# Tag Archives: Metrizable spaces

## The product of first countable spaces

All spaces under consideration are Hausdorff. First countable spaces are those spaces where there is a countable local base at every point in the space. This is quite a strong property. For example, every first countable space that is also compact has a cap on its cardinality and the cap is the cardinality of the real line (the continuum). See The cardinality of compact first countable spaces, I in this blog. In fact, if the compact and first countable space is uncountable, it has cardinality continuum (see The cardinality of compact first countable spaces, III). Any metric space (or metrizable space) is first countable. In this post, we discuss the product of first countable spaces. In this regard, first countable spaces and metrizable spaces behave similarly. We show that the product of countably many first countable spaces is first countable while the product of uncountably many first countable is not first countable. For more information on the product topology, see [2].

The Product Space
Consider a collection of sets $A_\alpha$ where $\alpha \in S$. Let $W=\bigcup \limits_{\alpha \in S} A_\alpha$. The product $\prod \limits_{\alpha \in S} A_\alpha$ is the set of all functions $f:S \mapsto W$ such that for each $\alpha \in S$, $f(\alpha) \in A_\alpha$. If the index set $S=\left\{1,2,\cdots,n\right\}$ is finite, the functions $f$ can be regarded as sequences $(f_1,f_2,\cdots,f_n)$ where each $f_i \in A_i$. If the index set $S=\mathbb{N}$, we can think of elements $f$ of the product as the sequence $(f_1,f_2,\cdots)$ where each $f_i \in A_i$. In general we can regard $f \in \prod \limits_{\alpha \in S} A_\alpha$ as functions $f:S \mapsto W$ or as sequences $f=(f_\alpha)_{\alpha \in S}$.

Consider the topological spaces $X_\alpha$ where $\alpha \in S$. Let $X=\prod \limits_{\alpha \in S} X_\alpha$ be the product as defined above. The product space of the spaces $X_\alpha$ is $X$ with the topology defined in the following paragraph.

Let $\tau_\alpha$ be the topology of each space $X_\alpha$, $\alpha \in S$. Consider $Y=\prod \limits_{\alpha \in S} O_\alpha$ where for each $\alpha \in S$, $O_\alpha \in \tau_\alpha$ (i.e. $O_\alpha$ is open in $X_\alpha$) and $O_\alpha=X_\alpha$ for all but finitely many $\alpha \in S$. The set of all such sets $Y$ is a base for a topology on the product $X=\prod \limits_{\alpha \in S} X_\alpha$. This topology is called the product topology of the spaces $X_\alpha$, $\alpha \in S$.

To more effectively work with product spaces, we consider a couple of equivalent bases that we can define for the product topology. Let $\mathcal{B}_\alpha$ be a base for the space $X_\alpha$. Consider $B=\prod \limits_{\alpha \in S} B_\alpha$ such that there is a finite set $F \subset S$ where $B_\alpha \in \mathcal{B}_\alpha$ for each $\alpha \in F$ and $B_\alpha=X_\alpha$ for all $\alpha \in S-F$. The set of all such sets $B$ is an equivalent base for the product topology.

Another equivalent base is defined using the projection maps. For each $\alpha \in S$, consider the map $\pi_\alpha:\prod \limits_{\beta \in S} X_\beta \mapsto X_\alpha$ such that $\pi_\alpha(f)=f_\alpha$ for each $f$ in the product. In words, the function $\pi_\alpha$ maps each point in the product space to its $\alpha^{th}$ coordinate. The mapping $\pi_\alpha$ is called the $\alpha^{th}$ projection map. For each set $U \subset X_\alpha$, $\pi_\alpha^{-1}(U)$ is the following set:

$\pi_\alpha^{-1}(U)=\left\{f \in \prod \limits_{\beta \in S} X_\beta: \pi_\alpha(f)=f_\alpha \in U\right\}$

Consider sets of the form $\bigcap \limits_{\alpha \in F} \pi_\alpha^{-1}(U_\alpha)$ where $F \subset S$ is finite and $U_\alpha$ is open in $X_\alpha$ for each $\alpha \in F$. The set of all such sets is another equivalent base for the product topology. If we only require that each $U_\alpha \in \mathcal{B}_\alpha$, a predetermined base for the coordinate space $X_\alpha$, we also obtain an equivalent base for the product topology.

Countable Product of First Countable Spaces
For $i=1,2,3,\cdots$, let $X_i$ be a first countable space. We show that $X=\prod \limits_{i=1}^{\infty}X_i$ is a first countable space.

