# A Space with G-delta Diagonal that is not Submetrizable

The property of being submetrizable implies having a $G_\delta$-diagonal. There are several other properties lying between these two properties (see [1]). Before diving into these other properties, it may be helpful to investigate a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The diagonal of a space $X$ is the set $\Delta=\left\{(x,x): x \in X \right\}$, a subset of the square $X \times X$. An interesting property is when the diagonal of a space is a $G_\delta$-set in $X \times X$ (the space is said to have a $G_\delta$-diagonal). Any compact space or a countably compact space with this property must be metrizable (see compact and countably compact space). A space $(X,\tau)$ is said to be submetrizable if there is a topology $\tau^*$ that can be defined on $X$ such that $(X,\tau^*)$ is a metrizable space and $\tau^* \subset \tau$. In other words, a submetrizable space is a space that has a coarser (weaker) metrizable topology. Every submetrizable space has a $G_\delta$-diagonal. Note that when $X$ has a weaker metric topology, the diagonal $\Delta$ is always a $G_\delta$-set in the metric square $X \times X$ (hence in the square in the original topology). The property of having a $G_\delta$-diagonal is strictly weaker than the property of having a weaker metric topology. In this post, we discuss the Mrowka space, which is a classic example of a space with a $G_\delta$-diagonal that is not submetrizable.

The Mrowka space (also called Psi space) was discussed previously in this blog (see this post). For the sake of completeness, the example is defined here.

First, we define some basic notions. Let $\omega$ be the first infinite ordinal (or more conveniently the set of all nonnegative integers). Let $\mathcal{A}$ be a family of infinite subsets of $\omega$. The family $\mathcal{A}$ is said to be an almost disjoint family if for each two distinct $A,B \in \mathcal{A}$, $A \cap B$ is finite. An almost disjoint family $\mathcal{A}$ is said to be a maximal almost disjoint family if $B$ is an infinite subset of $\omega$ such that $B \notin \mathcal{A}$, then $B \cap A$ is infinite for some $A \in \mathcal{A}$. In other words, if you put one more set into a maximal almost disjoint family, it ceases to be almost disjoint.

A natural question is whether there is an uncountable almost disjoint family of subsets of $\omega$. In fact, there is one whose cardinality is continuum (the cardinality of the real line). To see this, identify $\omega$ with $\mathbb{Q}=\lbrace{r_0,r_1,r_2,...}\rbrace$ (the set of all rational numbers). Let $\mathbb{P}=\mathbb{R}-\mathbb{Q}$ be the set of all irrational numbers. For each $x \in \mathbb{P}$, choose a subsequence of $\mathbb{Q}$ consisting of distinct elements that converges to $x$ (in the Euclidean topology). Then the family of all such sequences of rational numbers would be an almost disjoint family. By a Zorn’s Lemma argument, this almost disjoint family is contained within a maximal almost disjoint family. Thus we also have a maximal almost disjoint family of cardinality continuum. On the other hand, there is no countably infinite maximal almost disjoint family of subsets of $\omega$ (see this post).

Let $\mathcal{A}$ be an infinite almost disjoint family of subsets of $\omega$. We now define a Mrowka space (or $\Psi$-space), denoted by $\Psi(\mathcal{A})$. The underlying set is $\Psi(\mathcal{A})=\mathcal{A} \cup \omega$. Points in $\omega$ are isolated. For $A \in \mathcal{A}$, a basic open set is of the form $\lbrace{A}\rbrace \cup (A-F)$ where $F \subset \omega$ is finite. It is straightforward to verify that $\Psi(\mathcal{A})$ is Hausdorff, first countable and locally compact. It has a countable dense set of isolated points. Note that $\mathcal{A}$ is an infinite discrete and closed set in the space $\Psi(\mathcal{A})$. Thus $\Psi(\mathcal{A})$ is not countably compact.

We would like to point out that the definition of a Mrowka space $\Psi(\mathcal{A})$ only requires that the family $\mathcal{A}$ is an almost disjoint family and does not necessarily have to be maximal. For the example discribed in the title, $\mathcal{A}$ needs to be a maximal almost disjoint family of subsets of $\omega$.

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Example
Let $\mathcal{A}$ be a maximal almost disjoint family of subsets of $\omega$. Then $\Psi(\mathcal{A})$ as defined above is a space in which there is a $G_\delta$-diagonal that is not submetrizable.

Note that $\Psi(\mathcal{A})$ is pseudocompact (proved in this post). Because there is no countable maximal almost disjoint family of subsets of $\omega$, $\mathcal{A}$ must be an uncountable in addition to being a closed and discrete subspace of $\Psi(\mathcal{A})$ (thus the space is not Lindelof). Since $\Psi(\mathcal{A})$ is separable and is not Lindelof, $\Psi(\mathcal{A})$ is not metrizable. Any psuedocompact submetrizable space is metrizable (see Theorem 4 in this post). Thus $\Psi(\mathcal{A})$ must not be submetrizable.

On the other hand, any $\Psi$-space $\Psi(\mathcal{A})$ (even if $\mathcal{A}$ is not maximal) is a Moore space. It is well known that any Moore space has a $G_\delta$-diagonal. The remainder of this post has a brief discussion of Moore space.

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Moore Space

A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$ and each open set $U \subset X$ with $x \in U$, there is some $n$ such that any open set in $\mathcal{D}_n$ containing the point $x$ is contained in $U$. A developable space is one that has a development. A Moore space is a regular developable space.

Suppose that $X$ is a Moore space. We show that $X$ has a $G_\delta$-diagonal. That is, we wish to show that $\Delta=\left\{(x,x): x \in X \right\}$ is a $G_\delta$-set in $X \times X$.

Let $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ be a development. For each $n$, let $U_n=\bigcup \lbrace{V \times V:V \in \mathcal{D}_n}\rbrace$. Clearly $\Delta \subset \bigcap_{n<\omega} U_n$. Let $(x,y) \in \bigcap_{n<\omega} U_n$. For each $n$, $(x,y) \in V_n \times V_n$ for some $V_n \in \mathcal{D}_n$. We claim that $x=y$. Suppose that $x \ne y$. By the definition of development, there exists some $m$ such that every open set in $\mathcal{D}_m$ containing the point $x$ has to be a subset of $X-\left\{y \right\}$. Then $V_m \subset X-\left\{y \right\}$, which contradicts $y \in V_m$. Thus we have $\Delta = \bigcap_{n<\omega} U_n$.

The remaining thing to show is that $\Psi(\mathcal{A})$ is a Moore space. For each positive integer $n$, let $F_n=\left\{0,1,\cdots,n-1 \right\}$ and let $F_0=\varnothing$. The development is defined by $\lbrace{\mathcal{E}_n}\rbrace_{n<\omega}$, where for each $n$, $\mathcal{E}_n$ consists of open sets of the form $\lbrace{A}\rbrace \cup (A-F_n)$ where $A \in \mathcal{A}$ plus any singleton $\left\{j \right\}$ ($j \in \omega$) that has not been covered by the sets $\lbrace{A}\rbrace \cup (A-F_n)$.

Reference

1. Arhangel’skii, A. V., Buzyakova, R. Z., The rank of the diagonal and submetrizability, Commentationes Mathematicae Universitatis Carolinae, Vol. 47 (2006), No. 4, 585-597.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.