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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,T) be a topological space and give X×X the product topology (The product set ∏i∈IXi 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 ΔX (The diagonal ΔX⊆X×X, the diagonal map δX, and the pairing ⟨f,g⟩ of two maps) is closed in X×X:

X Hausdorff  ⟺  ΔX=ΔX‾ in X×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 ΔX back along a continuous pairing, and the graph result is a specialization of that argument.

Facts & Assumptions

Given: A topological space (X,T), the product X×X with the product topology, and the diagonal ΔX={ z∈X×X:z0=z1 }.

[A1]

X is Hausdorff when for all x≠y in X there are open U∋x and V∋y with U∩V=∅ (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L1]

Proof

technique · direct
1.1

Assume X is Hausdorff and let z∈X×X with z∉ΔX, so that z0≠z1; by [A1] there are open U∋z0 and V∋z1 with U∩V=∅.

A1
1.2

Assume ΔX is closed and let x,y∈X with x≠y; then z:=(x,y) satisfies z∉ΔX=ΔX‾, the equality holding by [L1] since ΔX is closed.

L1
2.1

The box U×V of step 1.1 is a basic open set containing z, and (U×V)∩ΔX=∅: a point w of the intersection would satisfy w0=w1 with w0∈U and w1∈V, putting w0 in U∩V=∅.

step 1.1A2
2.2

By [L1] applied to the basis of [A2], step 1.2 supplies a basic open box U×V with z∈U×V and (U×V)∩ΔX=∅; so x∈U and y∈V.

step 1.2A2L1
3.1

From step 2.1 and [L1], z∉ΔX‾ for every z∉ΔX; hence ΔX‾⊆ΔX, and with [L2] this gives ΔX‾=ΔX, so ΔX is closed.

step 1.1step 2.1L1L2
3.2

The sets U and V of step 2.2 are disjoint: if t∈U∩V then (t,t) lies in U×V and in ΔX, contradicting (U×V)∩ΔX=∅.

step 2.2
4.1

Step 3.1 shows that X Hausdorff implies ΔX closed, and steps 2.2 and 3.2 show that ΔX 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

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