# 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

# An example of a normal but not Lindelof Cp(X)

In this post, we discuss an example of a function space $C_p(X)$ that is normal and not Lindelof (as indicated in the title). Interestingly, much more can be said about this function space. In this post, we show that there exists a space $X$ such that

• $C_p(X)$ is collectionwise normal and not paracompact,
• $C_p(X)$ is not Lindelof but contains a dense Lindelof subspace,
• $C_p(X)$ is not first countable but is a Frechet space,
• As a corollary of the previous point, $C_p(X)$ cannot contain a copy of the compact space $\omega_1+1$,
• $C_p(X)$ is homeomorphic to $C_p(X)^\omega$,
• $C_p(X)$ is not hereditarily normal,
• $C_p(X)$ is not metacompact.

A short and quick description of the space $X$ is that $X$ is the one-point Lindelofication of an uncountable discrete space. As shown below, the function space $C_p(X)$ is intimately related to a $\Sigma$-product of copies of real lines. The results listed above are merely an introduction to this wonderful example and are derived by examining the $\Sigma$-products of copies of real lines. Deep results about $\Sigma$-product of real lines abound in the literature. The references listed at the end are a small sample. Example 3.2 in [2] is another interesting illustration of this example.

We now define the domain space $X=L_\tau$. In the discussion that follows, the Greek letter $\tau$ is always an uncountable cardinal number. Let $D_\tau$ be a set with cardinality $\tau$. Let $p$ be a point not in $D_\tau$. Let $L_\tau=D_\tau \cup \left\{p \right\}$. Consider the following topology on $L_\tau$:

• Each point in $D_\tau$ an isolated point, and
• open neighborhoods at the point $p$ are of the form $L_\tau-K$ where $K \subset D_\tau$ is countable.

It is clear that $L_\tau$ is a Lindelof space. The Lindelof space $L_\tau$ is sometimes called the one-point Lindelofication of the discrete space $D_\tau$ since it is a Lindelof space that is obtained by adding one point to a discrete space.

Consider the function space $C_p(L_\tau)$. See this post for general information on the pointwise convergence topology of $C_p(Y)$ for any completely regular space $Y$.

All the facts about $C_p(X)=C_p(L_\tau)$ mentioned at the beginning follow from the fact that $C_p(L_\tau)$ is homeomorphic to the $\Sigma$-product of $\tau$ many copies of the real lines. Specifically, $C_p(L_\tau)$ is homeomorphic to the following subspace of the product space $\mathbb{R}^\tau$.

$\Sigma_{\alpha<\tau}\mathbb{R}=\left\{ x \in \mathbb{R}^\tau: x_\alpha \ne 0 \text{ for at most countably many } \alpha<\tau \right\}$

Thus understanding the function space $C_p(L_\tau)$ is a matter of understanding a $\Sigma$-product of copies of the real lines. First, we establish the homeomorphism and then discuss the properties of $C_p(L_\tau)$ indicated above.

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The Homeomorphism

For each $f \in C_p(L_\tau)$, it is easily seen that there is a countable set $C \subset D_\tau$ such that $f(p)=f(y)$ for all $y \in D_\tau-C$. Let $W_0=\left\{f \in C_p(L_\tau): f(p)=0 \right\}$. Then each $f \in W_0$ has non-zero values only on a countable subset of $D_\tau$. Naturally, $W_0$ and $\Sigma_{\alpha<\tau}\mathbb{R}$ are homeomorphic.

We claim that $C_p(L_\tau)$ is homeomorphic to $W_0 \times \mathbb{R}$. For each $f \in C_p(L_\tau)$, define $h(f)=(f-f(p),f(p))$. Here, $f-f(p)$ is the function $g \in C_p(L_\tau)$ such that $g(x)=f(x)-f(p)$ for all $x \in L_\tau$. Clearly $h(f)$ is well-defined and $h(f) \in W_0 \times \mathbb{R}$. It can be readily verified that $h$ is a one-to-one map from $C_p(L_\tau)$ onto $W_0 \times \mathbb{R}$. It is not difficult to verify that both $h$ and $h^{-1}$ are continuous.

We use the notation $X_1 \cong X_2$ to mean that the spaces $X_1$ and $X_2$ are homeomorphic. Then we have:

$C_p(L_\tau) \ \cong \ W_0 \times \mathbb{R} \ \cong \ (\Sigma_{\alpha<\tau}\mathbb{R}) \times \mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$

Thus $C_p(L_\tau) \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R}$. This completes the proof that $C_p(L_\tau)$ is topologically the $\Sigma$-product of $\tau$ many copies of the real lines.

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Looking at the $\Sigma$-Product

Understanding the function space $C_p(L_\tau)$ is now reduced to the problem of understanding a $\Sigma$-product of copies of the real lines. Most of the facts about $\Sigma$-products that we need have already been proved in previous blog posts.

In this previous post, it is established that the $\Sigma$-product of separable metric spaces is collectionwise normal. Thus $C_p(L_\tau)$ is collectionwise normal. The $\Sigma$-product of spaces, each of which has at least two points, always contains a closed copy of $\omega_1$ with the ordered topology (see the lemma in this previous post). Thus $C_p(L_\tau)$ contains a closed copy of $\omega_1$ and hence can never be paracompact (and thus not Lindelof).

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Consider the following subspace of the $\Sigma$-product $\Sigma_{\alpha<\tau}\mathbb{R}$:

$\sigma_\tau=\left\{ x \in \Sigma_{\alpha<\tau}\mathbb{R}: x_\alpha \ne 0 \text{ for at most finitely many } \alpha<\tau \right\}$

In this previous post, it is shown that $\sigma_\tau$ is a Lindelof space. Though $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is not Lindelof, it has a dense Lindelof subspace, namely $\sigma_\tau$.

