In a previous post, I showed that any compact space with a countable network is metrizable. Another classic metrization theorem for compact spaces is that any compact space with a diagonal is metrizable (). The theroem I try to prove is: for a compact space , is perfectly normal if and only if has a diagonal if and only if is metrizable. My proof is based on the notion of diagonal. Every compact space with a diagonal has a diagonal, which allows us to define a countable base. The theorem discussed here had been generalized (see the comment at the end of this post). All spaces are at least Hausdorff.
Let be a space. The set is called the diagonal of the space . The space has a diagonal if is a set in .
Let be a collection of subsets of and let . Define . A sequence of open covers of is called a diagonal sequence of if for each , . Lemma 1 shows that a space has a diagonal if and only if it has a diagonal sequence. This lemma is due to Ceder ().
Another notion we need is that of the diagonal. The space has a diagonal if there is a diagonal sequence such that for each , . Such a diagonal sequence is called a diagonal sequence. The notion of diagonal is due to R. E. Hodel (). Lemma 2 below shows that any compact space with a diagonal has a diagonal.
Lemma 1. The space has a diagonal if and only if it has a diagonal sequence.
Proof. Suppose that where each is open in . Let denote the topology on . For each , let . We claim that for each , . Obviously . Let . For each , where and . Thus for each . This implies and . Thus . We have established that is a -diagonal sequence of .
Suppose is a diagonal sequence of . For each , let . Since , . To show the set inclusion for the other direction, let . For each , for some . This implies that for each . It follows that . Thus .
Lemma 2. If is compact and has a diagonal, then has a diagonal. Furthermore, each open cover in the diagonal sequence is finite.
Proof. Let be the diagonal sequence obtained in Lemma 1. We inductively define , another diagonal sequence.
Using the compactness of , obtain a finite subcollection of such that is a cover of . Here’s how I obtain . For each , choose an open set and an open set such that and . Choose open set such that and . Let be a finite subcollection of such that is a cover of . Continue the inductive process and we produce a sequence of open covers satisfying the following two claims.
For each , .
Because each open cover is chosen to be a subcover or a refinement of the open cover , we have . Since (from the definition of diagonal sequence), we have .
For each , .
We only need to show . Let for each . Because is finite, for some with . Each such for some . Thus implies . By Claim 1, .
We have shown that has a diagonal by producing a diagonal sequence .
Theorem. If is compact and has a diagonal, then is metrizable.
Proof. Let be the diagonal sequence obtained in Lemma 2. Furthermore, each is a finite open cover. Let . The collection satisfies the properties stated in the following two claims.
For each with , there is a such that and .
Since , for some . Then there is some such that and . In fact, for any with , it must be the case that .
Let . Then is a countable base for .
To see Claim 4, note that is a cover of and is closed under finite intersections. This makes is base for a topology. To show that this base generate the same topology on , let where is open. Then is compact. For each , let such that and . We can choose such that is a cover of . Then .
With Claims 3 and 4, the theorem is established.
Corollary. Let be a compact space. The following conditions are equivalent.
- is perfectly normal.
- has a diagonal.
- has a countable base.
Proof. and are obvious. follows from the theorem.
Examples. Based on the corollary, any non-metrizable compact Hausdorff space does not have a diagonal. One handy example is the uncountable product of the unit interval where . Both with the lexicographic order and the double arrow space are compact and non-metrizable (thus have no diagonal). I discussed these two spaces in a previous post.
Comment. The notion of diagonal plays an important role in metrization theorems. The theorem for compact space with diagonal had long been generalized. For example, in  Chaber had shown that any countably compact space with a diagonal is metrizable. In  and , it was shown that any paracompact space with a diagonal is submetrizable. The theorem proved in this post would simply be a corollary of this result. In upcoming posts, I plan to discuss some of these theorems as well as explore the connection of submetrizability and various diagonal properties.
- Borges, C. R. On stratifiable spaces, Pacific J. Math., 17 (1966), 1-16.
- Ceder, J. G. Some generalizations of metric spaces, Pacific J. Math., 11 (1961), 105-125.
- Chaber, Conditions which imply compactness in countably compact spaces, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998.
- Hodel, R., E., Moore spaces and spaces, Pacific J. Math., 38, (1971), 641-652.
- Okuyama, A., On metrizability of M-spaces, Proc. Japan. Acad., 40, 176-179.
- Sneider, V., Continuous images of Souslin and Borel sets: metrization theorems, Dokl. Acad. Nauk USSR, 50 (1945), 77-79.
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