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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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The cofinite topology on an infinite set is T1T_1 but neither Hausdorff nor regular nor normal

Example

Let XX be an infinite set — that is, a set that is not finite (Finite, countably infinite, countable, uncountable) — and give it the cofinite topology Tcof={}{UX:XU is finite}\mathcal{T}_{\mathrm{cof}} = \{\varnothing\} \cup \{\, U \subseteq X : X \setminus U \text{ is finite} \,\} (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), whose closed sets are XX together with the finite subsets of XX. Then:

  1. (X,Tcof)(X, \mathcal{T}_{\mathrm{cof}}) is T1T_1 (T0T_0 (Kolmogorov) and T1T_1 (Frechet) spaces).
  2. No two nonempty open sets are disjoint. Consequently the space 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), not regular (Regular spaces and T3T_3 spaces, with the source disagreement over whether regularity includes T1T_1 stated explicitly) and not normal (Normal spaces and T4T_4 spaces, with the source disagreement over whether normality includes T1T_1 stated explicitly).

So the cofinite topology on an infinite set satisfies T1T_1 and T0T_0 and fails every axiom above them. It is the standard witness that T1T_1 is strictly weaker than the Hausdorff condition, and that is how it is used on the main page (FALSE: every T1T_1 space is Hausdorff); here it is pushed further, to show that T1T_1 implies neither of the two axioms that sit above T2T_2 either.

Facts & Assumptions

Given: An infinite set XX with the cofinite topology Tcof\mathcal{T}_{\mathrm{cof}}, and points x,y,zXx, y, z \in X.

[A1]

UTcofU \in \mathcal{T}_{\mathrm{cof}} exactly when U=U = \varnothing or XUX \setminus U is finite; the closed sets are XX and the finite subsets of XX; and a union of two finite sets is finite (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, facts (i) and (ii) of that item, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L3]

A set with at most one element is finite, being equinumerous with 00 or with 11; so an infinite set has at least three distinct points, since a set with at most two elements is finite as a union of two sets each with at most one (Finite, countably infinite, countable, uncountable, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, fact (ii)).

Verification

technique · direct
1.1

TcofTcof\mathcal{T}_{\mathrm{cof}} \subseteq \mathcal{T}_{\mathrm{cof}}, so (X,Tcof)(X,\mathcal{T}_{\mathrm{cof}}) is T1T_1, which is claim 1.

L1
1.2

XX contains three distinct points xx, yy, zz.

L3
1.3

Let U,VU, V be nonempty open sets and suppose UV=U \cap V = \varnothing; then X=(XU)(XV)X = (X \setminus U) \cup (X \setminus V) is a union of two finite sets, hence finite by [A1], contradicting the hypothesis that XX is infinite. So no two nonempty open sets are disjoint, which is the first half of claim 2.

A1assume-hyp
2.1

xyx \ne y, and any open UxU \ni x and open VyV \ni y are nonempty, hence meet by step 1.3; so the space is not Hausdorff by [L2].

step 1.2step 1.3L2
2.2

{y}\{y\} is closed by [A1] and x{y}x \notin \{y\}; any open UxU \ni x and open V{y}V \supseteq \{y\} are nonempty, hence meet by step 1.3; so the space is not regular by [L2].

step 1.2step 1.3A1L2L4
2.3

{y}\{y\} and {z}\{z\} are disjoint nonempty closed sets by [A1] and step 1.2; any open U{y}U \supseteq \{y\} and open V{z}V \supseteq \{z\} are nonempty, hence meet by step 1.3; so the space is not normal by [L2].

step 1.2step 1.3A1L2L4
3.1

Steps 2.1, 2.2 and 2.3 complete claim 2, and step 1.1 is claim 1.

step 1.1step 2.1step 2.2step 2.3

Remarks

Depends on

Used by

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Sources