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A space $Y$ is first countable if there exists a countable local base at each point $y \in Y$. A space $Y$ is a Frechet space (or is Frechet-Urysohn) if for each $y \in Y$, if $y \in \overline{A}$ where $A \subset Y$, then there exists a sequence $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $A$ such that the sequence converges to $y$. Clearly, any first countable space is a Frechet space. The converse is not true (see Example 1 in this previous post).

For any uncountable cardinal number $\tau$, the product $\mathbb{R}^\tau$ is not first countable. In fact, any dense subspace of $\mathbb{R}^\tau$ is not first countable. In particular, the $\Sigma$-product $\Sigma_{\alpha<\tau}\mathbb{R}$ is not first countable. In this previous post, it is shown that the $\Sigma$-product of first countable spaces is a Frechet space. Thus $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is a Frechet space.

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As a corollary of the previous point, $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ cannot contain a homeomorphic copy of any space that is not Frechet. In particular, it cannot contain a copy of any compact space that is not Frechet. For example, the compact space $\omega_1+1$ is not embeddable in $C_p(L_\tau)$. The interest in compact subspaces of $C_p(L_\tau) \cong \Sigma_{\alpha<\tau}\mathbb{R}$ is that any compact space that is topologically embeddable in a $\Sigma$-product of real lines is said to be Corson compact. Thus any Corson compact space is a Frechet space.

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It can be readily verified that

$\Sigma_{\alpha<\tau}\mathbb{R} \ \cong \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \Sigma_{\alpha<\tau}\mathbb{R} \ \times \ \cdots \ \text{(countably many times)}$

Thus $C_p(L_\tau) \cong C_p(L_\tau)^\omega$. In particular, $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$ due to the following observation:

$C_p(L_\tau) \times C_p(L_\tau) \cong C_p(L_\tau)^\omega \times C_p(L_\tau)^\omega \cong C_p(L_\tau)^\omega \cong C_p(L_\tau)$

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As a result of the peculiar fact that $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$, it can be concluded that $C_p(L_\tau)$, though normal, is not hereditarily normal. This follows from an application of Katetov’s theorem. The theorem states that if $Y_1 \times Y_2$ is hereditarily normal, then either $Y_1$ is perfectly normal or every countably infinite subset of $Y_2$ is closed and discrete (see this previous post). The function space $C_p(L _\tau)$ is not perfectly normal since it contains a closed copy of $\omega_1$. On the other hand, there are plenty of countably infinite subsets of $C_p(L _\tau)$ that are not closed and discrete. As a Frechet space, $C_p(L _\tau)$ has many convergent sequences. Each such sequence without the limit is a countably infinite set that is not closed and discrete. As an example, let $\left\{x_1,x_2,x_3,\cdots \right\}$ be an infinite subset of $D_\tau$ and consider the following:

$C=\left\{f_n: n=1,2,3,\cdots \right\}$

where $f_n$ is such that $f_n(x_n)=n$ and $f_n(x)=0$ for each $x \in L_\tau$ with $x \ne x_n$. Note that $C$ is not closed and not discrete since the points in $C$ converge to $g \in \overline{C}$ where $g$ is the zero-function. Thus $C_p(L_\tau) \cong C_p(L_\tau) \times C_p(L_\tau)$ is not hereditarily normal.

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It is well known that collectionwise normal metacompact space is paracompact (see Theorem 5.3.3 in [4] where metacompact is referred to as weakly paracompact). Since $C_p(L_\tau)$ is collectionwise normal and not paracompact, $C_p(L_\tau)$ can never be metacompact.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Bella, A., Masami, S., Tight points of Pixley-Roy hyperspaces, Topology Appl., 160, 2061-2068, 2013.
3. Corson, H. H., Normality in subsets of product spaces, Amer. J. Math., 81, 785-796, 1959.
4. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A theorem about CCC spaces

It is a well known result in general topology that in any regular space with the countable chain condition, paracompactness and the Lindelof property are equivalent. The proof of this result hinges on one theorem about the spaces with the countable chain condition. In this post we are to put the spotlight on this theorem (Theorem 1 below) and then use it to prove a few results. These results indicate that in a space with the countable chain condition with some weaker covering property is either Lindelof or paracompact.

This post is centered on a theorem about the CCC property (Theorem 1 and Theorem 1a below). So it can be considered as a continuation of a previous post on CCC called Some basic properties of spaces with countable chain condition. The results that are derived from Theorem 1 are also found in [2]. But the theorem concerning CCC is only a small part of that paper among several other focuses. In this post, the exposition is to explain several interesting theorems that are derived from Theorem 1. One of the theorems is the statement that every locally compact metacompact perfectly normal space is paracompact, a theorem originally proved by Arhangelskii (see Theorem 11 below).

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CCC Spaces

All spaces under consideration are at least $T_1$ and regular. A space $X$ is said to have the countable chain condition (to have the CCC for short) if $\mathcal{U}$ is a disjoint collection of non-empty open subsets of $X$ (meaning that for any $A,B \in \mathcal{U}$ with $A \ne B$, we have $A \cap B=\varnothing$), then $\mathcal{U}$ is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of making a statement or stating a result, if $X$ has the CCC, we also say that $X$ is a CCC space or $X$ is CCC.

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The theorem of CCC spaces we want to discuss has to do with collections of open sets that are “nice”. We first define what we mean by nice. Let $\mathcal{A}$ be a collection of non-empty subsets of the space $X$. The collection $\mathcal{A}$ is said to be point-finite (point-countable) if each point of $X$ belongs to only finitely (countably) many sets in $\mathcal{A}$.

