# An exercise gleaned from the proof of a theorem on pseudocompact space

Filling in the gap is something that is done often when following a proof in a research paper or other published work. In fact this is necessary since it is not feasible for authors to prove or justify every statement or assertion in a proof (or define every term). The gap could be a basic result or could be an older result from another source. If the gap is a basic result or a basic fact that is considered folklore, it may be OK to put it on hold in the interest of pursuing the main point. Then come back later to fill the gap. In any case, filling in gaps is a great learning opportunity. In this post, we focus on one such example of filling in the gap. The example is from the book called Topological Function Spaces by A. V. Arkhangelskii [1].

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Pseudocompactness

The exercise we wish to highlight deals with continuous one-to-one functions defined on pseudocompact spaces. We first give a brief backgrounder on pseudocompact spaces with links to earlier posts.

All spaces considered are Hausdorff spaces. A space $X$ is a pseudocompact space if every continuous real-valued function defined on $X$ is bounded, i.e., if $f:X \rightarrow \mathbb{R}$ is a continuous function, then $f(X)$ is a bounded set in the real line. Compact spaces are pseudocompact. In fact, it is clear from definitions that

$\text{compact} \Longrightarrow \text{countably compact} \Longrightarrow \text{pseudocompact}$

None of the implications can be reversed. An example of a pseudocompact space that is not countably compact is the space $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$ (see here for the details). Some basic results on pseudocompactness focus on the conditions to add in order to turn a pseudocompact space into countably compact or even compact. For example, for normal spaces, pseudocompact implies countably compact. This tells us that when looking for pseudocompact space that is not countably compact, do not look among normal spaces. Another interesting result is that pseudocompact + metacompact implies compact. Likewise, when looking for pseudocompact space that is not compact, look among non-metacompact spaces. On the other hand, this previous post discusses when a pseudocompact space is metrizable. Another two previous posts also discuss pseudocompactness (see here and here).

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The exercise

Consider Theorem II.6.2 part (c) in pp. 76-77 in [1]. We do not state the theorem because it is not the focus here. Instead, we focus on an assertion in the proof of Theorem II.6.2.

The exercise that we wish to highlight is stated in Theorem 2 below. Theorem 1 is a standard result about continuous one-to-one functions defined on compact spaces and is stated here to contrast with Theorem 2.

Theorem 1
Let $Y$ be a compact space. Let $g: Y \rightarrow Z$ be a one-to-one continuous function from $Y$ onto a space $Z$. Then $g$ is a homeomorphism.

Theorem 2
Let $Y$ be a pseudocompact space. Let $g: Y \rightarrow Z$ be a one-to-one continuous function from $Y$ onto $Z$ where $Z$ is a separable and metrizable space. Then $g$ is a homeomorphism.

Theorem 1 says that any continuous one-to-one map from a compact space onto another compact space is a homeomorphism. To show a given map between two compact spaces is a homeomorphism, we only need to show that it is continuous in one direction. Theorem 2, the statement used in the proof of Theorem II.6.2 in [1], says that the standard result for compact spaces can be generalized to pseudocompactness if the range space is nice.

The proof of Theorem II.6.2 part (c) in [1] quoted [2] as a source for the assertion in our Theorem 2. Here, we leave both Theorem 1 and Theorem 2 as exercise. One way to prove Theorem 2 is to show that whenever there exists a map $g$ as described in Theorem 2, the domain $Y$ must be compact. Then Theorem 1 will finish the job.

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Reference

1. Arkhangelskii A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Arkhangelskii A. V., Ponomarev V. I., Fundamental of general topology: problems and exercises, Reidel, 1984. (Translated from the Russian).

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$\copyright \ 2015 \text{ by Dan Ma}$

# Products of compact spaces with countable tightness

In the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.

The following theorems are the main results in this post.

Theorem 1
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is compact, then $X \times Y$ is countably tight.

Theorem 2
The product of finitely many compact countably tight spaces is countably tight.

Theorem 3
Suppose that $X_1, X_2, X_3, \cdots$ are countably many compact spaces such that each $X_i$ has at least two points. If each $X_i$ is a countably tight space, then the product space $\prod_{i=1}^\infty X_i$ is countably tight.

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Finite products

Before proving Theorem 1 and Theorem 2, we prove the following results.

Theorem 4
Let $f:Y_1 \rightarrow Y_2$ be a continuous and closed map from the space $Y_1$ onto the space $Y_2$. Suppose that the space $Y_2$ is countably tight and that each fiber of the map $f$ is countably tight. Then the space $Y_1$ is countably tight.

Proof of Theorem 4
Let $x \in Y_1$ and $x \in \overline{A}$ where $A \subset Y_1$. We proceed to find a countable $W \subset Y_1$ such that $x \in \overline{W}$. Choose $y \in Y_2$ such that $y=f(x)$.

Let $M$ be the fiber of the map $f$ at the point $y$, i.e. $M=f^{-1}(y)$. By assumption, $M$ is countably tight. Call a point $w \in M$ countably reached by $A$ if there is some countable $C \subset A$ such that $w \in \overline{C}$. Let $G$ be the set of all points in $M$ that are countably reached by $A$. We claim that $x \in \overline{G}$.

Let $U \subset Y_1$ be open such that $x \in U$. Because the space $Y_1$ is regular, choose open $V \subset U$ such that $x \in V$ and $\overline{V} \subset U$. Then $V \cap A \ne \varnothing$. Furthermore, $x \in \overline{V \cap A}$. Let $C=f(V \cap A)$. By the continuity of $f$, we have $y \in \overline{C}$. Since $Y_2$ is countably tight, there exists some countable $D \subset C$ such that $y \in \overline{D}$. Choose a countable $E \subset V \cap A$ such that $f(E)=D$. It follows that $y \in \overline{f(E)}$.

We show that that $\overline{E} \cap M \ne \varnothing$. Since $E \subset \overline{E}$, we have $f(E) \subset f(\overline{E})$. Note that $f(\overline{E})$ is a closed set since $f$ is a closed map. Thus $\overline{f(E)} \subset f(\overline{E})$. As a result, $y \in f(\overline{E})$. Then $y=f(t)$ for some $t \in \overline{E}$. We have $t \in \overline{E} \cap M$.

By the definition of the set $G$, we have $\overline{E} \cap M \subset G$. Furthermore, $\overline{E} \cap M \subset \overline{V} \subset U$. Note that the arbitrary open neighborhood $U$ of $x$ contains points of $G$. This establishes the claim that $x \in \overline{G}$.

Since $M$ is a fiber of $f$, $M$ is countably tight by assumption. Choose some countable $T \subset G$ such that $x \in \overline{T}$. For each $t \in T$, choose a countable $W_t \subset A$ with $t \in \overline{W_t}$. Let $W=\bigcup_{t \in T} W_t$. Note that $W \subset A$ and $W$ is countable with $x \in \overline{W}$. This establishes the space $Y_1$ is countably tight at $x \in Y_1$. $\blacksquare$

Lemma 5
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a compact space, then $f$ is a closed map.

Proof of Lemma 5
Let $A$ be a closed subset of $X \times Y$. Suppose that $f(A)$ is not closed. Let $y \in \overline{f(A)}-f(A)$. It follows that no point of $X \times \left\{y \right\}$ belongs to $A$. For each $x \in X$, choose open subset $O_x$ of $X \times Y$ such that $(x,y) \in O_x$ and $O_x \cap A=\varnothing$. The set of all $O_x$ is an open cover of the compact space $X \times \left\{y \right\}$. Then there exist finitely many $O_x$ that cover $X \times \left\{y \right\}$, say $O_{x_i}$ for $i=1,2,\cdots,n$.

