Too many web development posts! Time for some topology!
As an exercise I’ve worked through the classical result that if a space \(X\) is compact Hausdorff and has a \(G_\delta\)-diagonal \(\Delta\), then \(X\) is metrizable. It relies on Urysohn’s theorem that regular second-countable spaces are metrizable, and since compact Hausdorff spaces are normal*, I’ll just show that the space is second countable.
When I went to compare, most of proofs I found online and in texts like Engleking were a lot less direct (involving characterizations of \(G_\delta\) diagonals using covers), so I hope someone gets some use out of this fairly straight-forward attack (due to a hint from my advisor Gary Gruenhage).
Definitions
We’ll assume you’re good on most of the terms, but \(G_\delta\) sets are countable intersections (from the \(\delta\)) of open sets (from the \(G\)). The diagonal of a space is what you expect:
\[\Delta = \{\langle x,x \rangle : x \in X \} \subseteq X^2\]Proof
Start by assuming that \(\Delta = \bigcap_{n<\omega} G_n\) where \(G_{n+1} \subseteq G_n\). Since \(X^2\) is normal (not because \(X\) is, but because \(X^2\) is also compact Hausdorff) and \(\Delta\) is closed, we can set \(U_0=G_0\), and \(\Delta \subseteq U_{n+1} \subseteq \overline{U_{n+1}}\subseteq U_n \cap G_{n+1}\).
It follows that \(\Delta \subseteq \bigcap_{n<\omega} U_n \subseteq \bigcap_{n<\omega} \overline{U_n} \subseteq \bigcap_{n<\omega} G_n = \Delta\)
These \(U_n\) have a special property: for every \(x\in X\) and every open neighborhood \(A\) of \(x\), there is some \(n<\omega\) such that
\[U_n[x]=\{y:\langle x,y\rangle \in U_n\}\subseteq A\]To see this, suppose not: then we may choose \(x_n \in U_n[x]\setminus A\). Since \(x_n \in \overline{U_m[x]}\) for \(n\geq m\), it follows that \(x_n\) must converge in \(\bigcap_{m<\omega} \overline{U_m[x]} = \{x\}\). But since \(x_n\) converges to \(x\), its neighborhood \(A\) must contain all but finitely many \(x_n\), a contradiction.
Now, since \(U_n\) is open, we can choose open \(B_{x,n}\) for each \(x\in X\) such that \((B_{x,n})^2\subseteq U_n\). Since \(X\) is compact, let \(\mathcal{B}_n\) be a finite subcover for \(\{B_{x,n}: x\in X\}\) for each \(n<\omega\).
We note that \(\mathcal{B} = \bigcup_{n<\omega} \mathcal{B}_n\) is countable, and we claim it is a base for \(X\). Let \(A\) be a neighborhood of \(x\); then choose \(U_n\) such that \(U_n[x]\subseteq A\). We need only take \(B_{z,n}\) from the cover \(\mathcal{B}_n\) of \(X\) such that \(x\in B_{z,n}\), and note that \((B_{z,n})^2\subseteq U_n\) implies \(x\in B_{z,n}\subseteq U_n[x] \subseteq A\). \(\Box\)
*Not too bad: you can easily show that in Hausdorff spaces, not just points can be separated by open sets, but all compact sets can be separated by open sets. All closed subsets of compact spaces are compact, so that’s that.