Let $\mathbb{N}$ be the set of positive integers. For each $n \in \mathbb{N}$, let $[n]=\left\{1,2,\cdots,n\right\}$ and let $\mathbb{N}^{[n]}$ be the set of all functions $t:[n] \mapsto \mathbb{N}$. For each $i$ and each $x \in X_i$, let $\mathcal{B}_x(i)=\left\{B_x(i,j): j \in \mathbb{N}\right\}$ be a countable local base at $x$.

Let $f \in X=\prod \limits_{i=1}^{\infty}X_i$. We wish to define a countable local base at $f$. For each $n \in \mathbb{N}$, define $W_n$ to be:

$W_n=\left\{\prod \limits_{i=1}^n B_{f(i)}(i,t(i)):t \in \mathbb{N}^{[n]}\right\}$

Let $W$ be the set of all subsets of the product space $X$ of the following form:

$\prod \limits_{j=1}^{\infty}V_j$ where there is some $n \in \mathbb{N}$ such that $\prod \limits_{j=1}^{n}V_j \in W_n$ and for all $j>n$, $V_j=X_j$.

Each $W_n$ is countable and $W$ is essentially the union of all the $W_n$. Thus $W$ is countable. We claim that $W$ is a local base at $f$. Let $O \subset X$ be an open set containing $f$. We can assume that $O=\prod \limits_{i=1}^\infty O_i$ where there is some $n \in \mathbb{N}$ such that for each $i \le n$, $O_i$ is open in $X_i$ and for $i>n$, $O_i=X_i$.

For each $i \le n$, $f(i) \in O_i$. Choose some $B_{f(i)}(i,t(i))$ such that $f(i) \in B_{f(i)}(i,t(i)) \subset O_i$. Let $V=\prod \limits_{i=1}^{\infty} V_i$ such that $\prod \limits_{i=1}^{n} V_i=\prod \limits_{i=1}^n B_{f(i)}(i,t(i))$ and $V_i=X_i$ for all $i>n$. Then $V \in W$ and $f \in V \subset O$. This completes the proof that $X=\prod \limits_{i=1}^{\infty}X_i$ is a first countable space.

Uncountable Product
Let $S$ be an uncountable index set. For $\alpha \in S$, let $X_\alpha$. We want to avoid the situation that all but countably many $X_\alpha$ are one-point space. So we assume each coordinate space $X_\alpha$ has at least two points, say, $p_\alpha$ and $q_\alpha$ with $p_\alpha \ne q_\alpha$. We show that $X=\prod \limits_{\alpha \in S}X_\alpha$ is not first countable.

Let $f \in \prod \limits_{\alpha \in S}X_\alpha$. Let $U_1,U_2, \cdots$ be open subsets of the product space such that for each $i$, $f \in U_i$. We show that there is some open set $O$ such that $f \in O$ and each $U_i \nsubseteq O$. For each $i$, there is a basic open set $B_i=\bigcap \limits_{\alpha \in F_i} \pi_\alpha^{-1}(U_{\alpha,i})$ such that $f \in B_i \subset U_i$.

Let $F=F_1 \cup F_2 \cup \cdots$. Since $S$ is uncountable and $F$ is countable, choose $\gamma \in S-F$. Since $X_\gamma$ has at least two points $p_\gamma$ and $q_\gamma$, choose one of them that is different from $f_\gamma$, say, $p_\gamma$. Choose two disjoint open subsets $M_1$ and $M_2$ of $X_\gamma$ such that $f_\gamma \in M_1$ and $p_\gamma \in M_2$ of $X_\gamma$. Let $O=\prod \limits_{\alpha \in S}O_\alpha$ such that $O_\gamma=M_1$ and $O_\alpha=X_\alpha$ for all $\alpha \ne \gamma$. We have $f \in O$. For each $i$, there is $g_i \in B_i \subset U_i$ such that $g_i(\gamma)=p_\gamma$. Thus each $g_i \notin O$. Thus there is no countable local base at $f$. Thus any product space with uncountably many factors, each of which has at least two points, is never first countable.

Reference

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

## 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.

## The Evaluation Map

The evaluation map is a useful tool for embedding a space $X$ into a product space. In this post we demonstrate that any Tychonoff space $X$ can be embedded into a cube $I^{\mathcal{K}}$ where $I$ is the unit interval $[0,1]$ and $\mathcal{K}$ is some cardinal. Any regular space with a countable base (second-countable space) can also be embedded into the Hilbert cube $I^{\omega}$ (Urysohn’s metrization theorem). The evaluation map also plays an important role in the theory of Cech-Stone compactification.

