LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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 space is normal if and only if every closed $A$ inside an open $U$ admits an open $V$ with $A \subseteq V \subseteq \overline{V} \subseteq U$

Statement

Let $(X, \mathcal{T})$ 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. The following two conditions are equivalent.

(a) $X$ is normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).
(b) For every closed $A \subseteq X$ and every open $U$ with $A \subseteq U$ there is an open $V$ with $A \subseteq V \subseteq \overline{V} \subseteq U .$

In particular, in a normal space any two disjoint closed sets $A$ and $D$ admit an open $V \supseteq A$ with $\overline{V} \cap D = \varnothing$ : apply (b) to $A$ and the open set $X \setminus D$ . That corollary is the form in which normality is used later on this page.

Facts & Assumptions

Given: A topological space $(X,\mathcal{T})$ , a closed set $A$ , an open set $U$ with $A \subseteq U$ , and disjoint closed sets $A_0, B_0$ .

[A1]

$X$ is normal when disjoint closed sets admit disjoint open supersets (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly).

[L1]

$\overline{V}$ 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).

[L2]

A set is closed exactly when its complement is open, and complementation reverses inclusion (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 $A$ be closed with $A \subseteq U$ and $U$ open; then $B := X \setminus U$ is closed by [L2] and $A \cap B = \varnothing$ , so [A1] gives disjoint open $V \supseteq A$ and $W \supseteq B$ .

A1L2assume-hyp

1.2

Assume (b), and let $A_0, B_0$ be disjoint closed sets; then $U_0 := X \setminus B_0$ is open by [L2] and contains $A_0$ , so (b) gives an open $V_0$ with $A_0 \subseteq V_0 \subseteq \overline{V_0} \subseteq U_0$ .

L2assume-hyp

2.1

Under step 1.1: $V \subseteq X \setminus W$ , since $V \cap W = \varnothing$ , and $X \setminus W$ is closed by [L2], so $\overline{V} \subseteq X \setminus W$ by [L1]; and $X \setminus W \subseteq X \setminus B = U$ because $B \subseteq W$ and complementation reverses inclusion.

step 1.1L1L2

2.2

Under step 1.2: put $W_0 := X \setminus \overline{V_0}$ , which is open by [L1] and [L2]; then $V_0 \cap W_0 = \varnothing$ because $V_0 \subseteq \overline{V_0}$ , and $B_0 = X \setminus U_0 \subseteq X \setminus \overline{V_0} = W_0$ because $\overline{V_0} \subseteq U_0$ .

step 1.2L1L2

3.1

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

step 2.1

3.2

Step 2.2 gives disjoint open $V_0 \supseteq A_0$ and $W_0 \supseteq B_0$ , so (b) implies (a) by [A1].

step 2.2A1

4.1

Steps 3.1 and 3.2 make (a) and (b) equivalent.

step 3.1step 3.2

5.1

For the final assertion, let $A$ and $D$ be disjoint closed sets in a normal $X$ ; then $X \setminus D$ is open by [L2] and contains $A$ , so (b) gives an open $V$ with $A \subseteq V \subseteq \overline{V} \subseteq X \setminus D$ , whence $\overline{V} \cap D = \varnothing$ .

step 4.1L2∎

Remarks

The name. Statement (b) is the shrinking form: an open set containing a closed set can be shrunk so that even its closure stays inside. It is the exact analogue for closed sets of the clause of 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 $x \in V \subseteq \overline{V} \subseteq U$ that shrinks an open set around a point.
Iterating (b) is what proves Urysohn's lemma, by indexing a family of open sets by the dyadic rationals; that iteration is a dependent choice and is not performed on this page (Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order). The single application above is choice free.
Nothing here uses a separation axiom. In particular $A$ and $D$ may be empty, and the corollary reads correctly in that case with $V = \varnothing$ or $V = X$ respectively.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 10 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

Normal space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §33 (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 normal if and only if every closed AAA inside an open UUU admits an open VVV with A⊆V⊆V‾⊆UA \subseteq V \subseteq \overline{V} \subseteq UA⊆V⊆V⊆U