# Drawing Sorgenfrey continuous functions

The Sorgenfrey line is a well known topological space. It is the real number line with open intervals defined as sets of the form $[a,b)$. Though this is a seemingly small tweak, it generates a vastly different space than the usual real number line. In this post, we look at the Sorgenfrey line from the continuous function perspective, in particular, the continuous functions that map the Sorgenfrey line into the real number line. In the process, we obtain insight into the space of continuous functions on the Sorgenfrey line.

The Sorgenfrey Line

Let $\mathbb{R}$ denote the real number line. The usual open intervals are of the form $(a,b)=\left\{x \in \mathbb{R}: a. The union of such open intervals is called an open set. If more than one topologies are considered on the real line, these open sets are referred to as the usual open sets or Euclidean open sets (on the real line). The open intervals $(a,b)$ form a base for the usual topology on the real line. One important fact abut the usual open sets is that the usual open sets can be generated by the intervals $(a,b)$ where both end points are rational numbers. Thus the usual topology on the real line is said to have a countable base.

Now tweak the usual topology by calling sets of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x open intervals. Then form open sets by taking unions of all such open intervals. The collection of such open sets is called the Sorgenfrey topology (on the real line). The real number line $\mathbb{R}$ with the Sorgenfry topology is called the Sorgenfrey line, denoted by $\mathbb{S}$. The Sorgenfrey line has been discussed in this blog, starting with this post. This post examines continuous functions from $\mathbb{S}$ into the real line. In the process, we gain insight on the space of continuous functions defined on $\mathbb{S}$.

Note that any usual open interval $(a,b)$ is the union of intervals of the form $[c,d)$. Thus any usual (Euclidean) open set is an open set in the Sorgenfrey line. Thus the usual topology (on the real line) is contained in the Sorgenfrey topology, i.e. the usual topology is a weaker (coarser) topology.

Let $C(\mathbb{R})$ be the set of all continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}$ where the domain is the real number line with the usual topology. Let $C(\mathbb{S})$ be the set of all continuous functions $f:\mathbb{S} \rightarrow \mathbb{R}$ where the domain is the Sorgenfrey line. In both cases, the range is always the number line with the usual topology. Based on the preceding paragraph, any continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ is also continuous with respect to the Sorgenfrey line, i.e. $C(\mathbb{R}) \subset C(\mathbb{S})$.

Pictures of Continuous Functions

Consider the following two continuous functions.

Figure 1 – CDF of the standard normal distribution

Figure 2 – CDF of the uniform distribution

The first one (Figure 1) is the cumulative distribution function (CDF) of the standard normal distribution. The second one (Figure 2) is the CDF of the uniform distribution on the interval $(0,a)$ where $a>0$. Both of these are continuous in the usual Euclidean topology (in the domain). Such graphs would make regular appearance in a course on probability and statistics. They also show up in a calculus course as an everywhere differentiable curve (Figure 1) and as a differentiable curve except at finitely many points (Figure 2). Both of these functions can also be regarded as continuous functions on the Sorgenfrey line.

Consider a function that is continuous in the Sorgenfrey line but not continuous in the usual topology.

Figure 3 – Right continuous function

Figure 3 is a function that maps the interval $(-\infty,0)$ to -1 and maps the interval $[0,\infty)$ to 1. It is not continuous in the usual topology because of the jump at $x=0$. But it is a continuous function when the domain is considered to be the Sorgenfrey line. Because of the open intervals being $[a,b)$, continuous functions defined on the Sorgenfrey line are right continuous.

The cumulative distribution function of a discrete probability distribution is always right continuous, hence continuous in the Sorgenfrey line. Here’s an example.

Figure 4 – CDF of a discrete uniform distribution

Figure 4 is the CDF of the uniform distribution on the finite set $\left\{0,1,2,3,4 \right\}$, where each point has probability 0.2. There is a jump of height 0.2 at each of the points from 0 to 4. Figure 3 and Figure 4 are step functions. As long as the left point of a step is solid and the right point is hollow, the step functions are continuous on the Sorgenfrey line.

The take away from the last four figures is that the real-valued continuous functions defined on the Sorgenfrey line are right continuous and that step functions (with the left point solid and the right point hollow) are Sorgenfrey continuous.

A Family of Sorgenfrey Continuous Functions

The four examples of continuous functions shown above are excellent examples to illustrate the Sorgenfrey topology. We now introduce a family of continuous functions $f_a:\mathbb{S} \rightarrow \mathbb{R}$ for $0. These continuous functions will lead to additional insight on the function space whose domain space is the Sorgenfrey line.

For any $0, the following gives the definition and the graph of the function $f_a$.

$\displaystyle f_a(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ -\infty

Figure 5 – a family of Sorgenfrey continuous functions

Function Space on the Sorgenfrey Line

This is the place where we switch the focus to function space. The set $C(\mathbb{S})$ is a subset of the product space $\mathbb{R}^\mathbb{R}$. So we can consider $C(\mathbb{S})$ as a topological space endowed with the topology inherited as a subspace of $\mathbb{R}^\mathbb{R}$. This topology on $C(\mathbb{S})$ is called the pointwise convergence topology and $C(\mathbb{S})$ with the product subspace topology is denoted by $C_p(\mathbb{S})$. See here for comments on how to work with the pointwise convergence topology.

For the present discussion, all we need is some notation on a base for $C_p(\mathbb{S})$. For $x \in \mathbb{S}$, and for any open interval $(a,b)$ (open in the usual topology of the real number line), let $[x,(a,b)]=\left\{h \in C_p(\mathbb{S}): h(x) \in (a, b) \right\}$. Then the collection of intersections of finitely many $[x,(a,b)]$ would form a base for $C_p(\mathbb{S})$.

The following is the main fact we wish to establish.

The function space $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum. In particular, the set $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$.

The above result will derive several facts on the function space $C_p(\mathbb{S})$, which are discussed in a section below. More interestingly, the proof of the fact that $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$ is based purely on the definition of the functions $f_a$ and the Sorgenfrey topology. The proof given below does not use any deep or high powered results from function space theory. So it should be a nice exercise on the Sorgenfrey topology.

I invite readers to either verify the fact independently of the proof given here or follow the proof closely. Lots of drawing of the functions $f_a$ on paper will be helpful in going over the proof. In this one instance at least, drawing continuous functions can help gain insight on function spaces.

Working out the Proof

The following diagram was helpful to me as I worked out the different cases in showing the discreteness of the family $F=\left\{f_a: 0. The diagram is a valuable aid in convincing myself that a given case is correct.

Figure 6 – A comparison of three Sorgenfrey continuous functions

Now the proof. First, $F$ is relatively discrete in $C_p(\mathbb{S})$. We show that for each $a$, there is an open set $O$ containing $f_a$ such that $O$ does not contain $f_w$ for any $w \ne a$. To this end, let $O=[a,V_1] \cap [-a,V_2]$ where $V_1$ and $V_2$ are the open intervals $V_1=(0.9,1.1)$ and $V_2=(-0.1,0.1)$. With Figure 6 as an aid, it follows that for $0, $f_b \notin O$ and for $a, $f_c \notin O$.

The open set $O=[a,V_1] \cap [-a,V_2]$ contains $f_a$, the function in the middle of Figure 6. Note that for $0, $f_b(-a)=1$ and $f_b(-a) \notin V_2$. Thus $f_b \notin O$. On the other hand, for $a, $f_c(a)=0$ and $f_c(a) \notin V_1$. Thus $f_c \notin O$. This proves that the set $F$ is a discrete subspace of $C_p(\mathbb{S})$ relative to $F$ itself.

Now we show that $F$ is closed in $C_p(\mathbb{S})$. To this end, we show that

for each $g \in C_p(\mathbb{S})$, there is an open set $U$ containing $g$ such that $U$ contains at most one point of $F$.

Actually, this has already been done above with points $g$ that are in $F$. One thing to point out is that the range of $f_a$ is $\left\{0,1 \right\}$. As we consider $g \in C_p(\mathbb{S})$, we only need to consider $g$ that maps into $\left\{0,1 \right\}$. Let $g \in C_p(\mathbb{S})$. The argument is given in two cases regarding the function $g$.

Case 1. There exists some $a \in (0,1)$ such that $g(a) \ne g(-a)$.

We assume that $g(a)=0$ and $g(-a)=1$. Then for all $0, $f_b(a)=1$ and for all $a, $f_c(-a)=0$. Let $U=[a,(-0.1,0.1)] \cap [-a,(0.9,1.1)]$. Then $g \in U$ and $U$ contains no $f_b$ for any $0 and $f_c$ for any $a. To help see this argument, use Figure 6 as a guide. The case that $g(a)=1$ and $g(-a)=0$ has a similar argument.

Case 2. For every $a \in (0,1)$, we have $g(a) = g(-a)$.

Claim. The function $g$ is constant on the interval $(-1,1)$. Suppose not. Let $0 such that $g(a) \ne g(b)$. Suppose that $0=g(b) < g(a)=1$. Consider $W=\left\{w. Clearly the number $a$ is an upper bound of $W$. Let $u \le a$ be a least upper bound of $W$. The function $g$ has value 1 on the interval $(u,a)$. Otherwise, $u$ would not be the least upper bound of the set $W$. There is a sequence of points $\left\{x_n \right\}$ in the interval $(b,u)$ such that $x_n \rightarrow u$ from the left such that $g(x_n)=0$ for all $n$. Otherwise, $u$ would not be the least upper bound of the set $W$.