The Evaluation Map
Let $X$ be a space. Let $\displaystyle Y=\Pi_{\alpha \in A}Y_\alpha$ be a product space. For each $y \in Y$, we use the notation $y=\langle y_\alpha \rangle_{\alpha \in A}$ to denote a point in the product space $Y$. Suppose we have a family of continuous functions $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$ where $f_\alpha:X \rightarrow Y_\alpha$ for each $\alpha$. Define a mapping that maps each $x \in X$ to the point $\langle f_\alpha(x) \rangle_{\alpha \in A} \in Y$. This mapping is called the evaluation map of the family of continuous functions $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$ and is denoted by $E_{\mathcal{F}}$.

The family of continuous functions $\mathcal{F}$ is said to separate points if for any two distinct points $x,y \in X$, there is a function $f \in \mathcal{F}$ such that $f(x) \neq f(y)$. The family of continuous functions $\mathcal{F}$ is said to separate points from closed sets if for each point $x \in X$ and for each closed set $C \subset X$ with $x \notin C$, there is a function $f \in \mathcal{F}$ such that $f(x) \notin \overline{f(C)}$.

Theorem 1. Given an evaluation map $E_{\mathcal{F}}$ as defined above, the following conditions hold.

1. The mapping $E_{\mathcal{F}}$ is continuous.
2. If the family of continuous functions $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$ separates points, then$E_{\mathcal{F}}$ is a one-to-one map.
3. If the family of continuous functions $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$ separates points from closed sets, then $E_{\mathcal{F}}$ is a homeomorphism from $X$ into the product space $\displaystyle Y=\Pi_{\alpha \in A}Y_\alpha$.

In this post, basic open sets in the product space $\displaystyle Y=\Pi_{\alpha \in A}Y_\alpha$ are of the form $\bigcap_{\alpha \in W} [\alpha,V_\alpha]$ where $W \subset A$ is finite, for each $\alpha \in W$, $V_\alpha$ is an open set in $Y_\alpha$ and $[\alpha,V_\alpha]=\lbrace{y \in Y:y_\alpha \in V_\alpha}\rbrace$.

Proof of 1. We show that $E_{\mathcal{F}}$ is continuous at each $x \in X$. Let $x \in X$. Let $h=\langle f_\alpha(x) \rangle_{\alpha \in A}$ and let $h \in V \cap E_{\mathcal{F}}(X)$ where $V=\bigcap_{\alpha \in W} [\alpha,V_\alpha]$ is a basic open set. Consider $U=\bigcap_{\alpha \in W} f_\alpha^{-1}(V_\alpha)$. It is easy to verify that $x \in U$ and $E_{\mathcal{F}}(U) \subset V\cap E_{\mathcal{F}}(X)$.

Proof of 2. Let $x,y \in X$ be distinct points. There is $\alpha \in A$  such that $f_\alpha(x) \neq f_\alpha(y)$. Clearly, $E_{\mathcal{F}}(x)= \langle f_\beta(x) \rangle_{\beta \in A} \neq E_{\mathcal{F}}(y)=\langle f_\beta(y) \rangle_{\beta \in A}$.

Proof of 3. Note that by condition 2 in this theorem, the map $E_{\mathcal{F}}$ is one-to-one. It suffices to show that $E_{\mathcal{F}}$ is an open map. Let $U \subset X$ be open. We show that $E_{\mathcal{F}}(U)$ is open in $E_{\mathcal{F}}(X)$. To this end, let $\langle f_\alpha(x) \rangle_{\alpha \in A} \in E_{\mathcal{F}}(U)$. Then $x \in U$. Since $\mathcal{F}$ separates points from closed sets, there is some $\beta$ such that $f_\beta(x) \notin \overline{f_\beta(X-U)}$. Let $V_\beta=Y_\beta-\overline{f_\beta(X-U)}$. Then $\langle f_\alpha(x) \rangle_{\alpha \in A} \in [\beta,V_\beta] \cap E_{\mathcal{F}}(X)=W_\beta$. We show that $W_\beta \subset E_{\mathcal{F}}(U)$. For each $\langle f_\alpha(y) \rangle_{\alpha \in A} \in W_\beta$, we have $f_\beta(y) \notin \overline{f_\beta(X-U)}$. If $y \notin U$, then $f_\beta(y) \in f_\beta(X-U)$, a contradiction. So we have $y \in U$ and this means that $\langle f_\alpha(y) \rangle_{\alpha \in A} \in E_{\mathcal{F}}(U)$. It follows that $W_\beta \subset E_{\mathcal{F}}(U)$.

