June | 2017 | Dan Ma's Topology Blog

Once it is known that a topological space is not metrizable, it is natural to ask, from a metrizability standpoint, which subspaces are metrizable, e.g. whether every compact subspace is metrizable. This post discusses several classes of spaces in which every compact subspace is metrizable. Though the goal here is not to find a complete characterization of such spaces, this post discusses several classes of spaces and various examples that have this property. The effort brings together many interesting basic and well known facts. Thus the notion “every compact subspace is metrizable” is an excellent learning opportunity.

Several Classes of Spaces

The notion “every compact subspace is metrizable” is a very broad class of spaces. It includes well known spaces such as Sorgenfrey line, Michael line and the first uncountable ordinal $\omega_1$ (with the order topology) as well as Moore spaces. Certain function spaces are in the class “every compact subspace is metrizable”. The following diagram is a good organizing framework.

$\displaystyle \begin{aligned} &1. \ \text{Metrizable} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&2. \ \text{Submetrizable} \Longleftarrow 5. \ \exists \ \text{countable network} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&3. \ \exists \ G_\delta \text{ diagonal} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&4. \ \text{Every compact subspace is metrizable} \end{aligned}$

Let $(X, \tau)$ be a space. It is submetrizable if there is a topology $\tau_1$ on the set $X$ such that $\tau_1 \subset \tau$ and $(X, \tau_1)$ is a metrizable space. The topology $\tau_1$ is said to be weaker (coarser) than $\tau$ . Thus a space $X$ is submetrizable if it has a weaker metrizable topology.

Let $\mathcal{N}$ be a set of subsets of the space $X$ . $\mathcal{N}$ is said to be a network for $X$ if for every open subset $O$ of $X$ and for each $x \in O$ , there exists $N \in \mathcal{N}$ such that $x \in N \subset O$ . Having a network that is countable in size is a strong property (see here for a discussion on spaces with a countable network).

The diagonal of the space $X$ is the subset $\Delta=\left\{(x,x): x \in X \right\}$ of the square $X \times X$ . The space $X$ has a $G_\delta$ -diagonal if $\Delta$ is a $G_\delta$ -subset of $X \times X$ , i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$ .

The implication $1 \Longrightarrow 2$ is clear. For $5 \Longrightarrow 2$ , see Lemma 1 in this previous post on countable network. The implication $2 \Longrightarrow 3$ is left as an exercise. To see $3 \Longrightarrow 4$ , let $K$ be a compact subset of $X$ . The property of having a $G_\delta$ -diagonal is hereditary. Thus $K$ has a $G_\delta$ -diagonal. According to a well known result, any compact space with a $G_\delta$ -diagonal is metrizable (see here).

None of the implications in the diagram is reversible. The first uncountable ordinal $\omega_1$ is an example for $4 \not \Longrightarrow 3$ . This follows from the well known result that any countably compact space with a $G_\delta$ -diagonal is metrizable (see here). The Mrowka space is an example for $3 \not \Longrightarrow 2$ (see here). The Sorgenfrey line is an example for both $2 \not \Longrightarrow 5$ and $2 \not \Longrightarrow 1$ .

To see where the examples mentioned earlier are placed, note that Sorgenfrey line and Michael line are submetrizable, both are submetrizable by the usual Euclidean topology on the real line. Each compact subspace of the space $\omega_1$ is countable and is thus contained in some initial segment $[0,\alpha]$ which is metrizable. Any Moore space has a $G_\delta$ -diagonal. Thus compact subspaces of a Moore space are metrizable.

Function Spaces

We now look at some function spaces that are in the class “every compact subspace is metrizable.” For any Tychonoff space (completely regular space) $X$ , $C_p(X)$ is the space of all continuous functions from $X$ into $\mathbb{R}$ with the pointwise convergence topology (see here for basic information on pointwise convergence topology).

Theorem 1
Suppose that $X$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.

