular when for every closed $C$ and every $x \notin C$ there are disjoint open $U_{0} ∋ x$ and $V_{0} \supseteq C$ (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly).

[L1]

$N$ is a neighbourhood of $x$ exactly when some open $W$ satisfies $x \in W \subseteq N$ ; a set is open exactly when it is a neighbourhood of each of its points (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open).

[L2]

$\overline{V}$ is the smallest closed superset of $V$ : it is closed, contains $V$ , and is contained in every closed set containing $V$ (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, claim 2, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L3]

$int (K)$ is the largest open subset of $K$ , and $x \in int (K)$ exactly when $K$ is a neighbourhood of $x$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L4]

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

1.1

Assume (a) and let $U$ be open with $x \in U$ ; then $C : = X ∖ U$ is closed by [L4] and $x \notin C$ , so [A1] gives disjoint open $V ∋ x$ and $W \supseteq C$ .

A1L4assume-hyp

1.2

Assume (b) and let $N$ be a neighbourhood of $x$ ; fix an open $U$ with $x \in U \subseteq N$ by [L1], and let $V$ be as in (b), so $x \in V \subseteq \overline{V} \subseteq U \subseteq N$ .

L1assume-hyp

1.3

Assume (c) and let $C$ be closed with $x \notin C$ ; then $X ∖ C$ is open by [L4] and contains $x$ , hence is a neighbourhood of $x$ by [L1], so (c) gives a closed neighbourhood $K$ of $x$ with $K \subseteq X ∖ C$ .

L1L4assume-hyp

2.1

Under step 1.1: $V \subseteq X ∖ W$ , since $V$ and $W$ are disjoint, and $X ∖ W$ is closed by [L4], so $\overline{V} \subseteq X ∖ W$ by [L2]; and $X ∖ W \subseteq X ∖ C = U$ because $C \subseteq W$ .

step 1.1L2L4

2.2

Under step 1.2: $\overline{V}$ is a closed set containing the open $V ∋ x$ , so it is a neighbourhood of $x$ by [L1], and it is a closed neighbourhood of $x$ contained in $N$ .

step 1.2L1L2

2.3

Under step 1.3: put $V_{0} : = int (K)$ , which is open and contains $x$ by [L3] since $K$ is a neighbourhood of $x$ ; and put $W_{0} : = X ∖ K$ , which is open by [L4] since $K$ is closed.

step 1.3L3L4

3.1

Step 2.1 gives $x \in V \subseteq \overline{V} \subseteq U$ with $V$ open, so (a) implies (b).

step 2.1

3.2

Step 2.2 gives, for every neighbourhood $N$ of $x$ , a closed neighbourhood of $x$ inside $N$ , so (b) implies (c).

step 2.2

3.3

Under step 2.3: $V_{0} \cap W_{0} = int (K) \cap (X ∖ K) = \emptyset$ because $int (K) \subseteq K$ by [L3], and $C \subseteq X ∖ K = W_{0}$ because $K \subseteq X ∖ C$ ; so $V_{0}$ and $W_{0}$ are disjoint open sets containing $x$ and $C$ respectively, and (c) implies (a).

step 2.3A1L3

4.1

By steps 3.1, 3.2 and 3.3 the three conditions (a), (b) and (c) are equivalent.

step 3.1step 3.2step 3.3∎

Remarks

Clause (b) is the working form. Every application of regularity below uses it in the shape "shrink an open set around a point so that even its closure stays inside", which is what makes regularity behave like a one-sided version of the normality shrinking lemma proved later on this page.
Clause (c) is what makes a clopen basis decisive. If a space has a basis of clopen sets then the basic sets containing a point are closed neighbourhoods of it and form a neighbourhood base (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open), so (c) holds and the space is regular with no further work. That is exactly the route by which the ordinal spaces later on this page are shown to be regular.
No separation hypothesis is used anywhere above. Points need not be closed, and the lemma is a statement about regularity alone; combining it with $T_{1}$ is the separate step that produces $T_{3}$ .

Depends on

Used by

Dependency tree · two levels

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Sources

Regular space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §31 (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 31: The Separation Axioms (East Tennessee State University) (standard reference, not scraped)

A space is regular if and only if every point has a neighbourhood base of closed neighbourhoods, if and only if x∈Ux∈U open gives an open VV with x∈V⊆V‾⊆Ux∈V⊆V⊆U