# A note about sigma-product of compact spaces

This is a basic discussion on $\Sigma-$products of compact spaces. Let $\displaystyle Y=\Pi_{\alpha \in S} X_\alpha$ be a product space and let $p \in Y$. The $\Sigma-$product of the spaces $\lbrace{X_\alpha}\rbrace_{\alpha \in S}$ about the fixed point $p$ is the following subspace of the product space $Y$:

$\displaystyle \Sigma_{\alpha \in S} X_\alpha=\lbrace{y \in Y:\lvert \lbrace{\alpha \in S:y(\alpha) \neq p(\alpha)}\rbrace \lvert \thinspace \leq \omega}\rbrace$.

The $\Sigma-$products had been introduced in this blog. In a previous post, it was shown that the $\Sigma-$product of uncountably many separable spaces is an example of a space that has the countable chain condition (ccc) that is not separable. In this post, we discuss the $\Sigma-$product of compact spaces. First, the $\Sigma-$product of uncountably many spaces can never be paracompact because it always contains a closed copy of the first uncountable ordinal $\omega_1$. Second, the $\Sigma-$product of compact spaces may not be normal. Our example is that $\displaystyle \Sigma_{\alpha \in S} X_\alpha$ is not normal where each $X_\alpha=\omega_1+1=[0,\omega_1]$. While it may be too much to expect that the Tychonoff theorem (the product space is compact if and only if each factor is compact) would have a counterpart for $\Sigma-$products, we do have a proof that the $\Sigma-$product of compact spaces is countably compact. This provides another example of a countably compact space that is non-compact. It is known that the $\Sigma-$product of metric spaces is normal ([1] and [2]).

When the index set $S$ is countable, the $\Sigma-$product is the entire product space $\displaystyle Y=\Pi_{\alpha \in S} X_\alpha$. When all but countably many $X_\alpha$ are one-point sets, the $\Sigma-$product can also be treated as just a product space of countably many factors. So we like to avoid these two situations by assuming that there are uncountably many spaces $X_\alpha$ and each $X_\alpha$ has at least two points. When this is the case, the $\Sigma-$product is a proper subspace of the associated product space. Such $\Sigma-$products are called proper.

For each $\beta \in S$, define $\pi_\beta:Y \rightarrow X_\beta$ by $\pi_\beta(f)=f(\beta)$ for each $\text{f}$ in the product space $\Pi X_\alpha$. The map $\pi_\beta$ maps each point in the product space to its $\beta^{th}$ coordinate and is called the projection map of $\Pi X_\alpha$ on $X_\beta$. Open sets of the form $\bigcap_{\alpha \in F} \pi^{-1}_\alpha(V_\alpha)$, where $F \subset S$ is finite and for each $\alpha \in F$, $V_\alpha$ is open in $X_\alpha$, form a base in the product topology.

Observation. Given $\displaystyle X=\Sigma_{\alpha \in S}X_\alpha$ and $T \subset S$, it is easy to verify that $\displaystyle X=\Sigma_{\alpha \in S}X_\alpha$ is homeomorphic to $\displaystyle \Sigma_{\alpha \in T}X_\alpha \times \Sigma_{\alpha \in S-T}X_\alpha$. We will make use of this observation in the rest of this note.

Lemma. Let $\lbrace{X_\alpha}\rbrace_{\alpha<\omega_1}$ be a family of spaces each of which has at least two points. Let $p \in \Pi_{\alpha<\omega_1}X_\alpha$. Then $\displaystyle X=\Sigma_{\alpha<\omega_1}X_\alpha$ (about the fixed point $p$) contains a closed copy of the first uncountable ordinal $\omega_1$. Thus any proper $\Sigma-$product can never be paracompact.

Proof. For each $\alpha<\omega_1$, choose a point $t_\alpha \in X_\alpha$ such that $t_\alpha \neq p(\alpha)$. Consider the point $q \in \Pi_{\alpha<\omega_1}X_\alpha$ such that $q(\alpha)=t_\alpha$ for each $\alpha<\omega_1$. Since each $X_\alpha$ is Hausdorff, choose disjoint open sets $U_\alpha,V_\alpha \subset X_\alpha$ containing $q(\alpha)$ and $p(\alpha)$ respectively.

We can think of elements of $X$ as all the functions $f$ such that $f(\alpha) \in X_\alpha$ for each $\alpha<\omega_1$ and $f(\alpha)=p(\alpha)$ for all but countably many $\alpha$. For each $\beta<\omega_1$, consider the function $f_\beta$ such that $f_\beta(\gamma)=q(\gamma)$ for all $\gamma<\beta$ and $f_\beta(\gamma)=p(\gamma)$ for all $\gamma \ge \beta$. In other words, $f_\beta$ agrees with $\text{q}$ on the initial segment $[0,\beta)$ and $f_\beta$ agrees with $p$ on the final segment $[\beta,\omega_1)$. Then $W=\lbrace{f_\beta:\beta<\omega_1}\rbrace$ is a closed copy of $\omega_1$.

