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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Under dependent choice a locally compact Hausdorff space is completely regular, hence Tychonoff

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$ -indexed chain). If $(X,\mathcal{T})$ is locally compact (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space) and Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), then $X$ is completely regular (Completely regular spaces and Tychonoff ( $T_{3\frac{1}{2}}$ ) spaces), and hence, being Hausdorff, $X$ is Tychonoff.

The proof passes through the one-point compactification (The one-point (Alexandroff) compactification $X^{*} = X \cup \{\infty\}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ ) rather than through a hereditary property of regularity or complete regularity: none is used or needed.

Facts & Assumptions

Given: A locally compact Hausdorff space $(X,\mathcal{T})$ , a closed set $C \subseteq X$ , and a point $x_0 \in X \setminus C$ .

[A1]

$X$ is locally compact (Locally compact topological space: every point has a compact neighbourhood; and what this says in a metric space) and Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L1]

The one-point compactification $X^{*} = X \cup \{\infty\}$ of a locally compact Hausdorff space $X$ : its open sets are the open sets of $X$ together with the sets $X^{*} \setminus K$ for $K$ a closed compact subset of $X$ (The one-point (Alexandroff) compactification $X^{*} = X \cup \{\infty\}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ ); consequently its closed sets are $\{\, F \cup \{\infty\} : F \text{ closed in } X \,\}$ together with $\{\, K : K \text{ closed compact in } X \,\}$ , the complements of the two families of open sets.

[L2]

$X^{*}$ is compact and contains $X$ as an open subspace (so the subspace topology $X$ inherits from $X^{*}$ is its own topology $\mathcal{T}$ ); and $X^{*}$ is Hausdorff, since $X$ is locally compact and Hausdorff ( $X^{*}$ is compact and contains $X$ as an open subspace; $X$ is dense in $X^{*}$ exactly when $X$ is not compact; and $X^{*}$ is Hausdorff exactly when $X$ is locally compact and Hausdorff).

[L3]

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

[L4]

In a $T_1$ space every singleton is closed (A space is $T_1$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology).

[L5]

Urysohn's lemma, clause 1: assuming DC, a normal space's disjoint closed sets admit a continuous $[0,1]$ -valued separating function (Urysohn's lemma, under the axiom of dependent choice: in a normal space two disjoint closed sets are separated by a continuous function into $[0,1]$ , and conversely such a space is normal).

[L6]

If $g : X^{*} \to Y$ is continuous and $X \subseteq X^{*}$ carries the subspace topology, then $g|_X$ is continuous (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[L7]

Completely regular: for closed $C$ and $x_0 \notin C$ , a continuous $f : X \to [0,1]$ with $f(x_0)=1$ and $f \equiv 0$ on $C$ (Completely regular spaces and Tychonoff ( $T_{3\frac{1}{2}}$ ) spaces).

Proof

technique · direct

1.1

By [L2], $X^{*}$ is compact and Hausdorff; by [L3], $X^{*}$ is regular and normal, hence $T_3$ and $T_4$ , that is normal and $T_1$ .

A1L2L3

1.2

$C \cup \{\infty\}$ is closed in $X^{*}$ : $C$ is closed in $X$ (given), so $C \cup \{\infty\}$ is one of the sets $F \cup \{\infty\}$ of [L1] with $F=C$ .

givenL1

1.3

$\{x_0\}$ and $C \cup \{\infty\}$ are disjoint: $x_0 \in X$ , so $x_0 \ne \infty$ , and $x_0 \notin C$ (given).

given

1.4

For $x \ne y$ in $X$ , Hausdorffness (given, [A1]) supplies disjoint open $U \ni x$ , $V \ni y$ ; then $y \notin U$ (else $y \in U \cap V = \varnothing$ ) and $x \notin V$ similarly, so $X$ is $T_1$ ( $T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces).

2.1

By step 1.1 ( $T_1$ ) and [L4], $\{x_0\} \subseteq X \subseteq X^{*}$ is closed in $X^{*}$ .

step 1.1L4

3.1

By step 1.1 ( $X^{*}$ normal), steps 2.1, 1.2 and 1.3, and [L5], fix a continuous $g : X^{*} \to [0,1]$ with $C \cup \{\infty\} \subseteq g^{-1}(\{0\})$ and $\{x_0\} \subseteq g^{-1}(\{1\})$ .

step 1.1step 2.1step 1.2step 1.3L5choose

4.1

By [L6] and [L2] ( $X$ a subspace of $X^{*}$ with its own topology), $f := g|_X : X \to [0,1]$ is continuous. For $x \in C$ : $x \in C \cup \{\infty\}$ , so $f(x)=g(x)=0$ ; and $f(x_0) = g(x_0) = 1$ , since $x_0 \in \{x_0\} \subseteq g^{-1}(\{1\})$ .

step 3.1L2L6

5.1

Since $C$ and $x_0 \notin C$ were arbitrary, step 4.1 exhibits, for every closed $C \subseteq X$ and $x_0 \in X \setminus C$ , a continuous $f : X \to [0,1]$ with $f(x_0)=1$ , $f \equiv 0$ on $C$ ; by [L7], $X$ is completely regular.

step 4.1L7

6.1

By steps 5.1 and 1.4, $X$ is completely regular and $T_1$ , that is Tychonoff.

step 5.1step 1.4∎

Remarks

Only two facts about $X^{*}$ are used: that it is compact Hausdorff (so normal, via A compact Hausdorff space is regular and normal, hence $T_3$ and $T_4$ ), and that $X$ sits inside it as an open subspace with its own topology, so that a Urysohn function on $X^{*}$ restricts to one on $X$ with no further argument. No property of $X^{*}$ beyond these two, and no hereditary behaviour of regularity, complete regularity or normality, is used anywhere in the proof.
The choice principle is the one already inside Urysohn's lemma, applied once, inside the compact Hausdorff space $X^{*}$ ; nothing above performs a further selection.

Depends on

Used by

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Sources

Locally compact space (Wikipedia) (standard reference, not scraped)
Alexandroff extension (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §33, 38 (standard reference, not scraped)
Tychonoff space (Wikipedia) (standard reference, not scraped)