# Compact Spaces With G-delta Diagonals

In a previous post, I showed that any compact space with a countable network is metrizable. Another classic metrization theorem for compact spaces is that any compact space with a $G_\delta-$diagonal is metrizable ([6]). The theroem I try to prove is: for a compact space $X$, $X^2$ is perfectly normal if and only if $X$ has a $G_\delta-$diagonal if and only if $X$ is metrizable. My proof is based on the notion of $G^*_\delta-$diagonal. Every compact space with a $G_\delta-$diagonal has a $G^*_\delta-$diagonal, which allows us to define a countable base. The theorem discussed here had been generalized (see the comment at the end of this post). All spaces are at least Hausdorff.

Let $X$ be a space. The set $\Delta=\lbrace{(x,x):x \in X}\rbrace$ is called the diagonal of the space $X$. The space $X$ has a $G_\delta-$diagonal if $\Delta$ is a $G_\delta-$set in $X^2$.

Let $\mathcal{G}$ be a collection of subsets of $X$ and let $x \in X$. Define $st(x,\mathcal{G})=\bigcup \lbrace{G \in \mathcal{G}:x \in G}\rbrace$. A sequence $\lbrace{\mathcal{G}_n}\rbrace_{n<\omega}$ of open covers of $X$ is called a $G_\delta-$diagonal sequence of $X$ if for each $x \in X$, $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n)$. Lemma 1 shows that a space has a $G_\delta-$diagonal if and only if it has a $G_\delta-$diagonal sequence. This lemma is due to Ceder ([2]).

Another notion we need is that of the $G^*_\delta-$diagonal. The space $X$ has a $G^*_\delta-$diagonal if there is a $G_\delta-$diagonal sequence $\lbrace{\mathcal{G}_n}\rbrace_{n<\omega}$ such that for each $x \in X$, $\lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{G}_n)}$. Such a $G_\delta-$diagonal sequence is called a $G^*_\delta-$diagonal sequence. The notion of $G^*_\delta-$diagonal is due to R. E. Hodel ([4]). Lemma 2 below shows that any compact space with a $G_\delta-$diagonal has a $G^*_\delta-$diagonal.

Lemma 1. The space $X$ has a $G_\delta-$diagonal if and only if it has a $G_\delta-$diagonal sequence.

Proof. $\Rightarrow$ Suppose that $\Delta=\bigcap_{n<\omega}U_n$ where each $U_n$ is open in $X^2$. Let $\tau$ denote the topology on $X$. For each $n$, let $\mathcal{G}_n=\lbrace{V \in \tau:V \times V \subset U_n}\rbrace$. We claim that for each $x \in X$, $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n)$. Obviously $\lbrace{x}\rbrace \subset \bigcap_{n<\omega} st(x,\mathcal{G}_n)$. Let $y \in \bigcap_{n<\omega} st(x,\mathcal{G}_n)$. For each $n$, $y \in V_n$ where $V_n \in \mathcal{G}_n$ and $x \in V_n$. Thus $(x,y) \in V_n \times V_n \subset U_n$ for each $n$. This implies $(x,y) \in \Delta$ and $x=y$. Thus $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n)$. We have established that $\{ \mathcal{G}_n \}$ is a $G_\delta$-diagonal sequence of $X$.

$\Leftarrow$ Suppose $\lbrace{\mathcal{G}_n}\rbrace$ is a $G_\delta-$diagonal sequence of $X$. For each $n$, let $U_n=\bigcup \lbrace{V \times V:V \in \mathcal{G}_n}\rbrace$. Since $\bigcap_n st(x,\mathcal{G}_n)=\{ x \}$, $\Delta \subset \bigcap_{n<\omega} U_n$. To show the set inclusion for the other direction, let $(x,y) \in \bigcap_{n<\omega} U_n$. For each $n$, $(x,y) \in V_n \times V_n$ for some $V_n \in \mathcal{G}_n$. This implies that $y \in st(x,\mathcal{G}_n)$ for each $n$. It follows that $y=x$. Thus $\Delta = \bigcap_{n<\omega} U_n$.

Lemma 2. If $X$ is compact and has a $G_\delta-$diagonal, then $X$ has a $G^*_\delta-$diagonal. Furthermore, each open cover in the $G_\delta-$diagonal sequence is finite.

Proof. Let $\lbrace{\mathcal{G}_n}\rbrace_{n<\omega}$ be the $G_\delta-$diagonal sequence obtained in Lemma 1. We inductively define $\lbrace{\mathcal{H}_n}\rbrace_{n<\omega}$, another $G_\delta-$diagonal sequence.

Using the compactness of $X$, obtain a finite subcollection $\mathcal{H}_0$ of $\mathcal{G}_0$ such that $\mathcal{H}_0$ is a cover of $X$. Here’s how I obtain $\mathcal{H}_1$. For each $x \in X$, choose an open set $G_x \in \mathcal{G}_1$ and an open set $H_x \in \mathcal{H}_0$ such that $x \in G_x$ and $x \in H_x$. Choose open set $V_x$ such that $x \in V_x$ and $\overline{V_x} \subset G_x \cap H_x$. Let $\mathcal{H}_1$ be a finite subcollection of $\lbrace{V_x:x \in X}\rbrace$ such that $\mathcal{H}_1$ is a cover of $X$. Continue the inductive process and we produce a sequence of open covers $\lbrace{\mathcal{H}_n}\rbrace_{n<\omega}$ satisfying the following two claims.

