Compact Spaces With G-delta Diagonals

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 G_\delta-diagonal is metrizable ([6]). The theroem I try to prove is: for a compact space X, X^2 is perfectly normal if and only if X has a G_\delta-diagonal if and only if X is metrizable. My proof is based on the notion of G^*_\delta-diagonal. Every compact space with a G_\delta-diagonal has a G^*_\delta-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 X be a space. The set \Delta=\lbrace{(x,x):x \in X}\rbrace is called the diagonal of the space X. The space X has a G_\delta-diagonal if \Delta is a G_\delta-set in X^2.

Let \mathcal{G} be a collection of subsets of X and let x \in X. Define st(x,\mathcal{G})=\bigcup \lbrace{G \in \mathcal{G}:x \in G}\rbrace. A sequence \lbrace{\mathcal{G}_n}\rbrace_{n<\omega} of open covers of X is called a G_\delta-diagonal sequence of X if for each x \in X, \lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n). Lemma 1 shows that a space has a G_\delta-diagonal if and only if it has a G_\delta-diagonal sequence. This lemma is due to Ceder ([2]).

Another notion we need is that of the G^*_\delta-diagonal. The space X has a G^*_\delta-diagonal if there is a G_\delta-diagonal sequence \lbrace{\mathcal{G}_n}\rbrace_{n<\omega} such that for each x \in X, \lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{G}_n)}. Such a G_\delta-diagonal sequence is called a G^*_\delta-diagonal sequence. The notion of G^*_\delta-diagonal is due to R. E. Hodel ([4]). Lemma 2 below shows that any compact space with a G_\delta-diagonal has a G^*_\delta-diagonal.

Lemma 1. The space X has a G_\delta-diagonal if and only if it has a G_\delta-diagonal sequence.

Proof. \Rightarrow Suppose that \Delta=\bigcap_{n<\omega}U_n where each U_n is open in X^2. Let \tau denote the topology on X. For each n, let \mathcal{G}_n=\lbrace{V \in \tau:V \times V \subset U_n}\rbrace. We claim that \lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n). Obviously \lbrace{x}\rbrace \subset \bigcap_{n<\omega} st(x,\mathcal{G}_n). Let y \in \bigcap_{n<\omega} st(x,\mathcal{G}_n). For each n, y \in V_n where V_n \in \mathcal{G}_n and x \in V_n. Thus (x,y) \in V_n \times V_n \subset U_n. This implies (x,y) \in \Delta and x=y.

\Leftarrow Suppose \lbrace{\mathcal{G}_n}\rbrace a G_\delta-diagonal sequence. For each n, let U_n=\bigcup \lbrace{V \times V:V \in \mathcal{G}_n}\rbrace. Clearly \Delta \subset \bigcap_{n<\omega} U_n. Let (x,y) \in \bigcap_{n<\omega} U_n. For each (x,y) \in V_n \times V_n for some V_n \in \mathcal{G}_n. This implies y \in \bigcap_{n<\omega} st(x,\mathcal{G}_n)=\lbrace{x}\rbrace. It follows that y=x, completing the proof of Lemma 1.

Lemma 2. If X is compact and has a G_\delta-diagonal, then X has a G^*_\delta-diagonal. Furthermore, each open cover in the G_\delta-diagonal sequence is finite.

Proof. Let \lbrace{\mathcal{G}_n}\rbrace_{n<\omega} be the G_\delta-diagonal sequence obtained in Lemma 1. We inductively define \lbrace{\mathcal{H}_n}\rbrace_{n<\omega}, another G_\delta-diagonal sequence.

Using the compactness of X, obtain a finite subcollection \mathcal{H}_0 of \mathcal{G}_0 such that \mathcal{H}_0 is a cover of X. Here’s how I obtain \mathcal{H}_1. For each x \in X, choose an open set G_x \in \mathcal{G}_1 and an open set H_x \in \mathcal{H}_0 such that x \in G_x and x \in H_x. Choose open set V_x such that x \in V_x and \overline{V_x} \subset G_x \cap H_x. Let \mathcal{H}_1 be a finite subcollection of \lbrace{V_x:x \in X}\rbrace such that \mathcal{H}_1 is a cover of X. Continue the inductive process and we produce a sequence of open covers \lbrace{\mathcal{H}_n}\rbrace_{n<\omega} satisfying the following two claims.

Claim 1
For each x \in X, \lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{H}_n).