It follows that $g(u)=1$. Otherwise, the function $g$ is not continuous at $u$. Now consider the 6 points $-a<-u<-b. By the assumption in Case 2, $g(u)=g(-u)=1$ and $g(b)=g(-b)=0$. Since $g(x_n)=0$ for all $n$, $g(-x_n)=0$ for all $n$. Note that $-x_n \rightarrow -u$ from the right. Since $g$ is right continuous, $g(-u)=0$, contradicting $g(-u)=1$. Thus we cannot have $0=g(b) < g(a)=1$.

Now suppose we have $1=g(b) > g(a)=0$ where $0. Consider $W=\left\{w. Clearly $W$ has an upper bound, namely the number $a$. Let $u \le a$ be a least upper bound of $W$. The function $g$ has value 0 on the interval $(u,a)$. Otherwise, $u$ would not be the least upper bound of the set $W$. There is a sequence of points $\left\{x_n \right\}$ in the interval $(b,u)$ such that $x_n \rightarrow u$ from the left such that $g(x_n)=1$ for all $n$. Otherwise, $u$ would not be the least upper bound of the set $W$.

It follows that $g(u)=0$. Otherwise, the function $g$ is not continuous at $u$. Now consider the 6 points $-a<-u<-b. By the assumption in Case 2, $g(u)=g(-u)=0$ and $g(b)=g(-b)=1$. Since $g(x_n)=1$ for all $n$, $g(-x_n)=1$ for all $n$. Note that $-x_n \rightarrow -u$ from the right. Since $g$ is right continuous, $g(-u)=1$, contradicting $g(-u)=0$. Thus we cannot have $1=g(b) > g(a)=0$.

The claim that the function $g$ is constant on the interval $(-1,1)$ is established. To wrap up, first assume that the function $g$ is 1 on the interval $(-1,1)$. Let $U=[0,(0.9,1.1)]$. It is clear that $g \in U$. It is also clear from Figure 5 that $U$ contains no $f_a$. Now assume that the function $g$ is 0 on the interval $(-1,1)$. Since $g$ is Sorgenfrey continuous, it follows that $g(-1)=0$. Let $U=[-1,(-0.1,0.1)]$. It is clear that $g \in U$. It is also clear from Figure 5 that $U$ contains no $f_a$.

We have established that the set $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$.

What does it Mean?

The above argument shows that the set $F$ is a closed an discrete subspace of the function space $C_p(\mathbb{S})$. We have the following three facts.

 Three Results $C_p(\mathbb{S})$ is separable. $C_p(\mathbb{S})$ is not hereditarily separable. $C_p(\mathbb{S})$ is not a normal space.

To show that $C_p(\mathbb{S})$ is separable, let’s look at one basic helpful fact on $C_p(X)$. If $X$ is a separable metric space, e.g. $X=\mathbb{R}$, then $C_p(X)$ has quite a few nice properties (discussed here). One is that $C_p(X)$ is hereditarily separable. Thus $C_p(\mathbb{R})$, the space of real-valued continuous functions defined on the number line with the pointwise convergence topology, is hereditarily separable and thus separable. Recall that continuous functions in $C_p(\mathbb{R})$ are also Soregenfrey line continuous. Thus $C_p(\mathbb{R})$ is a subspace of $C_p(\mathbb{S})$. The space $C_p(\mathbb{R})$ is also a dense subspace of $C_p(\mathbb{S})$. Thus the space $C_p(\mathbb{S})$ contains a dense separable subspace. It means that $C_p(\mathbb{S})$ is separable.

Secondly, $C_p(\mathbb{S})$ is not hereditarily separable since the subspace $F=\left\{f_a: 0 is a closed and discrete subspace.

Thirdly, $C_p(\mathbb{S})$ is not a normal space. According to Jones’ lemma, any separable space with a closed and discrete subspace of cardinality of continuum is not a normal space (see Corollary 1 here). The subspace $F=\left\{f_a: 0 is a closed and discrete subspace of the separable space $C_p(\mathbb{S})$. Thus $C_p(\mathbb{S})$ is not normal.

Remarks

The topology of the Sorgenfrey line is vastly different from the usual topology on the real line even though the the Sorgenfrey topology is obtained by a seemingly small tweak from the usual topology. The real line is a metric space while the Sorgenfrey line is not metrizable. The real number line is connected while the Sorgenfrey line is not. The countable power of the real number line is a metric space and thus a normal space. On the other hand, the Sorgenfrey line is a classic example of a normal space whose square is not normal. See here for a basic discussion of the Sorgenfrey line.

The pictures of Sorgenfrey continuous functions demonstrated here show that the real number line and the Sorgenfrey line are also very different from a function space perspective. The function space $C_p(\mathbb{R})$ has a whole host of nice properties: normal, Lindelof (hence paracompact and collectionwise normal), hereditarily Lindelof (hence hereditarily normal), hereditarily separable, and perfectly normal (discussed here).

Though separable, the function space $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum, making it not hereditarily separable and not normal.

For more information about $C_p(X)$ in general and $C_p(\mathbb{S})$ in particular, see [1] and [2]. A different proof that $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum can be found in Problem 165 in [2].

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Tkachuk V. V., A $C_p$-Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

# Normality in Cp(X)

Any collectionwise normal space is a normal space. Any perfectly normal space is a hereditarily normal space. In general these two implications are not reversible. In function spaces $C_p(X)$, the two implications are reversible. There is a normal space that is not countably paracompact (such a space is called a Dowker space). If a function space $C_p(X)$ is normal, it is countably paracompact. Thus normality in $C_p(X)$ is a strong property. This post draws on Dowker’s theorem and other results, some of them are previously discussed in this blog, to discuss this remarkable aspect of the function spaces $C_p(X)$.

Since we are discussing function spaces, the domain space $X$ has to have sufficient quantity of real-valued continuous functions, e.g. there should be enough continuous functions to separate the points from closed sets. The ideal setting is the class of completely regular spaces (also called Tychonoff spaces). See here for a discussion on completely regular spaces in relation to function spaces.

Let $X$ be a completely regular space. Let $C(X)$ be the set of all continuous functions from $X$ into the real line $\mathbb{R}$. When $C(X)$ is endowed with the pointwise convergence topology, the space is denoted by $C_p(X)$ (see here for further comments on the definition of the pointwise convergence topology).

When Function Spaces are Normal

Let $X$ be a completely regular space. We discuss these four facts of $C_p(X)$:

1. If the function space $C_p(X)$ is normal, then $C_p(X)$ is countably paracompact.
2. If the function space $C_p(X)$ is hereditarily normal, then $C_p(X)$ is perfectly normal.
3. If the function space $C_p(X)$ is normal, then $C_p(X)$ is collectionwise normal.
4. Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ has countable extent, i.e. every closed and discrete subset of $X$ is countable, implying that $X$ is collectionwise normal.

Fact #1 and Fact #2 rely on a representation of $C_p(X)$ as a product space with one of the factors being the real line. For $x \in X$, let $Y_x=\left\{f \in C_p(X): f(x)=0 \right\}$. Then $C_p(X) \cong Y_x \times \mathbb{R}$. This representation is discussed here.

Another useful tool is Dowker’s theorem, which essentially states that for any normal space $W$, the space $W$ is countably paracompact if and only if $W \times C$ is normal for all compact metric space $C$ if and only if $W \times [0,1]$ is normal. For the full statement of the theorem, see Theorem 1 in this previous post, which has links to the proofs and other discussion.

To show Fact #1, suppose that $C_p(X)$ is normal. Immediately we make use of the representation $C_p(X) \cong Y_x \times \mathbb{R}$ where $x \in X$. Since $Y_x \times \mathbb{R}$ is normal, $Y_x \times [0,1]$ is also normal. By Dowker’s theorem, $Y_x$ is countably paracompact. Note that $Y_x$ is a closed subspace of the normal $C_p(X)$. Thus $Y_x$ is also normal.

One more helpful tool is Theorem 5 in in this previous post, which is like an extension of Dowker’s theorem, which states that a normal space $W$ is countably paracompact if and only if $W \times T$ is normal for any $\sigma$-compact metric space $T$. This means that $Y_x \times \mathbb{R} \times \mathbb{R}$ is normal.

We want to show $C_p(X) \cong Y_x \times \mathbb{R}$ is countably paracompact. Since $Y_x \times \mathbb{R} \times \mathbb{R}$ is normal (based on the argument in the preceding paragraph), $(Y_x \times \mathbb{R}) \times [0,1]$ is normal. Thus according to Dowker’s theorem, $C_p(X) \cong Y_x \times \mathbb{R}$ is countably paracompact.

For Fact #2, a helpful tool is Katetov’s theorem (stated and proved here), which states that for any hereditarily normal $X \times Y$, one of the factors is perfectly normal or every countable subset of the other factor is closed (in that factor).

To show Fact #2, suppose that $C_p(X)$ is hereditarily normal. With $C_p(X) \cong Y_x \times \mathbb{R}$ and according to Katetov’s theorem, $Y_x$ must be perfectly normal. The product of a perfectly normal space and any metric space is perfectly normal (a proof is found here). Thus $C_p(X) \cong Y_x \times \mathbb{R}$ is perfectly normal.

The proof of Fact #3 is found in Problems 294 and 295 of [2]. The key to the proof is a theorem by Reznichenko, which states that any dense convex normal subspace of $[0,1]^X$ has countable extent, hence is collectionwise normal (problem 294). See here for a proof that any normal space with countable extent is collectionwise normal (see Theorem 2). The function space $C_p(X)$ is a dense convex subspace of $[0,1]^X$ (problem 295). Thus if $C_p(X)$ is normal, then it has countable extent and hence collectionwise normal.

Fact #4 says that normality of the function space imposes countable extent on the domain. This result is discussed in this previous post (see Corollary 3 and Corollary 5).

Remarks

The facts discussed here give a flavor of what function spaces are like when they are normal spaces. For further and deeper results, see [1] and [2].