Proof
The proof here actually shows more than is stated in the theorem. We show that $C_p(X)$ is submetrizable by a separable metric topology. Let $Y$ be a countable dense subspace of $X$ . Then $C_p(Y)$ is metrizable and separable since it is a subspace of the separable metric space $\mathbb{R}^{\omega}$ . Thus $C_p(Y)$ has a countable base. Let $\mathcal{E}$ be a countable base for $C_p(Y)$ .

Let $\pi:C_p(X) \longrightarrow C_p(Y)$ be the restriction map, i.e. for each $f \in C_p(X)$ , $\pi(f)=f \upharpoonright Y$ . Since $\pi$ is a projection map, it is continuous and one-to-one and it maps $C_p(X)$ into $C_p(Y)$ . Thus $\pi$ is a continuous bijection from $C_p(X)$ into $C_p(Y)$ . Let $\mathcal{B}=\left\{\pi^{-1}(E): E \in \mathcal{E} \right\}$ .

We claim that $\mathcal{B}$ is a base for a topology on $C_p(X)$ . Once this is established, the proof of the theorem is completed. Note that $\mathcal{B}$ is countable and elements of $\mathcal{B}$ are open subsets of $C_p(X)$ . Thus the topology generated by $\mathcal{B}$ is coarser than the original topology of $C_p(X)$ .

For $\mathcal{B}$ to be a base, two conditions must be satisfied – $\mathcal{B}$ is a cover of $C_p(X)$ and for $B_1,B_2 \in \mathcal{B}$ , and for $f \in B_1 \cap B_2$ , there exists $B_3 \in \mathcal{B}$ such that $f \in B_3 \subset B_1 \cap B_2$ . Since $\mathcal{E}$ is a base for $C_p(Y)$ and since elements of $\mathcal{B}$ are preimages of elements of $\mathcal{E}$ under the map $\pi$ , it is straightforward to verify these two points. $\square$

Theorem 1 is actually a special case of a duality result in $C_p$ function space theory. More about this point later. First, consider a corollary of Theorem 1.

Corollary 2
Let $X=\prod_{\alpha<c} X_\alpha$ where $c$ is the cardinality continuum and each $X_\alpha$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.

The key fact for Corollary 2 is that the product of continuum many separable spaces is separable (this fact is discussed here). Theorem 1 is actually a special case of a deep result.

Theorem 3
Suppose that $X=\prod_{\alpha<\kappa} X_\alpha$ is a product of separable spaces where $\kappa$ is any infinite cardinal. Then every compact subspace of $C_p(X)$ is metrizable.

Theorem 3 is a much more general result. The product of any arbitrary number of separable spaces is not separable if the number of factors is greater than continuum. So the proof for Theorem 1 will not work in the general case. This result is Problem 307 in [2].

A Duality Result

Theorem 1 is stated in a way that gives the right information for the purpose at hand. A more correct statement of Theorem 1 is: $X$ is separable if and only if $C_p(X)$ is submetrizable by a separable metric topology. Of course, the result in the literature is based on density and weak weight.

The cardinal function of density is the least cardinality of a dense subspace. For any space $Y$ , the weight of $Y$ , denoted by $w(Y)$ , is the least cardinaility of a base of $Y$ . The weak weight of a space $X$ is the least $w(Y)$ over all space $Y$ for which there is a continuous bijection from $X$ onto $Y$ . Thus if the weak weight of $X$ is $\omega$ , then there is a continuous bijection from $X$ onto some separable metric space, hence $X$ has a weaker separable metric topology.

There is a duality result between density and weak weight for $X$ and $C_p(X)$ . The duality result:

The density of $X$ coincides with the weak weight of $C_p(X)$ and the weak weight of $X$ coincides with the density of $C_p(X)$ . These are elementary results in $C_p$ -theory. See Theorem I.1.4 and Theorem I.1.5 in [1].

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Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Tkachuk V. V., A $C_p$ -Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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