False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $S = \{a,b\}$ , $a \ne b$ , with topology $\mathcal{T}_{\mathrm{Sier}} = \{\varnothing, \{b\}, S\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L1]

The closed sets of $S$ are the complements of $\mathcal{T}_{\mathrm{Sier}}$ : $S \setminus \varnothing = S$ , $S \setminus \{b\} = \{a\}$ , $S \setminus S = \varnothing$ ; so the closed sets are $\{S, \{a\}, \varnothing\}$ (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 $a \ne b$ and $\mathcal{T}_{\mathrm{Sier}} = \{\varnothing,\{b\},S\}$ ; by [L1] its closed sets are $\{S,\{a\},\varnothing\}$ .

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 $A=\varnothing$ or $B=\varnothing$ . If $A=\varnothing$ , take $U:=\varnothing \supseteq A$ and $V:=S \supseteq B$ ; if $B=\varnothing$ , take $U:=S \supseteq A$ and $V:=\varnothing \supseteq B$ . Either way $U,V$ are open and $U \cap V = \varnothing$ .

step 1.1L1L2algebra

2.2

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

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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 104 results over 27 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

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