Fact #1 is essentially driven by Dowker’s theorem. It follows from the theorem that whenever the product space $X \times Y$ is normal, one of the factor must be countably paracompact if the other factor has a non-trivial convergent sequence (see Theorem 2 in this previous post). As a result, there is no Dowker space that is a $C_p(X)$. No pathology can be found in $C_p(X)$ with respect to finding a Dowker space. In fact, not only $C_p(X) \times C$ is normal for any compact metric space $C$, it is also true that $C_p(X) \times T$ is normal for any $\sigma$-compact metric space $T$ when $C_p(X)$ is normal.

The driving force behind Fact #2 is Katetov’s theorem, which basically says that the hereditarily normality of $X \times Y$ is a strong statement. Coupled with the fact that $C_p(X)$ is of the form $Y_x \times \mathbb{R}$, Katetov’s theorem implies that $Y_x \times \mathbb{R}$ is perfectly normal. The argument also uses the basic fact that perfectly normality is preserved when taking product with metric spaces.

There are examples of normal but not collectionwise normal spaces (e.g. Bing’s Example G). Resolution of the question of whether normal but not collectionwise normal Moore space exists took extensive research that spanned decades in the 20th century (the normal Moore space conjecture). The function $C_p(X)$ is outside of the scope of the normal Moore space conjecture. The function space $C_p(X)$ is usually not a Moore space. It can be a Moore space only if the domain $X$ is countable but then $C_p(X)$ would be a metric space. However, it is still a powerful fact that if $C_p(X)$ is normal, then it is collectionwise normal.

On the other hand, a more interesting point is on the normality of $X$. Suppose that $X$ is a normal Moore space. If $C_p(X)$ happens to be normal, then Fact #4 says that $X$ would have to be collectionwise normal, which means $X$ is metrizable. If the goal is to find a normal Moore space $X$ that is not collectionwise normal, the normality of $C_p(X)$ would kill the possibility of $X$ being the example.

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Tkachuk V. V., A $C_p$-Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

# Looking for spaces in which every compact subspace is metrizable

Once it is known that a topological space is not metrizable, it is natural to ask, from a metrizability standpoint, which subspaces are metrizable, e.g. whether every compact subspace is metrizable. This post discusses several classes of spaces in which every compact subspace is metrizable. Though the goal here is not to find a complete characterization of such spaces, this post discusses several classes of spaces and various examples that have this property. The effort brings together many interesting basic and well known facts. Thus the notion “every compact subspace is metrizable” is an excellent learning opportunity.

Several Classes of Spaces

The notion “every compact subspace is metrizable” is a very broad class of spaces. It includes well known spaces such as Sorgenfrey line, Michael line and the first uncountable ordinal $\omega_1$ (with the order topology) as well as Moore spaces. Certain function spaces are in the class “every compact subspace is metrizable”. The following diagram is a good organizing framework.

\displaystyle \begin{aligned} &1. \ \text{Metrizable} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&2. \ \text{Submetrizable} \Longleftarrow 5. \ \exists \ \text{countable network} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&3. \ \exists \ G_\delta \text{ diagonal} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&4. \ \text{Every compact subspace is metrizable} \end{aligned}

Let $(X, \tau)$ be a space. It is submetrizable if there is a topology $\tau_1$ on the set $X$ such that $\tau_1 \subset \tau$ and $(X, \tau_1)$ is a metrizable space. The topology $\tau_1$ is said to be weaker (coarser) than $\tau$. Thus a space $X$ is submetrizable if it has a weaker metrizable topology.

Let $\mathcal{N}$ be a set of subsets of the space $X$. $\mathcal{N}$ is said to be a network for $X$ if for every open subset $O$ of $X$ and for each $x \in O$, there exists $N \in \mathcal{N}$ such that $x \in N \subset O$. Having a network that is countable in size is a strong property (see here for a discussion on spaces with a countable network).

The diagonal of the space $X$ is the subset $\Delta=\left\{(x,x): x \in X \right\}$ of the square $X \times X$. The space $X$ has a $G_\delta$-diagonal if $\Delta$ is a $G_\delta$-subset of $X \times X$, i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$.

The implication $1 \Longrightarrow 2$ is clear. For $5 \Longrightarrow 2$, see Lemma 1 in this previous post on countable network. The implication $2 \Longrightarrow 3$ is left as an exercise. To see $3 \Longrightarrow 4$, let $K$ be a compact subset of $X$. The property of having a $G_\delta$-diagonal is hereditary. Thus $K$ has a $G_\delta$-diagonal. According to a well known result, any compact space with a $G_\delta$-diagonal is metrizable (see here).

None of the implications in the diagram is reversible. The first uncountable ordinal $\omega_1$ is an example for $4 \not \Longrightarrow 3$. This follows from the well known result that any countably compact space with a $G_\delta$-diagonal is metrizable (see here). The Mrowka space is an example for $3 \not \Longrightarrow 2$ (see here). The Sorgenfrey line is an example for both $2 \not \Longrightarrow 5$ and $2 \not \Longrightarrow 1$.

To see where the examples mentioned earlier are placed, note that Sorgenfrey line and Michael line are submetrizable, both are submetrizable by the usual Euclidean topology on the real line. Each compact subspace of the space $\omega_1$ is countable and is thus contained in some initial segment $[0,\alpha]$ which is metrizable. Any Moore space has a $G_\delta$-diagonal. Thus compact subspaces of a Moore space are metrizable.

Function Spaces

We now look at some function spaces that are in the class “every compact subspace is metrizable.” For any Tychonoff space (completely regular space) $X$, $C_p(X)$ is the space of all continuous functions from $X$ into $\mathbb{R}$ with the pointwise convergence topology (see here for basic information on pointwise convergence topology).

Theorem 1
Suppose that $X$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.

Proof
The proof here actually shows more than is stated in the theorem. We show that $C_p(X)$ is submetrizable by a separable metric topology. Let $Y$ be a countable dense subspace of $X$. Then $C_p(Y)$ is metrizable and separable since it is a subspace of the separable metric space $\mathbb{R}^{\omega}$. Thus $C_p(Y)$ has a countable base. Let $\mathcal{E}$ be a countable base for $C_p(Y)$.

Let $\pi:C_p(X) \longrightarrow C_p(Y)$ be the restriction map, i.e. for each $f \in C_p(X)$, $\pi(f)=f \upharpoonright Y$. Since $\pi$ is a projection map, it is continuous and one-to-one and it maps $C_p(X)$ into $C_p(Y)$. Thus $\pi$ is a continuous bijection from $C_p(X)$ into $C_p(Y)$. Let $\mathcal{B}=\left\{\pi^{-1}(E): E \in \mathcal{E} \right\}$.

We claim that $\mathcal{B}$ is a base for a topology on $C_p(X)$. Once this is established, the proof of the theorem is completed. Note that $\mathcal{B}$ is countable and elements of $\mathcal{B}$ are open subsets of $C_p(X)$. Thus the topology generated by $\mathcal{B}$ is coarser than the original topology of $C_p(X)$.

For $\mathcal{B}$ to be a base, two conditions must be satisfied – $\mathcal{B}$ is a cover of $C_p(X)$ and for $B_1,B_2 \in \mathcal{B}$, and for $f \in B_1 \cap B_2$, there exists $B_3 \in \mathcal{B}$ such that $f \in B_3 \subset B_1 \cap B_2$. Since $\mathcal{E}$ is a base for $C_p(Y)$ and since elements of $\mathcal{B}$ are preimages of elements of $\mathcal{E}$ under the map $\pi$, it is straightforward to verify these two points. $\square$

Theorem 1 is actually a special case of a duality result in $C_p$ function space theory. More about this point later. First, consider a corollary of Theorem 1.

Corollary 2
Let $X=\prod_{\alpha where $c$ is the cardinality continuum and each $X_\alpha$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.

The key fact for Corollary 2 is that the product of continuum many separable spaces is separable (this fact is discussed here). Theorem 1 is actually a special case of a deep result.

Theorem 3
Suppose that $X=\prod_{\alpha<\kappa} X_\alpha$ is a product of separable spaces where $\kappa$ is any infinite cardinal. Then every compact subspace of $C_p(X)$ is metrizable.

Theorem 3 is a much more general result. The product of any arbitrary number of separable spaces is not separable if the number of factors is greater than continuum. So the proof for Theorem 1 will not work in the general case. This result is Problem 307 in [2].

A Duality Result

Theorem 1 is stated in a way that gives the right information for the purpose at hand. A more correct statement of Theorem 1 is: $X$ is separable if and only if $C_p(X)$ is submetrizable by a separable metric topology. Of course, the result in the literature is based on density and weak weight.

The cardinal function of density is the least cardinality of a dense subspace. For any space $Y$, the weight of $Y$, denoted by $w(Y)$, is the least cardinaility of a base of $Y$. The weak weight of a space $X$ is the least $w(Y)$ over all space $Y$ for which there is a continuous bijection from $X$ onto $Y$. Thus if the weak weight of $X$ is $\omega$, then there is a continuous bijection from $X$ onto some separable metric space, hence $X$ has a weaker separable metric topology.

There is a duality result between density and weak weight for $X$ and $C_p(X)$. The duality result:

The density of $X$ coincides with the weak weight of $C_p(X)$ and the weak weight of $X$ coincides with the density of $C_p(X)$. These are elementary results in $C_p$-theory. See Theorem I.1.4 and Theorem I.1.5 in [1].

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Tkachuk V. V., A $C_p$-Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

Countably paracompact spaces are discussed in a previous post. The discussion of countably paracompactness in the previous post is through discussing Dowker’s theorem. In this post, we discuss a few more facts that can be derived from Dowker’s theorem.