Let $W=\bigcup_{i=1}^n O_{x_i}$. We have $X \times \left\{y \right\} \subset W$. Since $X$ is compact, we can then use the Tube Lemma which implies that there exists open $G \subset Y$ such that $X \times \left\{y \right\} \subset X \times G \subset W$. It follows that $G \cap f(A) \ne \varnothing$. Choose $t \in G \cap f(A)$. Then for some $x \in X$, $(x,t) \in A$. Since $t \in G$, $(x,t) \in W$, implying that $W \cap A \ne \varnothing$, a contradiction. Thus $f(A)$ must be a closed set in $Y$. This completes the proof of the lemma. $\blacksquare$

Proof of Theorem 1
Let $X$ be the factor that is compact. let $f: X \times Y \rightarrow Y$ be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space $Y$ is countably tight by assumption. Each fiber of the projection map $f$ is of the form $X \times \left\{y \right\}$ where $y \in Y$, which is countably tight. Then use Theorem 4 to establish that $X \times Y$ is countably tight. $\blacksquare$

Proof of Theorem 2
This is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight. $\blacksquare$

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Countable products

Our proof to establish that the product space $\prod_{i=1}^\infty X_i$ is countably tight is an indirect one and makes use of two non-trivial results. We first show that $\omega_1 \times \prod_{i=1}^\infty X_i$ is a closed subspace of a $\Sigma$-product that is normal. It follows from another result that the second factor $\prod_{i=1}^\infty X_i$ is countably tight. We now present all the necessary definitions and theorems.

Consider a product space $Y=\prod_{\alpha<\kappa} Y_\alpha$ where $\kappa$ is an infinite cardinal number. Fix a point $p \in Y$. The $\Sigma$-product of the spaces $Y_\alpha$ with $p$ as the base point is the following subspace of the product space $Y=\prod_{\alpha<\kappa} Y_\alpha$:

$\displaystyle \Sigma_{\alpha<\kappa} Y_\alpha=\left\{y \in \prod_{\alpha<\kappa} Y_\alpha: y_\alpha \ne p_\alpha \text{ for at most countably many } \alpha < \kappa \right\}$

The definition of the space $\Sigma_{\alpha<\kappa} Y_\alpha$ depends on the base point $p$. The discussion here is on properties of $\Sigma_{\alpha<\kappa} Y_\alpha$ that hold regardless of the choice of base point. If the factor spaces are indexed by a set $A$, the notation is $\Sigma_{\alpha \in A} Y_\alpha$.

If all factors $Y_\alpha$ are identical, say $Y_\alpha=Z$ for all $\alpha$, then we use the notation $\Sigma_{\alpha<\kappa} Z$ to denote the $\Sigma$-product. Once useful fact is that if there are $\omega_1$ many factors and each factor has at least 2 points, then the space $\omega_1$ can be embedded as a closed subspace of the $\Sigma$-product.

Theorem 6
For each $\alpha<\omega_1$, let $Y_\alpha$ be a space with at least two points. Then $\Sigma_{\alpha<\omega_1} Y_\alpha$ contains $\omega_1$ as a closed subspace. See Exercise 3 in this previous post.

Now we discuss normality of $\Sigma$-products. This previous post shows that if each factor is a separable metric space, then the $\Sigma$-product is normal. It is also well known that if each factor is a metric space, the $\Sigma$-product is normal. The following theorem handles the case where each factor is a compact space.

Theorem 7
For each $\alpha<\kappa$, let $Y_\alpha$ be a compact space. Then the $\Sigma$-product $\Sigma_{\alpha<\kappa} Y_\alpha$ is normal if and only if each factor $Y_\alpha$ is countably tight.

Theorem 7 is Theorem 7.5 in page 821 of [1]. Theorem 7.5 in [1] is stated in a more general setting where each factor of the $\Sigma$-product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.

Theorem 8
Let $Y$ be a compact space. Then the product space $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight.

We now prove Theorem 3.

Proof of Theorem 3
Let $\omega_1=\cup \left\{A_n: n \in \omega \right\}$, where for each $n$, $\lvert A_n \lvert=\omega_1$ and that $A_n \cap A_m=\varnothing$ if $n \ne m$. For each $n=1,2,3,\cdots$, let $S_n=\Sigma_{\alpha \in A_n} X_n$. By Theorem 7, each $S_n$ is normal. Let $S_0=\Sigma_{\alpha \in A_0} X_1$, which is also normal. By Theorem 6, the space $\omega_1$ of countable ordinals is a closed subspace of $S_0$. Let $T=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots$. We have the following derivation.

\displaystyle \begin{aligned} T&=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots \\&\subset S_0 \times S_1 \times S_2 \times S_3 \times \cdots \\&\cong W=\Sigma_{\alpha<\omega_1} W_\alpha \end{aligned}

Recall that $\omega_1=\cup \left\{A_n: n \in \omega \right\}$. The space $W=\Sigma_{\alpha<\omega_1} W_\alpha$ is defined such that for each $n \ge 1$ and for each $\alpha \in A_n$, $W_\alpha=X_n$. Furthermore, for $n=0$, for each $\alpha \in A_0$, let $W_\alpha=X_1$. Thus $W$ is a $\Sigma$-product of compact countably tight spaces and is thus normal by Theorem 7. The space $T=\omega_1 \times \prod_{n=1}^\infty X_n$ is a closed subspace of the normal space $W$. By Theorem 8, the product space $\prod_{n=1}^\infty X_n$ must be countably tight. $\blacksquare$

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Remarks

Theorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let $X=X_1 \times X_2 \times \cdots \times X_n$.

\displaystyle \begin{aligned} X&=X_1 \times X_2 \times \cdots \times X_n \\&\cong X_1 \times X_2 \times \cdots \times X_n \times \left\{t_{n+1} \right\} \times \left\{t_{n+2} \right\} \times \cdots \\&\subset X_1 \times X_2 \times \cdots \times X_n \times X_{n+1} \times X_{n+2} \times \cdots \end{aligned}

In the above derivation, $t_m$ is a point of $X_m$ for all $m >n$. When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.

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Exercise

Exercise 1
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a countably compact space and $Y$ is a Frechet space, then $f$ is a closed map.

Exercise 2
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is a countably compact space and the other space is a Frechet space, then $X \times Y$ is countably tight.

Exercise 2 is a variation of Theorem 1. One factor is weakened to “countably compact”. However, the other factor is strengthened to “Frechet”.

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Reference

1. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Normality in the powers of countably compact spaces

Let $\omega_1$ be the first uncountable ordinal. The topology on $\omega_1$ we are interested in is the ordered topology, the topology induced by the well ordering. The space $\omega_1$ is also called the space of all countable ordinals since it consists of all ordinals that are countable in cardinality. It is a handy example of a countably compact space that is not compact. In this post, we consider normality in the powers of $\omega_1$. We also make comments on normality in the powers of a countably compact non-compact space.