Now we define what we mean by “nice” collection of open sets. The collection $\mathcal{A}$ is said to be locally finite (locally countable) at a point $x \in X$ if there exists an open set $O \subset X$ with $x \in O$ such that $O$ meets at most finitely (countably) many sets in $\mathcal{A}$. The collection $\mathcal{A}$ is said to be locally finite (locally countable) if it is locally finite (locally countable) at each $x \in X$.

The property of being a separable space implies the CCC. The reverse is not true. However the CCC property is still a very strong property. The CCC property is equivalent to the property that if a collection of non-empty open sets is “nice” on a dense set of points, then the collection of open sets is a countable collection. The following is a precise statement.

Theorem 1

Let $X$ be a CCC space. Then if $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that the following set

$D(\mathcal{U})=\left\{x \in X: \mathcal{U} \text{ is locally-countable at } x \right\}$

is dense in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

The collections of open sets in the above theorem do not have to be open covers. However, if they are open covers, the theorem can tie CCC spaces with some covering properties. As long as the space has the CCC, any open cover that is locally-countable on a dense set must be countable. Looking at it in the contrapositive angle, in a CCC space, any uncountable open cover is not locally-countable in some open set.

Proof of Theorem 1
Let $\mathcal{U}$ be a collection of open subsets of $X$ such that the set $D(\mathcal{U})$ as defined above is dense in the open subspace $\bigcup \mathcal{U}$. We show that $\mathcal{U}$ is countable. Suppose not.

For each $U \in \mathcal{U}$, since $U \cap D(\mathcal{U}) \ne \varnothing$, we can choose a non-empty open set $f(U) \subset U$ such that $f(U)$ has non-empty intersection with only countably many sets in $\mathcal{U}$. Let $\mathcal{U}_f$ be the following collection:

$\mathcal{U}_f=\left\{f(U): U \in \mathcal{U} \right\}$

For $H,K \in \mathcal{U}_f$, by a chain from $H$ to $K$, we mean a finite collection

$\left\{W_1,W_2,\cdots,W_n \right\} \subset \mathcal{U}_f$

such that $H=W_1$, $K=W_n$ and $W_j \cap W_{j+1} \ne \varnothing$ for any $1 \le j . For each open set $W \in \mathcal{U}_f$, define $\mathcal{C}(W)$ and $\mathcal{E}(W)$ as follows:

$\mathcal{C}(W)=\left\{V \in \mathcal{U}_f: \text{there exists a chain from } W \text{ to } V \right\}$

$\mathcal{E}(W)=\bigcup \mathcal{C}(W)$

One observation we make is that for $W_1,W_2 \in \mathcal{U}_f$, if $\mathcal{E}(W_1) \cap \mathcal{E}(W_2) \ne \varnothing$, then $\mathcal{C}(W_1)=\mathcal{C}(W_2)$ and $\mathcal{E}(W_1)=\mathcal{E}(W_2)$. So the distinct $\mathcal{E}(W)$ are pairwise disjoint. Because the space $X$ has the CCC, there can be only countably many distinct open sets $\mathcal{E}(W)$. Thus there can be only countably many distinct collections $\mathcal{C}(W)$.

Note that each $\mathcal{C}(W)$ is a countable collection of open sets. Each $V \in \mathcal{U}_f$ meets only countably many open sets in $\mathcal{U}$. So each $V \in \mathcal{U}_f$ can meet only countably many sets in $\mathcal{U}_f$, since for each $V \in \mathcal{U}_f$, $V \subset U$ for some $U \in \mathcal{U}$. Thus for each $W \in \mathcal{U}_f$, in considering all finite-length chain starting from $W$, there can be only countably many open sets in $\mathcal{U}_f$ that can be linked to $W$. Thus $\mathcal{C}(W)$ must be countable. In taking the union of all $\mathcal{C}(W)$, we get back the collection $\mathcal{U}_f$. Thus we have:

$\mathcal{U}_f=\bigcup \limits_{W \in \mathcal{U}_f} \mathcal{C}(W)$

Because the space $X$ is CCC, there are only countably many distinct collections $\mathcal{C}(W)$ in the above union. Each $\mathcal{C}(W)$ is countable. So $\mathcal{U}_f$ is a countable collection of open sets.

Furthermore, each $U \in \mathcal{U}$ contains at least one set in $\mathcal{U}_f$. From the way we choose sets in $\mathcal{U}_f$, we see that for each $V \in \mathcal{U}_f$, $V=f(U) \subset U$ for at most countably many $U \in \mathcal{U}$. The argument indicates that we have a one-to-countable mapping from $\mathcal{U}_f$ to $\mathcal{U}$. Thus the original collection $\mathcal{U}$ must be countable. $\blacksquare$

The property in Theorem 1 is actually equivalent to the CCC property. Just that the proof of Theorem 1 represents the hard direction that needs to be proved. Theorem 1 can be expanded to be the following theorem.

Theorem 1a

Let $X$ be a space. Then the following conditions are equivalent.

1. The space $X$ has the CCC.
2. If $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that the following set

$D(\mathcal{U})=\left\{x \in X: \mathcal{U} \text{ is locally-countable at } x \right\}$

is dense in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

3. If $\mathcal{U}$ is a collection of non-empty open subsets of $X$ such that $\mathcal{U}$ is locally-countable at every point in the open subspace $\bigcup \mathcal{U}$, then $\mathcal{U}$ must be countable.

The direction $1 \rightarrow 2$ has been proved above. The directions $2 \rightarrow 3$ and $3 \rightarrow 1$ are straightforward.

