# A note on metrization theorems for compact spaces

In a previous post (Metrization Theorems for Compact Spaces), three classic metrization theorems for compact spaces are discussed. The three theorems are: any Hausdorff compact space $X$ is metrizable if any of the following holds:

1. $X$ has a countable network,
2. $X$ has a $G_\delta$ diagonal,
3. $X$ has a point countable base.

The metrization results for conditions 2 and 3 hold for countably compact spaces as well. See the following posts:

In this post, we discuss another metrization theorem for compact spaces. We show that a compact Hausdorff space $X$ is metrizable if and only if the function space $C_p(X)$ is separable.

Let’s discuss the function space. Let $Y$ by any Tychonoff space and let $\mathbb{R}$ be the set of all real numbers. Let $C(Y,\mathbb{R})$ be the set of all real-valued continuous functions defined on $Y$. For any $A \subset Y$ and for any $V \subset \mathbb{R}$, define $[A,V]=\lbrace{f \in C(Y,\mathbb{R}): f(A) \subset V}\rbrace$. If we restrict $A$ to $\lbrace{x}\rbrace$ and restrict $V$ to open sets, then the set of all $[A,V]$ is a subbase for a topology on $C(Y,\mathbb{R})$. This topology is called the pointwise convergence topology. The function space $C(Y,\mathbb{R})$ with this topology is denoted by $C_p(Y)$.

It is a theorem that $C_p(Y)$ is separable if and only if $Y$ has a weaker topology that forms a separable metric space. The result on compact spaces is a corollary of this theorem.

Theorem. Let $Y$ be a Tychonoff space with $\tau$ being the topology. The following conditions are equivalent:

1. $C_p(Y)$ is separable.
2. There is a topology $\tau_1 \subset \tau$ such that $(Y, \tau_1)$ is a separable metric space.

Proof
$1 \Rightarrow 2$. Let $D \subset C_p(Y)$ be a countable dense subspace. Let $\mathcal{V}$ be the class of all bounded open intervals of $\mathbb{R}$ with rational endpoints. Consider $\mathcal{B}=\lbrace{f^{-1}(V): f \in D, V \in \mathcal{V}}\rbrace$. Note that $\mathcal{B}$ is a subbase for a topology $\tau_1$ on $Y$. Since $\mathcal{B}$ is countable, the topology $\tau_1$ has a countable base and is thus separable and metrizable.

$2 \Rightarrow 1$. Let $\tau_1 \subset \tau$ be a topology. Suppose that $\tau_1$ is generated by a countable base $\mathcal{U}$. As in $1 \Rightarrow 2$, let $\mathcal{V}$ be the class of all bounded open intervals of $\mathbb{R}$ with rational endpoints. Let $\mathcal{N}$ be the class of all finite intersections of the sets in the following collection of sets.

$\displaystyle \lbrace{[A,V]: A \in \mathcal{U},V \in \mathcal{V}}\rbrace$

Note that $\mathcal{N}$ is countable. For each $W \in \mathcal{N}$, choose $f_W \in W$. We claim that $\lbrace{f_W: W \in \mathcal{N}}\rbrace$ is a countable dense set of $C_p(Y)$. To see this, let $T=\bigcap_{i \le n} [A_i,V_i]$ be a basic open set in $C_p(Y)$ where $A_i=\lbrace{x_i}\rbrace$. Fix $f \in T$. For each $i$, choose $O_i \in \mathcal{U}$ such that $x_i \in O_i$ and $f(O_i) \subset V_i$. Then $W=\bigcap_{i \le n} [O_i,V_i] \in \mathcal{N}$. Now, we have $f_W \in T$.

Corollary. Let $X$ be a compact Hausdorff space. Then the following conditions are equivalent:

1. $X$ is metrizable.
2. $C_p(X)$ is separable.

Proof.
$1 \Rightarrow 2$. This follows from $2 \Rightarrow 1$ in the above theorem.