Let $\omega$ be the first infinite ordinal. It is well known that $\omega^{\omega_1}$, the product space of $\omega_1$ many copies of $\omega$, is not normal (a proof can be found in this earlier post). This means that any product space $\prod_{\alpha<\kappa} X_\alpha$, with uncountably many factors, is not normal as long as each factor $X_\alpha$ contains a countable discrete space as a closed subspace. Thus in order to discuss normality in the product space $\prod_{\alpha<\kappa} X_\alpha$, the interesting case is when each factor is infinite but contains no countable closed discrete subspace (i.e. no closed copies of $\omega$). In other words, the interesting case is that each factor $X_\alpha$ is a countably compact space that is not compact (see this earlier post for a discussion of countably compactness). In particular, we would like to discuss normality in $X^{\kappa}$ where $X$ is a countably non-compact space. In this post we start with the space $X=\omega_1$ of the countable ordinals. We examine $\omega_1$ power $\omega_1^{\omega_1}$ as well as the countable power $\omega_1^{\omega}$. The former is not normal while the latter is normal. The proof that $\omega_1^{\omega}$ is normal is an application of the normality of $\Sigma$-product of the real line.

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The uncountable product

Theorem 1
The product space $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal.

Theorem 1 follows from Theorem 2 below. For any space $X$, a collection $\mathcal{C}$ of subsets of $X$ is said to have the finite intersection property if for any finite $\mathcal{F} \subset \mathcal{C}$, the intersection $\cap \mathcal{F} \ne \varnothing$. Such a collection $\mathcal{C}$ is called an f.i.p collection for short. It is well known that a space $X$ is compact if and only collection $\mathcal{C}$ of closed subsets of $X$ satisfying the finite intersection property has non-empty intersection (see Theorem 1 in this earlier post). Thus any non-compact space has an f.i.p. collection of closed sets that have empty intersection.

In the space $X=\omega_1$, there is an f.i.p. collection of cardinality $\omega_1$ using its linear order. For each $\alpha<\omega_1$, let $C_\alpha=\left\{\beta<\omega_1: \alpha \le \beta \right\}$. Let $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$. It is a collection of closed subsets of $X=\omega_1$. It is an f.i.p. collection and has empty intersection. It turns out that for any countably compact space $X$ with an f.i.p. collection of cardinality $\omega_1$ that has empty intersection, the product space $X^{\omega_1}$ is not normal.

Theorem 2
Let $X$ be a countably compact space. Suppose that there exists a collection $\mathcal{C}=\left\{C_\alpha: \alpha < \omega_1 \right\}$ of closed subsets of $X$ such that $\mathcal{C}$ has the finite intersection property and that $\mathcal{C}$ has empty intersection. Then the product space $X^{\omega_1}$ is not normal.

Proof of Theorem 2
Let’s set up some notations on product space that will make the argument easier to follow. By a standard basic open set in the product space $X^{\omega_1}=\prod_{\alpha<\omega_1} X$, we mean a set of the form $O=\prod_{\alpha<\omega_1} O_\alpha$ such that each $O_\alpha$ is an open subset of $X$ and that $O_\alpha=X$ for all but finitely many $\alpha<\omega_1$. Given a standard basic open set $O=\prod_{\alpha<\omega_1} O_\alpha$, the notation $\text{Supp}(O)$ refers to the finite set of $\alpha$ for which $O_\alpha \ne X$. For any set $M \subset \omega_1$, the notation $\pi_M$ refers to the projection map from $\prod_{\alpha<\omega_1} X$ to the subproduct $\prod_{\alpha \in M} X$. Each element $d \in X^{\omega_1}$ can be considered a function $d: \omega_1 \rightarrow X$. By $(d)_\alpha$, we mean $(d)_\alpha=d(\alpha)$.

For each $t \in X$, let $f_t: \omega_1 \rightarrow X$ be the constant function whose constant value is $t$. Consider the following subspaces of $X^{\omega_1}$.

$H=\prod_{\alpha<\omega_1} C_\alpha$

$\displaystyle K=\left\{f_t: t \in X \right\}$

Both $H$ and $K$ are closed subsets of the product space $X^{\omega_1}$. Because the collection $\mathcal{C}$ has empty intersection, $H \cap K=\varnothing$. We show that $H$ and $K$ cannot be separated by disjoint open sets. To this end, let $U$ and $V$ be open subsets of $X^{\omega_1}$ such that $H \subset U$ and $K \subset V$.

Let $d_1 \in H$. Choose a standard basic open set $O_1$ such that $d_1 \in O_1 \subset U$. Let $S_1=\text{Supp}(O_1)$. Since $S_1$ is the support of $O_1$, it follows that $\pi_{S_1}^{-1}(\pi_{S_1}(d_1)) \subset O_1 \subset U$. Since $\mathcal{C}$ has the finite intersection property, there exists $a_1 \in \bigcap_{\alpha \in S_1} C_\alpha$.

Define $d_2 \in H$ such that $(d_2)_\alpha=a_1$ for all $\alpha \in S_1$ and $(d_2)_\alpha=(d_1)_\alpha$ for all $\alpha \in \omega_1-S_1$. Choose a standard basic open set $O_2$ such that $d_2 \in O_2 \subset U$. Let $S_2=\text{Supp}(O_2)$. It is possible to ensure that $S_1 \subset S_2$ by making more factors of $O_2$ different from $X$. We have $\pi_{S_2}^{-1}(\pi_{S_2}(d_2)) \subset O_2 \subset U$. Since $\mathcal{C}$ has the finite intersection property, there exists $a_2 \in \bigcap_{\alpha \in S_2} C_\alpha$.

Now choose a point $d_3 \in H$ such that $(d_3)_\alpha=a_2$ for all $\alpha \in S_2$ and $(d_3)_\alpha=(d_2)_\alpha$ for all $\alpha \in \omega_1-S_2$. Continue on with this inductive process. When the inductive process is completed, we have the following sequences:

• a sequence $d_1,d_2,d_3,\cdots$ of point of $H=\prod_{\alpha<\omega_1} C_\alpha$,
• a sequence $S_1 \subset S_2 \subset S_3 \subset \cdots$ of finite subsets of $\omega_1$,
• a sequence $a_1,a_2,a_3,\cdots$ of points of $X$

such that for all $n \ge 2$, $(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_{n-1}$ and $\pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$. Let $A=\left\{a_1,a_2,a_3,\cdots \right\}$. Either $A$ is finite or $A$ is infinite. Let’s examine the two cases.

Case 1
Suppose that $A$ is infinite. Since $X$ is countably compact, $A$ has a limit point $a$. That means that every open set containing $a$ contains some $a_n \ne a$. For each $n \ge 2$, define $y_n \in \prod_{\alpha< \omega_1} X$ such that

• $(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$,
• $(y_n)_\alpha=a$ for all $\alpha \in \omega_1-S_n$

From the induction step, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$. Let $t=f_a \in K$, the constant function whose constant value is $a$. It follows that $t$ is a limit of $\left\{y_1,y_2,y_3,\cdots \right\}$. This means that $t \in \overline{U}$. Since $t \in K \subset V$, $U \cap V \ne \varnothing$.

Case 2
Suppose that $A$ is finite. Then there is some $m$ such that $a_m=a_j$ for all $j \ge m$. For each $n \ge 2$, define $y_n \in \prod_{\alpha< \omega_1} X$ such that

• $(y_n)_\alpha=(d_n)_\alpha=a_{n-1}$ for all $\alpha \in S_n$,
• $(y_n)_\alpha=a_m$ for all $\alpha \in \omega_1-S_n$

As in Case 1, we have $y_n \in \pi_{S_n}^{-1}(\pi_{S_n}(d_n)) \subset U$ for all $n$. Let $t=f_{a_m} \in K$, the constant function whose constant value is $a_m$. It follows that $t=y_n$ for all $n \ge m+1$. Thus $U \cap V \ne \varnothing$.

