ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-05 (claude-sonnet-5)

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.

The particular-point topology is $T_{0}$ , it is not $T_{1}$ and not regular once the set has at least two points, and it is not normal once the set has at least three

Example

Let $X$ be a set, fix $p \in X$ , and give $X$ the particular-point topology $T_{p} = {\emptyset} \cup {U \subseteq X : p \in U}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose closed sets are $X$ together with the subsets of $X$ that do not contain $p$ . Then:

$(X, T_{p})$ is $T_{0}$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces), for every $X$ and every $p$ .
If $X$ has at least two points then $(X, T_{p})$ is not $T_{1}$ , and hence not Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).
If $X$ has at least two points then $(X, T_{p})$ is not regular (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly).
If $X$ has at least three points then $(X, T_{p})$ is not normal (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

Clause 4 needs one point more than clause 3, and the extra point is not slack: on a two-point set the particular-point topology is Sierpinski space, which is normal (Sierpinski space is $T_{0}$ and normal but neither $T_{1}$ nor regular: normality without $T_{1}$ implies nothing). So this family separates the two failures, and it shows that "not regular" and "not normal" begin at different sizes.

Facts & Assumptions

Given: A set $X$ , a point $p \in X$ , the topology $T_{p}$ above, and points $x, y \in X$ .

[A1]

$U \in T_{p}$ exactly when $U = \emptyset$ or $p \in U$ ; the closed sets are $X$ and the subsets not containing $p$ (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).

[L1]

$T_{0}$ : some open set contains exactly one of two distinct points. $T_{1}$ : every singleton is closed. Every Hausdorff space is $T_{1}$ ( $T_{0}$ (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, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L2]

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

Verification

technique · direct

1.1

Every nonempty open set contains $p$ , by [A1].

1.2

Let $x \neq y$ in $X$ . If $x = p$ then ${p}$ is open by [A1], contains $x$ and not $y$ ; if $y = p$ the same argument applies with the roles exchanged; and if neither is $p$ then ${x, p}$ is open by [A1], contains $x$ and not $y$ . So $(X, T_{p})$ is $T_{0}$ , which is claim 1.

A1L1

1.3

Suppose $X$ has at least two points, so that $X ∖ {p} \neq \emptyset$ . Then ${p} \neq X$ , and ${p}$ contains $p$ , so ${p}$ is not closed by [A1]; hence $(X, T_{p})$ is not $T_{1}$ by [L1], and not Hausdorff, which is claim 2.

A1L1assume-hyp

1.4

Suppose $X$ has at least three points and fix $x, y \in X$ with $x \neq y$ , $x \neq p$ and $y \neq p$ . Then ${x}$ and ${y}$ are disjoint nonempty closed sets by [A1].

A1assume-hyp

2.1

Under step 1.3 fix $x \in X$ with $x \neq p$ ; then ${x}$ does not contain $p$ , so ${x}$ is closed by [A1], and $p \notin {x}$ .

step 1.3A1

2.2

Under step 1.4: any open $U \supseteq {x}$ and open $V \supseteq {y}$ are nonempty, hence both contain $p$ by step 1.1, so $U \cap V \neq \emptyset$ and $(X, T_{p})$ is not normal, which is claim 4.

step 1.1step 1.4L2

3.1

Under step 1.3: any open $U$ with $p \in U$ and any open $V$ with ${x} \subseteq V$ are both nonempty, so both contain $p$ by step 1.1 and $U \cap V \neq \emptyset$ . Hence the point $p$ and the closed set ${x}$ cannot be separated and $(X, T_{p})$ is not regular, which is claim 3.

step 1.1step 2.1L2

4.1

Steps 1.2, 1.3, 3.1 and 2.2 are claims 1 to 4.

step 1.2step 1.3step 3.1step 2.2∎

Remarks

Closures here are as large as they can be. For $A \neq \emptyset$ with $p \notin A$ one has $\overline{A} = A$ , since $A$ is already closed; but $\overline{{p}} = X$ , because the only closed set containing $p$ is $X$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set). So the particular point is dense, and a single point can be dense in a space with any number of points at all.
Every nonempty open set contains $p$ , which is the one fact behind clauses 3 and 4 alike. It makes the space as far from Hausdorff as possible while still distinguishing points: no two nonempty open sets are ever disjoint.
Sierpinski space is the case of two points. With $X = {p, x}$ the open sets are $\emptyset$ , ${p}$ and $X$ , which is The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies's Sierpinski topology with the open point named $p$ ; so clause 4 genuinely needs its extra hypothesis.

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

Particular point topology (Wikipedia) (standard reference, not scraped)
Separation axiom (Wikipedia) (standard reference, not scraped)
L. Steen and J. Seebach, Counterexamples in Topology, §8 (standard reference, not scraped)
Sierpiński space (Wikipedia) (standard reference, not scraped)

The particular-point topology is T0T0​, it is not T1T1​ and not regular once the set has at least two points, and it is not normal once the set has at least three