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Dowker’s Theorem

Essentially, Dowker’s theorem is the statement that for a normal space $X$, the space $X$ is countably paracompact if any only if $X \times Y$ is normal for any infinite compact metric space. The following is the full statement of Dowker’s theorem. The long list of equivalent conditions is important for applications in various scenarios.

Theorem 1 (Dowker’s Theorem)
Let $X$ be a normal space. The following conditions are equivalent.

1. The space $X$ is countably paracompact.
2. Every countable open cover of $X$ has a point-finite open refinement.
3. If $\left\{U_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$, there exists an open refinement $\left\{V_n: n=1,2,3,\cdots \right\}$ such that $\overline{V_n} \subset U_n$ for each $n$.
4. The product space $X \times Y$ is normal for any infinite compact metric space $Y$.
5. The product space $X \times [0,1]$ is normal where $[0,1]$ is the closed unit interval with the usual Euclidean topology.
6. The product space $X \times S$ is normal where $S$ is a non-trivial convergent sequence with the limit point. Note that $S$ can be taken as a space homeomorphic to $\left\{1,\frac{1}{2},\frac{1}{3},\cdots \right\} \cup \left\{0 \right\}$ with the Euclidean topology.
7. For each sequence $\left\{A_n \subset X: n=1,2,3,\cdots \right\}$ of closed subsets of $X$ such that $A_1 \supset A_2 \supset A_3 \supset \cdots$ and $\cap_n A_n=\varnothing$, there exist open sets $B_1,B_2,B_3,\cdots$ such that $A_n \subset B_n$ for each $n$ such that $\cap_n B_n=\varnothing$.

A Dowker space is any normal space that is not countably paracompact. The notion of Dowker space was motivated by Dowker’s theorem since such a space would be a normal space $X$ for which $X \times [0,1]$ is not normal. The search for such a space took about 20 years from 1951 when C. H. Dowker proved the theorem to 1971 when M. E. Rudin constructed a ZFC example of a Dowker space.

Theorem 1 (Dowker’s theorem) is proved here and is further discussed in this previous post on countably paracompact space. The statement appears in Condition 6 here is not found in the previous version of the theorem. However, no extra effort is required to support it. Condition 5 trivially implies condition 6. The proof of condition 5 implying condition 7 (the proof of 4 implies 5 shown here) only requires that the product of $X$ and a convergent sequence is normal. So inserting condition 6 does not require extra proof.

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Getting More from Dowker’s Theorem

As a result of Theorem 1, normal countably paracompact spaces are productive in normality with respect to compact metric spaces (condition 4 in Dowker’s theorem as stated above). Another way to look at condition 4 is that the normality in the product $X \times Y$ is a strong property. Whenever the product $X \times Y$ is normal, we know that each factor is normal. Dowker’s theorem tells us that whenever $X \times Y$ is normal and one of the factor is a compact metric space such as the unit interval $[0,1]$, the other factor is countably paracompact. The fact can be extended. Even if the factors are not metric spaces, as long as one of the factors has a non-discrete point with “countable” tightness, normality of the product confers countably paracompactness on one of the factors. The following two theorems make this clear.

Theorem 2
Suppose that the product $X \times Y$ is normal. If one of the factor contains a non-trivial convergent sequence, then the other factor is countably paracompact.

Proof of Theorem 2
Suppose $Y$ contains a non-trivial convergent sequence. Let this sequence be denoted by $S =\left\{ x_n:n=1,2,3,\cdots \right\} \cup \left\{x \right\}$ such that the point $x$ is the limit point. Since $X \times Y$ is normal, both $X$ and $Y$ are normal and that $X \times S$ is normal. By Theorem 1, $X$ is countably paracompact. $\square$

Theorem 3
Suppose that the product $X \times Y$ is normal. If one of the factor contains a countable subset that is non-discrete, then the other factor is countably paracompact.

Proof of Theorem 3
To discuss this fact, we need to turn to the generalized Dowker’s theorem, which is Theorem 2 in this previous post. We will not re-state the theorem. The crucial direction is $7 \longrightarrow 4$ in that theorem. To avoid confusion, we call these two conditions A7 and A4. The following are the conditions.

A7

The product $X \times Y$ is a normal space for some space $Y$ containing a non-discrete subspace of cardinality $\kappa$.

A4

For each decreasing family $\left\{F_\alpha: \alpha<\kappa \right\}$ of closed subsets of $X$ such that $\bigcap_{\alpha<\kappa} F_\alpha=\varnothing$, there exists a family $\left\{G_\alpha: \alpha<\kappa \right\}$ of open subsets of $X$ satisfying $\bigcap_{\alpha<\kappa} G_\alpha=\varnothing$ and $F_\alpha \subset G_\alpha$ for all $\alpha<\kappa$.

Actually the proof in the previous post shows that A7 implies another condition that is equivalent to A4 for any infinite cardinal $\kappa$. In particular, A7 $\longrightarrow$ A4 would hold for the countably infinite $\kappa=\omega$. Note that under $\kappa=\omega$, A4 would be the same as condition 7 in Theorem 1 above.

Thus by Theorem 2 in this previous post for the countably infinite case and by Theorem 1 in this post, the theorem is established. $\square$

Remarks
In Theorem 2, the second factor $Y$ does not have to be a metric space. As long as it has a non-trivial convergent sequence, the normality of the product (a big if in some situation) implies countably paracompactness in the other factor.

Theorem 3 is essentially a corollary of the proof of Theorem 2 in the previous post. One way to look at Theorem 3 is that the normality of the product $X \times Y$ is a strong statement. If the product is normal and if one factor has a countable non-discrete subspace, then the other factor is countably paracompact. Another way to look at it is through the angle of Dowker spaces. By Dowker’s theorem (Theorem 1), the product of any Dowker space with any infinite compact metric space is not normal. The pathology is actually more severe. A Dowker space is severely lacking in ability to form normal product, as the following corollary makes clear.

Corollary 4
If $X$ is a Dowker space, then $X \times Y$ is not normal for any space $Y$ containing a non-discrete countable subspace.

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More Results

Two more results are discussed. According to Dowker’s theorem, the product of a countably paracompact space $X$ and any compact metric space is normal. In particular, $X \times [0,1]$ is normal. Theorem 5 is saying that with a little extra work, it can be shown that $X \times \mathbb{R}$ is normal. What makes this works is that the metric factor is $\sigma$-compact.

Theorem 5
Let $X$ be a normal space. The following conditions are equivalent.

1. The space $X$ is countably paracompact.
2. The product space $X \times Y$ is normal for any non-discrete $\sigma$-compact metric space $Y$.
3. The product space $X \times \mathbb{R}$ is normal where $\mathbb{R}$ is the real number line with the usual Euclidean topology.

Proof of Theorem 5
$1 \rightarrow 2$
Suppose that $X$ is countably paracompact. Let $Y=\bigcup_{j=1}^\infty Y_j$ where each $Y_j$ is compact. Since $Y$ is a $\sigma$-compact metric space, it is Lindelof. The Lindelof number and the weight agree in a metric space. Thus $Y$ has a countable base. According to Urysohn’s metrization theorem (discussed here), $Y$ can be embedded into the compact metric space $\prod_{j=1}^\infty W_j$ where each $W_j=[0,1]$. For convenience, we consider $Y$ as a subspace of $\prod_{j=1}^\infty W_j$. Furthermore, $X \times Y=\bigcup_{j=1}^\infty (X \times Y_j) \subset X \times\prod_{j=1}^\infty W_j$.

By Theorem 1, each $X \times Y_j$ is normal and that $X \times\prod_{j=1}^\infty W_j$ is normal. Note that $X \times Y$ is an $F_\sigma$-subset of the normal space $X \times\prod_{j=1}^\infty W_j$. Since normality is passed to $F_\sigma$-subsets, $X \times Y$ is normal.

Note. For a proof that $F_\sigma$-subsets of normal spaces are normal, see 2.7.2(b) on p. 112 of Englelking [1].

$2 \rightarrow 3$ is immediate.

$3 \rightarrow 1$
Suppose that $X \times \mathbb{R}$ is normal. Then $X \times [0,1]$ is normal since it is a closed subspace of $X \times \mathbb{R}$. By Theorem 1, $X$ is countably paracompact. $\square$

Theorem 6
Let $X$ be a normal space. Let $Y$ be a non-discrete $\sigma$-compact metric space. Then $X \times Y$ is a normal space if and only if $X \times Y$ is countably paracompact.

Proof of Theorem 6
Let $Y=\bigcup_{j=1}^\infty Y_j$ where each $Y_j$ is compact. As in the proof of Theorem 5, we use the compact metric space $\prod_{j=1}^\infty W_j$ where each $W_j=[0,1]$.

Suppose that $X \times Y$ is normal. Since $Y$ is a non-discrete metric space, $Y$ contains a countable non-discrete subspace. Then by either Theorem 2 or Theorem 3, $X$ is countably paracompact.

By Theorem 1, $X \times\prod_{j=1}^\infty W_j$ is normal. Note that $X \times \prod_{j=1}^\infty W_j \times [0,1]$ is normal since $(\prod_{j=1}^\infty W_j) \times [0,1]$ is a compact metric space. By Theorem 1 again, $X \times\prod_{j=1}^\infty W_j$ is countably paracompact.

As in the proof of Theorem 5, we can consider $Y$ as a subspace of $\prod_{j=1}^\infty W_j$. Furthermore, $X \times Y=\bigcup_{j=1}^\infty X \times Y_j \subset X \times\prod_{j=1}^\infty W_j$.

Note that $X \times Y$ is $F_\sigma$-subset of the countably paracompact space $X \times\prod_{j=1}^\infty W_j$. Since countably paracompactness is passed to $F_\sigma$-subsets, we conclude that $X \times Y$ is countably paracompact.