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Tying Theorem 1 to “Nice” Open Covers

One easy application of Theorem 1 is to tie it to locally-finite and locally-countable open covers. We have the following theorem.

Theorem 2

In any CCC space, any locally-countable open cover must be countable. Thus any locally-finite open cover must also be countable.

Theorem 2 gives the well known result that any CCC paracompact space is Lindelof (see Theorem 5 below). In fact, Theorem 2 gives the result that any CCC para-Lindelof space is Lindelof (see Theorem 6 below). A space $X$ is para-Lindelof if every open cover has a locally-countable open refinement.

Can Theorem 2 hold for point-finite covers (or point-countable covers)? The answer is no (see Example 1 below). With the additional property of having a Baire space, we have the following theorem.

Theorem 3

In any Baire space with the CCC, any point-finite open cover must be countable.

A Space $X$ is a Baire space if $U_1,U_2,U_3,\cdots$ are dense open subsets of $X$, then $\bigcap \limits_{j=1}^\infty U_j \ne \varnothing$. For more information about Baire spaces, see this previous post.
.

Proof of Theorem 3
Let $X$ be a Baire space with the CCC. Let $\mathcal{U}$ be a point-finite open cover of $X$. Suppose that $\mathcal{U}$ is uncountable. We show that this assumption with lead to a contradiction. Thus $\mathcal{U}$ must be countable.

By Theorem 1, there exists an open set $V \subset X$ such that $\mathcal{U}$ is not locally-countable at any point in $V$. For each positive integer $n$, let $H_n$ be the following:

$H_n=\left\{x \in V: x \text{ is in at most } n \text{ sets in } \mathcal{U} \right\}$

Note that $V=\bigcup \limits_{j=1}^\infty H_j$. Furthermore, each $H_n$ is a closed set in the space $V$. Since $X$ is a Baire space, every non-empty open subset of $X$ is of second category (i.e. it cannot be a union of countably many closed and nowhere dense sets). Thus it cannot be that each $H_n$ is nowhere dense in $V$. For some $n$, $H_n$ is not nowhere dense. There must exist some open $W \subset V$ such that $H_n \cap W$ is dense in $W$. Because $H_n$ is closed, $W \subset H_n$.

Choose $y \in W$. The point $y$ is in at most $n$ open sets in $\mathcal{U}$. Let $U_1,U_2,\cdots,U_m \in \mathcal{U}$ such that $y \in \bigcap \limits_{j=1}^m U_j$. Clearly $1 \le m \le n$. Let $U=W \cap U_1 \cap \cdots \cap U_m$. Note that $y \in U \subset H_n \subset V$.

Every point in $U$ belongs to at most $n$ many sets in $\mathcal{U}$ and already belong to $m$ sets in $\mathcal{U}$. So each point in $U$ can belong to at most $n-m$ additional open sets in $\mathcal{U}$. Consider the case $n-m=0$ and the case $n-m>0$. We show that each case leads to a contradiction.

Suppose that $n-m=0$. Then each point of $U$ can only meet $n$ open sets in $\mathcal{U}$, namely $U_1,U_2,\cdots,U_m$. This contradicts that $\mathcal{U}$ is not locally-countable at points in $U \subset V$.

Suppose that $k=n-m>0$. Let $\mathcal{U}^*=\mathcal{U}-\left\{U_1,\cdots,U_m \right\}$. Let $\mathcal{M}$ be the following collection:

$\mathcal{M}=\left\{U \cap \bigcap \limits_{O \in M} O \ne \varnothing: M \subset \mathcal{U}^* \text{ and } \lvert M \lvert=k \right\}$

Each element of $\mathcal{M}$ is an open subset of $U$ that is the intersection of exactly $n$ many open sets in $\mathcal{U}$. So $\mathcal{M}$ is a collection of pairwise disjoint open sets. The open set $U$ as a topological space has the CCC. So $\mathcal{M}$ is at most countable. Thus the open set $U$ meets at most countably many open sets in $\mathcal{U}$, contradicting that $\mathcal{U}$ is not locally-countable at points in $U \subset V$.

Both cases $n-m=0$ and $n-m>0$ lead to contradiction. So $\mathcal{U}$ must be countable. The proof to Theorem 3 is completed. $\blacksquare$

As a corollary to Theorem 3, we have the result that every Baire CCC metacompact space is Lindelof.

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Some Applications of Theorems 2 and 3

In proving paracompactness in some of the theorems, we need a theorem involving the concept of star-countable open cover. A collection $\mathcal{A}$ of subsets of a space $X$ is said to be star-finite (star-countable) if for each $A \in \mathcal{A}$, only finitely (countably) many sets in $\mathcal{A}$ meets $A$, i.e., the following set

$\left\{B \in \mathcal{A}: B \cap A \ne \varnothing \right\}$

is finite (countable). The proof of the following theorem can be found in Engleking (see the direction (iv) implies (i) in the proof of Theorem 5.3.10 on page 326 in [1]).

Theorem 4

If every open cover of a regular space $X$ has a star-countable open refinement, then $X$ is paracompact.

As indicated in the above section, Theorem 2 and Theorem 3 have some obvious applications. We have the following theorems.

Theorem 5

Let $X$ be a CCC space. Then $X$ is paracompact if and only of $X$ is Lindelof.

Proof of Theorem 5
The direction $\Longleftarrow$ follows from the fact that any regular Lindelof space is paracompact.

The direction $\Longrightarrow$ follows from Theorem 2. $\blacksquare$

Theorem 6

Every CCC para-Lindelof space is Lindelof.

Proof of Theorem 6
This also follows from Theorem 2. $\blacksquare$

Theorem 7

Every Baire CCC metacompact space is Lindelof.

