# Pseudocompact spaces with regular G-delta diagonals

This post complements two results discussed in two previous blog posts concerning $G_\delta$-diagonal. One result is that any compact space with a $G_\delta$-diagonal is metrizable (see here). The other result is that the compactness in the first result can be relaxed to countably compactness. Thus any countably compact space with a $G_\delta$-diagonal is metrizable (see here). The countably compactness in the second result cannot be relaxed to pseudocompactness. The Mrowka space is a pseudocompact space with a $G_\delta$-diagonal that is not submetrizable, hence not metrizable (see here). However, if we strengthen the $G_\delta$-diagonal to a regular $G_\delta$-diagonal while keeping pseudocompactness fixed, then we have a theorem. We prove the following theorem.

Theorem 1
If the space $X$ is pseudocompact and has a regular $G_\delta$-diagonal, then $X$ is metrizable.

All spaces are assumed to be Hausdorff and completely regular. The assumption of completely regular is crucial. The proof of Theorem 1 relies on two lemmas concerning pseudocompact spaces (one proved in a previous post and one proved here). These two lemmas work only for completely regular spaces.

The proof of Theorem 1 uses a metrization theorem. The best metrization to use in this case is Moore metrization theorem (stated below). The result in Theorem 1 is found in [2].

First some basics. Let $X$ be a space. The diagonal of the space $X$ is the set $\Delta=\{ (x,x): x \in X \}$. When the diagonal $\Delta$, as a subset of $X \times X$, is a $G_\delta$-set, i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$, the space $X$ is said to have a $G_\delta$-diagonal.

The space $X$ is said to have a regular $G_\delta$-diagonal if the diagonal $\Delta$ is a regular $G_\delta$-set in $X \times X$, i.e. $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$. If $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$, then $\Delta=\bigcap_{n=1}^\infty \overline{U_n}=\bigcap_{n=1}^\infty U_n$. Thus if a space has a regular $G_\delta$-diagonal, it has a $G_\delta$-diagonal. We will see that there exists a space with a $G_\delta$-diagonal that fails to be a regular $G_\delta$-diagonal.

The space $X$ is a pseudocompact space if for every continuous function $f:X \rightarrow \mathbb{R}$, the image $f(X)$ is a bounded set in the real line $\mathbb{R}$. Pseudocompact spaces are discussed in considerable details in this previous post. We will rely on results from this previous post to prove Theorem 1.

The following lemma is used in proving Theorem 1.

Lemma 2
Let $X$ be a pseudocompact space. Suppose that $O_1,O_2,O_2,\cdots$ is a decreasing sequence of non-empty open subsets of $X$ such that $\bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}$ for some point $x \in X$. Then $\{ O_n \}$ is a local base at the point $x$.

Proof of Lemma 2
Let $O_1,O_2,O_2,\cdots$ be a decreasing sequence of open subsets of $X$ such that $\bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}$. Let $U$ be open in $X$ with $x \in U$. If $O_n \subset U$ for some $n$, then we are done. Suppose that $O_n \not \subset U$ for each $n$.

Choose open $V$ with $x \in V \subset \overline{V} \subset U$. Consider the sequence $\{ O_n \cap (X-\overline{V}) \}$. This is a decreasing sequence of non-empty open subsets of $X$. By Theorem 2 in this previous post, $\bigcap \overline{O_n \cap (X-\overline{V})} \ne \varnothing$. Let $y$ be a point in this non-empty set. Note that $y \in \bigcap_{n=1}^\infty \overline{O_n}$. This means that $y=x$. Since $x \in \overline{O_n \cap (X-\overline{V})}$ for each $n$, any open set containing $x$ would contain a point not in $\overline{V}$. This is a contradiction since $x \in V$. Thus it must be the case that $x \in O_n \subset U$ for some $n$. $\square$

The following metrization theorem is useful in proving Theorem 1.

Theorem 3 (Moore Metrization Theorem)
Let $X$ be a space. Then $X$ is metrizable if and only if the following condition holds.

There exists a decreasing sequence $\mathcal{B}_1,\mathcal{B}_2,\mathcal{B}_3,\cdots$ of open covers of $X$ such that for each $x \in X$, the sequence $\{ St(St(x,\mathcal{B}_n),\mathcal{B}_n):n=1,2,3,\cdots \}$ is a local base at the point $x$.

For any family $\mathcal{U}$ of subsets of $X$, and for any $A \subset X$, the notation $St(A,\mathcal{U})$ refers to the set $\cup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}$. In other words, it is the union of all sets in $\mathcal{U}$ that contain points of $A$. The set $St(A,\mathcal{U})$ is also called the star of the set $A$ with respect to the family $\mathcal{U}$. If $A=\{ x \}$, we write $St(x,\mathcal{U})$ instead of $St(\{ x \},\mathcal{U})$. The set $St(St(x,\mathcal{B}_n),\mathcal{B}_n)$ indicated in Theorem 3 is the star of the set $St(x,\mathcal{B}_n)$ with respect to the open cover $\mathcal{B}_n$.

Theorem 3 follows from Theorem 1.4 in [1], which states that for any $T_0$-space $X$, $X$ is metrizable if and only if there exists a sequence $\mathcal{G}_1, \mathcal{G}_2, \mathcal{G}_3,\cdots$ of open covers of $X$ such that for each open $U \subset X$ and for each $x \in U$, there exist an open $V \subset X$ and an integer $n$ such that $x \in V$ and $St(V,\mathcal{G}_n) \subset U$.