$2 \Rightarrow 1$. This follows from $1 \Rightarrow 2$ in the above theorem. Note that any compact Hausdorff space cannot have a strictly weaker (or coarser) Hausdorff topology. Thus if a compact Hausdorff space has a weaker metrizable topology, it must be metrizable.

# On Spaces That Can Never Be Dowker

A Dowker space is a normal space $X$ for which the product with the closed unit interval $[0,1]$ is not normal. In 1951, Dowker characterized Dowker’s spaces as those spaces that are normal but not countably paracompact ([1]). Soon after, spaces that are normal but not countably paracompact became known as Dowker spaces. In 1971, M. E. Rudin ([2]) constructed a ZFC example of a Dowker’s space. But this Dowker’s space is large. It has cardinality $(\omega_\omega)^\omega$ and is pathological in many ways. Thus the search for “nice” Dowker’s spaces continued. The Dowker’s spaces being sought were those with additional properties such as having various cardinal functions (e.g. density, character and weight) countable. Many “nice” Dowker’s spaces had been constructed using various additional set-theoretic assumptions. In 1996, Balogh constructed a first “small” Dowker’s space (cardinaltiy continuum) without additional set-theoretic axioms beyond ZFC ([4]). Rudin’s survey article is an excellent reference for Dowker’s spaces ([3]).

In this note, I make several additional observations on Dowker’s spaces. In this previous post, I presented a proof of the Dowker’s theorem characterizing the normal spaces for which the product with the unit interval is normal (see the statement of the Dowker’s theorem below). In another post, I showed that perfectly normal spaces can never be Dowker’s spaces. Based on the Dowker’s theorem, several other classes of spaces are easily seen as not Dowker.

Dowker’s Theorem. For a normal space $X$, the following conditions are equivalent.

1. The space $X$ is countably paracompact.
2. The product $X \times Y$ is normal for any infinite compact metric space $Y$.
3. The product $X \times [0,1]$ is normal.
4. For each sequence of closed subsets $\lbrace{A_0,A_1,A_2,...}\rbrace$ of $X$ such that $A_0 \supset A_1 \supset A_2 \supset ...$ and $\bigcap_{n<\omega} A_n=\phi$, there is open sets $U_n \supset A_n$ for each $n$ such that $\bigcap_{n<\omega} U_n=\phi$.

Observations. If $X$ is perfectly compact, then it can be shown that it is countably paracompact by showing that it satisfies condition 4 in the Dowker’s theorem (there is a proof in this blog). Thus there are no perfectly normal Dowker’s spaces. There are no countably compact Dowker’s spaces since any countably compact space is countably paracompact. This can also be seen using condition 4 above. In a countably compact space, any decreasing nested sequence of closed sets has non-empty intersection and thus condition 4 is satisfied vacuously. Furthermore, all metric spaces, compact spaces, regular Lindelof spaces cannot be Dowker since these spaces are paracomapct.

Normal Moore spaces are perfectly normal. Thus there are no Dowker’s spaces that are Moore spaces. Note that a space is perfectly normal if it is normal and if every closed set is $G_\delta$. We show that in a Moore space, every closed set is $G_\delta$. Let $\lbrace{\mathcal{O}_n:n \in \omega}\rbrace$ be a development for the regular space $X$. Let $A$ be a closed set in $X$. We show that $A$ is a $G_\delta-$ set in $X$. For each $n$, let $U_n=\lbrace{O \in \mathcal{O}_n:O \bigcap A \neq \phi}\rbrace$. Obviously, $A \subset \bigcap_n U_n$. Let $x \in \bigcap_n U_n$. If $x \notin A$, there is some $n$ such that for each $O \in \mathcal{O}_n$ with $x \in O$, we have $O \subset X-A$. Since $x \in \bigcap_n U_n$, $x \in O$ for some $O \in \mathcal{O}_n$ and $O \cap A \neq \phi$, a contradiction. Thus we have $A=\bigcap_n U_n$.