Both cases show that $U \cap V \ne \varnothing$. This completes the proof the product space $X^{\omega_1}$ is not normal. $\blacksquare$

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The countable product

Theorem 3
The product space $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is normal.

Proof of Theorem 3
The proof here actually proves more than normality. It shows that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is collectionwise normal, which is stronger than normality. The proof makes use of the $\Sigma$-product of $\kappa$ many copies of $\mathbb{R}$, which is the following subspace of the product space $\mathbb{R}^{\kappa}$.

$\Sigma(\kappa)=\left\{x \in \mathbb{R}^{\kappa}: x(\alpha) \ne 0 \text{ for at most countably many } \alpha<\kappa \right\}$

It is well known that $\Sigma(\kappa)$ is collectionwise normal (see this earlier post). We show that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$ is a closed subspace of $\Sigma(\kappa)$ where $\kappa=\omega_1$. Thus $\omega_1^{\omega}$ is collectionwise normal. This is established in the following claims.

Claim 1
We show that the space $\omega_1$ is embedded as a closed subspace of $\Sigma(\omega_1)$.

For each $\beta<\omega_1$, define $f_\beta:\omega_1 \rightarrow \mathbb{R}$ such that $f_\beta(\gamma)=1$ for all $\gamma<\beta$ and $f_\beta(\gamma)=0$ for all $\beta \le \gamma <\omega_1$. Let $W=\left\{f_\beta: \beta<\omega_1 \right\}$. We show that $W$ is a closed subset of $\Sigma(\omega_1)$ and $W$ is homeomorphic to $\omega_1$ according to the mapping $f_\beta \rightarrow W$.

First, we show $W$ is closed by showing that $\Sigma(\omega_1)-W$ is open. Let $y \in \Sigma(\omega_1)-W$. We show that there is an open set containing $y$ that contains no points of $W$.

Suppose that for some $\gamma<\omega_1$, $y_\gamma \in O=\mathbb{R}-\left\{0,1 \right\}$. Consider the open set $Q=(\prod_{\alpha<\omega_1} Q_\alpha) \cap \Sigma(\omega_1)$ where $Q_\alpha=\mathbb{R}$ except that $Q_\gamma=O$. Then $y \in Q$ and $Q \cap W=\varnothing$.

So we can assume that for all $\gamma<\omega_1$, $y_\gamma \in \left\{0, 1 \right\}$. There must be some $\theta$ such that $y_\theta=1$. Otherwise, $y=f_0 \in W$. Since $y \ne f_\theta$, there must be some $\delta<\gamma$ such that $y_\delta=0$. Now choose the open interval $T_\theta=(0.9,1.1)$ and the open interval $T_\delta=(-0.1,0.1)$. Consider the open set $M=(\prod_{\alpha<\omega_1} M_\alpha) \cap \Sigma(\omega_1)$ such that $M_\alpha=\mathbb{R}$ except for $M_\theta=T_\theta$ and $M_\delta=T_\delta$. Then $y \in M$ and $M \cap W=\varnothing$. We have just established that $W$ is closed in $\Sigma(\omega_1)$.

Consider the mapping $f_\beta \rightarrow W$. Based on how it is defined, it is straightforward to show that it is a homeomorphism between $\omega_1$ and $W$.

Claim 2
The $\Sigma$-product $\Sigma(\omega_1)$ has the interesting property it is homeomorphic to its countable power, i.e.

$\Sigma(\omega_1) \cong \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \ \ \ \ \ \ \ \ \ \ \ \text{(countably many times)}$.

Because each element of $\Sigma(\omega_1)$ is nonzero only at countably many coordinates, concatenating countably many elements of $\Sigma(\omega_1)$ produces an element of $\Sigma(\omega_1)$. Thus Claim 2 can be easily verified. With above claims, we can see that

$\displaystyle \omega_1^{\omega}=\omega_1 \times \omega_1 \times \omega_1 \times \cdots \subset \Sigma(\omega_1) \times \Sigma(\omega_1) \times \Sigma(\omega_1) \cdots \cong \Sigma(\omega_1)$

Thus $\omega_1^{\omega}$ is a closed subspace of $\Sigma(\omega_1)$. Any closed subspace of a collectionwise normal space is collectionwise normal. We have established that $\omega_1^{\omega}$ is normal. $\blacksquare$

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The normality in the powers of $X$

We have established that $\prod_{\alpha<\omega_1} \omega_1=\omega_1^{\omega_1}$ is not normal. Hence any higher uncountable power of $\omega_1$ is not normal. We have also established that $\prod_{\alpha<\omega} \omega_1=\omega_1^{\omega}$, the countable power of $\omega_1$ is normal (in fact collectionwise normal). Hence any finite power of $\omega_1$ is normal. However $\omega_1^{\omega}$ is not hereditarily normal. One of the exercises below is to show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Theorem 2 can be generalized as follows:

Theorem 4
Let $X$ be a countably compact space has an f.i.p. collection $\mathcal{C}$ of closed sets such that $\bigcap \mathcal{C}=\varnothing$. Then $X^{\kappa}$ is not normal where $\kappa=\lvert \mathcal{C} \lvert$.

The proof of Theorem 2 would go exactly like that of Theorem 2. Consider the following two theorems.

Theorem 5
Let $X$ be a countably compact space that is not compact. Then there exists a cardinal number $\kappa$ such that $X^{\kappa}$ is not normal and $X^{\tau}$ is normal for all cardinal number $\tau<\kappa$.

By the non-compactness of $X$, there exists an f.i.p. collection $\mathcal{C}$ of closed subsets of $X$ such that $\bigcap \mathcal{C}=\varnothing$. Let $\kappa$ be the least cardinality of such an f.i.p. collection. By Theorem 4, that $X^{\kappa}$ is not normal. Because $\kappa$ is least, any smaller power of $X$ must be normal.

Theorem 6
Let $X$ be a space that is not countably compact. Then $X^{\kappa}$ is not normal for any cardinal number $\kappa \ge \omega_1$.

Since the space $X$ in Theorem 6 is not countably compact, it would contain a closed and discrete subspace that is countable. By a theorem of A. H. Stone, $\omega^{\omega_1}$ is not normal. Then $\omega^{\omega_1}$ is a closed subspace of $X^{\omega_1}$.

Thus between Theorem 5 and Theorem 6, we can say that for any non-compact space $X$, $X^{\kappa}$ is not normal for some cardinal number $\kappa$. The $\kappa$ from either Theorem 5 or Theorem 6 is at least $\omega_1$. Interestingly for some spaces, the $\kappa$ can be much smaller. For example, for the Sorgenfrey line, $\kappa=2$. For some spaces (e.g. the Michael line), $\kappa=\omega$.

Theorems 4, 5 and 6 are related to a theorem that is due to Noble.

Theorem 7 (Noble)
If each power of a space $X$ is normal, then $X$ is compact.

A proof of Noble’s theorem is given in this earlier post, the proof of which is very similar to the proof of Theorem 2 given above. So the above discussion the normality of powers of $X$ is just another way of discussing Theorem 7. According to Theorem 7, if $X$ is not compact, some power of $X$ is not normal.

The material discussed in this post is excellent training ground for topology. Regarding powers of countably compact space and product of countably compact spaces, there are many topics for further discussion/investigation. One possibility is to examine normality in $X^{\kappa}$ for more examples of countably compact non-compact $X$. One particular interesting example would be a countably compact non-compact $X$ such that the least power $\kappa$ for non-normality in $X^{\kappa}$ is more than $\omega_1$. A possible candidate could be the second uncountable ordinal $\omega_2$. By Theorem 2, $\omega_2^{\omega_2}$ is not normal. The issue is whether the $\omega_1$ power $\omega_2^{\omega_1}$ and countable power $\omega_2^{\omega}$ are normal.