Note. For a proof that countably paracompactness is passed to $F_\sigma$-subsets, see the proof that paracompactness is passed to $F_\sigma$-subsets in this previous post. Just apply the same proof but start with a countable open cover.

For the other direction, suppose that $X \times Y$ is countably paracompact. Since $X \times \left\{y \right\}$ is a closed subspace of $Y$ with $y \in Y$ and is a copy of $X$, $X$ is countably paracompact. Then by Theorem 5, $X \times Y$ is a normal space. $\square$

Remarks
Theorem 5 seems like an extension of Theorem 1. But the amount of extra work is very little. So normal countably paracompact spaces are productive with not just compact metric spaces but also with $\sigma$-compact metric spaces. The $\sigma$-compactness is absolutely crucial. The product of a normal countably paracompact space with a metric space does not have to be normal. For example, the Michael line $\mathbb{M}$ is paracompact and thus countably paracompact. The product of $\mathbb{M}$ and metric space is not necessarily normal (discussed here). However, the product of $\mathbb{M}$ and $\mathbb{R}$ or other $\sigma$-compact metric space is normal.

Recall that a space is called a Dowker space if it is normal and not countably paracompact. For the type of product $X \times Y$ discussed in Theorem 6, it cannot be Dowker (if it is normal, it is countably paracompact). The two notions are the same with such product $X \times Y$. Theorem 6 actually holds for a wider class than indicated. The following is Corollary 4.3 in [2].

Theorem 7
Let $X$ be a normal space. Let $Y$ be a non-discrete metric space. Then $X \times Y$ is a normal space if and only if $X \times Y$ is countably paracompact.

So $\sigma$-compactness is not necessary for Theorem 6. However, when the metric factor is $\sigma$-compact, the proof is simplified considerably. For the full proof, see Corollary 4.3 in [2].

Among the products $X \times Y$, the two notions of normality and countably paracompactness are the same as long as one factor is normal and the other factor is a non-discrete metric space. For such product, determining normality is equivalent to determining countably paracompactness, a covering property. In showing countably paracompactness, a shrinking property as well as a condition about decreasing sequence of closed sets being expanded by open sets (see Theorem 4 and Theorem 5 in this previous post) can be used.

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Reference

1. Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
2. 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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$\copyright$ 2017 – Dan Ma

# The product of locally compact paracompact spaces

It is well known that when $X$ and $Y$ are paracompact spaces, the product space $X \times Y$ is not necessarily normal. Classic examples include the product of the Sorgenfrey line with itself (discussed here) and the product of the Michael line and the space of irrational numbers (discussed here). However, if one of the paracompact factors is “compact”, the product can be normal or even paracompact. This post discusses several classic results along this line. All spaces are Hausdorff and regular.

Suppose that $X$ and $Y$ are paracompact spaces. We have the following results:

1. If $Y$ is a compact space, then $X \times Y$ is paracompact.
2. If $Y$ is a $\sigma$-compact space, then $X \times Y$ is paracompact.
3. If $Y$ is a locally compact space, then $X \times Y$ is paracompact.
4. If $Y$ is a $\sigma$-locally compact space, then $X \times Y$ is paracompact.

The proof of the first result makes uses the tube lemma. The second result is a corollary of the first. The proofs of both results are given here. The third result is a corollary of the fourth result. We give a proof of the fourth result.

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Proof of the Fourth Result

The fourth result indicated above is restated as Theorem 2 below. It is a theorem of K. Morita [1]. This is one classic result on product of paracompact spaces. After proving the theorem, comments are made about interesting facts and properties that follow from this result. Theorem 2 is also Theorem 3.22 in chapter 18 in the Handbook of Set-Theoretic Topology [2].

A space $W$ is a locally compact space if for each $w \in W$, there is an open subset $O$ of $W$ such that $w \in O$ and $\overline{O}$ is compact. When we say $Y$ is a $\sigma$-locally compact space, we mean that $Y=\bigcup_{j=1}^\infty Y_j$ where each $Y_j$ is a locally compact space. In proving the result discussed here, we also assume that each $Y_j$ is a closed subspace of $Y$. The following lemma will be helpful.

Lemma 1
Let $Y$ be a paracompact space. Suppose that $Y$ is $\sigma$-locally compact. Then there exists a cover $\mathcal{C}=\bigcup_{j=1}^\infty \mathcal{C}_j$ of $Y$ such that each $\mathcal{C}_j$ is a locally finite family consisting of compact sets.

Proof of Lemma 1
Let $Y=\bigcup_{n=1}^\infty Y_n$ such that each $Y_n$ is closed and is locally compact. Fix an integer $n$. For each $y \in Y_n$, let $O_{n,y}$ be an open subset of $Y_n$ such that $y \in O_{n,y}$ and $\overline{O_{n,y}}$ is compact (the closure is taken in $Y_n$). Consider the open cover $\mathcal{O}=\left\{ O_{n,y}: y \in Y_j \right\}$ of $Y_n$. Since $Y_n$ is a closed subspace of $Y$, $Y_n$ is also paracompact. Let $\mathcal{V}=\left\{ V_{n,y}: y \in Y_j \right\}$ be a locally finite open cover of $Y_n$ such that $\overline{V_{n,y}} \subset O_{n,y}$ for each $y \in Y_n$ (again the closure is taken in $Y_n$). Each $\overline{V_{n,y}}$ is compact since $\overline{V_{n,y}} \subset O_{n,y} \subset \overline{O_{n,y}}$. Let $\mathcal{C}_n=\left\{ \overline{V_{n,y}}: y \in Y_n \right\}$.

We claim that $\mathcal{C}_n$ is a locally finite family with respect to the space $Y$. For each $y \in Y-Y_n$, $Y-Y_n$ is an open set containing $y$ that intersects no set in $\mathcal{C}_n$. For each $y \in Y_n$, there is an open set $O \subset Y_n$ that meets only finitely many sets in $\mathcal{C}_n$. Extend $O$ to an open subset $O_1$ of $Y$. That is, $O_1$ is an open subset of $Y$ such that $O=O_1 \cap Y_n$. It is clear that $O_1$ can only meets finitely many sets in $\mathcal{C}_n$.

Then $\mathcal{C}=\bigcup_{j=1}^\infty \mathcal{C}_j$ is the desired $\sigma$-locally finite cover of $Y$. $\square$

Theorem 2
Let $X$ be any paracompact space and let $Y$ be any $\sigma$-locally compact paracompact space. Then $X \times Y$ is paracompact.

Proof of Theorem 2
By Lemma 1, let $\mathcal{C}=\bigcup_{n=1}^\infty \mathcal{C}_n$ be a $\sigma$-locally finite cover of $Y$ such that each $\mathcal{C}_n$ consists of compact sets. To show that $X \times Y$ is paracompact, let $\mathcal{U}$ be an open cover of $X \times Y$. For each $C \in \mathcal{C}$ and for each $x \in X$, the set $\left\{ x \right\} \times C$ is obviously compact.

Fix $C \in \mathcal{C}$ and fix $x \in X$. For each $y \in C$, the point $(x,y) \in U_{y}$ for some $U_{y} \in \mathcal{U}$. Choose open $H_y \subset X$ and open $K_y \subset Y$ such that $(x,y) \in H_y \times K_y \subset U_{x,y}$. Letting $y$ vary, the open sets $H_y \times K_y$ cover the compact set $\left\{ x \right\} \times C$. Choose finitely many open sets $H_y \times K_y$ that also cover $\left\{ x \right\} \times C$. Let $H(C,x)$ be the intersection of these finitely many $H_y$. Let $\mathcal{K}(C,x)$ be the set of these finitely many $K_y$.

To summarize what we have obtained in the previous paragraph, for each $C \in \mathcal{C}$ and for each $x \in X$, there exists an open subset $H(C,x)$ containing $x$, and there exists a finite set $\mathcal{K}(C,x)$ of open subsets of $Y$ such that

• $C \subset \bigcup \mathcal{K}(C,x)$,
• for each $K \in \mathcal{K}(C,x)$, $H(C,x) \times K \subset U$ for some $U \in \mathcal{U}$.

For each $C \in \mathcal{C}$, the set of all $H(C,x)$ is an open cover of $X$. Since $X$ is paracompact, for each $C \in \mathcal{C}$, there exists a locally finite open cover $\mathcal{L}_C=\left\{L(C,x): x \in X \right\}$ such that $L(C,x) \subset H(C,x)$ for all $x$. Consider the following families of open sets.

$\mathcal{E}_n=\left\{L(C,x) \times K: C \in \mathcal{C}_n \text{ and } x \in X \text{ and } K \in \mathcal{K}(C,x) \right\}$

$\mathcal{E}=\bigcup_{n=1}^\infty \mathcal{E}_n$

We claim that $\mathcal{E}$ is a $\sigma$-locally finite open refinement of $\mathcal{U}$. First, show that $\mathcal{E}$ is an open cover of $X \times Y$. Let $(a,b) \in X \times Y$. Then for some $n$, $b \in C$ for some $C \in \mathcal{C}_n$. Furthermore, $a \in L(C,x)$ for some $x \in X$. The information about $C$ and $x$ are detailed above. For example, $C \subset \bigcup \mathcal{K}(C,x)$. Thus there exists some $K \in \mathcal{K}(C,x)$ such that $b \in K$. We now have $(a,b) \in L(C,x) \times K \in \mathcal{E}_n$.

Next we show that $\mathcal{E}$ is a refinement of $\mathcal{U}$. Fix $L(C,x) \times K \in \mathcal{E}_n$. Immediately we see that $L(C,x) \subset H(C,x)$. Since $K \in \mathcal{K}(C,x)$, $H(C,x) \times K \subset U$ for some $U \in \mathcal{U}$. Then $L(C,x) \times K \subset U$.

