a space X is Hausdorff if and only if Δ⁢(X) is closed


Theorem.

A space X is HausdorffPlanetmathPlanetmath if and only if

{(x,x)∈X×X∣x∈X}

is closed in X×X under the product topology.

Proof.

First, some preliminaries: Recall that the diagonal map Δ:X→X×X is defined as x⟼Δ(x,x). Also recall that in a topologyMathworldPlanetmath generated by a basis (like the product topology), a set Y is open if and only if, for every point y∈Y, there’s a basis element B with y∈B⊂Y. Basis elements for X×X have the form U×V where U,V are open sets in X.

Now, suppose that X is Hausdorff. We’d like to show its image under Δ is closed. We can do that by showing that its complement Δ⁢(X)c is open. Δ⁢(X) consists of points with equal coordinates, so Δ⁢(X)c consists of points (x,y) with x and y distinct.

For any (x,y)∈Δ⁢(X)c, the Hausdorff condition gives us disjoint open U,V⊂X with x∈U,y∈V. Then U×V is a basis element containing (x,y). U and V have no points in common, so U×V contains nothing in the image of the diagonal map: U×V is contained in Δ⁢(X)c. So Δ⁢(X)c is open, making Δ⁢(X) closed.

Now let’s suppose Δ⁢(X) is closed. Then Δ⁢(X)c is open. Given any (x,y)∈Δ⁢(X)c, there’s a basis element U×V with (x,y)∈U×V⊂Δ⁢(X)c. U×V lying in Δ⁢(X)c implies that U and V are disjoint.

If we have x≠y in X, then (x,y) is in Δ⁢(X)c. The basis element containing (x,y) gives us open, disjoint U,V with x∈U,y∈V. X is Hausdorff, just like we wanted. ∎

Title a space X is Hausdorff if and only if Δ⁢(X) is closed
Canonical name ASpacemathnormalXIsHausdorffIfAndOnlyIfDeltaXIsClosed
Date of creation 2013-03-22 14:20:47
Last modified on 2013-03-22 14:20:47
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 9
Author mathcam (2727)
Entry type Proof
Classification msc 54D10
Related topic DiagonalEmbedding
Related topic T2Space
Related topic ProductTopology
Related topic SeparatedScheme