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Exercises

Exercise 1
Show that $\omega_1 \times \omega_1$ is not hereditarily normal.

Exercise 2
Show that the mapping $f_\beta \rightarrow W$ in Claim 3 in the proof of Theorem 3 is a homeomorphism.

Exercise 3
The proof of Theorem 3 shows that the space $\omega_1$ is a closed subspace of the $\Sigma$-product of the real line. Show that $\omega_1$ can be embedded in the $\Sigma$-product of arbitrary spaces.

For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Let $p \in \prod_{\alpha<\omega_1} X_\alpha$. The $\Sigma$-product of the spaces $X_\alpha$ is the following subspace of the product space $\prod_{\alpha<\omega_1} X_\alpha$.

$\Sigma(X_\alpha)=\left\{x \in \prod_{\alpha<\omega_1} X_\alpha: x(\alpha) \ne p(\alpha) \ \text{for at most countably many } \alpha<\omega_1 \right\}$

The point $p$ is the center of the $\Sigma$-product. Show that the space $\Sigma(X_\alpha)$ contains $\omega_1$ as a closed subspace.

Exercise 4
Find a direct proof of Theorem 3, that $\omega_1^{\omega}$ is normal.

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$\copyright \ 2015 \text{ by Dan Ma}$

# The product of uncountably many factors is never hereditarily normal

The space $Y=\prod_{\alpha<\omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1}$ is the product of $\omega_1$ many copies of the two-element set $\left\{0,1 \right\}$ where $\omega_1$ is the first uncountable ordinal. It is a compact space by Tychonoff’s theorem. It is a normal space since every compact Hausdorff space is normal. A space is hereditarily normal if every subspace is normal. Is the space $Y$ hereditarily normal? In this post, we give two proofs that it is not hereditarily normal. It then follows that any product space $\prod X_\alpha$ cannot be hereditarily normal as long as there are uncountably many factors and every factor has at least two point.

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The connection with a theorem of Katetov

It turns out that there is a connection with a theorem of Katetov. For any compact space, knowing hereditary normality of the first several self product spaces can reveal a great deal of information about the compact space. More specifically, for any compact space $X$, knowing whether $X$, $X^2$ and $X^3$ are hereditarily normal can tell us whether $X$ is metrizable. If all three are hereditarily normal, then $X$ is metrizable. If one of the three self products is not hereditarily normal, then $X$ is not metrizable. This fact is based on a theorem of Katetov (see this previous post). The space $Y=\left\{0,1 \right\}^{\omega_1}$ is not metrizable since it is not first countable (see Problem 1 below). Thus one of its first three self products must fail to be hereditarily normal.

These two proofs are not direct proof in the sense that a non-normal subspace is not explicitly produced. Instead the proofs use other theorem or basic but important background results. One of the two proofs (#2) uses a theorem of Katetov on hereditarily normal spaces. The other proof (#1) uses the fact that the product of uncountably many copies of a countable discrete space is not normal. We believe that these two proofs and the required basic facts are an important training ground for topology. We list out these basic facts as exercises. Anyone who wishes to fill in the gaps can do so either by studying the links provided or by consulting other sources.

The theorem of Katetov mentioned earlier provides a great exercise – for any non-metrizable compact space $X$, determine where the hereditary normality fails. Does it fail in $X$, $X^2$ or $X^3$? This previous post examines a small list of compact non-metrizable spaces. In all the examples in this list, the hereditary normality fails in $X$ or $X^2$. The space $Y=\left\{0,1 \right\}^{\omega_1}$ can be added to this list. All the examples in this list are defined using no additional set theory axioms beyond ZFC. A natural question: does there exist an example of compact non-metrizable space $X$ such that the hereditary normality holds in $X^2$ and fails in $X^3$? It turns out that this was a hard problem and the answer is independent of ZFC. This previous post provides a brief discussion and has references for the problem.

All spaces under consideration are Hausdorff spaces.

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Exercises

Problem 1
Let $X$ be a compact space. Show that $X$ is normal.

Problem 2
For each $\alpha<\omega_1$, let $A_\alpha$ be a set with cardinality $\le \omega_1$. Show that $\lvert \bigcup_{\alpha<\omega_1} A_\alpha \lvert \le \omega_1$.

Problem 2 holds for any infinite cardinal, not just $\omega_1$. One reference for Problem 2 is Lemma 10.21 on page 30 of Set Theorey, An Introduction to Independence Proofs by Kenneth Kunen.

Problem 3
For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Show that for every point $p \in \prod_{\alpha<\omega_1} X_\alpha$, there does not exist a countable base at the point $p$. In other words, the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not first countable at every point. It follows that product space $\prod_{\alpha<\omega_1} X_\alpha$ is not metrizable.

Problem 4
In any space, a $G_\delta$-set is a set that is the intersection of countably many open sets. When a singleton set $\left\{ x \right\}$ is a $G_\delta$-set, we say the point $x$ is a $G_\delta$-point. For each $\alpha<\omega_1$, let $X_\alpha$ be a space with at least two points. Show that every point $p$ in the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not a $G_\delta$-point.

Note that Problem 4 implies Problem 3.

For Problem 3 and Problem 4, use the fact that there are uncountably many factors and that a basic open set in the product space is of the form $\prod_{\alpha<\omega_1} O_\alpha$ and that it has only finitely many coordinates at which $O_\alpha \ne X_\alpha$.

Problem 5
For each $\alpha<\omega_1$, let $X_\alpha=\left\{0,1,2,\cdots \right\}$ be the set of non-negative integers with the discrete topology. Show that the product space $\prod_{\alpha<\omega_1} X_\alpha$ is not normal.

See here for a discussion of Problem 5.

Problem 6
Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. Show that $Y$ has a countably infinite subspace

$W=\left\{y_0,y_1,y_2,y_3\cdots \right\}$

such that $W$ is relatively discrete. In other words, $W$ is discrete in the subspace topology of $W$. However $W$ is not discrete in the product space $Y$ since $Y$ is compact.

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Proof #1

Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. We show that $Y$ is not hereditarily normal.

Note that the product space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$ can be written as the product of $\omega_1$ many copies of itself:

$\displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

The fact (1) follows from the fact that the union of $\omega_1$ many pairwise disjoint sets, each of which has cardinality $\omega_1$, has cardinality $\omega_1$ (see Problem 2). The space $\left\{0,1 \right\}^{\omega_1}$ has a countably infinite subspace that is relatively discrete (see Problem 6). In other words, it has a subspace that is homemorphic to $\omega=\left\{0,1,2,\cdots \right\}$ where $\omega$ has the discrete topology. Thus the following is homeomorphic to a subspace of $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$.

$\displaystyle \omega^{\omega_1} = \omega \times \omega \times \omega \times \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

By Problem 5, the space $\omega^{\omega_1}$ is not normal. Hence the compact space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$ contains the non-normal space $\omega^{\omega_1}$ and is thus not hereditarily normal. $\blacksquare$

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Proof #2

Let $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. We show that $Y$ is not hereditarily normal. This proof uses a theorem of Katetov, discussed in this previous post and stated below.

Theorem 1
If $X_1 \times X_2$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

• The factor $X_1$ is perfectly normal.
• Every countable and infinite subset of the factor $X_2$ is closed.

