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

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FALSE: Every normal space is completely regular

Statement

FALSE. Every normal space is completely regular.

This is exactly why 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 carries the hypothesis $T_{1}$ : normality alone, without $T_{1}$ , gives no separation property above itself.

Facts & Assumptions

Given: Sierpinski space $S = {a, b}$ , $a \neq b$ , with topology $T_{S i e r} = {\emptyset, {b}, S}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L1]

The closed sets of $S$ are the complements of $T_{S i e r}$ : $S ∖ \emptyset = S$ , $S ∖ {b} = {a}$ , $S ∖ S = \emptyset$ ; so the closed sets are ${S, {a}, \emptyset}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L2]

$S$ is normal when disjoint closed subsets of $S$ admit disjoint open supersets (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

[L3]

$S$ is regular when a point and a closed set not containing it admit disjoint open neighbourhoods (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly).

[L4]

Every completely regular space is regular (Every completely regular space is regular, and every Tychonoff space is $T_{3}$ , Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces).

Refutation

technique · constructive

1.1

Let $S = {a, b}$ with $a \neq b$ and $T_{S i e r} = {\emptyset, {b}, S}$ ; by [L1] its closed sets are ${S, {a}, \emptyset}$ .

givenL1construct

2.1

$S$ is normal: let $A, B$ be disjoint closed subsets of $S$ . The nonempty closed sets are ${a}$ and $S$ , and ${a} \subseteq S$ , so any two nonempty closed sets of $S$ meet at $a$ ; hence disjointness of $A, B$ forces $A = \emptyset$ or $B = \emptyset$ . If $A = \emptyset$ , take $U : = \emptyset \supseteq A$ and $V : = S \supseteq B$ ; if $B = \emptyset$ , take $U : = S \supseteq A$ and $V : = \emptyset \supseteq B$ . Either way $U, V$ are open and $U \cap V = \emptyset$ .

step 1.1L1L2algebra

2.2

$S$ is not regular: $b \notin {a}$ , since $a \neq b$ , and ${a}$ is closed by step 1.1. Every open set containing $a$ equals $S$ , since among $\emptyset, {b}, S$ only $S$ contains $a$ ; so any open $V \supseteq {a}$ has $V = S$ , and any open $U ∋ b$ then satisfies $U \cap V = U \cap S = U \neq \emptyset$ , since $b \in U$ . So no disjoint open $U ∋ b$ , $V \supseteq {a}$ exist, and $S$ is not regular.

step 1.1L1L3

3.1

By [L4], complete regularity implies regularity; by step 2.2, $S$ is not regular, so $S$ is not completely regular. With step 2.1, $S$ is a normal space that is not completely regular, refuting the statement.

step 2.1step 2.2L4discharge-construct∎

Remarks

The failure is exactly the missing $T_{1}$ . Sierpinski space is $T_{0}$ (the open set ${b}$ distinguishes $a$ from $b$ ) but not $T_{1}$ : the singleton ${b}$ is not among the closed sets ${S, {a}, \emptyset}$ of step 1.1, so $b$ has no closed singleton. Consequently 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 never claims anything about a normal space that is not $T_{1}$ .
Every disjoint closed pair in $S$ is separated for a trivial reason. Step 2.1 never invokes Urysohn's lemma or any function; normality here has nothing to do with continuous functions, because the only disjoint pairs available involve $\emptyset$ .

Depends on

Used by

Sierpinski space is normal and not completely regular, so the T₁ hypothesis in the Urysohn corollary is not decoration Example

Dependency tree · two levels

33 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

Sierpinski space (Wikipedia) (standard reference, not scraped)
Normal space (Wikipedia) (standard reference, not scraped)