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 ([6]). 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 ([2]).

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 ([4]). 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.

*Claim 1*

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 .

*Claim 2*

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.

*Claim 3*

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 .

*Claim 4*

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 [3] Chaber had shown that any countably compact space with a diagonal is metrizable. In [1] and [5], 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.

**Reference**

- 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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