First, $Y$ can be written as the product of two copies of itself:

$\displaystyle \left\{0,1 \right\}^{\omega_1} \cong \left\{0,1 \right\}^{\omega_1} \times \left\{0,1 \right\}^{\omega_1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

This is because the union of two disjoints sets, each of which has cardinality $\omega_1$, has carinality $\omega_1$. Note that the countably infinite subset $W$ from Problem 6 is not a closed subset of $Y$. If it were, the compact space $Y$ would contain an infinite set with no limit point. Thus the second condition of Theorem 1 is not satisfied. If $Y \cong Y \times Y$ were to be hereditarily normal, then the first condition must be satisfied, i.e. $Y$ is perfectly normal (meaning that $Y$ is normal and that every closed subset of it is a $G_\delta$-set). However, Problem 4 indicates that no point in $Y$ can be a $G_\delta$ point. Therefore $Y$ cannot be hereditarily normal. $\blacksquare$

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Corollary

The product of uncountably many spaces, each one of which has at least two points, contains a homeomorphic copy of the space $\displaystyle Y=\left\{0,1 \right\}^{\omega_1}$. Thus such a product space can never be hereditarily normal. We state this more formally below.

Theorem 2
Let $\kappa$ be any uncountable cardinal. For each $\alpha<\kappa$, let $X_\alpha$ be a space with at least two points. Then $\prod_{\alpha<\kappa} X_\alpha$ is not hereditarily normal.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Looking for non-normal subspaces of the square of a compact X

A theorem of Katetov states that if $X$ is compact with a hereditarily normal cube $X^3$, then $X$ is metrizable (discussed in this previous post). This means that for any non-metrizable compact space $X$, Katetov’s theorem guarantees that some subspace of the cube $X^3$ is not normal. Where can a non-normal subspace of $X^3$ be found? Is it in $X$, in $X^2$ or in $X^3$? In other words, what is the “dimension” in which the hereditary normality fails for a given compact non-metrizable $X$ (1, 2 or 3)? Katetov’s theorem guarantees that the dimension must be at most 3. Out of curiosity, we gather a few compact non-metrizable spaces. They are discussed below. In this post, we motivate an independence result using these examples.

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Katetov’s theorems

First we state the results of Katetov for reference. These results are proved in this previous post.

Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

• The factor $X$ is perfectly normal.
• Every countable and infinite subset of the factor $Y$ is closed.

Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.

Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.

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Examples of compact non-metrizable spaces

The set-theoretic result presented here is usually motivated by looking at Theorem 3. The question is: Can $X^3$ in Theorem 3 be replaced by $X^2$? We take a different angle of looking at some standard compact non-metric spaces and arrive at the same result. The following is a small listing of compact non-metrizable spaces. Each example in this list is defined in ZFC alone, i.e. no additional axioms are used beyond the generally accepted axioms of set theory.

1. One-point compactification of the Tychonoff plank.
2. One-point compactification of $\psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$.
3. The first compact uncountable ordinal, i.e. $\omega_1+1$.
4. The one-point compactification of an uncountable discrete space.
5. Alexandroff double circle.
6. Double arrow space.
7. Unit square with the lexicographic order.

Since each example in the list is compact and non-metrizable, the cube of each space must not be hereditarily normal according to Theorem 3 above. Where does the hereditary normality fail? For #1 and #2, $X$ is a compactification of a non-normal space and thus not hereditarily normal. So the dimension for the failure of hereditary normality is 1 for #1 and #2.

For #3 through #7, $X$ is hereditarily normal. For #3 through #5, each $X$ has a closed subset that is not a $G_\delta$ set (hence not perfectly normal). In #3 and #4, the non-$G_\delta$-set is a single point. In #5, the the non-$G_\delta$-set is the inner circle. Thus the compact space $X$ in #3 through #5 is not perfectly normal. By Theorem 2, the dimension for the failure of hereditary normality is 2 for #3 through #5.

For #6 and #7, each $X^2$ contains a copy of the Sorgenfrey plane. Thus the dimension for the failure of hereditary normality is also 2 for #6 and #7.

In the small sample of compact non-metrizable spaces just highlighted, the failure of hereditary normality occurs in “dimension” 1 or 2. Naturally, one can ask:

Question. Is there an example of a compact non-metrizable space $X$ such that the failure of hereditary nornmality occurs in “dimension” 3? Specifically, is there a compact non-metrizable $X$ such that $X^2$ is hereditarily normal but $X^3$ is not hereditarily normal?

Such a space $X$ would be an example to show that the condition “$X^3$ is hereditarily normal” in Theorem 3 is necessary. In other words, the hypothesis in Theorem 3 cannot be weakened if the example just described were to exist.

The above list of compact non-metrizable spaces is a small one. They are fairly standard examples for compact non-metrizable spaces. Could there be some esoteric example out there that fits the description? It turns out that there are such examples. In [1], Gruenhage and Nyikos constructed a compact non-metrizable $X$ such that $X^2$ is hereditarily normal. The construction was done using MA + not CH (Martin’s Axiom coupled with the negation of the continuum hypothesis). In that same paper, they also constructed another another example using CH. With the examples from [1], one immediate question was whether the additional set-theoretic axioms of MA + not CH (or CH) was necessary. Could a compact non-metrizable $X$ such that $X^2$ is hereditarily normal be still constructed without using any axioms beyond ZFC, the generally accepted axioms of set theory? For a relatively short period of time, this was an open question.

In 2001, Larson and Todorcevic [3] showed that it is consistent with ZFC that every compact $X$ with hereditarily normal $X^2$ is metrizable. In other words, there is a model of set theory that is consistent with ZFC in which Theorem 3 can be improved to assuming $X^2$ is hereditarily normal. Thus it is impossible to settle the above question without assuming additional axioms beyond those of ZFC. This means that if a compact non-metrizable $X$ is constructed without using any axiom beyond ZFC (such as those in the small list above), the hereditary normality must fail at dimension 1 or 2. Numerous other examples can be added to the above small list. Looking at these ZFC examples can help us appreciate the results in [1] and [3]. These ZFC examples are excellent training ground for general topology.

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Reference

1. Gruenhage G., Nyikos P. J., Normality in $X^2$ for Compact $X$, Trans. Amer. Math. Soc., Vol 340, No 2 (1993), 563-586
2. Katetov M., Complete normality of Cartesian products, Fund. Math., 35 (1948), 271-274
3. Larson P., Todorcevic S., KATETOV’S PROBLEM, Trans. Amer. Math. Soc., Vol 354, No 5 (2001), 1783-1791

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$\copyright \ 2015 \text{ by Dan Ma}$

# When a product space is hereditarily normal

When the spaces $X$ and $Y$ are normal spaces, the product space $X \times Y$ is not necessarily normal. Even if one of the factors is metrizable, there is still no guarantee that the product is normal. So it is possible that the normality of each of the factors $X$ and $Y$ can have no influence on the normality of the product $X \times Y$. The dynamics in the other direction are totally different. When the product $X \times Y$ is hereditarily normal, the two factors $X$ and $Y$ are greatly impacted. In this post, we discuss a theorem of Katetov, which shows that the hereditary normality of the product can impose very strict conditions on the factors, which lead to many interesting results. This theorem also leads to an interesting set-theoretic result, and thus can possibly be a good entry point to the part of topology that deals with consistency and independence results – statements that cannot be proved true or false based on the generally accepted axioms of set theory (ZFC). In this post, we discuss Katetov’s theorem and its consequences. In the next post, we discuss examples that further motivate the set-theoretic angle.