Proof of Theorem 7
Let $X$ be a Baire CCC metacompact space. Let $\mathcal{U}$ be an open cover of $X$. By metacompactness, let $\mathcal{V}$ be a point-finite open refinement of $\mathcal{U}$. By Theorem 3, $\mathcal{V}$ must be countable. $\blacksquare$

Theorem 8

Every Baire CCC hereditarily metacompact space is hereditarily Lindelof.

Proof of Theorem 8
Let $X$ be a Baire CCC hereditarily metacompact space. To show that $X$ is hereditarily Lindelof, it suffices to show that every non-empty open subset is Lindelof. Let $Y \subset X$ be open. Then $Y$ has the CCC and is also metacompact. Being a Baire space is hereditary with respect to open subspaces. So $Y$ is a Baire space too. By Theorem 7, $Y$ is Lindelof. $\blacksquare$

Theorem 9

Every locally CCC regular para-Lindelof space is paracompact.

Proof of Theorem 9
A space is locally CCC if every point has an open neighborhood that has the CCC. Let $X$ be a regular space that is locally CCC and para-Lindelof. Let $\mathcal{U}$ be an open cover of $X$. Using the locally CCC assumption and by taking a refinement of $\mathcal{U}$ if necessary, we can assume that each open set in $\mathcal{U}$ has the CCC. By the para-Lindelof assumption, let $\mathcal{V}$ be a locally-countable open refinement of $\mathcal{U}$. So each open set in $\mathcal{V}$ has the CCC too.

Now we show that $\mathcal{V}$ is star-countable. Let $V \in \mathcal{V}$. Let $\mathcal{G}$ be the following collection:

$\mathcal{G}=\left\{V \cap W: W \in \mathcal{V} \right\}$

which is is open cover of $V$. Within the subspace $V$, $\mathcal{G}$ is a locally-countable open cover. By Theorem 2, $\mathcal{G}$ must be countable. The collection $\mathcal{G}$ represents all the open sets in $\mathcal{V}$ that have non-empty intersection with $V$. Thus only countably many open sets in $\mathcal{V}$ can meet $V$. So $\mathcal{V}$ is a star-countable open refinement of $\mathcal{U}$. By Theorem 4, $X$ is paracompact. $\blacksquare$

Theorem 10

Every locally CCC regular metacompact Baire space is paracompact.

Proof of Theorem 10
Let $X$ be a regular space that is locally CCC and is a metacompact Baire space. Let $\mathcal{U}$ be an open cover of $X$. Using the locally CCC assumption and by taking a refinement of $\mathcal{U}$ if necessary, we can assume that each open set in $\mathcal{U}$ has the CCC. By the metacompact assumption, let $\mathcal{V}$ be a point-finite open refinement of $\mathcal{U}$. So each open set in $\mathcal{V}$ has the CCC too. Each open set in $\mathcal{V}$ is also a Baire space.

Now we show that $\mathcal{V}$ is star-countable. Let $V \in \mathcal{V}$. Let $\mathcal{G}$ be the following collection:

$\mathcal{G}=\left\{V \cap W: W \in \mathcal{V} \right\}$

which is is open cover of $V$. Within the subspace $V$, $\mathcal{G}$ is a point-finite open cover. By Theorem 3, $\mathcal{G}$ must be countable. The collection $\mathcal{G}$ represents all the open sets in $\mathcal{V}$ that have non-empty intersection with $V$. Thus only countably many open sets in $\mathcal{V}$ can meet $V$. So $\mathcal{V}$ is a star-countable open refinement of $\mathcal{U}$. By Theorem 4, $X$ is paracompact. $\blacksquare$

Theorem 11

Every locally compact metacompact perfectly normal space is paracompact.

Proof of Theorem 11
This follows from Theorem 10 after we prove the following two points:

• Any locally compact space is a Baire space.
• Any perfect locally compact space is locally CCC.

To see the first point, let $Y$ be a locally compact space. Let $W_1,W_2,W_3,\cdots$ be dense open sets in $Y$. Let $y \in Y$ and let $W \subset Y$ be open such that $y \in W$ and $\overline{W}$ is compact. We show that $W$ contains a point that belongs to all $W_n$. Let $X_1=W \cap W_1$, which is open and non-empty. Next choose non-empty open $X_2$ such that $\overline{X_2} \subset X_1$ and $X_2 \subset W_2$. Next choose non-empty open $X_3$ such that $\overline{X_3} \subset X_2$ and $X_3 \subset W_3$. Continue in this manner, we have a sequence of open sets $X_1,X_2,X_3,\cdots$ such that for each $n$, $\overline{X_{n+1}} \subset X_n$ and $\overline{X_n}$ is compact. The intersection of all the $X_n$ is non-empty. The points in the intersection must belong to each $W_n$.

To see the second point, let $Y$ be a locally compact space such that every closed set is a $G_\delta$-set. Suppose that $Y$ is not locally CCC at $y \in Y$. Let $U \subset Y$ be open such that $y \in U$ and $\overline{U}$ is compact. Then $U$ must not have the CCC. Let $\left\{U_\alpha: \alpha<\omega_1 \right\}$ be a pairwise disjoint collection of open subsets of $U$. Let $O=\bigcup \limits_{\alpha<\omega_1} U_\alpha$ and let $C=Y-O$.

Let $C=\bigcap \limits_{n=1}^\infty V_n$ where each $V_n$ is open in $Y$ and $V_{n+1} \subset V_n$ for each integer $n$. For each $\alpha<\omega_1$, pick $y_\alpha \in U_\alpha$. For each $y_\alpha$, there is some integer $f(\alpha)$ such that $y_\alpha \notin V_{f(\alpha)}$. So there must exist some integer $n$ such that $A=\left\{y_\alpha: f(\alpha)=n \right\}$ is uncountable.