There are other classes of spaces that can never be Dowker. We point these out without proof. For example, there are no linearly ordered Dowker’s spaces and there are no monotonically normal Dowker’s spaces (see Rudin’s survey article [3]).

Reference

1. Dowker, C. H., On Countably Paracompact Spaces, Canad. J. Math. 3, (1951) 219-224.
2. Rudin, M. E., A normal space $X$ for which $X \times I$ is not normal, Fund. Math., 73 (1971), 179-186.
3. Rudin, M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
4. Balogh, Z., A small Dowker space in ZFC, Proc. Amer. Math. Soc., 124 (1996), 2555-2560.

# Countably Compact Spaces with G-delta Diagonals

It is a classic result in general topology that any compact space with a $G_\delta-$diagonal is metrizable ([3]). This theorem also holds for countably compact spaces (due to Chaber in [2]). The goal of this post is to present a proof of this theorem. We prove that if $X$ is countably compact and has a $G_\delta-$diagonal, then $X$ is compact and thus metrizable. All spaces are at least Hausdorff. This post has a discussion on the theorem on compact spaces with $G_\delta-$diagonal. This post has a discussion on some metrizaton theorems for compact spaces.

If $\mathcal{T}$ is a collection of subsets of a space $X$, then for each $x \in X$, define $st(x,\mathcal{T})=\bigcup\lbrace{T \in \mathcal{T}:x \in T}\rbrace$. A sequence of open covers $\lbrace{\mathcal{T}_n:n \in \omega}\rbrace$ of the space $X$ is a $G_\delta-$diagonal sequence for $X$ if for each $x \in X$, we have $\lbrace{x}\rbrace=\bigcap_{n<\omega} st(x,\mathcal{T}_n)$. We use the following lemma (due to Ceder, [1]). This lemma was proved in this previous post.

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

Theorem. Let $X$ be a countably compact space that has a $G_\delta-$diagonal. Then $X$ is compact.

Proof. Let $X$ be a countably compact space. Let $\lbrace{\mathcal{T}_n:n \in \omega}\rbrace$ be a $G_\delta-$diagonal sequence for $X$. If $X$ is Lindelof, then we are done. Suppose we have an open cover $\mathcal{V}$ of $X$ that has no countable subcover. From this open cover $\mathcal{V}$, we derive a contradiction.

We inductively, for each $\alpha < \omega_1$, choose a point $x_\alpha \in X$ and an integer $m(\alpha) \in \omega$ with the following properties:

For each $\alpha < \omega_1$,

1. $x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace$, and
2. the open cover $\mathcal{V}$ does not have a countable subcollection that covers $X-\bigcup_{\beta \leq \alpha} st(x_\beta,\mathcal{T}_{m(\beta)})$.

To start off, choose $x_0 \in X$. There is an integer $m(0) \in \omega$ such that no countable subcollection of $\mathcal{V}$ covers $X-st(x_0,\mathcal{T}_{m(0)})$. Suppose this integer $m(0)$ does not exist. Then for each $n \in \omega$, we have a countable $\mathcal{V}_n \subset \mathcal{V}$ such that $\mathcal{V}_n$ covers $X-st(x_0,\mathcal{T}_n)$. Then $\bigcup_{n<\omega} \mathcal{V}_n$ would be a countable subcollection of $\mathcal{V}$ that covers $X-\lbrace{x_0}\rbrace$. This would mean that $\mathcal{V}$ has a countable subcover of $X$.

Suppose that $\lbrace{x_\beta:\beta<\alpha}\rbrace$ and $\lbrace{m(\beta):\beta<\alpha}\rbrace$ have been chosen such that conditions (1) and (2) are satisfied for each $\beta<\alpha$. We have the following claim. Proving this claim allows us to choose $x_\alpha$ and $m(\alpha)$.

