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 (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.
Let be a space. It is submetrizable if there is a topology on the set such that and is a metrizable space. The topology is said to be weaker (coarser) than . Thus a space is submetrizable if it has a weaker metrizable topology.
Let be a set of subsets of the space . is said to be a network for if for every open subset of and for each , there exists such that . 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 is the subset of the square . The space has a -diagonal if is a -subset of , i.e. is the intersection of countably many open subsets of .
The implication is clear. For , see Lemma 1 in this previous post on countable network. The implication is left as an exercise. To see , let be a compact subset of . The property of having a -diagonal is hereditary. Thus has a -diagonal. According to a well known result, any compact space with a -diagonal is metrizable (see here).
None of the implications in the diagram is reversible. The first uncountable ordinal is an example for . This follows from the well known result that any countably compact space with a -diagonal is metrizable (see here). The Mrowka space is an example for (see here). The Sorgenfrey line is an example for both and .
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 is countable and is thus contained in some initial segment which is metrizable. Any Moore space has a -diagonal. Thus compact subspaces of a Moore space are metrizable.
We now look at some function spaces that are in the class “every compact subspace is metrizable.” For any Tychonoff space (completely regular space) , is the space of all continuous functions from into with the pointwise convergence topology (see here for basic information on pointwise convergence topology).
Suppose that is a separable space. Then every compact subspace of is metrizable.
The proof here actually shows more than is stated in the theorem. We show that is submetrizable by a separable metric topology. Let be a countable dense subspace of . Then is metrizable and separable since it is a subspace of the separable metric space . Thus has a countable base. Let be a countable base for .
Let be the restriction map, i.e. for each , . Since is a projection map, it is continuous and one-to-one and it maps into . Thus is a continuous bijection from into . Let .
We claim that is a base for a topology on . Once this is established, the proof of the theorem is completed. Note that is countable and elements of are open subsets of . Thus the topology generated by is coarser than the original topology of .
For to be a base, two conditions must be satisfied – is a cover of and for , and for , there exists such that . Since is a base for and since elements of are preimages of elements of under the map , it is straightforward to verify these two points.
Theorem 1 is actually a special case of a duality result in function space theory. More about this point later. First, consider a corollary of Theorem 1.
Let where is the cardinality continuum and each is a separable space. Then every compact subspace of 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.
Suppose that is a product of separable spaces where is any infinite cardinal. Then every compact subspace of 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 .
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: is separable if and only if 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 , the weight of , denoted by , is the least cardinaility of a base of . The weak weight of a space is the least over all space for which there is a continuous bijection from onto . Thus if the weak weight of is , then there is a continuous bijection from onto some separable metric space, hence has a weaker separable metric topology.
There is a duality result between density and weak weight for and . The duality result:
- Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
- Tkachuk V. V., A -Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.