Because each open cover \mathcal{H}_n is chosen to be a subcover or a refinement of the open cover \mathcal{G}_n, we have st(x,\mathcal{H}_n) \subset st(x,\mathcal{G}_n). Since \lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{G}_n) (from the definition of G_\delta-diagonal sequence), we have \lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{H}_n).

Claim 2
For each x \in X, \lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)}.

We only need to show \bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)} \subset \lbrace{x}\rbrace. Let y \in \overline{st(x,\mathcal{H}_n)} for each n. Because \mathcal{H}_n is finite, y \in \overline{V} for some V \in \mathcal{H}_n with x \in V. Each such \overline{V} \subset U for some U \in \mathcal{H}_{n-1}. Thus y \in \overline{st(x,\mathcal{H}_n)} implies y \in st(x,\mathcal{H}_{n-1}). By Claim 1, y=x.

We have shown that X has a G^*_\delta-diagonal by producing a G^*_\delta-diagonal sequence \lbrace{\mathcal{H}_n}\rbrace_{n<\omega}.

Theorem. If X is compact and has a G_\delta-diagonal, then X is metrizable.

Proof. Let \lbrace{\mathcal{H}_n}\rbrace_{n<\omega} be the G^*_\delta-diagonal sequence obtained in Lemma 2. Furthermore, each \mathcal{H}_n is a finite open cover. Let \mathcal{H}=\bigcup_{n<\omega} \mathcal{H}_n. The collection \mathcal{H} satisfies the properties stated in the following two claims.

Claim 3
For each x,y \in X with x \neq y, there is a U \in \mathcal{H} such that x \in U and y \notin \overline{U}.

Since \lbrace{x}\rbrace=\bigcap_{n<\omega} \overline{st(x,\mathcal{H}_n)}, y \notin \overline{st(x,\mathcal{H}_n)} for some n. Then there is some U \in \mathcal{H}_n such that x \in U and y \notin \overline{U}.

Claim 4
Let \mathcal{B}=\lbrace{X-\overline{\bigcup F}: F \subset \mathcal{H} \phantom{x} and \phantom{x} \vert F \lvert < \omega}\rbrace. Then \mathcal{B} is a countable base for X.

To see Claim 4, note that \mathcal{B} is a cover of X and is closed under finite intersections. This makes \mathcal{B} is base for a topology. To show that this base generate the same topology on X, let y \in U \subset X where U is open. Then X-U is compact. For each x \in X-U, let V_x \in \mathcal{H} such that x \in V_x and y \notin \overline{V_x}. We can choose F=\lbrace{V_{x(0)},...,V_{x(n)}}\rbrace such that F is a cover of X-U. Then y \in X-\overline{\bigcup F} \subset U.

With Claims 3 and 4, the theorem is established.

Corollary. Let X be a compact space. The following conditions are equivalent.

  1. X^2 is perfectly normal.
  2. X has a G_\delta-diagonal.
  3. X has a countable base.

Proof. 1 \rightarrow 2 and 3 \rightarrow 1 are obvious. 2 \rightarrow 3 follows from the theorem.

Examples. Based on the corollary, any non-metrizable compact Hausdorff space does not have a G_\delta-diagonal. One handy example is the uncountable product of the unit interval I^{\omega_1} where I=[0,1]. Both I \times I with the lexicographic order and the double arrow space are compact and non-metrizable (thus have no G_\delta-diagonal). I discussed these two spaces in a previous post.

Comment. The notion of G_\delta-diagonal plays an important role in metrization theorems. The theorem for compact space with G_\delta-diagonal had long been generalized. For example, in [3] Chaber had shown that any countably compact space with a G_\delta- diagonal is metrizable. In [1] and [5], it was shown that any paracompact space with a G_\delta-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 G_\delta-diagonal properties.


  1. Borges, C. R. On stratifiable spaces, Pacific J. Math., 17 (1966), 1-16.
  2. Ceder, J. G. Some generalizations of metric spaces, Pacific J. Math., 11 (1961), 105-125.
  3. Chaber, Conditions which imply compactness in countably compact spaces, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998.
  4. Hodel, R., E., Moore spaces and w \Delta-spaces, Pacific J. Math., 38, (1971), 641-652.
  5. Okuyama, A., On metrizability of M-spaces, Proc. Japan. Acad., 40, 176-179.
  6. Sneider, V., Continuous images of Souslin and Borel sets: metrization theorems, Dokl. Acad. Nauk USSR, 50 (1945), 77-79.

One thought on “Compact Spaces With G-delta Diagonals

  1. Pingback: Looking for spaces in which every compact subspace is metrizable | Dan Ma's Topology Blog

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