Claim. No countable subcollection of $\mathcal{V}$ covers $X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)})$.

Suppose we do have a countable $\mathcal{W} \subset \mathcal{V}$ such that $\mathcal{W}$ covers $X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)})$. Then $\mathcal{S}=\lbrace{st(x_\beta,\mathcal{T}_m(\beta)):\beta < \alpha}\rbrace \cup \mathcal{W}$ is a countable open cover of $X$ and thus has a finite subcover $\mathcal{F}$. Let $\delta$ be the largest ordinal $<\alpha$ such that $st(x_\delta,\mathcal{T}_m(\delta))$ is in this finite subcover $\mathcal{F}$. Then $\mathcal{W}$ is a counntable subcollection of $\mathcal{V}$ that covers $X-\bigcup_{\beta \leq \delta} st(x_\beta,\mathcal{T}_{m(\beta)})$. This violates condition (2) above for the ordinal $\delta$. This proves the claim.

Now, pick $x_\alpha \in X-\bigcup\lbrace{st(x_\beta,\mathcal{T}_{m(\beta)}): \beta < \alpha}\rbrace$. There must be some integer $m(\alpha) \in \omega$ such that conditon (2) above is satisfied for $\alpha$. If not, for each $n \in \omega$, there is some countable $\mathcal{V}_n \subset \mathcal{V}$ such that $\mathcal{V}_n$ covers $X-\bigcup_{\beta \leq \alpha} st(x_\beta,\mathcal{T}_n)$. Then $\bigcup_{n<\omega} \mathcal{V}_n$ would be a countable subcollection of $\mathcal{V}$ that covers $X-\biggl(\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)}) \biggr) \bigcup \lbrace{x_\alpha}\rbrace$. This would mean that $\mathcal{V}$ has a countable subcover of $X-\bigcup_{\beta<\alpha} st(x_\beta,\mathcal{T}_{m(\beta)})$. This violates the above claim. Now the induction process is completed.

To conclude the proof of the theorem, note that there is some $n \in \omega$ and there is some uncountable $D \subset \omega_1$ such that for each $\alpha \in D$, $n=m(\alpha)$. Then $Y=\lbrace{x_\alpha:\alpha \in D}\rbrace$ is an uncountable closed and discrete set in $X$. Note that each open set in $\mathcal{T}_n$ contains at most one point of $Y$. Thus $X$ must be Lindelof. With $X$ being countably compact, $X$ is compact.

Reference

1. Ceder, J. G. Some generalizations of metric spaces, Pacific J. Math., 11 (1961), 105-125.
2. Chaber, Conditions which imply compactness in countably compact spaces, Bull. Acad. Pol. Sci. Ser. Math., 24 (1976), 993-998.
3. Sneider, V., Continuous images of Souslin and Borel sets: metrization theorems, Dokl. Acad. Nauk USSR, 50 (1945), 77-79.

# Metrization Theorems for Compact Spaces

In this blog I have already presented two metrization theorems for compact spaces: (1) any compact space with a countable network is metrizable (see the post), (2) any compact space with a $G_\delta-$diagonal is metrizable (see the post). I now present another classic theorem: any countably compact space with a point-countable base is metrizable. This theorem is a classic result of Miscenko ([1]). All spaces are at least Hausdorff and regular. We have the following three metrization theorems for compact spaces. In subsequent posts, I will discuss generalizations of these theorems and discuss related concepts.

Thoerem 1. Any compact space with a countable network is metrizable.
The proof is in this post.

Thoerem 2. Any compact space with a $G_\delta-\text{diagonal}$ is metrizable.
The proof is in this post.

Thoerem 3. Any countably compact space with a point-countable base is metrizable.

A base $\mathcal{B}$ for a space $X$ is a point-countabe base if every point in $X$ belongs to at most countably elements of $\mathcal{B}$.