Some Applications

A space $X$ is a Tychonoff space (also known as completely regular space) if for each $x \in X$ and for each closed set $C \subset X$ where $x \notin C$, there is a continuous function $f:X \rightarrow I$ such that $f(x)=0$ and $f(y)=1$ for all $y \in C$. The following is a corollary to theorem 1.

Corollary 1. Any Tychonoff space can be embedded in a cube $I^{\mathcal{K}}$.

Proof. Let $\mathcal{F}$ be the family of all continuous functions from the Tychonoff space $X$ into the unit interval $I$. By the definition of Tychonoff space, $\mathcal{F}$ separates points from closed sets. By theorem 1, the evaluation map $E_{\mathcal{F}}$ is a homeomorphism from $X$ into the cube $I^{\mathcal{K}}$ where $\mathcal{K}=\lvert \mathcal{F} \lvert$.

We now turn our attention to regular second countable space. Having a countable base has many strong properties, one of which is that it can be embedded into the Hilbert Cube $I^{\omega}=I^{\aleph_0}$. Before we prove this, observe that any regular space with a countable base is a regular Lindelof space. Furthermore, the property of having a countable base is hereditary. Thus a regular space with a countable base is hereditarily Lindelof (hence perfectly normal). The Vendenisoff Theorem states that in a perfectly normal space, every closed set is a zero-set (i.e. every open set is a cozero-set). So we make use of this theorem to obtain continuous functions that separate points from closed sets. There is a proof of The Vendenisoff Theorem in this blog. A set $Z \subset X$ is a zero-set in the space $X$ if there is a continuous function $f:X \rightarrow I$ such that $f^{-1}(0)=Z$. A set $W \subset X$ is a cozero-set if $X-W$ is a zero-set. We are now ready to prove one part of the Urysohn’s metrization theorem.

Urysohn’s metrization theorem. The following conditions are equivalent.

1. The space $X$ is a regular space with a countable base.
2. The space $X$ can be embedded into the Hilbert cube $I^{\aleph_0}$.
3. The space $X$ is a separable metric space.

We prove the direction $1 \Rightarrow 2$. Let $\lbrace{B_0,B_1,B_2,...}\rbrace$ be a countable base for the regular space $X$. Based on the preceding discussion, $X$ is perfectly normal. By the Vendenisoff Theorem, for each $n$$X-B_n$ is a zero-set. Thus for each $n$, there is a continuous function $f_n:X \rightarrow I$ such that $f_n^{-1}(0)=X-B_n$ and $f_n^{-1}((0,1])=B_n$. Let $\mathcal{F}=\lbrace{f_0,f_1,f_2,...}\rbrace$. It is easy to verify that $\mathcal{F}$ separates points from closed sets. Thus the evaluation map $E_{\mathcal{F}}$ is a homeomorphism from $X$ into $I^{\aleph_0}$.

## 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.

## A Note About Countably Compact Spaces

This is a discussion on several additional conditions that would turn a countably compact space into a compact space. For example, a countably compact space having a $G_\delta-$ diagonal is compact (proved in this post). Each of the following properties, if possessed by a countably compact space, would lead to compactness: (1) having a $G_\delta-$ diagonal, (2) being metrizable, (3) being a Moore space, (4) being paracompact, and (5) being metacompact. All spaces are at least Hausdorff. We have the following theorem. Some relevant definitions and links to posts in this blog are given below. For any terms that are not defined here, see Engelking ([1]).

Theorem. Let $X$ be a countably compact space. If $X$ possesses any one of the following conditions, then $X$ is compact.

1. Having a $G_\delta-$ diagonal.
2. Being a metrizable space.
3. Being a Moore space.
4. Being a paracompact space.
5. Being a metacompact space.

The proof of 1 has already been presented in another post in this blog. Since metrizable spaces are Moore spaces, between 2 and 3 we only need to prove 3. Between 4 and 5, we only need to prove 5 (since paracompact compact spaces are metacompact).

Proof of 3. A Moore space is a regular space that has a development (see this post for the definition). In this post, I showed that a space $X$ has a $G_\delta-$diagonal if and only it has a $G_\delta-$diagonal sequence. It is easy to verify that the development for a Moore space is a $G_\delta-$diagonal sequence. Thus any Moore space has a $G_\delta-$diagonal and any countably compact Moore space is compact (and metrizable). Saying in another way, in the class of Moore spaces, countably compactness is equivalent to compactness.