A subset $W$ of a space $X$ is said to be a $G_\delta$-set in $X$ if $W$ is the intersection of countably many open subsets of $X$. A space $X$ is perfectly normal if it is normal and that every closed subset of $X$ is a $G_\delta$-set. Some authors use other statements to characterize perfect normality (here is one such characterization). Perfect normality implies hereditarily normal (see Theorem 6 in this previous post). The implication cannot be reversed. Katetov’s theorem implies that the hereditary normality of the product $X \times Y$ will in many cases make one or both of the factors perfectly normal. Thus the hereditary normality in the product $X \times Y$ is a very strong property.

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Katetov’s theorems

Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

• The factor $X$ is perfectly normal.
• Every countable and infinite subset of the factor $Y$ is closed.

Proof of Theorem 1
The strategy we use is to define a subspace of $X \times Y$ that is not normal after assuming that none of the two conditions is true. So assume that $X$ has a closed subspace $W$ that is not a $G_\delta$-set and assume that $T=\left\{t_n: n=1,2,3,\cdots \right\}$ is an infinite subset of $Y$ that is not closed. Let $p \in Y$ be a limit point of $T$ such that $p \notin T$. The candidate for a non-normal subspace of $X \times Y$ is:

$M=X \times Y-W \times \left\{p \right\}$

Note that $M$ is an open subspace of $X \times Y$ since it is the result of subtracting a closed set from $X \times Y$. The following are the two closed sets that demonstrate that $M$ is not normal.

$H=W \times (Y-\left\{p \right\})$

$K=(X-W) \times \left\{p \right\}$

It is clear that $H$ and $K$ are closed subsets of $M$. Let $U$ and $V$ be open subsets of $M$ such that $H \subset U$ and $K \subset V$. We show that $U \cap V \ne \varnothing$. To this end, define $U_j=\left\{x \in X: (x,t_j) \in U \right\}$ for each $j$. It follows that for each $j$, $W \subset U_j$. Furthermore each $U_j$ is an open subspace of $X$. Thus $W \subset \bigcap_j U_j$. Since $W$ is not a $G_\delta$-set in $X$, there must exist $t \in \bigcap_j U_j$ such that $t \notin W$. Then $(t, p) \in K$ and $(t, p) \in V$.

Since $V$ is open in the product $X \times Y$, choose open sets $A \subset X$ and $B \subset Y$ such that $(t,p) \in A \times B$ and $A \times B \subset V$. With $p \in B$, there exists some $j$ such that $t_j \in B$. First, $(t,t_j) \in V$. Since $t \in U_j$, $(t,t_j) \in U$. Thus $U \cap V \ne \varnothing$. This completes the proof that the subspace $M$ is not normal and that $X \times Y$ is not hereditarily normal. $\blacksquare$

Let’s see what happens in Theorem 1 when both factors are compact. If both $X$ and $Y$ are compact and if $X \times Y$ is hereditarily normal, then both $X$ and $Y$ must be perfect normal. Note that in any infinite compact space, not every countably infinite subset is closed. Thus if compact spaces satisfy the conclusion of Theorem 1, they must be perfectly normal. Hence we have the following theorem.

Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.

Moe interestingly, Theorem 1 leads to a metrization theorem for compact spaces.

Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.

Proof of Theorem 3
Suppose that $X^3$ is hereditarily normal. By Theorem 2, the compact spaces $X^2$ and $X$ are perfectly normal. In particular, the following subset of $X^2$ is a $G_\delta$-set in $X^2$.

$\Delta=\left\{(x,x): x \in X \right\}$

The set $\Delta$ is said to be the diagonal of the space $X$. It is a well known result that any compact space whose diagonal is a $G_\delta$-set in the square is metrizable (discussed here). $\blacksquare$

The results discussed here make it clear that hereditary normality in product spaces is a very strong property. One obvious question is whether Theorem 3 can be improved by assuming only the hereditary normality of $X^2$. This was indeed posted by Katetov himself. This leads to the discussion in the next post.

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Reference

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

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$\copyright \ 2015 \text{ by Dan Ma}$

# Compact metrizable scattered spaces

A scattered space is one in which there are isolated points found in every subspace. Specifically, a space $X$ is a scattered space if every non-empty subspace $Y$ of $X$ has a point $y \in Y$ such that $y$ is an isolated point in $Y$, i.e. the singleton set $\left\{y \right\}$ is open in the subspace $Y$. A handy example is a space consisting of ordinals. Note that in a space of ordinals, every non-empty subset has an isolated point (e.g. its least element). In this post, we discuss scattered spaces that are compact metrizable spaces.

Here’s what led the author to think of such spaces. Consider Theorem III.1.2 found on page 91 of Arhangelskii’s book on topological function space [1], which is Theorem 1 stated below:

Thereom 1
For any compact space $X$, the following conditions are equivalent:

• The function space $C_p(X)$ is a Frechet-Urysohn space.
• The function space $C_p(X)$ is a k space.
• $X$ is a scattered space.

Let’s put aside the Frechet-Urysohn property and the k space property for the moment. For any Hausdorff space $X$, let $C(X)$ be the set of all continuous real-valued functions defined on the space $X$. Since $C(X)$ is a subspace of the product space $\mathbb{R}^X$, a natural topology that can be given to $C(X)$ is the subspace topology inherited from the product space $\mathbb{R}^X$. Then $C_p(X)$ is simply the set $C(X)$ with the product subspace topology (also called the pointwise convergence topology).

Let’s say the compact space $X$ is countable and infinite. Then the function space $C_p(X)$ is metrizable since it is a subspace of $\mathbb{R}^X$, a product of countably many lines. Thus the function space $C_p(X)$ has the Frechet-Urysohn property (being metrizable implies Frechet-Urysohn). This means that the compact space $X$ is scattered. The observation just made is a proof that any infinite compact space that is countable in cardinality must be scattered. In particular, every infinite compact and countable space must have an isolated point. There must be a more direct proof of this same fact without taking the route of a function space. The indirect argument does not reveal the essential nature of compact metric spaces. The essential fact is that any uncountable compact metrizable space contains a Cantor set, which is as unscattered as any space can be. Thus the only scattered compact metrizable spaces are the countable ones.

The main part of the proof is the construction of a Cantor set in a compact metrizable space (Theorem 3). The main result is Theorem 4. In many settings, the construction of a Cantor set is done in the real number line (e.g. the middle third Cantor set). The construction here is in a more general setting. But the idea is still the same binary division process – the splitting of a small open set with compact closure into two open sets with disjoint compact closure. We also use that fact that any compact metric space is hereditarily Lindelof (Theorem 2).

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Compact metrizable spaces

We first define some notions before looking at compact metrizable spaces in more details. Let $X$ be a space. Let $A \subset X$. Let $p \in X$. We say that $p$ is a limit point of $A$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$. So the notion of limit point here is from a topology perspective and not from a metric perspective. In a topological space, a limit point does not necessarily mean that it is the limit of a convergent sequence (however, it does in a metric space). The proof of the following theorem is straightforward.

Theorem 2
Let $X$ be a hereditarily Lindelof space (i.e. every subspace of $X$ is Lindelof). Then for any uncountable subset $A$ of $X$, all but countably many points of $A$ are limit points of $A$.

We now discuss the main result.

Theorem 3
Let $X$ be a compact metrizable space such that every point of $X$ is a limit point of $X$. Then there exists an uncountable closed subset $C$ of $X$ such that every point of $C$ is a limit point of $C$.