Proof of Theorem 3. Let $\mathcal{B}$ be a point-countable base for the countably compact space $X$. We show that $X$ is separable. Once we have a countable dense subset, the base $\mathcal{B}$ has to be a countable base. So we inductively define a sequence of countable sets $\lbrace{D_0,D_1,...}\rbrace$ such that $D=\bigcup_{n<\omega}D_n$ is dense in $X$.

Let $D_0=\lbrace{x_0}\rbrace$ be a one-point set to start with. For $n>0$, let $E_n=\bigcup_{i. Let $\mathcal{B}_n=\lbrace{B \in \mathcal{B}:B \cap E_n \neq \phi}\rbrace$. For each finite $T \subset \mathcal{B}_n$ such that $X - \bigcap T \neq \phi$, choose a point $x(T) \in X - \bigcup T$. Let $D_n$ be the union of $E_n$ and the set of all points $x(T)$. Let $D=\bigcup_{n<\omega}D_n$.

We claim that $\overline{D}=X$. Suppose we have $x \in X-\overline{D}$. Let $\mathcal{A}=\lbrace{B \in \mathcal{B}:B \cap D \neq \phi \phantom{X} \text{and} \thinspace x \notin B}\rbrace$. We know that $\mathcal{A}$ is countable since every element of $\mathcal{A}$ contains points of the countable set $D$. We also know that $\mathcal{A}$ is an open cover of $\overline{D}$. By the countably compactness of $\overline{D}$, we can find a finite $T \subset \mathcal{A}$ such that $\overline{D} \subset \bigcup T$. The finite set $T$ must have appeared during the induction process of selecting points for $D_n$ for some $n$ (i.e. $T \subset \mathcal{B}_n$). So a point $x(T)$ has been chosen such that $x(T) \notin \bigcup T$ (thus we have $x(T) \in D_n \subset \overline{D}$). On the other hand, since $\overline{D} \subset \bigcup T$, we observe that $x(T) \notin \overline{D}$, producing a contradiction. Thus the countable set $D$ is dense in $X$, making the point-countable base $\mathcal{B}$ a countable base.

Reference

1. Miscenko, A., Spaces with a point-countable base, Dokl. Acad. Nauk SSSR, 144 (1962), 985-988. (English translation: Soviet Math. Dokl. 3 (1962), 1199-1202).

# 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.

Reference

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.

# Perfectly Normal Spaces Can Never Be Dowker Spaces

The Dowker’s theorem states that for a normal space $X$, $X \times [0,1]$ is normal if and only if $X$ is countably paracompact. Since this theorem was published, any normal space that is not countably paracompact became known as Dowker space. There are classes of spaces that can never be Dowker spaces (e.g. metrizable spaces, paracompact spaces, compact spaces and Lindelof spaces). In [Katetov], it was shown that there are no perfectly normal Dowker spaces. My blog has a proof of the Dowker’s theorem (see the proof here). For more background on Dowker’s spaces, see the survey article [Rudin]. Dowker’s theorem was published in [Dowker].

Theorem. If $X$ is perfectly normal, then $X$ is countably paracompact.

To prove this theorem, we use the following characterization of countably paracompactness (you can find a proof here).

Lemma. Let $X$ be a normal space. Then $X$ is countably paracompact if and only if for each sequence $\lbrace{A_n:n \in \omega}\rbrace$ of closed subsets of $X$ such that $A_0 \supset A_1 \supset ...$ and $\bigcap_n A_n=\phi$, there exist open sets $B_n \supset A_n$ such that $\bigcap_n B_n=\phi$.

Proof of Theorem. Suppose $X$ is perfectly normal. Let $A_0 \supset A_1 \supset ...$ be a sequence of closed sets such that $\bigcap_n A_n=\phi$. For each $n$, let $A_n=\bigcap_{i<\omega} U_{n,i}$ where each $U_{n,i}$ is open in $X$. For each $n$, define $B_n=\bigcap_{i \leq j \leq n}U_{j,i}$. Clearly, $B_n \supset A_n$. It is easy to see that $\bigcap_n B_n=\phi$. Note that all the open sets $U_{n,j}$ are used in defining the sequence $B_0,B_1,B_2,\cdots$. Thus $\bigcap_n B_n \neq \phi$ would imply $\bigcap_n A_n \neq \phi$.