Claim 1. The set $W$ is closed in $X$.

We show $X-W$ is open. Let $y \in X-W$. If for some $\alpha \in \omega_1$, $y(\alpha) \in X_\alpha-\lbrace{p(\alpha),q(\alpha)}\rbrace$, choose open $O \subset X_\alpha$ such that $y(\alpha) \in O$ and $O$ does not contain both $p(\alpha),q(\alpha)$. Then $y \in \pi^{-1}_\alpha(O)$ and $\pi^{-1}_\alpha(O) \cap W =\phi$. Now we can assume that for each $\alpha \in \omega_1$, we have $y(\alpha) \in \lbrace{p(\alpha),q(\alpha)}\rbrace$. Since $y \notin W$, there are $\alpha<\beta<\omega_1$ such that $y(\alpha)=p(\alpha)$ and $y(\beta)=q(\beta)$. Then $y \in \pi^{-1}_\alpha(V_\alpha) \cap \pi^{-1}_\beta(U_\beta)=B$ and $B \cap W=\phi$.

Claim 2. The mapping $\alpha \rightarrow f_\alpha$ is a homeomorphism between $\omega_1$ and $W$.

First show that the mapping is continuous.  Let $\bigcap_{\alpha \in F} \pi^{-1}_\alpha(O_\alpha)$ be an open set containing $f_\beta$ where $F \subset \omega_1$ is finite and $O_\alpha \subset X_\alpha$ is open for each $\alpha \in F$. Let $V=(\gamma,\beta]$ be an open interval in the order topology of $\omega_1$ such that $V$ misses $F-\lbrace{\beta}\rbrace$. For each $\delta \in V$, it is clear that $f_\delta \in \bigcap_{\alpha \in F} \pi^{-1}_\alpha(O_\alpha)$. Now, we show that the inverse is continuous. Consider the open interval $(\gamma,\beta]$. Note that $f_\beta \in \pi^{-1}_\gamma(U_\gamma) \cap \pi^{-1}_\beta(V_\beta)=O$. For each $f_\delta \in O$, we have $\delta \in (\gamma,\beta]$.

Example. The space $\displaystyle Z=\Sigma_{\alpha<\omega_1}Z_\alpha$ where each $Z_\alpha=\omega_1+1=[0,\omega_1]$ is not normal.

Let $p \in (\omega_1+1)^{\omega_1}$. Consider the $\Sigma-$product $\displaystyle Z=\Sigma_{\alpha<\omega_1}Z_\alpha$ about the fixed point $p$ where each $Z_\alpha=\omega_1+1=[0,\omega_1]$. Based on the observation above, $\displaystyle Z=\Sigma_{\alpha<\omega_1}Z_\alpha$ is homeomorphic to $\displaystyle Z_1=\Sigma_{0<\alpha<\omega_1}Z_\alpha \times (\omega_1+1)$. In turn $Z_1$ is homeomorphic to $\displaystyle Z_2=\Sigma_{\alpha<\omega_1}Z_\alpha \times (\omega_1+1)=Z \times (\omega_1+1)$. Based on the lemma, $Z$ contains a closed copy of $\omega_1$. It can be shown that $\omega_1 \times (\omega_1+1)$ is not normal (see a proof in this blog). Thus $Z \cong Z \times (\omega_1+1)$ contains the non-normal closed subspace $\omega_1 \times (\omega_1+1)$ and is thus not normal.

Theorem 1. Let $\lbrace{X_\alpha}\rbrace_{\alpha \in S}$ be a family of compact spaces. Then $\displaystyle X=\Sigma_{\alpha \in S}X_\alpha$ is countably compact.

Proof. Let $A \subset X=\Sigma_{\alpha \in S}X_\alpha$ be a countably infinite set. We show that $A$ has an accumulation point in $X$. For each $f \in A$, let $T_f \subset S$ be the countable set on which $f \neq$ the fixed point $p$. Since $A$ is countable, $T=\bigcup_{f \in A}T_f$ is countable. We can consider $A$ as a subspace of $Z=\Pi_{\alpha \in T}X_\alpha$. Since $Z$ is compact, $A$ has an accumulation point $\text{q}$ in $Z$. Extend $\text{q}$ by letting $q=p$ on $S-T$. It follows that $q \in X$ is an accumulation point of $A$.