Proof of Theorem 3
Note that any compact metrizable space is a complete metric space. Consider a complete metric $\rho$ on the space $X$. One fact that we will use is that if there is a sequence of closed sets $X \supset H_1 \supset H_2 \supset H_3 \supset \cdots$ such that the diameters of the sets $H$ (based on the complete metric $\rho$) decrease to zero, then the sets $H_n$ collapse to one point.

The uncountable closed set $C$ we wish to define is a Cantor set, which is constructed from a binary division process. To start, pick two points $p_0,p_1 \in X$ such that $p_0 \ne p_1$. By assumption, both points are limit points of the space $X$. Choose open sets $U_0,U_1 \subset X$ such that

• $p_0 \in U_0$,
• $p_1 \in U_1$,
• $K_0=\overline{U_0}$ and $K_1=\overline{U_1}$,
• $K_0 \cap K_1 = \varnothing$,
• the diameters for $K_0$ and $K_1$ with respect to $\rho$ are less than 0.5.

Note that each of these open sets contains infinitely many points of $X$. Then we can pick two points in each of $U_0$ and $U_1$ in the same manner. Before continuing, we set some notation. If $\sigma$ is an ordered string of 0’s and 1’s of length $n$ (e.g. 01101 is a string of length 5), then we can always extend it by tagging on a 0 and a 1. Thus $\sigma$ is extended as $\sigma 0$ and $\sigma 1$ (e.g. 01101 is extended by 011010 and 011011).

Suppose that the construction at the $n$th stage where $n \ge 1$ is completed. This means that the points $p_\sigma$ and the open sets $U_\sigma$ have been chosen such that $p_\sigma \in U_\sigma$ for each length $n$ string of 0’s and 1’s $\sigma$. Now we continue the picking for the $(n+1)$st stage. For each $\sigma$, an $n$-length string of 0’s and 1’s, choose two points $p_{\sigma 0}$ and $p_{\sigma 1}$ and choose two open sets $U_{\sigma 0}$ and $U_{\sigma 1}$ such that

• $p_{\sigma 0} \in U_{\sigma 0}$,
• $p_{\sigma 1} \in U_{\sigma 1}$,
• $K_{\sigma 0}=\overline{U_{\sigma 0}} \subset U_{\sigma}$ and $K_{\sigma 1}=\overline{U_{\sigma 1}} \subset U_{\sigma}$,
• $K_{\sigma 0} \cap K_{\sigma 1} = \varnothing$,
• the diameters for $K_{\sigma 0}$ and $K_{\sigma 1}$ with respect to $\rho$ are less than $0.5^{n+1}$.

For each positive integer $m$, let $C_m$ be the union of all $K_\sigma$ over all $\sigma$ that are $m$-length strings of 0’s and 1’s. Each $C_m$ is a union of finitely many compact sets and is thus compact. Furthermore, $C_1 \supset C_2 \supset C_3 \supset \cdots$. Thus $C=\bigcap \limits_{m=1}^\infty C_m$ is non-empty. To complete the proof, we need to show that

• $C$ is uncountable (in fact of cardinality continuum),
• every point of $C$ is a limit point of $C$.

To show the first point, we define a one-to-one function $f: \left\{0,1 \right\}^N \rightarrow C$ where $N=\left\{1,2,3,\cdots \right\}$. Note that each element of $\left\{0,1 \right\}^N$ is a countably infinite string of 0’s and 1’s. For each $\tau \in \left\{0,1 \right\}^N$, let $\tau \upharpoonright n$ denote the string of the first $n$ digits of $\tau$. For each $\tau \in \left\{0,1 \right\}^N$, let $f(\tau)$ be the unique point in the following intersection:

$\displaystyle \bigcap \limits_{n=1}^\infty K_{\tau \upharpoonright n} = \left\{f(\tau) \right\}$

This mapping is uniquely defined. Simply conceptually trace through the induction steps. For example, if $\tau$ are 01011010…., then consider $K_0 \supset K_{01} \supset K_{010} \supset \cdots$. At each next step, always pick the $K_{\tau \upharpoonright n}$ that matches the next digit of $\tau$. Since the sets $K_{\tau \upharpoonright n}$ are chosen to have diameters decreasing to zero, the intersection must have a unique element. This is because we are working in a complete metric space.

It is clear that the map $f$ is one-to-one. If $\tau$ and $\gamma$ are two different strings of 0’s and 1’s, then they must differ at some coordinate, then from the way the induction is done, the strings would lead to two different points. It is also clear to see that the map $f$ is reversible. Pick any point $x \in C$. Then the point $x$ must belong to a nested sequence of sets $K$‘s. This maps to a unique infinite string of 0’s and 1’s. Thus the set $C$ has the same cardinality as the set $\left\{0,1 \right\}^N$, which has cardinality continuum.

To see the second point, pick $x \in C$. Suppose $x=f(\tau)$ where $\tau \in \left\{0,1 \right\}^N$. Consider the open sets $U_{\tau \upharpoonright n}$ for all positive integers $n$. Note that $x \in U_{\tau \upharpoonright n}$ for each $n$. Based on the induction process described earlier, observe these two facts. This sequence of open sets has diameters decreasing to zero. Each open set $U_{\tau \upharpoonright n}$ contains infinitely many other points of $C$ (this is because of all the open sets $U_{\tau \upharpoonright k}$ that are subsets of $U_{\tau \upharpoonright n}$ where $k \ge n$). Because the diameters are decreasing to zero, the sequence of $U_{\tau \upharpoonright n}$ is a local base at the point $x$. Thus, the point $x$ is a limit point of $C$. This completes the proof. $\blacksquare$

Theorem 4
Let $X$ be a compact metrizable space. It follows that $X$ is scattered if and only if $X$ is countable.

Proof of Theorem 4
$\Longleftarrow$
In this direction, we show that if $X$ is countable, then $X$ is scattered (the fact that can be shown using the function space argument pointed out earlier). Here, we show the contrapositive: if $X$ is not scattered, then $X$ is uncountable. Suppose $X$ is not scattered. Then every point of $X$ is a limit point of $X$. By Theorem 3, $X$ would contain a Cantor set $C$ of cardinality continuum.

$\Longrightarrow$
In this direction, we show that if $X$ is scattered, then $X$ is countable. We also show the contrapositive: if $X$ is uncountable, then $X$ is not scattered. Suppose $X$ is uncountable. By Theorem 2, all but countably many points of $X$ are limit points of $X$. After discarding these countably many isolated points, we still have a compact space. So we can just assume that every point of $X$ is a limit point of $X$. Then by Theorem 3, $X$ contains an uncountable closed set $C$ such that every point of $C$ is a limit point of $C$. This means that $X$ is not scattered. $\blacksquare$

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Remarks

A corollary to the above discussion is that the cardinality for any compact metrizable space is either countable (including finite) or continuum (the cardinality of the real line). There is nothing in between or higher than continuum. To see this, the cardinality of any Lindelof first countable space is at most continuum according to a theorem in this previous post (any compact metric space is one such). So continuum is an upper bound on the cardinality of compact metric spaces. Theorem 3 above implies that any uncountable compact metrizable space has to contain a Cantor set, hence has cardinality continuum. So the cardinality of a compact metrizable space can be one of two possibilities – countable or continuum. Even under the assumption of the negation of the continuum hypothesis, there will be no uncountable compact metric space of cardinality less than continuum. On the other hand, there is only one possibility for the cardinality of a scattered compact metrizable, which is countable.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2015 \text{ by Dan Ma}$