The remaining point to make is that each $\mathcal{E}_n$ is a locally finite family of open subsets of $X \times Y$. Let $(a,b) \in X \times Y$. Since $\mathcal{C}_n$ is locally finite in $Y$, there exists some open $Q \subset Y$ such that $b \in Q$ and $Q$ meets only finitely many sets in $\mathcal{C}_n$, say $C_1,C_2,\cdots,C_m$. Recall that $\mathcal{L}_{C_j}$ is the set of all $L(C_j,x)$ and is locally finite. Thus there exists an open $O \subset X$ such that $a \in O$ and $O$ meets only finitely many sets in each $\mathcal{L}_{C_j}$ where $j=1,2,\cdots,m$. Thus the open set $O$ meets only finitely many sets $L(C,x)$ for finitely many $C \in \mathcal{C}_n$ and finitely many $x \in X$. These finitely many $C$ and $x$ lead to finitely many $K$. Thus it follows that $O \times Q$ meets only finitely many sets $L(C,x) \times K$ in $\mathcal{E}_n$. Thus $\mathcal{E}_n$ is locally finite.

What has been established is that every open cover of $X \times Y$ has a $\sigma$-locally finite open refinement. This fact is equivalent to paracompactness (according to Theorem 1 in this previous post). This concludes the proof of the theorem. $\square$

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Productively Paracompact Spaces

Consider this property for a space $X$.

(*) The space $X$ satisfies the property that $X \times Y$ is a paracompact space for every paracompact space $Y$.

Such a space can be called a productively paracompact space (for some reason, this term is not used in the literature).

According to the four results stated at the beginning, any space in any one of the following four classes

1. Compact spaces.
2. $\sigma$-compact spaces.
3. Locally compact paracompact spaces.
4. $\sigma$-locally compact paracompact spaces.

satisfies this property. Both the Michael line and the space of the irrational numbers are examples of paracompact spaces that do not have this productively paracompact property. According to comments made on page 799 [2], the theorem of Morita (Theorem 2 here) triggered extensive research to investigate this class of spaces. The class of spaces is broader than the four classes listed here. For example, the productively paracompact spaces also include the closed images of locally compact paracompact spaces. The handbook [2] has more references.

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Normal P-Spaces

Consider this property.

(**) The space $X$ satisfies the property that $X \times Y$ is a normal space for every metric space $Y$.

These spaces can be called productively normal spaces with respect to metric spaces. They go by another name. Morita defined the notion of P-spaces and proved that a space $X$ is a normal P-space if and only if the product of $X$ with any metric space is normal.

Since the class of metric spaces contain the paracompact spaces, any space has property (*) would have property (**), i.e. a normal P-space.Thus any locally compact paracompact space is a normal P-space. Any $\sigma$-locally compact paracompact space is a normal P-space. If a paracompact space has any one of the four “compact” properties discussed here, it is a normal P-space.

Other examples of normal P-spaces are countably compact normal spaces (see here) and perfectly normal spaces (see here).

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Looking at Diagrams

Let’s compare these classes of spaces: productively paracompact spaces (the spaces satisfying property (*)), normal P-spaces and paracompact spaces. We have the following diagram.

Diagram 1

$\displaystyle \begin{array}{ccccc} \text{ } &\text{ } & \text{Productively Paracompact} & \text{ } & \text{ } \\ \text{ } & \swarrow & \text{ } & \searrow & \text{ } \\ \text{Paracompact} &\text{ } & \text{ } & \text{ } & \text{Normal P-space} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \end{array}$

Clearly productively paracompact implies paracompact. As discussed in the previous section, productively paracompact implies normal P. If a space $X$ is such that the product of $X$ with every paracompact space is paracompact, then the product of $X$ with every metric space is paracompact and hence normal.

However, the arrows in Diagram 1 are not reversible. The Michael line mentioned at the beginning will shed some light on this point. Here’s the previous post on Michael line. Let $\mathbb{M}$ be the Michael line. Let $\mathbb{P}$ be the space of the irrational numbers. The space $\mathbb{M}$ would be a paracompact space that is not productively paracompact since its product with $\mathbb{P}$ is not normal, hence not paracompact.

On the other hand, the space of irrational numbers $\mathbb{P}$ is a normal P-space since it is a metric space. But it is not productively paracompact since its product with the Michael line $\mathbb{M}$ is not normal, hence not paracompact.

The two classes of spaces at the bottom of Diagram 1 do not relate. The Michael line $\mathbb{M}$ is a paracompact space that is not a normal P-space since its product with $\mathbb{P}$ is not normal. Normal P-space does not imply paracompact. Any space that is normal and countably compact is a normal P-space. For example, the space $\omega_1$, the first uncountable ordinal, with the ordered topology is normal and countably compact and is not paracompact.

There are other normal P-spaces that are not paracompact. For example, Bing’s Example H is perfectly normal and not paracompact. As mentioned in the previous section, any perfectly normal space is a normal P-space.

The class of spaces whose product with every paracompact space is paracompact is stronger than both classes of paracompact spaces and normal P-spaces. It is a strong property and an interesting class of spaces. It is also an excellent topics for any student who wants to dig deeper into paracompact spaces.

Let’s add one more property to Diagram 1.

Diagram 2

$\displaystyle \begin{array}{ccccc} \text{ } &\text{ } & \text{Productively Paracompact} & \text{ } & \text{ } \\ \text{ } & \swarrow & \text{ } & \searrow & \text{ } \\ \text{Paracompact} &\text{ } & \text{ } & \text{ } & \text{Normal P-space} \\ \text{ } & \searrow & \text{ } & \swarrow & \text{ } \\ \text{ } &\text{ } & \text{Normal Countably Paracompact} & \text{ } & \text{ } \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \end{array}$

All properties in Diagram 2 except for paracompact are productive. Normal countably paracompact spaces are productive. According to Dowker’s theorem, the product of any normal countably paracompact space with any compact metric space is normal (see Theorem 1 in this previous post). The last two arrows in Diagram 2 are also not reversible.

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Reference

1. Morita K., On the Product of Paracompact Spaces, Proc. Japan Acad., Vol. 39, 559-563, 1963.
2. 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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$\copyright$ 2017 – Dan Ma

# The product of a perfectly normal space and a metric space is perfectly normal

The previous post gives a positive result for normality in product space. It shows that the product of a normal countably compact space and a metric space is always normal. In this post, we discuss another positive result, which is the following theorem.

Main Theorem
If $X$ is a perfectly normal space and $Y$ is a metric space, then $X \times Y$ is a perfectly normal space.

As a result of this theorem, perfectly normal spaces belong to a special class of spaces called P-spaces. K. Morita defined the notion of P-space and he proved that a space $Y$ is a Normal P-space if and only if $X \times Y$ is normal for every metric space $X$ (see the section below on P-spaces). Thus any perfectly normal space is a Normal P-space.

All spaces under consideration are Hausdorff. A subset $A$ of the space $X$ is a $G_\delta$-subset of the space $X$ if $A$ is the intersection of countably many open subsets of $X$. A subset $B$ of the space $X$ is an $F_\sigma$-subset of the space $X$ if $B$ is the union of countably many closed subsets of $X$. Clearly, a set $A$ is a $G_\delta$-subset of the space $X$ if and only if $X-A$ is an $F_\sigma$-subset of the space $X$.

A space $X$ is said to be a perfectly normal space if $X$ is normal with the additional property that every closed subset of $X$ is a $G_\delta$-subset of $X$ (or equivalently every open subset of $X$ is an $F_\sigma$-subset of $X$).

The perfect normality has a characterization in terms of zero-sets and cozero-sets. A subset $A$ of the space $X$ is said to be a zero-set if there exists a continuous function $f: X \rightarrow [0,1]$ such that $A=f^{-1}(0)$, where $f^{-1}(0)=\left\{x \in X: f(x)=0 \right\}$. A subset $B$ of the space $X$ is a cozero-set if $X-B$ is a zero-set, or more explicitly if there is a continuous function $f: X \rightarrow [0,1]$ such that $B=\left\{x \in X: f(x)>0 \right\}$.

It is well known that the space $X$ is perfectly normal if and only if every closed subset of $X$ is a zero-set, equivalently every open subset of $X$ is a cozero-set. See here for a proof of this result. We use this result to show that $X \times Y$ is perfectly normal.

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The Proof

Let $X$ be a perfectly normal space and $Y$ be a metric space. Since $Y$ is a metric space, let $\mathcal{B}=\bigcup_{j=1}^\infty \mathcal{B}_j$ be a base for $Y$ such that each $\mathcal{B}_j$ is locally finite. We show that $X \times Y$ is perfectly normal. To that end, we show that every open subset of $X \times Y$ is a cozero-set. Let $U$ be an open subset of $X \times Y$.

For each $(x,y) \in X \times Y$, there exists open $O_{x,y} \subset X$ and there exists $B_{x,y} \in \mathcal{B}$ such that $(x,y) \in O_{x,y} \times B_{x,y} \subset U$. Then $U$ is the union of all sets $O_{x,y} \times B_{x,y}$. Observe that $B_{x,y} \in \mathcal{B}_{j}$ for some integer $j$. For each $B \in \mathcal{B}$ such that $B=B_{x,y}$ for some $(x,y) \in X \times Y$, let $O(B)$ be the union of all corresponding open sets $O_{x,y}$ for all applicable $(x,y)$.

For each positive integer $j$, let $\mathcal{W}_j$ be the collection of all open sets $O(B) \times B$ such that $B \in \mathcal{B}_j$ and $B=B_{x,y}$ for some $(x,y) \in X \times Y$. Let $\mathcal{V}_j=\cup \mathcal{W}_j$. As a result, $U=\bigcup_{j=1}^\infty \mathcal{V}_j$.