The set $A$ is an infinite subset of the compact set $\overline{U}$. So $A$ has a limit point, say $p$ (also called cluster point). Clearly $p \notin O$. So $p \in C$. In particular, $p \in V_n$. Then $V_n$ contains some points of $A$. But for any $y_\alpha \in A$, $y_\alpha \notin V_n=V_{f(\alpha)}$, a contradiction. So $Y$ must be locally CCC at each $y \in Y$. $\blacksquare$

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Some Examples

Example 1
A CCC space $X$ with an uncountable point-finite open covers. This example demonstrates that in Theorem 2, locally-finite or locally-countable cannot be replaced by point-finite. Consider the following product space:

$Y=\prod \limits_{\alpha < \omega_1} \left\{0,1 \right\}=\left\{0,1 \right\}^{\omega_1}$

i.e, the product space of $\omega_1$ many copies of the two-point discrete space $\left\{0,1 \right\}$. Let $X$ be the set of all points $h \in Y$ such that $h(\alpha)=1$ for only finitely many $\alpha<\omega_1$.

The product space $Y$ is the product of separable spaces, hence has the CCC. The space $X$ is dense in $Y$. Hence $X$ has the CCC. For each $\alpha<\omega_1$, define $U_\alpha$ as follows:

$U_\alpha=\left\{h \in X: h(\alpha)=1 \right\}$

Then $\left\{U_\alpha:\alpha<\omega_1 \right\}$ is a point-finite open cover of $X$. Of course, $X$ in this example is not a Baire space. $\blacksquare$

The following three examples center around the four properties in Theorem 7 (Baire + CCC + metacompact imply Lindelof). These examples show that each property in the hypothesis is crucial.

Example 2
A separable non-Lindelof space that is a Baire space. This example shows that the metacompact assumption is crucial for Theorem 7.

The example is the Sorgenfrey plane $S \times S$ where $S$ is the real line with the Sorgenfrey topology (generated by the half-open intervals of the form $[a,b)$). It is well known that $S \times S$ is not Lindelof. The Sorgenfrey plane is Baire and is separable (hence CCC). Furthermore, $S \times S$ is not metacompact (if it were, it would be Lindelof by Theorem 7). $\blacksquare$

Example 3
A non-Lindelof metacompact Baire space $M$. This example shows that the CCC assumption in Theorem 7 is necessary.

This space $M$ is the subspace of Bing’s Example G that has finite support (defined and discussed in the post A subspace of Bing’s example G. It is normal and not collectionwise normal (hence cannot be Lindelof) and metacompact. The space $M$ does not have CCC since it has uncountably many isolated points. Any space with a dense set of isolated points is a Baire space. Thus the space $M$ is also a Baire space. $\blacksquare$

Example 4
A non-Lindelof CCC metacompact non-Baire space $W$. This example shows that the Baire space assumption in Theorem 7 is necessary.

Let $W$ be the set of all non-empty finite subsets of the real line with the Pixley-Roy topology. Note that $W$ is non-Lindelof and has the CCC and is metacompact. Of course it is not Baire. For more information on Pixley-Roy spaces, see the post called Pixley-Roy hyperspaces. For the purpose of this example, the Pixley-Roy space can be built on any uncountable separable metrizable space.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Tall, F. D., The Countable Chain Condition Versus Separability – Applications of Martin’s Axiom, Gen. Top. Appl., 4, 315-339, 1974.

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$\copyright \ 2014 \text{ by Dan Ma}$

# One way to find collectionwise normal spaces

Collectionwise normality is a property that is weaker than paracompactness and stronger than normality (see the implications below). Normal spaces need not be collectionwise normal. Bing’s Example G is an example of a normal and not collectionwise normal space (see the blog post “Bing’s Example G”). We discuss one instance when normal spaces are collectionwise normal, giving a way to obtain collectionwise normal spaces that are not paracompact.

$\text{ }$
$\text{paracompact} \Longrightarrow \text{collectionwise normal} \Longrightarrow \text{normal}$

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Collectionwise Normal Spaces

A normal space is one in which any two disjoint closed sets can be separated by disjoint open sets. By induction, in a normal space any finite number of disjoint closed sets can be separated by disjoint open sets. Of course, the inductive reasoning cannot be carried over to the case of infinitely many disjoint closed sets. In the real line with the usual topology, the singleton sets $\left\{x \right\}$, where $x$ is rational, are disjoint closed sets that cannot be simultaneously separated by disjoint open sets. In order to separate an infinite collection of disjoint closed sets, it makes sense to restrict on the type of collections of closed sets. A space $X$ is collectionwise normal if every discrete collection of closed subsets of $X$ can be separated by pairwise disjoint open subsets of $X$. The following is a more specific definition.

Definition
A space $X$ is collectionwise normal if for every discrete collection $\mathcal{A}$ of closed subsets of $X$, there exists a pairwise disjoint collection $\mathcal{U}=\left\{U_A: A \in \mathcal{A} \right\}$ of open subsets of $X$ such that $A \subset U_A$ for each $A \in \mathcal{A}$.

For more details about the definitions of collectionwise normality, see “Definitions of Collectionwise Normal Spaces”. The implications displayed above are repeated below. None of the arrows is reversible.

$\text{ }$
$\text{paracompact} \Longrightarrow \text{collectionwise normal} \Longrightarrow \text{normal}$

As indicated above, Bing’s Example G is an example of a normal and not collectionwise normal space (see the blog post “Bing’s Example G”). The propositions in the next section can be used to obtain collectionwise normal spaces that are not paracompact.

