TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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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

Statement

Let $(X, T)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $T_{c o f}$ be the cofinite topology on the set $X$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). The following four conditions are equivalent.

(a) $X$ is $T_{1}$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).
(b) ${x}$ is closed for every $x \in X$ .
(c) $F$ is closed for every finite $F \subseteq X$ (Finite, countably infinite, countable, uncountable).
(d) $T_{c o f} \subseteq T$ , that is, the topology of $X$ is finer than the cofinite topology on the same set.

Condition (d) says that the cofinite topology is the coarsest $T_{1}$ topology on any set: it is $T_{1}$ by the equivalence, and every $T_{1}$ topology on that set contains it.

Facts & Assumptions

Given: A topological space $(X, T)$ , the cofinite topology $T_{c o f}$ on the same set $X$ , points $x, y \in X$ and a finite subset $F \subseteq X$ .

[A1]

$X$ is $T_{1}$ when for all $x \neq y$ there are open $U, V$ with $x \in U$ , $y \notin U$ , $y \in V$ and $x \notin V$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).

[L1]

A set is closed exactly when its complement is open; $\emptyset$ and $X$ are open and closed; and a union of two closed sets is closed by (C3), hence so is a union of finitely many by iterating (C3) (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L2]

A set is open exactly when it is a neighbourhood of each of its points, that is, exactly when each of its points lies in an open subset of it (Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, consequence 4).

[L3]

The cofinite topology on $X$ consists of $\emptyset$ together with the sets whose complement in $X$ is finite; its closed sets are $X$ together with the finite subsets of $X$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L4]

A finite set is one equinumerous with a natural number, so a finite $F$ may be listed as $F = {x_{0}, \dots, x_{n - 1}}$ for some $n \in N$ , the case $n = 0$ being $F = \emptyset$ (Finite, countably infinite, countable, uncountable).

Proof

technique · direct

1.1A1

(a) implies (b): fix $x \in X$ and let $y \in X ∖ {x}$ ; then $y \neq x$ , so [A1] supplies an open $V$ with $y \in V$ and $x \notin V$ , whence $y \in V \subseteq X ∖ {x}$ .

1.2L1L4

(b) implies (c): let $F \subseteq X$ be finite and list it as $F = {x_{0}, \dots, x_{n - 1}}$ by [L4], so that $F = {x_{0}} \cup \dots \cup {x_{n - 1}}$ ; for $n = 0$ this reads $F = \emptyset$ , which is closed by [L1].

1.3L1L3

(c) implies (d): let $U \in T_{c o f}$ ; if $U = \emptyset$ then $U \in T$ by [L1], and otherwise $X ∖ U$ is finite by [L3], hence closed by (c), hence $U$ is open.

1.4A1L3

(d) implies (a): let $x \neq y$ in $X$ ; the sets $X ∖ {y}$ and $X ∖ {x}$ have finite complements, so they lie in $T_{c o f}$ by [L3] and hence in $T$ by (d), and they witness the $T_{1}$ condition, since $x \in X ∖ {y}$ , $y \notin X ∖ {y}$ , $y \in X ∖ {x}$ and $x \notin X ∖ {x}$ .

2.1step 1.1L1L2

By step 1.1 the set $X ∖ {x}$ is a neighbourhood of each of its points, hence open by [L2], so ${x}$ is closed by [L1]; this completes the implication (a) implies (b).

2.2step 1.2L1

By step 1.2 and (b) the set $F$ is a union of $n$ closed sets, hence closed by [L1]; this completes the implication (b) implies (c).

3.1step 1.3step 1.4step 2.1step 2.2

The four implications of steps 2.1, 2.2, 1.3 and 1.4 close the cycle (a) implies (b) implies (c) implies (d) implies (a), so the four conditions are equivalent.

4.1step 3.1L3∎

In particular $T_{c o f}$ itself satisfies (d) with $T = T_{c o f}$ , so the cofinite topology on any set is $T_{1}$ by step 3.1, and by (d) it is contained in every $T_{1}$ topology on that set; this is the final assertion of the statement.

Remarks

The theorem is the reason $T_{1}$ is quoted as "points are closed". Every later use of $T_{1}$ on this page goes through clause (b): the $T_{1}$ hypothesis in $T_{3}$ and $T_{4}$ is used exactly to turn a point into a closed set so that regularity or normality applies to it.
Clause (c) is not a strengthening of clause (b). It follows from it by a finite union, and the finite union is genuinely finite: an arbitrary union of closed sets need not be closed, and in the cofinite topology on an infinite set no infinite proper subset is closed at all, although every singleton is.
Clause (d) locates the cofinite topology. It is the smallest $T_{1}$ topology on a given set, in the sense of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison's comparison order, and this is why it is the standard witness for a $T_{1}$ space that fails every stronger separation axiom; the witness is worked on the companion page.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A space is T1T1​ 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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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