In any topological space , a point is a point if the one-point set is the intersection of countably many open subsets of . It is well known that any compact Hausdorff space is first countable at every point, i.e., if a point of a compact space is a point, then there is a countable local base at that point. It is also well known that uncountable power of first countable spaces can fail to be first countable at every point. For example, no point of the compact space can be a point. In this post, we show that any Corson compact space has a dense set of point. Therefore, any Corson compact space is first countable on a dense set (see Corollary 4 below). However, it is not true that every Corson compact space has a dense metrizable subspace. See Theorem 9.14 in  for an example of a first countable Corson compact space with no dense metrizable subspace. A list of other blog posts on Corson compact spaces is given at the end of this post.
The fact that every Corson compact space has a dense first countable subspace is taken as a given in the literature. For one example, see chapter c-16 of . Even though Corollary 4 is a basic fact of Corson compact spaces, the proof involves much more than a direct application of the relevant definitions. The proof given here is intended to be an online resource for any one interested in knowing more about Corson compact spaces.
For any infinite cardinal number , the -product of many copies of is the following subspace of :
A compact space is said to be a Corson compact space if it can be embedded in for some infinite cardinal .
For each , let denote the support of the point , i.e., is the set of all such that .
Let be a Corson compact space. Then has a point.
Proof of Proposition 1
If is finite, then every point is isolated and is thus a point. Assume is infinite. Let be an infinite cardinal number such that . For , define if the following holds:
It is relatively straightforward to verify that the following three properties are satisfied:
- for all . (reflexivity)
- For all , if and , then . (antisymmetry)
- For all , if and , then . (transitivity)
Thus as defined here is a partial order on the compact space . Let such that is a chain with respect to , i.e., for all , or . We show that has an upper bound (in ) with respect to the partial order . We need this for an argument using Zorn’s lemma.
Let . For each , choose some such that and define . For all , define . Because is a chain, the point is well-defined. It is also clear that for all . If , then is a desired upper bound of . So assume . It follows that is a limit point of , i.e., every open set containing contains a point of different from . Hence is a limit point of too. Since is compact, , a contradiction. Thus it must be that . Thus every chain in the partially ordered set has an upper bound. By Zorn’s lemma, there exists at least one maximal element with respect to the partial order , i.e., there exists such that for all .
We now show that is a point in . Let . For each and for each positive integer , let be the open interval . For each positive integer , define the open set as follows:
Note that . Because is a maximal element, note that if such that for all , then it must be the case that . Thus if , then for all . We have .
Let be a compact space such that for every non-empty compact subspace of , there exists a point in . Then every non-empty open subset of contains a point.
Proof of Lemma 2
Let be a non-empty open subset of the compact space . If there exists such that is open in , then is a point. So assume that every point of is a non-isolated point of . By regularity, choose an open subset of such that . Continue in the same manner and obtain a decreasing sequence of open subsets of such that for each positive integer . Then is a non-empty closed subset of and thus compact. By assumption, has a point, say .
Then where each is open in . For each , let be open in such that . For each , let , which is open in . Then . This means that is a point in the compact space . Note that , the open set we start with. This completes the proof that every non-empty open subset of contains a point.
Let be a Corson compact space. Then has a dense set of points.
Proof of Proposition 3
Note that Corson compactness is hereditary with respect to closed sets. Thus every compact subspace of is also Corson compact. By Proposition 1, every compact subspace of has a point. By Lemma 2, has a dense set of points.
Every Corson compact space has a dense first countable subspace.
Blog posts on Corson compact spaces
- Basic topological properties of Corson compact spaces
- Every Corson compact space has a dense first countable subspace (this post)
- An example of a non-metrizable Corson compact space
- Sigma-products of separable metric spaces are monolithic
- A short note on monolithic spaces
- Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
- Todorcevic, S., Trees and Linearly Ordered Sets, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 235-293, 1984.