Comment. As a consequence of this theorem and the Dowker’s theorem, if $X$ is perfectly normal, then $X \times Y$ is normal for any compact metric space $Y$.

Reference
[Dowker]
Dowker, C. H. [1951], On Countably Paracompact Spaces, Canad. J. Math. 3, 219-224.

[Katetov]
Katetov, M., On real-valued functions in topological spaces, Fund. Math. 38 (1951), pp. 85-91.

[Rudin]
Rudin, M. E., [1984], Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 761-780.

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$\copyright \ 2009-2016 \text{ by Dan Ma}$
Revised November 23, 2016

# A Theorem About Hereditarily Normality

Let $Y$ be either the closed unit interval $\mathbb{I}=[0,1]$ or $[0,\omega]=\omega+1$. Let $X$ be any one of the following spaces:

$X=[0,\omega_1]=\omega_1+1$,
$X=[0,\omega_1)=\omega_1$,
$X=$ Michael Line,
$X=$ Sorgenfrey Line.

The cross product $X \times Y$ is normal in all four cases. The $X$ factor, in the order listed, is compact, countably compact, paracompact and Lindelof. Thus they are all countably paracompact. According to the Dowker’s theorem, the product of a normal space $X$ and a compact metric space $Y$ is normal if and only if $X$ is countably paracompact. My goal here is to show that the four cases of $X \times Y$ here cannot be hereditarily normal. This is from a theorem due to Katetov. We prove the following theorem.

Theorem. If $X \times Y$ is hereditarily normal, then either $X$ is perfectly normal or every countably infinite subspace of $Y$ is closed and discrete.

Proof. Suppose that $X$ is not perfectly normal and $Y$ has a countably infinite subset that is not closed and discrete. Let $H \subset X$ be a closed set that is not a $G_\delta-$set. Let $C=\lbrace{y_n:n \in \omega}\rbrace \subset Y$ be an infinite set with an accumulation point $y$. We assume that $y \notin C$.

We show that the open subspace $U=(X \times Y)-(H \times \lbrace{y}\rbrace)$ is not normal. To this end, let $A=H \times (Y-\lbrace{y}\rbrace)$ and $B=(X-H) \times \lbrace{y}\rbrace$. The sets $A$ and $B$ are disjoint closed subspaces of the open subspace $U$. Suppose we have disjoint open sets $S$ and $T$ such that $A \subset S$ and $B \subset T$.

For each $n \in \omega$, let $O_n=\lbrace{x \in X:(x,y_n) \in S}\rbrace$. Each $O_n$ is open and $H \subset O_n$. Thus $H \subset \bigcap_n O_n$. Let $x \in \bigcap_n O_n$. Then $(x,y) \in \overline{S}$. This means $x \in H$. If $x \notin H$, then $(x,y) \in B \subset T$ (which is impossible). So we have $H=\bigcap_n O_n$, indicating that $H$ is a $G_\delta-$set, and leading to a contradiction. So the subspace $U=(X \times Y)-(H \times \lbrace{y}\rbrace)$ is not normal.

Corollary. For countably compact spaces (in particular compact spaces) $X$ and $Y$, if the Cartesian product $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are prefectly normal.

Proof. Note that both factors, being countably compact, cannot have closed and discrete infinite subsets.

Comment. Note that the converse of the corollary is not true. Let both factors be the double arrow space, which is perfectly normal. But the square of the double arrow space contains a copy of the Sorgenfrey Plane, which is not normal.

Reference
[Katetov] Katetov, M., [1948] Complete normality of Cartesian products, Fund. Math., 36, 271-274.