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

1.1

(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 $y \in V \subseteq X ∖ {x}$ .

1.2

(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].

L1L4

1.3

(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.

L1L3

1.4

(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 $x \notin X ∖ {x}$ .

A1L3

2.1

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).

step 1.1L1L2

2.2

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).

step 1.2L1

3.1

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.

step 1.3step 1.4step 2.1step 2.2

4.1

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.

step 3.1L3∎

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.

Depends on

Used by

Dependency tree · two levels

18 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

T1 space (Wikipedia) (standard reference, not scraped)
Cofiniteness (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §17 (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 17: Closed Sets and Limit Points (East Tennessee State University) (standard reference, not scraped)

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