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A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology

Statement

Let $(X, \mathcal{T})$ be a topological space and give $X \times X$ the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Then $X$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) if and only if the diagonal $\Delta_X$ (The diagonal $\Delta_X \subseteq X \times X$ , the diagonal map $\delta_X$ , and the pairing $\langle f, g \rangle$ of two maps) is closed in $X \times X$ :

$X \text{ Hausdorff} \iff \Delta_X = \overline{\Delta_X} \text{ in } X \times X .$

The condition on the right is a single closedness statement about one subset of one space, with no quantifier over pairs of points visible in it; that is what makes the criterion useful. In particular, the closed agreement-set result below is obtained by pulling $\Delta_X$ back along a continuous pairing, and the graph result is a specialization of that argument.

Facts & Assumptions

Given: A topological space $(X,\mathcal{T})$ , the product $X \times X$ with the product topology, and the diagonal $\Delta_X = \{\, z \in X \times X : z_0 = z_1 \,\}$ .

[A1]

$X$ is Hausdorff when for all $x \ne y$ in $X$ there are open $U \ni x$ and $V \ni y$ with $U \cap V = \varnothing$ (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[A2]

The boxes $U \times V$ with $U, V \in \mathcal{T}$ form a basis for the product topology on $X \times X$ , the index set being $2$ (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, Basis and subbasis for a topology, and the topology generated by a family of sets, The diagonal $\Delta_X \subseteq X \times X$ , the diagonal map $\delta_X$ , and the pairing $\langle f, g \rangle$ of two maps).

[L1]

For a basis $\mathcal{B}$ of a space, a point lies in $\overline{A}$ if and only if every $B \in \mathcal{B}$ containing it meets $A$ ; and $A$ is closed if and only if $A = \overline{A}$ (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, claims 1(d) and 2, Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

[L2]

$A \subseteq \overline{A}$ for every subset $A$ of a space (Interior, closure, boundary, exterior, derived set and isolated point in a topological space, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Proof

technique · direct

1.1

Assume $X$ is Hausdorff and let $z \in X \times X$ with $z \notin \Delta_X$ , so that $z_0 \ne z_1$ ; by [A1] there are open $U \ni z_0$ and $V \ni z_1$ with $U \cap V = \varnothing$ .

1.2

Assume $\Delta_X$ is closed and let $x, y \in X$ with $x \ne y$ ; then $z := (x,y)$ satisfies $z \notin \Delta_X = \overline{\Delta_X}$ , the equality holding by [L1] since $\Delta_X$ is closed.

2.1

The box $U \times V$ of step 1.1 is a basic open set containing $z$ , and $(U \times V) \cap \Delta_X = \varnothing$ : a point $w$ of the intersection would satisfy $w_0 = w_1$ with $w_0 \in U$ and $w_1 \in V$ , putting $w_0$ in $U \cap V = \varnothing$ .

step 1.1A2

2.2

By [L1] applied to the basis of [A2], step 1.2 supplies a basic open box $U \times V$ with $z \in U \times V$ and $(U \times V) \cap \Delta_X = \varnothing$ ; so $x \in U$ and $y \in V$ .

step 1.2A2L1

3.1

From step 2.1 and [L1], $z \notin \overline{\Delta_X}$ for every $z \notin \Delta_X$ ; hence $\overline{\Delta_X} \subseteq \Delta_X$ , and with [L2] this gives $\overline{\Delta_X} = \Delta_X$ , so $\Delta_X$ is closed.

step 1.1step 2.1L1L2

3.2

The sets $U$ and $V$ of step 2.2 are disjoint: if $t \in U \cap V$ then $(t,t)$ lies in $U \times V$ and in $\Delta_X$ , contradicting $(U \times V) \cap \Delta_X = \varnothing$ .

step 2.2

4.1

Step 3.1 shows that $X$ Hausdorff implies $\Delta_X$ closed, and steps 2.2 and 3.2 show that $\Delta_X$ closed implies that any two distinct points of $X$ have disjoint open neighbourhoods, which by [A1] is the Hausdorff condition; the two implications are the theorem.

step 2.2step 3.1step 3.2A1∎

Remarks

The criterion is about the product topology on a binary product, and there the box basis and the product basis are the same family (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space), so the boxes tested in steps 2.1 and 2.2 are legitimately basic. No infinite product is formed anywhere in the argument, and the criterion says nothing about one.
Neither direction spends a choice principle. The forward direction produces one box from one Hausdorff separation of one named pair, and the backward direction reads one box out of the closure characterisation; there is no family to select from in either.
What the criterion does not say. It does not say that $\Delta_X$ is closed in $X \times X$ carrying some other topology, and it does not say that $\Delta_X$ is closed in $X$ — the latter is not even a statement, $\Delta_X$ being a subset of the square. The hypothesis that $X \times X$ carries the product topology is used at [A2] and cannot be dropped.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 60 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Hausdorff space (Wikipedia) (standard reference, not scraped)
Product topology (Wikipedia) (standard reference, not scraped)
Stacks Project, Topology, Lemma 5.3 (Tag 08ZD) (standard reference, not scraped)