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When Normal implies Collectionwise Normal

Being able to simultaneously separate any discrete collection of closed sets is stronger than the property of merely being able to separate finite collection of disjoint closed sets. It turns out that the stronger property of collectionwise normality is required only for separating uncountable discrete collections of closed sets. As the following lemma shows, normality is sufficient to separate any countable discrete collection of closed sets.

Lemma 1
Let $X$ be a normal space. Then for every discrete collection $\left\{C_1,C_2,C_3,\cdots \right\}$ of closed subsets of $X$, there exists a pairwise disjoint collection $\left\{O_1,O_2,O_3,\cdots \right\}$ of open subsets of $X$ such that $C_i \subset O_i$ for each $i$.

Proof of Lemma 1
Let $\left\{C_1,C_2,C_3,\cdots \right\}$ be a discrete collection of closed subsets of $X$. For each $i$, choose disjoint open sets $U_i$ and $V_i$ such that $C_i \subset U_i$ and $\cup \left\{C_j: j \ne i \right\} \subset V_i$. Let $O_1=U_1$. For each $i>1$, let $O_i=U_i \cap V_1 \cap \cdots \cap V_{i-1}$. It follows that $O_i \cap O_j = \varnothing$ for all $i \ne j$. It is also clear that for each $i$, $C_i \subset O_i$. $\blacksquare$

We have the following propositions.

Proposition 2
Let $X$ be a normal space. If all discrete collections of closed subsets of $X$ are at most countable, then $X$ is collectionwise normal.

Proposition 3
Let $X$ be a normal space. If all closed and discrete subsets of $X$ are at most countable (such a space is said to have countable extent), then $X$ is collectionwise normal.

Proposition 4
Any normal and countably compact space is collectionwise normal.

Proposition 2 follows from Lemma 1. As noted in Proposition 3, any space in which all closed and discrete subsets are countable is said to have countable extent. It is easy to verify that $X$ has countable extent if and only if all discrete collections of closed subsets of $X$ are at most countable. If $X$ is a countably compact space, then every infinite subset of $X$ has a limit point. Thus Proposition 4 follows from the fact that any countably compact space has countable extent.

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Paracompact Spaces

One way to find a collectionwise normal space that is not paracompact is to find a non-paracompact space that satisfies Propositions 3, 4 or 5. For example, $\omega_1$, the space of all countable ordinals with the order topology, is not paracompact. Since $\omega_1$ is normal and countably compact, it is collectionwise normal by Proposition 4. For a basic discussion of $\omega_1$ as a topological space, see “The First Uncountable Ordinal”.

As the following theorem shows, paracompact spaces are collectionwise normal. Thus the class of collectionwise normal spaces includes all metric spaces and paracompact spaces.

Theorem 5
If a space $X$ is paracompact, then $X$ is collectionwise normal.

Proof of Theorem 5
Let $X$ be a paracompact space. Let $\mathcal{A}$ be a discrete collection of closed subsets of $X$. For each $x \in X$, let $O_x$ be open such that $x \in O_x$ and $O_x$ meets at most one element of $\mathcal{A}$. Let $\mathcal{O}=\left\{O_x: x \in X \right\}$. By the paracompactness of $X$, $\mathcal{O}$ has a locally finite open refinement $\mathcal{V}$.

For each $A \in \mathcal{A}$, let $W_A=\cup \left\{\overline{V}: V \in \mathcal{V} \text{ and } \overline{V} \cap A=\varnothing \right\}$ and let $U_A=X-W_A$. Each $W_A$ is a closed set since $\mathcal{V}$ is locally finite. Thus each $U_A$ is open. Furthermore, for each $A \in \mathcal{A}$, $A \subset U_A$. It is easily checked that $\left\{U_A: A \in \mathcal{A} \right\}$ is pairwise disjoint. $\blacksquare$

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Reference

1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$
Revised June 16, 2019

# A subspace of Bing’s example G

Bing’s Example G is the first example of a topological space that is normal but not collectionwise normal (see [1]). Example G was an influential example from an influential paper. The Example G and its subspaces had been extensively studied. In addition to being normal and not collectionwise normal, Example G is not perfectly normal and not metacompact. See the previous post “Bing’s Example G” for a basic discussion of Example G. In this post we focus on one subspace of Example G examined by Michael in [3]. This subspace is normal, not collectionwise normal and not perfectly normal just like Example G. However it is metacompact. In [3], Michael proved that any metacompact collectionwise normal space is paracompact (metacompact was called pointwise paracompact in that paper). This subspace of Example G demonstrates that collectionwise normality in Michael’s theorem cannot be replaced by normality.

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Bing’s Example G

For a more detailed discussion of Bing’s Example G in this blog, see the blog post “Bing’s Example G”. For the sake of completeness, we repeat the definition of Example G. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. Let $F=2^Q$ be the set of all functions $f: Q \rightarrow 2=\left\{0,1 \right\}$. Obviously $2^Q$ is simply the Cartesian product of $\lvert Q \lvert$ many copies of the two-point discrete space $\left\{0,1 \right\}$, i.e., $\prod \limits_{q \in Q} \left\{0,1 \right\}$. For each $p \in P$, define the function $f_p: Q \rightarrow 2$ by the following:

$\forall q \in Q$, $f_p(q)=1$ if $p \in q$ and $f_p(q)=0$ if $p \notin q$

Let $F_P=\left\{f_p: p \in P \right\}$. Let $\tau$ be the set of all open subsets of $2^Q$ in the product topology. The following is another topology on $2^Q$:

$\tau^*=\left\{U \cup V: U \in \tau \text{ and } V \subset 2^Q \text{ with } V \cap F_P=\varnothing \right\}$

Bing’s Example G is the set $F=2^Q$ with the topology $\tau^*$. In other words, each $x \in F-F_P$ is made an isolated point and points in $F_P$ retain the usual product open sets.

