a space X is Hausdorff if and only if Δ(X) is closed
Theorem.
Proof.
First, some preliminaries: Recall that the diagonal map Δ:X→X×X is defined as xΔ⟼(x,x). Also recall that in a topology 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 |
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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 |