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


A space X is HausdorffPlanetmathPlanetmath if and only if


is closed in X×X under the product topology.


First, some preliminaries: Recall that the diagonal map Δ:XX×X is defined as xΔ(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 yY, there’s a basis element B with yBY. 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,VX with xU,yV. 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×VΔ(X)c. U×V lying in Δ(X)c implies that U and V are disjoint.

If we have xy in X, then (x,y) is in Δ(X)c. The basis element containing (x,y) gives us open, disjoint U,V with xU,yV. 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