Theorem. A space is Hausdorff iff its diagonal map is closed [math-000B]

2023-11-23 · Lîm Tsú-thuàn

The proposition says a space is Hausdorff if and only if its diagonal map Δ:X→X×X\Delta : X \to X \times X to product space is closed, which given an alternative definition. Notice that close means its complement Δ∁\Delta^\complement is open. The definition of Δ\Delta is

Δ={(x,x)∣x∈X} \Delta = \{ (x, x) \mid x \in X \}
Not hard to see below argument also applies to all finite XX-product spaces.

Proof [local-0]

(Hausdorff => diagonal map is closed) Hausdorff means every two different points has a pair of disjoint neighborhoods (U∈Nx,V∈Ny)(U \in \mathcal{N}_x, V \in \mathcal{N}_y), where x∈Ux \in U and y∈Vy \in V. Therefore, every pair (x,y)(x, y) not line on the diagonal has U×VU \times V cover them. The union of all these open sets U×VU \times V covers Δ∁\Delta^\complement, so Δ\Delta the complement of the union is closed.

(diagonal map is closed => Hausdorff) Since

Δ∁={(x,y)∣x,y∈X(x≠y)}\Delta^\complement = \{(x, y) \mid x, y \in X (x \ne y) \}

is open, which implies for all x,yx, y the pair (x,y)∈Δ∁(x, y) \in \Delta^\complement. Since N(x,y)=Nx×Ny\mathcal{N}_{(x, y)} = \mathcal{N}_x \times \mathcal{N}_y, this implies the fact that Δ∁∈Nx×Ny\Delta^\complement \in \mathcal{N}_x \times \mathcal{N}_y.

Also, for all U∈NxU \in \mathcal{N}_x and V∈NyV \in \mathcal{N}_y, it's natural that U×V⊆Δ∁U \times V \subseteq \Delta^\complement, since the open set U×VU \times V at most cover Δ∁\Delta^\complement.

Consider that reversely again, that means for all (x,x)∈Δ(x, x) \in \Delta, pair (x,x)∉U×V(x, x) \notin U \times V (i.e. U×VU \times V will not cover any part of diagonal), that implies U∩V=∅U \cap V = \emptyset as desired.