In a previous post (Metrization Theorems for Compact Spaces), three classic metrization theorems for compact spaces are discussed. The three theorems are: any Hausdorff compact space is metrizable if any of the following holds:
- has a countable network,
- has a diagonal,
- has a point countable base.
The metrization results for conditions 2 and 3 hold for countably compact spaces as well. See the following posts:
Countably Compact Spaces with G-delta Diagonals
Metrization Theorems for Compact Spaces
In this post, we discuss another metrization theorem for compact spaces. We show that a compact Hausdorff space is metrizable if and only if the function space is separable.
Let’s discuss the function space. Let by any Tychonoff space and let be the set of all real numbers. Let be the set of all real-valued continuous functions defined on . For any and for any , define . If we restrict to and restrict to open sets, then the set of all is a subbase for a topology on . This topology is called the pointwise convergence topology. The function space with this topology is denoted by .
It is a theorem that is separable if and only if has a weaker topology that forms a separable metric space. The result on compact spaces is a corollary of this theorem.
Theorem. Let be a Tychonoff space with being the topology. The following conditions are equivalent:
- is separable.
- There is a topology such that is a separable metric space.
Proof
. Let be a countable dense subspace. Let be the class of all bounded open intervals of with rational endpoints. Consider . Note that is a subbase for a topology on . Since is countable, the topology has a countable base and is thus separable and metrizable.
. Let be a topology. Suppose that is generated by a countable base . As in , let be the class of all bounded open intervals of with rational endpoints. Let be the class of all finite intersections of the sets in the following collection of sets.
Note that is countable. For each , choose . We claim that is a countable dense set of . To see this, let be a basic open set in where . Fix . For each , choose such that and . Then . Now, we have .
Corollary. Let be a compact Hausdorff space. Then the following conditions are equivalent:
- is metrizable.
- is separable.
Proof.
. This follows from in the above theorem.
. This follows from in the above theorem. Note that any compact Hausdorff space cannot have a strictly weaker (or coarser) Hausdorff topology. Thus if a compact Hausdorff space has a weaker metrizable topology, it must be metrizable.