Claim 1
For each $x \in X$, $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{H}_n)$.

Because each open cover $\mathcal{H}_n$ is chosen to be a subcover or a refinement of the open cover $\mathcal{G}_n$, we have $st(x,\mathcal{H}_n) \subset st(x,\mathcal{G}_n)$. Since $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n)$ (from the definition of $G_\delta-$diagonal sequence), we have $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{H}_n)$.

Claim 2
For each $x \in X$, $\lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)}$.

We only need to show $\bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)} \subset \lbrace{x}\rbrace$. Let $y \in \overline{st(x,\mathcal{H}_n)}$ for each $n$. Because $\mathcal{H}_n$ is finite, $y \in \overline{V}$ for some $V \in \mathcal{H}_n$ with $x \in V$. Each such $\overline{V} \subset U$ for some $U \in \mathcal{H}_{n-1}$. Thus $y \in \overline{st(x,\mathcal{H}_n)}$ implies $y \in st(x,\mathcal{H}_{n-1})$. By Claim 1, $y=x$.

We have shown that $X$ has a $G^*_\delta-$diagonal by producing a $G^*_\delta-$diagonal sequence $\lbrace{\mathcal{H}_n}\rbrace_{n<\omega}$.

Theorem. If $X$ is compact and has a $G_\delta-$diagonal, then $X$ is metrizable.

Proof. Let $\lbrace{\mathcal{H}_n}\rbrace_{n<\omega}$ be the $G^*_\delta-$diagonal sequence obtained in Lemma 2. Furthermore, each $\mathcal{H}_n$ is a finite open cover. Let $\mathcal{H}=\bigcup_{n<\omega} \mathcal{H}_n$. The collection $\mathcal{H}$ satisfies the properties stated in the following two claims.

Claim 3
For each $x,y \in X$ with $x \neq y$, there is a $U \in \mathcal{H}$ such that $x \in U$ and $y \notin \overline{U}$.

Since $\lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)}$, $y \notin \overline{st(x,\mathcal{H}_n)}$ for some $n$. Then there is some $U \in \mathcal{H}_n$ such that $x \in U$ and $y \notin \overline{U}$. In fact, for any $U \in \mathcal{H}_n$ with $x \in U$, it must be the case that $y \notin \overline{U}$.

Claim 4
Let $\mathcal{B}=\lbrace{X-\overline{\bigcup F}: F \subset \mathcal{H} \phantom{x} and \phantom{x} \vert F \lvert < \omega}\rbrace$. Then $\mathcal{B}$ is a countable base for $X$.

To see Claim 4, note that $\mathcal{B}$ is a cover of $X$ and is closed under finite intersections. This makes $\mathcal{B}$ is base for a topology. To show that this base generate the same topology on $X$, let $y \in U \subset X$ where $U$ is open. Then $X-U$ is compact. For each $x \in X-U$, let $V_x \in \mathcal{H}$ such that $x \in V_x$ and $y \notin \overline{V_x}$. We can choose $F=\lbrace{V_{x(0)},...,V_{x(n)}}\rbrace$ such that $F$ is a cover of $X-U$. Then $y \in X-\overline{\bigcup F} \subset U$.

With Claims 3 and 4, the theorem is established.

Corollary. Let $X$ be a compact space. The following conditions are equivalent.

1. $X^2$ is perfectly normal.
2. $X$ has a $G_\delta-$diagonal.
3. $X$ has a countable base.

Proof. $1 \rightarrow 2$ and $3 \rightarrow 1$ are obvious. $2 \rightarrow 3$ follows from the theorem.

Examples. Based on the corollary, any non-metrizable compact Hausdorff space does not have a $G_\delta-$diagonal. One handy example is the uncountable product of the unit interval $I^{\omega_1}$ where $I=[0,1]$. Both $I \times I$ with the lexicographic order and the double arrow space are compact and non-metrizable (thus have no $G_\delta-$diagonal). I discussed these two spaces in a previous post.

Comment. The notion of $G_\delta-$diagonal plays an important role in metrization theorems. The theorem for compact space with $G_\delta-$diagonal had long been generalized. For example, in [3] Chaber had shown that any countably compact space with a $G_\delta-$ diagonal is metrizable. In [1] and [5], it was shown that any paracompact space with a $G_\delta-$diagonal is submetrizable. The theorem proved in this post would simply be a corollary of this result. In upcoming posts, I plan to discuss some of these theorems as well as explore the connection of submetrizability and various $G_\delta-$diagonal properties.

Reference

1. Borges, C. R. On stratifiable spaces, Pacific J. Math., 17 (1966), 1-16.
2. Ceder, J. G. Some generalizations of metric spaces, Pacific J. Math., 11 (1961), 105-125.
3. Chaber, Conditions which imply compactness in countably compact spaces, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998.
4. Hodel, R., E., Moore spaces and $w \Delta-$spaces, Pacific J. Math., 38, (1971), 641-652.
5. Okuyama, A., On metrizability of M-spaces, Proc. Japan. Acad., 40, 176-179.
6. Sneider, V., Continuous images of Souslin and Borel sets: metrization theorems, Dokl. Acad. Nauk USSR, 50 (1945), 77-79.

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$\copyright$ 2009 – Dan Ma, Revised February 1, 2018