Proof of Theorem 1

Suppose $X$ is pseudocompact such that its diagonal $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$. We can assume that $U_1 \supset U_2 \supset \cdots$. For each $n \ge 1$, define the following:

$\mathcal{U}_n=\{ U \subset X: U \text{ open in } X \text{ and } U \times U \subset U_n \}$

Note that each $\mathcal{U}_n$ is an open cover of $X$. Also note that $\{ \mathcal{U}_n \}$ is a decreasing sequence since $\{ U_n \}$ is a decreasing sequence of open sets. We show that $\{ \mathcal{U}_n \}$ is a sequence of open covers of $X$ that satisfies Theorem 3. We establish this by proving the following claims.

Claim 1. For each $x \in X$, $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$.

To prove the claim, let $x \ne y$. There is an integer $n$ such that $(x,y) \notin \overline{U_n}$. Choose open sets $U$ and $V$ such that $(x,y) \in U \times V$ and $(U \times V) \cap \overline{U_n}=\varnothing$. Note that $(x,y) \notin U_k$ and $(U \times V) \cap U_n=\varnothing$.

We want to show that $V \cap St(x,\mathcal{U}_n)=\varnothing$, which implies that $y \notin \overline{St(x,\mathcal{U}_n)}$. Suppose $V \cap St(x,\mathcal{U}_n) \ne \varnothing$. This means that $V \cap W \ne \varnothing$ for some $W \in \mathcal{U}_n$ with $x \in W$. Then $(U \times V) \cap (W \times W) \ne \varnothing$. Note that $W \times W \subset U_n$. This implies that $(U \times V) \cap U_n \ne \varnothing$, a contradiction. Thus $V \cap St(x,\mathcal{U}_n)=\varnothing$. Since $y \in V$, $y \notin \overline{St(x,\mathcal{U}_n)}$. We have established that for each $x \in X$, $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$.

Claim 2. For each $x \in X$, $\{ St(x,\mathcal{U}_n) \}$ is a local base at the point $x$.

Note that $\{ St(x,\mathcal{U}_n) \}$ is a decreasing sequence of open sets such that $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$. By Lemma 2, $\{ St(x,\mathcal{U}_n) \}$ is a local base at the point $x$.

Claim 3. For each $x \in X$, $\bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}=\{ x \}$.

Let $x \ne y$. There is an integer $n$ such that $(x,y) \notin \overline{U_n}$. Choose open sets $U$ and $V$ such that $(x,y) \in U \times V$ and $(U \times V) \cap \overline{U_n}=\varnothing$. It follows that $(U \times V) \cap \overline{U_t}=\varnothing$ for all $t \ge n$. Furthermore, $(U \times V) \cap U_t=\varnothing$ for all $t \ge n$. By Claim 2, choose integers $i$ and $j$ such that $St(x,\mathcal{U}_i) \subset U$ and $St(y,\mathcal{U}_j) \subset V$. Choose an integer $k \ge \text{max}(n,i,j)$. It follows that $(St(x,\mathcal{U}_i) \times St(y,\mathcal{U}_j)) \cap U_k=\varnothing$. Since $\mathcal{U}_k \subset \mathcal{U}_i$ and $\mathcal{U}_k \subset \mathcal{U}_j$, it follows that $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k=\varnothing$.

We claim that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$. Suppose not. Choose $w \in St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)$. It follows that $w \in B$ for some $B \in \mathcal{U}_k$ such that $B \cap St(x,\mathcal{U}_k) \ne \varnothing$ and $B \cap St(y,\mathcal{U}_k) \ne \varnothing$. Furthermore $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap (B \times B)=\varnothing$. Note that $B \times B \subset U_k$. This means that $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k \ne \varnothing$, contradicting the fact observed in the preceding paragraph. It must be the case that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$.

Because there is an open set containing $y$, namely $St(y,\mathcal{U}_k)$, that contains no points of $St(St(x,\mathcal{U}_k), \mathcal{U}_k)$, $y \notin \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}$. Thus Claim 3 is established.

Claim 4. For each $x \in X$, $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n)) \}$ is a local base at the point $x$.

Note that $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \}$ is a decreasing sequence of open sets such that $\bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n))}=\{ x \}$. By Lemma 2, $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \}$ is a local base at the point $x$.

In conclusion, the sequence $\mathcal{U}_1,\mathcal{U}_2,\mathcal{U}_3,\cdots$ of open covers satisfies the properties in Theorem 3. Thus any pseudocompact space with a regular $G_\delta$-diagonal is metrizable. $\square$

Example

Any submetrizable space has a $G_\delta$-diagonal. The converse is not true. A classic example of a non-submetrizable space with a $G_\delta$-diagonal is the Mrowka space (discussed here). The Mrowka space is also called the psi-space since it is sometimes denoted by $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal family of almost disjoint subsets of $\omega$. Actually $\Psi(\mathcal{A})$ would be a family of spaces since $\mathcal{A}$ is any maximal almost disjoint family. For any maximal $\mathcal{A}$, $\Psi(\mathcal{A})$ is a pseudocompact non-submetrizable space that has a $G_\delta$-diagonal. This example shows that the requirement of a regular $G_\delta$-diagonal in Theorem 1 cannot be weakened to a $G_\delta$-diagonal. See here for a more detailed discussion of this example.

Reference

1. Gruenhage, G., Generalized Metric Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 423-501, 1984.
2. McArthur W. G., $G_\delta$-Diagonals and Metrization Theorems, Pacific Journal of Mathematics, Vol. 44, No. 2, 613-317, 1973.

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