Since both $X$ and $Y$ are perfectly normal, for each $O(B) \times B \in \mathcal{W}_j$, there exist continuous functions

$F_{O(B),j}: X \rightarrow [0,1]$

$G_{B,j}: Y \rightarrow [0,1]$

such that

$O(B)=\left\{x \in X: F_{O(B),j}(x) >0 \right\}$

$B=\left\{y \in Y: G_{B,j}(y) >0 \right\}$

Now define $H_j: X \times Y \rightarrow [0,1]$ by the following:

$\displaystyle H_j(x,y)=\sum \limits_{O(B) \times B \in \mathcal{W}_j} F_{O(B),j}(x) \ G_{B,j}(y)$

for all $(x,y) \in X \times Y$. Note that the function $H_j$ is well defined. Since $\mathcal{B}_j$ is locally finite in $Y$, $\mathcal{W}_j$ is locally finite in $X \times Y$. Thus $H_j(x,y)$ is obtained by summing a finite number of values of $F_{O(B),j}(x) \ G_{B,j}(y)$. On the other hand, it can be shown that $H_j$ is continuous for each $j$. Based on the definition of $H_j$, it can be readily verified that $H_j(x,y)>0$ for all $(x,y) \in \cup \mathcal{W}_j$ and $H_j(x,y)=0$ for all $(x,y) \notin \cup \mathcal{W}_j$.

Define $H: X \times Y \rightarrow [0,1]$ by the following:

$\displaystyle H(x,y)=\sum \limits_{j=1}^\infty \biggl[ \frac{1}{2^j} \ \frac{H_j(x,y)}{1+H_j(x,y)} \biggr]$

It is clear that $H$ is continuous. We claim that $U=\left\{(x,y) \in X \times Y: H(x,y) >0 \right\}$. Recall that the open set $U$ is the union of all $O(B) \times B \in \mathcal{W}_j$ for all $j$. Thus if $(x,y) \in \cup \mathcal{W}_j$ for some $j$, then $H(x,y)>0$ since $H_j(x,y)>0$. If $(x,y) \notin \cup \mathcal{W}_j$ for all $j$, $H(x,y)=0$ since $H_j(x,y)=0$ for all $j$. Thus the open set $U$ is an $F_\sigma$-subset of $X \times Y$. This concludes the proof that $X \times Y$ is perfectly normal. $\square$

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Remarks

The main theorem here is a classic result in general topology. An alternative proof is to show that any perfectly normal space is a P-space (definition given below). Then by Morita’s theorem, the product of any perfectly normal space and any metric space is normal (Theorem 1 below). For another proof that is elementary, see Lemma 7 in this previous post.

The notions of perfectly normal spaces and paracompact spaces are quite different. By the theorem discussed here, perfectly normal spaces are normally productive with metric spaces. It is possible for a paracompact space to have a non-normal product with a metric space. The classic example is the Michael line (discussed here).

On the other hand, there are perfectly normal spaces that are not paracompact. One example is Bing’s Example H, which is perfectly normal and not paracompact (see here).

Even though a perfectly normal space is normally productive with metric spaces, it cannot be normally productive in general. For each non-discrete perfectly normal space $X$, there exists a normal space $Y$ such that $X \times Y$ is not normal. This follows from Morita’s first conjecture (now a true statement). Morita’s first conjecture is discussed here.

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P-Space in the Sense of Morita

Morita defined the notion of P-spaces [1] and [2]. Let $\kappa$ be a cardinal number such that $\kappa \ge 1$. Let $\Gamma$ be the set of all finite ordered sequences $(\alpha_1,\alpha_2,\cdots,\alpha_n)$ where $n=1,2,\cdots$ and all $\alpha_i < \kappa$. Let $X$ be a space. The collection $\left\{F_\sigma \subset X: \sigma \in \Gamma \right\}$ is said to be decreasing if this condition holds: $\sigma =(\alpha_1,\alpha_2,\cdots,\alpha_n)$ and $\delta =(\alpha_1,\alpha_2,\cdots,\alpha_n, \cdots, \alpha_m)$ with $n imply that $F_{\delta} \subset F_{\sigma}$. The space $X$ is a P-space if for any cardinal $\kappa \ge 1$ and for any decreasing collection $\left\{F_\sigma \subset X: \sigma \in \Gamma \right\}$ of closed subsets of $X$, there exists open set $U_\sigma$ for each $\sigma \in \Gamma$ such that the following conditions hold:

• for all $\sigma \in \Gamma$, $F_\sigma \subset U_\sigma$,
• for any infinite sequence $(\alpha_1,\alpha_2,\cdots,\alpha_n,\cdots)$ where each each finite subsequence $\sigma_n=(\alpha_1,\alpha_2,\cdots,\alpha_n)$ is an element of $\Gamma$, if $\bigcap_{n=1}^\infty F_{\sigma_n}=\varnothing$, then $\bigcap_{n=1}^\infty U_{\sigma_n}=\varnothing$.

If $\kappa=1$ where $1=\left\{0 \right\}$. Then the index set $\Gamma$ defined above can be viewed as the set of all positive integers. As a result, the definition of P-space with $\kappa=1$ implies the a condition in Dowker’s theorem (see condition 6 in Theorem 1 here). Thus any space $X$ that is normal and a P-space is countably paracompact (or countably shrinking or that $X \times Y$ is normal for every compact metric space or any other equivalent condition in Dowker’s theorem). The following is a theorem of Morita.

Theorem 1 (Morita)
Let $X$ be a space. Then $X$ is a normal P-space if and only if $X \times Y$ is normal for every metric space $Y$.

In light of Theorem 1, both perfectly normal spaces and normal countably compact spaces are P-spaces (see here). According to Theorem 1 and Dowker’s theorem, it follows that any normal P-space is countably paracompact.

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Reference

1. Morita K., On the Product of a Normal Space with a Metric Space, Proc. Japan Acad., Vol. 39, 148-150, 1963. (article information; paper)
2. Morita K., Products of Normal Spaces with Metric Spaces, Math. Ann., Vol. 154, 365-382, 1964.

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$\copyright \ 2017 \text{ by Dan Ma}$

# The product of a normal countably compact space and a metric space is normal

It is well known that normality is not preserved by taking products. When nothing is known about the spaces $X$ and $Y$ other than the facts that they are normal spaces, there is not enough to go on for determining whether $X \times Y$ is normal. In fact even when one factor is a metric space and the other factor is a hereditarily paracompact space, the product can be non-normal (discussed here). This post discusses a productive scenario – the first factor is a normal space and second factor is a metric space with the first factor having the additional property that it is countably compact. In this scenario the product is always normal. This is a well known result in general topology. The goal here is to nail down a proof for use as future reference.

Main Theorem
Let $X$ be a normal and countably compact space. Then $X \times Y$ is a normal space for every metric space $Y$.

The proof of the main theorem uses the notion of shrinkable open covers.

Remarks
The main theorem is a classic result and is often used as motivation for more advanced results for products of normal spaces. Thus we would like to present a clear and complete proof of this classic result for anyone who would like to study the topics of normality (or the lack of) in product spaces. We found that some proofs of this result in the literature are hard to follow. In A. H. Stone’s paper [2], the result is stated in a footnote, stating that “it can be shown that the topological product of a metric space and a normal countably compact space is normal, though not necessarily paracompact”. We had seen several other papers citing [2] as a reference for the result. The Handbook [1] also has a proof (Corollary 4.10 in page 805), which we feel may not be the best proof to learn from. We found a good proof in [3] using the idea of shrinking of open covers.

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The Notion of Shrinking

The key to the proof is the notion of shrinkable open covers and shrinking spaces. Let $X$ be a space. Let $\mathcal{U}$ be an open cover of $X$. The open cover of $\mathcal{U}$ is said to be shrinkable if there is an open cover $\mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\}$ of $X$ such that $\overline{V(U)} \subset U$ for each $U \in \mathcal{U}$. When this is the case, the open cover $\mathcal{V}$ is said to be a shrinking of $\mathcal{U}$. If an open cover is shrinkable, we also say that the open cover can be shrunk (or has a shrinking). Whenever an open cover has a shrinking, the shrinking is indexed by the open cover that is being shrunk. Thus if the original cover is indexed, e.g. $\left\{U_\alpha: \alpha<\kappa \right\}$, then a shrinking has the same indexing, e.g. $\left\{V_\alpha: \alpha<\kappa \right\}$.

A space $X$ is a shrinking space if every open cover of $X$ is shrinkable. Every open cover of a paracompact space has a locally finite open refinement. With a little bit of rearranging, the locally finite open refinement can be made to be a shrinking (see Theorem 2 here). Thus every paracompact space is a shrinking space. For other spaces, the shrinking phenomenon is limited to certain types of open covers. In a normal space, every finite open cover has a shrinking, as stated in the following theorem.

Theorem 1
The following conditions are equivalent.

1. The space $X$ is normal.
2. Every point-finite open cover of $X$ is shrinkable.
3. Every locally finite open cover of $X$ is shrinkable.
4. Every finite open cover of $X$ is shrinkable.
5. Every two-element open cover of $X$ is shrinkable.

The hardest direction in the proof is $1 \Longrightarrow 2$, which is established in this previous post. The directions $2 \Longrightarrow 3 \Longrightarrow 4 \Longrightarrow 5$ are immediate. To see $5 \Longrightarrow 1$, let $H$ and $K$ be two disjoint closed subsets of $X$. By condition 5, the two-element open cover $\left\{X-H,X-K \right\}$ has a shrinking $\left\{U,V \right\}$. Then $\overline{U} \subset X-H$ and $\overline{V} \subset X-K$. As a result, $H \subset X-\overline{U}$ and $K \subset X-\overline{V}$. Since the open sets $U$ and $V$ cover the whole space, $X-\overline{U}$ and $X-\overline{V}$ are disjoint open sets. Thus $X$ is normal.

