CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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Under dependent choice a normal $T_{1}$ space is completely regular, so $T_{4} \Rightarrow T_{3 \frac{1}{2}}$ , and together with the implications already proved this is the whole classical chain

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). If $(X, T)$ is normal and $T_{1}$ , that is $T_{4}$ (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), then $X$ is completely regular (Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces). Since $X$ is also $T_{1}$ , $X$ is Tychonoff, and $T_{4} \Rightarrow T_{3 \frac{1}{2}}$ .

Combined with The implications proved on this page: perfectly normal gives completely normal under countable choice, and completely normal gives normal; normal with $T_{1}$ gives $T_{3}$ ; completely regular gives regular; regular with $T_{1}$ gives Urysohn, hence Hausdorff, hence $T_{1}$ , hence $T_{0}$ ; and metrizable gives every one of them, every arrow of

$T_{6} \Rightarrow T_{5} \Rightarrow T_{4} \Rightarrow T_{3 \frac{1}{2}} \Rightarrow T_{3} \Rightarrow T_{2 \frac{1}{2}} \Rightarrow T_{2} \Rightarrow T_{1} \Rightarrow T_{0}$

now holds: the first arrow under the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the arrow $T_{4} \Rightarrow T_{3 \frac{1}{2}}$ proved here under dependent choice, and every other arrow with no choice principle at all. No arrow of this chain is asserted to reverse.

Facts & Assumptions

Given: A normal, $T_{1}$ topological space $(X, T)$ , a closed set $C \subseteq X$ , and a point $x_{0} \in X ∖ C$ .

[A1]

$X$ is normal (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly) and $T_{1}$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).

[L1]

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, clause (b)).

[L2]

Urysohn's lemma, clause 1: assuming DC, if $X$ is normal and $P, Q \subseteq X$ are disjoint closed sets, there is a continuous $h : X \to [0, 1]$ with $P \subseteq h^{- 1} ({0})$ and $Q \subseteq h^{- 1} ({1})$ (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).

[L3]

$X$ is completely regular when for every closed $C$ and every $x_{0} \in X ∖ C$ there is 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).

[L4]

Clauses 3 and 4 of The implications proved on this page: perfectly normal gives completely normal under countable choice, and completely normal gives normal; normal with $T_{1}$ gives $T_{3}$ ; completely regular gives regular; regular with $T_{1}$ gives Urysohn, hence Hausdorff, hence $T_{1}$ , hence $T_{0}$ ; and metrizable gives every one of them: normal with $T_{1}$ implies $T_{3}$ ; completely regular implies regular, and Tychonoff implies $T_{3}$ ; and clauses 1, 2 and 5 give the remaining arrows of the displayed chain, clause 1 — perfectly normal implies completely normal, that is $T_{6} \Rightarrow T_{5}$ — under the Axiom of Countable Choice.

Proof

technique · direct

1.1

${x_{0}}$ is closed, since $X$ is $T_{1}$ by [A1].

A1L1

1.2

${x_{0}} \cap C = \emptyset$ , since $x_{0} \notin C$ .

given

2.1

By [A1] $X$ is normal, so [L2] applies to the disjoint closed sets $C$ and ${x_{0}}$ : there is a continuous $f : X \to [0, 1]$ with $C \subseteq f^{- 1} ({0})$ and ${x_{0}} \subseteq f^{- 1} ({1})$ , that is $f \equiv 0$ on $C$ and $f (x_{0}) = 1$ .

step 1.1step 1.2A1L2

3.1

Since $C$ and $x_{0} \notin C$ were arbitrary, step 2.1 exhibits, for every closed $C$ and every $x_{0} \in X ∖ C$ , a continuous $f : X \to [0, 1]$ with $f (x_{0}) = 1$ and $f \equiv 0$ on $C$ ; by [L3] this makes $X$ completely regular.

step 2.1L3

4.1

Since $X$ is also $T_{1}$ by [A1], $X$ is Tychonoff, so $T_{4} \Rightarrow T_{3 \frac{1}{2}}$ .

step 3.1A1

5.1

By [L4], $T_{3 \frac{1}{2}} \Rightarrow T_{3} \Rightarrow T_{2 \frac{1}{2}} \Rightarrow T_{2} \Rightarrow T_{1} \Rightarrow T_{0}$ and $T_{6} \Rightarrow T_{5} \Rightarrow T_{4}$ all hold, the arrow $T_{6} \Rightarrow T_{5}$ under countable choice; combined with step 4.1, every arrow of the displayed chain holds.

step 4.1L4∎

Remarks

This corollary supplies exactly the one arrow the published separation-axioms page could not reach. That page's own rem-separation-axiom-conventions names the missing arrow as normal $T_{1}$ implies completely regular and records that no rearrangement of material already on that page could supply it, since the implication is Urysohn's lemma. Nothing in this corollary revisits or amends that page; it only supplies, at a later point in the reading order, the theorem that page named as absent.
The chain above is not asserted to be a theorem of ZF. Its weakest link is this corollary's own dependent-choice hypothesis, and the first arrow separately costs countable choice; neither cost is removed by combining the arrows, and no clause of The implications proved on this page: perfectly normal gives completely normal under countable choice, and completely normal gives normal; normal with $T_{1}$ gives $T_{3}$ ; completely regular gives regular; regular with $T_{1}$ gives Urysohn, hence Hausdorff, hence $T_{1}$ , hence $T_{0}$ ; and metrizable gives every one of them is reproved here.

Depends on

Used by

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Urysohn's lemma (Wikipedia) (standard reference, not scraped)
Separation axiom (Wikipedia) (standard reference, not scraped)

Under dependent choice a normal T1T1​ space is completely regular, so T4⇒T312T4​⇒T321​​, and together with the implications already proved this is the whole classical chain