False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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FALSE: every normal space is Hausdorff, so the $T_{1}$ hypothesis in $T_{4}$ is redundant

Statement

False claim: every normal space (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly) is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not); equivalently, the $T_{1}$ hypothesis in the definition of $T_{4}$ is redundant.

The refutation is the indiscrete topology $T_{i n d} = {\emptyset, X}$ on a two-point set $X = {a, b}$ with $a \neq b$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). It is normal, because its only closed sets are $\emptyset$ and $X$ and no two nonempty closed sets are disjoint; and it is not Hausdorff, not $T_{1}$ and not even $T_{0}$ , because the only open set containing either point is $X$ .

Facts & Assumptions

Given: The two-point set $X = {a, b}$ with $a \neq b$ , carrying the indiscrete topology $T_{i n d} = {\emptyset, X}$ .

[A1]

A space is normal when any two disjoint closed sets have disjoint open supersets (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

[A2]

A space is Hausdorff when distinct points have disjoint open neighbourhoods, $T_{1}$ when each of two distinct points has an open set containing it and missing the other, and $T_{0}$ when some open set contains exactly one of them (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).

[L1]

The closed sets of $T_{i n d}$ are $\emptyset$ and $X$ (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).

Refutation

technique · direct

1.1

Let $A$ and $B$ be disjoint closed subsets of $X$ ; by [L1] each is $\emptyset$ or $X$ , and since $X \neq \emptyset$ they cannot both be $X$ .

1.2

If $A = \emptyset$ then $U : = \emptyset$ and $V : = X$ are disjoint open sets with $A \subseteq U$ and $B \subseteq V$ ; if $B = \emptyset$ then $U : = X$ and $V : = \emptyset$ do the same.

L1construct

1.3

The only open set containing $a$ is $X$ , and the only open set containing $b$ is $X$ , since $\emptyset$ contains neither.

2.1

By steps 1.1 and 1.2 every pair of disjoint closed sets is separated by disjoint open sets, so $(X, T_{i n d})$ is normal.

step 1.1step 1.2A1

2.2

By step 1.3 no open set contains exactly one of $a$ and $b$ , so $(X, T_{i n d})$ is not $T_{0}$ , hence not $T_{1}$ and not Hausdorff.

step 1.3A2

3.1

By step 2.1 the space is normal and by step 2.2 it is not Hausdorff, so the claim is false; and since it is not $T_{1}$ either, the $T_{1}$ hypothesis in the definition of $T_{4}$ is not redundant.

step 2.1step 2.2A1A2∎

Remarks

This is the reason this library does not build $T_{1}$ into the word normal. Normality on its own places a space nowhere in the hierarchy: the witness above is normal and fails the weakest axiom of all. Sierpinski space, on the companion page, is a second witness, normal and $T_{0}$ and not regular.
The same two-point space refutes more than this. It is also regular, completely regular, completely normal and perfectly normal, and still not $T_{0}$ ; the verification is on the companion page. So every unnumbered adjective on this page is compatible with the total failure of point separation, which is exactly what the numerals $T_{3}$ to $T_{6}$ are for.
What survives. With $T_{1}$ added, normality does give the whole descending chain (A normal $T_{1}$ space is regular, hence $T_{3}$ , hence Urysohn, Hausdorff, $T_{1}$ and $T_{0}$ ); the hypothesis is spent at one step, turning a point into a closed set.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 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

Normal space (Wikipedia) (standard reference, not scraped)
Trivial topology (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §32 (standard reference, not scraped)

FALSE: every normal space is Hausdorff, so the T1T1​ hypothesis in T4T4​ is redundant