In a normal space, all finite open covers are shrinkable. In general, an infinite open cover of a normal space may or may not be shrinkable. It turns out that finding a normal space with an infinite open cover that is not shrinkable is no trivial matter (see Dowker’s theorem in this previous post). However, if an open cover in a normal space is point-finite or locally finite, then it is shrinkable.

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Key Idea

We now discuss the key idea to the proof of the main theorem. Consider the product space $X \times Y$. Let $\mathcal{U}$ be an open cover of $X \times Y$. Let $M \subset Y$. The set $M$ is stable with respect to the open cover $\mathcal{U}$ if for each $x \in X$, there is an open set $O_x$ containing $x$ such that $O_x \times M \subset U$ for some $U \in \mathcal{U}$.

Let $\kappa$ be a cardinal number (either finite or infinite). A space $X$ is a $\kappa$-shrinking space if for each open cover $\mathcal{W}$ of $X$ such that the cardinality of $\mathcal{W}$ is $\le \kappa$, then $\mathcal{W}$ is shrinkable. According to Theorem 1, any normal space is 2-shrinkable.

Theorem 2
Let $\kappa$ be a cardinal number (either finite or infinite). Let $X$ be a $\kappa$-shrinking space. Let $Y$ be a paracompact space. Suppose that $\mathcal{U}$ is an open cover of $X \times Y$ such that the following two conditions are satisfied:

• Each point $y \in Y$ has an open set $V_y$ containing $y$ such that $V_y$ is stable with respect to $\mathcal{U}$.
• $\lvert \mathcal{U} \lvert = \kappa$.

Then $\mathcal{U}$ is shrinkable.

Proof of Theorem 2
Let $\mathcal{U}$ be any open cover of $X \times Y$ satisfying the hypothesis. We show that $\mathcal{U}$ has a shrinking.

For each $y \in Y$, obtain the open covers $\left\{G(U,y): U \in \mathcal{U} \right\}$ and $\left\{H(U,y): U \in \mathcal{U} \right\}$ of $X$ as follows. For each $U \in \mathcal{U}$, define the following:

$G(U,y)=\cup \left\{O: O \text{ is open in } X \text{ such that } O \times V_y \subset U \right\}$

Then $\left\{G(U,y): U \in \mathcal{U} \right\}$ is an open cover of $X$. Since $X$ is $\kappa$-shrinkable, there is an open cover $\left\{H(U,y): U \in \mathcal{U} \right\}$ of $X$ such that $\overline{H(U,y)} \subset G(U,y)$ for each $U \in \mathcal{U}$.

Now $\left\{V_y: y \in Y \right\}$ is an open cover of $Y$. By the paracompactness of $Y$, let $\left\{W_y: y \in Y \right\}$ be a locally finite open cover of $Y$ such that $\overline{W_y} \subset V_y$ for each $y \in Y$. For each $U \in \mathcal{U}$, define the following:

$W_U=\cup \left\{H(U,y) \times W_y: y \in Y \text{ such that } \overline{H(U,y) \times W_y} \subset U \right\}$

We claim that $\mathcal{W}=\left\{ W_U: U \in \mathcal{U} \right\}$ is a shrinking of $\mathcal{U}$. First it is a cover of $X \times Y$. Let $(x,t) \in X \times Y$. Then $t \in W_y$ for some $y \in Y$. There exists $U \in \mathcal{U}$ such that $x \in H(U,y)$. Note the following.

$\overline{H(U,y) \times W_y} \subset \overline{H(U,y)} \times \overline{W_y} \subset G(U,y) \times V_y \subset U$

This means that $H(U,y) \times W_y \subset W_U$. Since $(x,t) \in H(U,y) \times W_y$, $(x,t) \in W_U$. Thus $\mathcal{W}$ is an open cover of $X \times Y$.

Now we show that $\mathcal{W}$ is a shrinking of $\mathcal{U}$. Let $U \in \mathcal{U}$. To show that $\overline{W_U} \subset U$, let $(x,t) \in \overline{W_U}$. Let $L$ be open in $Y$ such that $t \in L$ and that $L$ meets only finitely many $W_y$, say for $y=y_1,y_2,\cdots,y_n$. Immediately we have the following relations.

$\forall \ i=1,\cdots,n, \ \overline{W_{y_i}} \subset V_{y_i}$

$\forall \ i=1,\cdots,n, \ \overline{H(U,y_i)} \subset G(U,y_i)$

$\forall \ i=1,\cdots,n, \ \overline{H(U,y_i) \times W_{y_i}} \subset \overline{H(U,y_i)} \times \overline{W_{y_i}} \subset G(U,y_i) \times V_{y_i} \subset U$

Then it follows that

$\displaystyle (x,t) \in \overline{\bigcup \limits_{j=1}^n H(U,y_j) \times W_{y_j}}=\bigcup \limits_{j=1}^n \overline{H(U,y_j) \times W_{y_j}} \subset U$

Thus $U \in \mathcal{U}$. This shows that $\mathcal{W}$ is a shrinking of $\mathcal{U}$. $\square$

Remark
Theorem 2 is the Theorem 3.2 in [3]. Theorem 2 is a formulation of Theorem 3.2 [3] for the purpose of proving Theorem 3 below.

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Main Theorem

Theorem 3 (Main Theorem)
Let $X$ be a normal and countably compact space. Let $Y$ be a metric space. Then $X \times Y$ is a normal space.

Proof of Theorem 3
Let $\mathcal{U}$ be a 2-element open cover of $X \times Y$. We show that $\mathcal{U}$ is shrinkable. This would mean that $X \times Y$ is normal (according to Theorem 1). To show that $\mathcal{U}$ is shrinkable, we show that the open cover $\mathcal{U}$ satisfies the two bullet points in Theorem 2.

Fix $y \in Y$. Let $\left\{B_n: n=1,2,3,\cdots \right\}$ be a base at the point $y$. Define $G_n$ as follows:

$G_n=\cup \left\{O \subset X: O \text{ is open such that } O \times B_n \subset U \text{ for some } U \in \mathcal{U} \right\}$

It is clear that $\mathcal{G}=\left\{G_n: n=1,2,3,\cdots \right\}$ is an open cover of $X$. Since $X$ is countably compact, choose $m$ such that $\left\{G_1,G_2,\cdots,G_m \right\}$ is a cover of $X$. Let $E_y=\bigcap_{j=1}^m B_j$. We claim that $E_y$ is stable with respect to $\mathcal{U}$. To see this, let $x \in X$. Then $x \in G_j$ for some $j \le m$. By the definition of $G_j$, there is some open set $O_x \subset X$ such that $x \in O_x$ and $O_x \times B_j \subset U$ for some $U \in \mathcal{U}$. Furthermore, $O_x \times E_y \subset O_x \times B_j \subset U$.

To summarize: for each $y \in Y$, there is an open set $E_y$ such that $y \in E_y$ and $E_y$ is stable with respect to the open cover $\mathcal{U}$. Thus the first bullet point of Theorem 2 is satisfied. The open cover $\mathcal{U}$ is a 2-element open cover. Thus the second bullet point of Theorem 2 is satisfied. By Theorem 2, the open cover $\mathcal{U}$ is shrinkable. Thus $X \times Y$ is normal. $\square$

Corollary 4
Let $X$ be a normal and pseudocompact space. Let $Y$ be a metric space. Then $X \times Y$ is a normal space.

The corollary follows from the fact that any normal and pseudocompact space is countably compact (see here).

Remarks
The proof of Theorem 3 actually gives a more general result. Note that the second factor only needs to be paracompact and that every point has a countable base (i.e. first countable). The first factor $X$ has to be countably compact. The shrinking requirement for $X$ is flexible – if open covers of a certain size for $X$ are shrinkable, then open covers of that size for the product are shrinkable. We have the following corollaries.

Corollary 5
Let $X$ be a $\kappa$-shrinking and countably compact space and let $Y$ be a paracompact first countable space. Then $X \times Y$ is a $\kappa$-shrinking space.

Corollary 6
Let $X$ be a shrinking and countably compact space and let $Y$ be a paracompact first countable space. Then $X \times Y$ is a shrinking space.

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Remarks

The main theorem (Theorem 3) says that any normal and countably compact space is productively normal with one class of spaces, namely the metric spaces. Thus if one wishes to find a non-normal product space with one factor being countably compact, the other factor must not be a metric space. For example, if $W=\omega_1$, the first uncountable ordinal with the ordered topology, then $W \times X$ is always normal for every metric $X$. For non-normal example, $W \times C$ is not normal for any compact space $C$ with uncountable tightness (see Theorem 1 in this previous post). Another example, $W \times L_{\omega_1}$ is not normal where $L_{\omega_1}$ is the one-point Lindelofication of a discrete space of cardinality $\omega_1$ (follows from Example 1 and Theorem 7 in this previous post).

Another comment is that normal countably paracompact spaces are examples of Normal P-spaces. K. Morita defined the notion of P-space and he proved that a space $Y$ is a Normal P-space if and only if $X \times Y$ is normal for every metric space $X$.

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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.
2. Stone A. H., Paracompactness and Product Spaces, Bull. Amer. Math. Soc., Vol. 54, 977-982, 1948. (paper)
3. Yang L., The Normality in Products with a Countably Compact Factor, Canad. Math. Bull., Vol. 41 (2), 245-251, 1998. (abstract, paper)

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$\copyright \ 2017 \text{ by Dan Ma}$