Proof of 5. A space $X$ is metacompact if every open cover of $X$ has a point-finite open refinement. Let $X$ be metacompact. Let $\mathcal{U}$ be an open cover of $X$. By the metacompactness, $\mathcal{U}$ has a point-finite open refinement $\mathcal{O}$. We are done if we can show $\mathcal{O}$ has a finite subcover. This finite subcover is obtained through the following claims.

Claim 1. There is a set $M \subset X$ such that $\lvert M \cap O \lvert \thinspace \leq 1$ for each $O \in \mathcal{O}$ and such that $M$ is maximal. That is, by adding an additional point $x \notin M$, $\lvert (M \cup \lbrace{x}\rbrace) \cap O \lvert \thinspace \ge 2$ for some $O \in \mathcal{O}$.

Such a set can be obtained by using the Zorn’s Lemma.

Claim 2. Let $\mathcal{W}=\lbrace{O \in \mathcal{O}:O \cap M \neq \phi}\rbrace$. We claim that $\mathcal{W}$ is an open cover of $X$.

To see this, let $x \in X$. If $x \in M$, then $x \in O$ for some $O \in \mathcal{W}$. If $x \notin M$, then by the maximality of $M$, $M \cup \lbrace{x}\rbrace$ intersects with some $O \in \mathcal{O}$ with at least 2 points. This means that $x$ and at least one point of $M$ are in $O$. Then $O \in \mathcal{W}$.

Since each open set in $\mathcal{W}$ contains at most one point of $M$, $M$ is a closed and discrete set in $X$. By the countably compactness of $X$, $M$ must be finite. Since each point of $M$ is in at most finitely many open sets in $\mathcal{O}$, $\mathcal{W}$ is finite. Thus $\mathcal{W}$ is a finite subcover of $\mathcal{O}$.

Reference

1. Engelking, R., General Topology, Revised and Completed Edition, 1989, Heldermann Verlag, Berlin.

## 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.

## Perfect Image of Separable Metric Spaces

In a previous post on countable network, it was shown that having a countable network is equivalent to being the continuous image of a separable metric space. Since there is an example of a non-metrizable space with countable netowrk, the continuous image of a separable metric space needs not be a separable metric space. However, the perfect image of a separable metrizable space is separable metrizable. First some definitions. A continuous mapping $f:X \rightarrow Y$ is a closed mapping if $f(H)$ is closed in $Y$ for any closed set $H \subset X$. A continuous surjection $f:X \rightarrow Y$ is a perfect mapping if $f$ is closed and $f^{-1}(y)$ is compact for each $y \in Y$.

Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. Because $f$ is a closed mapping, $f(X-B)$ is closed and $f(B)$ is open in $Y$ for each $B \in \mathcal{B}$. We show that $\mathcal{B}_f=\lbrace{f(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$. Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. We have $y \in f(B) \subset U$. Thus the topology on $Y$ can be generated by $\mathcal{B}_f$.

Update (11/24/2009):
The proof in the above paragraph is faulty. Thanks to Dave Milovich for pointing this out. Here’s the corrected proof.

Let me first prove a lemma.

Lemma. Let $f: X \rightarrow Y$ be a closed mapping and let $V \subset X$ be open. Then $f_*(V)=\lbrace{y \in Y:f^{-1}(y) \subset V}\rbrace$ is open in $Y$. Furthermore, $f_*(V) \subset f(V)$.

Proof of Lemma. Since $f$ is a closed mapping, $f(X-V)$ is closed. We claim that $f(X-V)=Y-f_*(V)$. It is clear that $f(X-V) \subset Y-f_*(V)$. To show that $Y-f_*(V) \subset f(X-V)$, let $z \in Y-f_*(V)$. Then $f^{-1}(z)$ cannot be a subset of $V$. Choose $x \in f^{-1}(z)-V$. Then we have $z=f(x) \in f(X-V)$. Thus $f(X-V)=Y-f_*(V)$ and $f_*(V)$ is open. It is straitforward to verify that $f_*(V) \subset f(V)$.

Now I prove that the perfect image of a separable metric space is a separable metric space. Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. We show that $\mathcal{B}_f=\lbrace{f_*(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$.

Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $x \in B_x$ and $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. Since $f^{-1}(y) \subset B$, we have $y \in f_*(B)$. We also have $f_*(B) \subset f(B) \subset U$. Thus the topology on $Y$ can be generated by the countable base $\mathcal{B}_f$.

## A Discussion About The Michael Line

The original post was a basic discussion of the Michael line. It was written back in Oct 2009 and is now replaced by the following newer posts.

“Michael Line Basics”

“Finite and Countable Products of the Michael Line”

“Bernstein Sets and the Michael Line”

“The Michael Line and the Continuum Hypothesis”