# If $X$ Hausdorff, compact with $G_{\delta }$ diagonal then $X$ has a countable basis

Suppose that $$X$$ is a Hausorff and compact space. Moreover, suppose that $$\Delta:=\{(x,x):x\in X\}$$ is a $$G_{\delta}$$ set. I need to show that $$X$$ has a countable basis.

As $$X$$ is Hausdorff, $$\Delta$$ is closed in $$X\times X$$, moreover, as $$X\times X$$ is compact and Hausdorff then it is a normal space. Now, as $$\triangle$$ is a $$G_{\delta}$$ closed set, there exists a continuous function $$f:X\times X\to \mathbb{R}$$ such that $$f^{-1}(\{0\})=\Delta$$ (It follows by Urysohn's lemma). Let $$\mathcal{B}:=\{B_n\mid n\in\omega\}$$ be a countable basis for $$\mathbb{R}$$ and consider the set $$\{f^{-1}(B_n)\mid n\in\omega\}$$. I claim that this set is a basis for $$X\times X$$.

As $$f$$ is continuous with compact domain and Hausdorff image, $$f$$ is a closed function. Let $$G\subseteq X\times X$$ be an open set. So $$\mathbb{R}\setminus f[(X\times X)\setminus G]$$ is a open set in $$\mathbb{R}$$. Let $$n\in\omega$$ such that $$B_n\subseteq \mathbb{R}\setminus f[(X\times X)\setminus G]$$, so we have that $$f^{-1}(B_n)\subseteq G$$. Then $$X\times X$$ has a countable basis, so $$X$$ has a countable basis too.

I think that there is a mistake in my proof because I don't use the fact that $$\Delta$$ is $$G_{\delta}$$ in a significant way. Can someone tell me where I am wrong?

• In my personal opinion $\Delta$(\Delta) looks better than $\triangle$(\triangle) and is shorter to write. Jun 14, 2019 at 2:14

The inverse images of the $$B_n$$ won't necessarily form a countable base for $$X \times X$$. So you need a better approach:

You can show that if $$\Delta$$ is a $$G_\delta$$ in $$X^2$$ and $$X$$ is compact Hausdorff then $$X$$ has a countable $$T_2$$-separating family of open sets (proof below). (Such a family $$\mathcal{O}$$ has the property that for all $$x \neq y$$ in $$X$$ we have $$O_1,O_2 \in \mathcal{O}$$ with $$x \in O_1, x \in O_2, O_1 \cap O_2=\emptyset$$) and this implies that $$X$$ has countable weight, by compactness: the topology $$\tau$$ generated by $$\mathcal{O}$$ is second countable and Hausdorff and a subset of the original topology $$\tau_X$$ on $$X$$ so the identity $$1_X: (X, \tau_X) \to (X,\tau)$$ is a continuous closed (compact to Hausdorff) bijection and so $$\tau_X=\tau$$ and we're done; a standard argument.

The family $$\mathcal{O}$$ can be found as follows: write $$\Delta = \bigcap_n U_n$$ where $$U_n \subseteq X^2$$ open. Fix $$n$$ and for each $$x \in X$$ find $$O^n_x$$ open in $$X$$ containing $$x$$ such that $$O^n_x \times O^n_x \subseteq U_n$$ by openness of $$U_n$$. Find an open $$V^n_x$$ containing $$x$$ such that $$\overline{V^n_x} \subseteq O^n_x$$ by regularity of $$X$$. Finitely many of these cover $$X$$ by compactness, so we have $$F_n \subseteq X$$ finite such that $$X = \bigcup \{V^n_x: x \in F_n\}$$. Now define

$$\mathcal{O} = \{V^n_x, X\setminus \overline{V^n_x}\mid x \in F_n, n \in \Bbb N\}$$

and check that $$\mathcal{O}$$ is (countable and) $$T_2$$-separating. (And so these form a countable subbase for $$X$$ as we saw in the final argument of the first paragraph.)

• Note that $\Delta = \cap_n U_n$ is used in the proof of $T_2$-separating in an essential way: ($x \neq y$ means $(x,y) \notin \Delta$ so $(x,y) \notin U_m$ for some $m$ etc. Jun 15, 2019 at 9:09

Existence of $$f$$ requires the fact that $$\Delta$$ is a $$G_{\delta}$$ set. If we arite $$\Delta$$ as the intersection of open sets $$U_1,U_2,...$$ then there exits $$f_n$$'s which are $$0$$ on $$\Delta$$ and $$1$$ on $$U_n^{c}$$ with $$0\leq f_n \leq 1$$ and $$f =\sum \frac 1 {2^{n}} f_n$$ would serve the purpose.

• Yes, I know that. But suppose that there exists a closed set $F\subseteq X\times X$ such that $F\cap\triangle=\emptyset$. So, there exists a continuous function $f:X\times X\to\mathbb{R}$ such that $f(\triangle)\subseteq\{0\}$ and $f(F)\subseteq\{1\}$. Then, with this $f$ I can use the same argument above for the proof. In this case, that $\triangle$ was be a $G_{\delta}$ set is not important. Jun 13, 2019 at 23:50
• @Gödel You have to make sure that $f=0$ only at points of $\Delta$. Your $f$ may vanish at points not in $\Delta$ also. Jun 13, 2019 at 23:52
• But why I need $f^{-1}(0)=\triangle$ in my proof? Jun 13, 2019 at 23:54
• Your third paragraph has mistakes. $f^{-1}(B_n)$ may be empty. You have to show that for any $u \in G$ there exists $n$ such that $u \in f^{-1}(B_n) \subset G$. Jun 14, 2019 at 0:03