TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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A compact Hausdorff space is regular and normal, hence $T_3$ and $T_4$

Statement

Let $X$ be a compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) topological space. Then:

$X$ is regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly);
$X$ is normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly);
$X$ is $T_1$ ( $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces), and hence $X$ is $T_3$ and $T_4$ .

Following Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly and Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly, regular and normal name the separation conditions alone and the numerals $T_3$ and $T_4$ name their conjunctions with $T_1$ ; claim 3 is what supplies the $T_1$ half, and it is stated separately for that reason.

Nothing stronger is claimed. In particular it is not asserted here that a compact Hausdorff space is completely regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly distinguishes the two conditions), and no continuous real-valued function is produced anywhere below.

Facts & Assumptions

Given: A compact Hausdorff topological space $X$ .

[A1]

$X$ is regular when for every closed $C \subseteq X$ and every $x \in X \setminus C$ there are disjoint open $U \ni x$ and $V \supseteq C$ ; the case $C = \varnothing$ is met by $U = X$ and $V = \varnothing$ , and $T_3$ is regular together with $T_1$ (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly, $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

[A2]

$X$ is normal when for all disjoint closed $A, B \subseteq X$ there are disjoint open $U \supseteq A$ and $V \supseteq B$ ; the cases $A = \varnothing$ and $B = \varnothing$ are met by $\varnothing$ together with $X$ , and $T_4$ is normal together with $T_1$ (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly, $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

[A3]

$X$ is a topological space, so a subset 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).

[L1]

A closed subspace of a compact space is compact (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[L2]

In a Hausdorff space, a point and a disjoint compact set have disjoint open neighbourhoods, and two disjoint compact sets have disjoint open neighbourhoods (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[L3]

Every Hausdorff space is $T_1$ (Every Urysohn space is Hausdorff, every Hausdorff space is $T_1$ and hence $T_0$ , and every regular $T_1$ space is Urysohn, claim 2, $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

Proof

technique · direct

1.1

Let $C \subseteq X$ be closed and let $x \in X \setminus C$ ; since $X$ is compact and $C$ is closed in $X$ , the subspace $C$ is compact, and $x$ does not lie in it.

A3L1

1.2

Let $A, B \subseteq X$ be closed with $A \cap B = \varnothing$ ; since $X$ is compact and both are closed in $X$ , both subspaces $A$ and $B$ are compact.

A3L1

1.3

$X$ is $T_1$ , being Hausdorff.

2.1

By [L2], applied to the point $x$ and the disjoint compact set $C$ of step 1.1, there are disjoint open $U \ni x$ and $V \supseteq C$ ; as $C$ and $x$ were arbitrary this is exactly the condition of [A1], so $X$ is regular, which is claim 1.

step 1.1A1L2

2.2

By [L2], applied to the two disjoint compact sets $A$ and $B$ of step 1.2, there are disjoint open $U \supseteq A$ and $V \supseteq B$ ; as $A$ and $B$ were arbitrary this is the condition of [A2], so $X$ is normal, which is claim 2.

step 1.2A2L2

3.1

By step 1.3 the space is $T_1$ ; with step 2.1 it is regular and $T_1$ , hence $T_3$ , and with step 2.2 it is normal and $T_1$ , hence $T_4$ . This is claim 3.

step 1.3step 2.1step 2.2A1A2

4.1

Steps 2.1, 2.2 and 3.1 are claims 1, 2 and 3, so a compact Hausdorff space is regular, normal, $T_3$ and $T_4$ .

step 2.1step 2.2step 3.1∎

Remarks

The whole content is that "closed" and "compact" coincide here, in the direction that is needed. Regularity asks a point to be separated from a closed set and normality asks two closed sets to be separated; compactness of the ambient space converts each closed set into a compact one, and the separation of compact sets in a Hausdorff space is what In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones supplies. No new separation argument is run.
Why compactness of $X$ is needed and not just of the sets separated. The hypothesis is used only through [L1], to know that an arbitrary closed subset of $X$ is compact. A Hausdorff space in which the sets to be separated happen to be compact is separated by [L2] alone and needs no hypothesis on the ambient space at all; what compactness of $X$ buys is that every closed set is such a set.
The degenerate cases are not a gap. If $C$ , $A$ or $B$ is empty the required open sets are named outright in [A1] and [A2], so the argument does not depend on any nonemptiness hidden in the compact-separation clauses.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 66 results over 15 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)
Compact space (Wikipedia) (standard reference, not scraped)
Separation axiom (Wikipedia) (standard reference, not scraped)
A. Hatcher, Topology Notes (standard reference, not scraped)

A compact Hausdorff space is regular and normal, hence T3T_3T3​ and T4T_4T4​