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Section 3.2 Compact Hausdorff topological spaces

Definition 3.2.1.

A topological space \(X\) is said to be Hausdorff if for any two points \(x,y\in X\text{,}\) there exist open neighbourhoods \(U\) of \(x\) and \(V\) of \(y\) such that \(U\cap V=\varnothing\text{.}\)

Any discrete topological space is Hausdorff.

Any subspace of Euclidean space is Hausdorff: for any two points \(x,y\in X\text{,}\) if \(r\coloneq d(x,y)/2\text{,}\) then the balls \(B(x,r)\) and \(B(y,r)\) are disjoint.

The indiscrete topology on a set \(X\) is not Hausdorff unless \(\# X\leq 1\text{.}\)

Any subspace of a Hausdorff space is Hausdorff.

If \((X,\tau)\) is Hausdorff, then for any topology \(\tau'\) on \(X\) that is finer than \(\tau\text{,}\) the topological space \((X,\tau')\) is Hausdorff as well.

This points in the opposite direction from compactness. Whereas Hausdorffitude is stable under passage to a finer topology, compactness is stable under passage to a coarser topology.

Suppose \(y\in X \smallsetminus K\text{.}\) We aim to prove that there is an open neighbourhood \(V\) of \(y\) that does not intersect \(K\text{.}\) For each point \(x\in K\text{,}\) select open neighbourhoods \(U_x\) of \(x\) and \(V_x\) of \(y\) such that \(U_x\cap V_x=\varnothing\text{.}\) Now \(\{U_x\cap K : x\in X\}\) is an open cover of \(K\text{,}\) so it contains a finite subcover \(\{U_{x_1}\cap K,\dots,U_{x_n}\cap K\}\text{.}\) Now \(V\coloneq V_{x_1}\cap\dots\cap V_{x_n}\) is an open neighbourhood of \(y\) that does not intersect \(K\text{.}\)

Suppose \(K\subseteq X\) a \(\tau'\)-closed subset, hence compact with the subspace topology. Since the identity is continuous \(\fromto{(X,\tau')}{(X,\tau)}\) and the continuous image of a compactum is compact, it follows that \(K\) is compact as a subspace of \((X,\tau)\text{.}\) The previous lemma now implies that \(K\) is \(\tau\)-closed.

For every pair of points \(x\in K\) and \(y\in L\text{,}\) select open neighbourhoods \(U_{x,y}\) of \(x\) and \(V_{x,y}\) of \(y\) such that \(U_{x,y}\cap V_{x,y}=\varnothing\text{.}\) For any \(y\in L\text{,}\) we obtain an open cover \(\{U_{x,y}\cap K : x\in K\}\) of \(K\text{;}\) it contains a finite subcover \(\{U_{x_1,y}\cap K,\dots,U_{x_m,y}\cap K\}\text{.}\) We form disjoint open sets

\begin{equation*} U_y\coloneq\bigcup_{i=1}^mU_{x_i,y}\text{ and } V_y\coloneq\bigcap_{i=1}^mV_{x_i,y} \end{equation*}

such that \(U_y\supseteq K\) and \(y\in V_y\text{.}\) Thus we obtain an open cover \(\{V_y\cap L : y\in L\}\) of \(L\text{;}\) it contains a finite subcover \(\{V_{y_1}\cap L,\dots,V_{y_n}\cap L\}\text{.}\) We form disjoint open sets

\begin{equation*} U\coloneq\bigcap_{j=1}^nU_{y_j}\text{ and } V\coloneq\bigcup_{j=1}^nV_{y_j} \end{equation*}

such that \(U\supseteq K\) and \(V\supseteq L\text{.}\)

Assume that \(X\) is Hausdorff, and let \(x\in X\text{.}\) Surely \(x\in I_x\text{,}\) and we claim that any point \(y\in X\smallsetminus\{x\}\) is not in \(I\text{.}\) Indeed, for any such point, one may find disjoint open neighbourhoods \(U\) of \(x\) and \(V\) of \(y\text{,}\) whence \(X \smallsetminus V\) is a closed neighbourhood of \(x\) not containing \(y\text{.}\) Thus \(I=\{x\}\text{.}\)

Conversely, suppose that, for any point \(x\in X\text{,}\) one has \(I_x=\{x\}\text{.}\) Suppose \(x\) and \(y\) two distinct points of \(X\text{.}\) Then since \(I_x=\{x\}\text{,}\) there exists a closed neighbourhood \(W\) of \(x\) not containing \(y\text{,}\) and now the interior \(\iota W\) and the complement \(X \smallsetminus W\) are disjoint open neighbourhoods of \(x\) and \(y\text{,}\) respectively.

The key point is that if \(U,V \subseteq X\) are subsets, then \(U \cap V = \varnothing\) if and only if \((U \times V) \cap \Delta_X = \varnothing\text{.}\)

Assume that \(X\) is Hausdorff. Let \((x,y) \in (X \times X) \smallsetminus \Delta_X\text{.}\) Choose an open neighborhood \(U\) of \(x\) and \(V\) of \(y\) such that \(U \cap V = \varnothing\text{.}\) Thus \(U \times V\) is an open neighborhood of \((x,y)\) in \((X \times X) \smallsetminus \Delta_X\text{.}\)

Conversely, assume that \(\Delta_X\) is closed in \(X \times X\text{.}\) Now let \(x, y \in X\) be points such that \(x \neq y\text{.}\) There exist open neighborhoods \(U\) of \(x\) and \(V\) of \(y\) such that \((U \times V) \subseteq (X \times X) \smallsetminus \Delta_X\text{.}\) It follows that \(U \cap V = \varnothing\text{.}\)

This follows from the fact that \(\Gamma_f = (f \times \id)^{-1}(\Delta_Y)\text{.}\)