Reference

1. Gul’ko, S. P., On the properties of sets lying in $\Sigma-$products, Dokl. Acad. Nauk. SSSR, 237, (1977) 505-508 (in Russian).
2. Rudin, M. E., $\Sigma-$products of metric spaces are normal, Preprint, 1977.

# 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 $E_{\mathcal{F}}:X \longrightarrow \Pi_{\alpha \in A}Y_\alpha$ as follows:

For each $x \in X$, $E_{\mathcal{F}}(x)$ is 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 $\mathcal{F}=\lbrace{f_\alpha:\alpha \in A}\rbrace$. If the family $\mathcal{F}$ is understood, we may skip the subscript and use $E$ to denote the evaluation map.

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 $E_{\mathcal{F}}:X \longrightarrow \Pi_{\alpha \in A}Y_\alpha$ 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 $f(x)=1$ and $f(y)=0$ 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 $I^{\aleph_0}$.

# The Vedenissoff Theorem

The purpose of this post is to present a proof of the Vedenissoff Theorem, a useful theorem that characterizes perfect normality. A space $X$ is perfectly normal if $X$ is normal and every closed subset of $X$ is a $G_\delta-$set. The Vedenissoff Theorem characterizes perfect normality in terms of zero sets and cozero sets. A subset $A$ of a space $X$ is a zero-set if there is a continuous $f:X \rightarrow I=[0,1]$ such that $A=f^{-1}(0)$. A set $A \subset X$ is a cozero-set if $X-A$ is a zero-set. The Vedenissoff Theorem were published in 1936 and 1940 (references found in [1]). All spaces are at least Hausdorff. The Urysohn’s Lemma is used as a basic tool to obtain continuous functions on a normal space and is stated without proof.

Urysohn’s Lemma. Let $X$ be a normal space. If $H \subset X$ and $K \subset X$ are disjoint closed sets, then there is a continuous $f:X \rightarrow I$ such that $f(x)=0$ for each $x \in H$ and $f(x)=1$ for each $x \in K$.

The Vedenissoff Theorem. Given a space $X$, the following conditions are equivalent.

1. The space $X$ is perfectly normal.
2. Every closed subset of $X$ is a zero-set.
3. Every open subset of $X$ is a cozero-set.
4. If $H \subset X$ and $K \subset X$ are disjoint closed sets, then there is a continuous $f:X \rightarrow I$ such that $H=f^{-1}(0)$ and $K=f^{-1}(1)$.

Proof. $1 \rightarrow 2$. Let $A \subset X$ be a closed set and let $A=\bigcap_{n<\omega}U_n$ where each $U_n$ is open. By the Urysohn’s Lemma, for each $n$, there is a continuous $f_n:X \rightarrow I$ such that $f_n(x)=0$ for each $x \in A$ and $f_n(y)=1$ for each $y \in X-U_n$. Then define a function $f:X \rightarrow I$ by $\displaystyle f(x)=\sum_{n<\omega} \frac{f_n(x)}{2^{n+1}}$. Clearly, $f(x)=0$ for each $x \in A$. For $x \notin A$, $x \in X-U_n$ for some $n$ and $f(x)> \frac{1}{2^n}$. Thus $A=f^{-1}(0)$. The remaining thing to show is that $f$ is continuous. Note that the sequence of continuous functions $\lbrace{g_n}\rbrace$ converges uniformly to the function $f$ where $\displaystyle g_n=\sum_{i=0}^n \frac{f_i}{2^{i+1}}$. This implies that the limit $f$ must be continuous.

$2 \rightarrow 3$ is clear.

$3 \rightarrow 4$. Let $H$ and $K$ be two disjoint closed subsets of $X$. There is a continuous $f:X \rightarrow I$ such that $f^{-1}(0)=H$. Likewise, there is a continuous $g:X \rightarrow I$ such that $g^{-1}(0)=K$. Define $\displaystyle h=\frac{f}{f+g}$. This is a continuous function. It is easy to verify that $h^{-1}(0)=H$ and $h^{-1}(1)=K$.

$4 \rightarrow 1$. Clearly, condition 4 implies that $X$ is normal. Now show that every closed set is $G_\delta$. Let $A$ be a proper closed subset (i.e., there is $z \in X-A$). Then we have continuous $f:X \rightarrow I$ such that $f^{-1}(0)=A$ and $f^{-1}(1)=\lbrace{z}\rbrace$. We have $\displaystyle A=\bigcap_{n<\omega}U_n$ where $U_n=f^{-1}([0,\frac{1}{2^{n+1}}))$.

Reference

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

# 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.