ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Sierpinski space is $T_{0}$ and normal but neither $T_{1}$ nor regular: normality without $T_{1}$ implies nothing

Example

Let $S = {a, b}$ with $a \neq b$ carry the Sierpinski topology $T_{S i e r} = {\emptyset, {b}, S}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), so that $b$ is the open point and $a$ the closed one. Then:

$S$ is $T_{0}$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces): the open set ${b}$ contains $b$ and not $a$ .
$S$ is not $T_{1}$ : the singleton ${b}$ is not 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). It is not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) either, and for an independent reason rather than as a consequence of the previous sentence: the only open set containing $a$ is $S$ itself, which also contains $b$ .
$S$ is normal (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly): its only closed sets are $\emptyset$ , ${a}$ and $S$ , and no two disjoint closed sets are both nonempty.
$S$ is not regular (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly): the point $b$ and the closed set ${a}$ have no disjoint open neighbourhoods, because the only open set containing $a$ is $S$ itself.

Sierpinski space is therefore the sharpest small witness that normality without $T_{1}$ implies nothing: it is normal, satisfies the weakest axiom $T_{0}$ , and fails the axiom immediately above it.

Facts & Assumptions

Given: The two-point set $S = {a, b}$ with $a \neq b$ and the topology $T_{S i e r} = {\emptyset, {b}, S}$ .

[A1]

$T_{S i e r}$ is a topology on $S$ and its members are exactly $\emptyset$ , ${b}$ and $S$ (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).

[A2]

A set is closed exactly when its complement is open (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L1]

$T_{0}$ : some open set contains exactly one of two distinct points. $T_{1}$ : each of two distinct points has an open set containing it and missing the other, equivalently every singleton is closed ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces, 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).

[L2]

Normal: disjoint closed sets have disjoint open supersets. Regular: a point and a closed set not containing it have disjoint open supersets (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly, Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly).

[L3]

Hausdorff: distinct points have disjoint open neighbourhoods (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

Verification

technique · direct

1.1

The closed subsets of $S$ are the complements of $\emptyset$ , ${b}$ and $S$ , namely $S$ , ${a}$ and $\emptyset$ .

A1A2

1.2

${b}$ is open and contains $b$ but not $a$ , so $S$ is $T_{0}$ , which is claim 1.

A1L1

1.3

The only open set containing $a$ is $S$ , by [A1], since $\emptyset$ contains nothing and ${b}$ does not contain $a$ .

2.1

${b}$ is not closed, since ${b}$ is not among the three closed sets of step 1.1, so $S$ is not $T_{1}$ by [L1]; and $S$ is not Hausdorff, since by step 1.3 the only open set containing $a$ is $S$ , which also contains $b$ , so $a$ and $b$ have no disjoint open neighbourhoods as [L3] would require. Together these are claim 2.

step 1.1step 1.3L1L3

2.2

No two disjoint closed subsets of $S$ are both nonempty: by step 1.1 the nonempty closed sets are ${a}$ and $S$ , and ${a} \cap {a} = {a}$ , ${a} \cap S = {a}$ and $S \cap S = S$ are all nonempty.

step 1.1

2.3

Suppose $U$ and $V$ are disjoint open sets with $b \in U$ and ${a} \subseteq V$ ; then $V = S$ by step 1.3, so $U \cap V = U \neq \emptyset$ , a contradiction. Hence $b$ and the closed set ${a}$ cannot be separated and $S$ is not regular by [L2], which is claim 4.

step 1.1step 1.3L2

3.1

Let $A, B$ be disjoint closed subsets of $S$ ; by step 2.2 one of them is $\emptyset$ , and then $\emptyset$ together with $S$ separates the pair, in the order matching which of $A$ and $B$ is empty. So $S$ is normal by [L2], which is claim 3.

step 2.2A1L2

4.1

Steps 1.2, 2.1, 3.1 and 2.3 are claims 1 to 4, so Sierpinski space is $T_{0}$ and normal and is neither $T_{1}$ nor regular.

step 1.2step 2.1step 3.1step 2.3∎

Remarks

Normality without $T_{1}$ does not imply regularity. This space is normal and not regular. Other combinations occur too: the indiscrete two-point space is regular and normal but not $T_{0}$ , and the cofinite topology on an infinite set is $T_{1}$ and neither regular nor normal, both worked on this page. This page leaves the converse open; it is refuted later by Assuming countable choice, refuted: every regular space is normal ↗. Only with $T_{1}$ does normality imply regularity (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).
Why "the closed point" is the right name for $a$ . ${a}$ is closed and ${b}$ is not, so the two points of $S$ are not interchangeable even though the set has only two elements; the labelling is fixed once and for all in The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies.
The space is $T_{0}$ and nothing more. It sits at the very bottom of the chain, and it is the standard reminder that the bottom of the chain is not empty: $T_{0}$ is a real condition, satisfied here and failed by the indiscrete topology on the same underlying set.

Depends on

Used by

The particular-point topology is T₀, it is not T₁ and not regular once the set has at least two points, and it is not normal once the set has at least three Example

Dependency tree · two levels

23 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)
Separation axiom (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §17 (standard reference, not scraped)
Particular point topology (Wikipedia) (standard reference, not scraped)

Sierpinski space is T0T0​ and normal but neither T1T1​ nor regular: normality without T1T1​ implies nothing