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Michael’s Subspace of Example G

For each $f \in F$, let $supp(f)$ be the support of $f$, i.e., $supp(f)=\left\{q \in Q:f(q) \ne 0 \right\}$. Michael in [3] considered the following subspace of $F$.

$M=F_P \cup \left\{f \in F: supp(f) \text{ is finite} \right\}$

Michael in [3] used the letter $G$ to denote the space $M$. We choose another letter to distinguish it from Example G. The subspace $M$ consists of all points $f_p \in F_P$ and all other $f \in F$ such that $f(q)=1$ for only finitely many $q \in Q$. The space $M$ is normal and not collectionwise Hausdorff (hence not collectionwise normal and not paracompact). By eliminating points $f \in F$ that have values of $1$ for infinitely many $q \in Q$, we obtain a subspace that is metacompact. We discuss the following points:

• The space $M$ is normal.
• The space $M$ is not collectionwise Hausdorff and hence not collectionwise normal.
• The space $M$ is metacompact.
• The space $M$ is not perfectly normal.

The space $M$ is normal since the space $F$ that is Example G is hereditarily normal (see the section called Bing’s Example G is Completely Normal in the post “Bing’s Example G”).

To show that the space $M$ is not collectionwise Hausdorff, it is helpful to first look at $M$ as a subspace of the product space $2^Q$. The product space $2^Q$ has the countable chain condition (CCC) since it is a product of separable spaces. Note that $M$ is dense in the product space $2^Q$. Thus $M$ as a subspace of the product space has the CCC.

In the space $M$, the set $F_P$ is still a closed and discrete set. In the space $M$, open sets containing points of $F_P$ are the same as product open sets in $2^Q$ relative to the set $M$. Since $M$ has CCC (as a subspace of the product space $2^Q$), $M$ cannot have uncountably many pairwise disjoint open sets containing points of $F_P$ (in either the product topology or the Example G subspace topology). It follows that $M$ is not collectionwise Hausdorff. If it were, there would be uncountably many pairwise disjoint product open sets separating points in $F_P$, which is not possible.

To see that $M$ is metacompact, let $\mathcal{U}$ be an open cover of $M$. For each $p \in P$, choose $U_p \in \mathcal{U}$ such that $f_p \in U_p$. For each $p \in P$, let $W_p=\left\{f \in M: f(\left\{p \right\})=1 \right\}$. Let $\mathcal{V}$ be the following:

$\mathcal{V}=\left\{U_p \cap W_p: p \in P \right\} \cup \left\{\left\{x \right\}: x \in M-F_P \right\}$

Note that $\mathcal{V}$ is a point-finite open refinement of $\mathcal{U}$. Each $U_p \cap W_p$ contains only one point of $F_P$, namely $f_p$. On the other hand, each $f \in M$ with finite support can belong to at most finitely many $U_p \cap W_p$.

The space $M$ is not perfectly normal. This point is alluded to in [3] by Michael and elsewhere in the literature, e.g. in Bing’s paper (see [1]) and in Engelking’s general topology text (see 5.53 on page 338 of [2]). In fact Michael indicated that one can obtain a perfectly normal example with the aforementioned properties using Example H defined in [1] instead of using the subspace $M$ defined here in this post.

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Reference

1. Bing, R. H., Metrization of Topological Spaces, Canad. J. Math., 3, 175-186, 1951.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
3. Michael, E., Point-finite and locally finite coverings, Canad. J. Math., 7, 275-279, 1955.
4. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

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# Cartesian Products of Two Paracompact Spaces – Continued

Consider the real line $\mathbb{R}$ with a topology finer than the usual topology obtained by isolating each point in $\mathbb{P}$ where $\mathbb{P}$ is the set of all irrational numbers. The real line with this finer topology is called the Michael line and we use $\mathbb{M}$ to denote this topological space. It is a classic result that $\mathbb{M} \times \mathbb{P}$ is not normal (see “Michael Line Basics”). Even though the Michael line $\mathbb{M}$ is paracompact (it is in fact hereditarily paracompact), $\mathbb{M}$ is not perfectly normal. Result 3 below will imply that the Michael line cannot be perfectly normal. Otherwise $\mathbb{M} \times \mathbb{P}$ would be paracompact (hence normal). Result 3 is the statement that if $X$ is paracompact and perfectly normal and Y is a metric space then $X \times Y$ is paracompact and perfectly normal. We also use this result to show that if $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof (see Result 4 below).

This post is a continuation of the post “Cartesian Products of Two Paracompact Spaces”. In that post, four results are listed. They are:

Result 1

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

Result 2

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Result 3

If $X$ is paracompact and perfectly normal and $Y$ is metrizable, then $X \times Y$ is paracompact and perfectly normal.

Result 4

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

Result 1 and Result 2 are proved in the previous post “Cartesian Products of Two Paracompact Spaces”. Result 3 and Result 4 are proved in this post. All spaces are assumed to be regular.

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Paracompact Spaces, Lindelof Spaces and Other Information

A paracompact space is one in which every open cover has a locally finite open refinement. The previous post “Cartesian Products of Two Paracompact Spaces” has a basic discussion on paracompact spaces. For the sake of completeness, we repeat here some of the results discussed in that post. A proof of Proposition 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).. For a proof of Proposition 2, see Theorem 3 in the previous post “Cartesian Products of Two Paracompact Spaces”. We provide a proof for Proposition 3.

Proposition 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$-locally finite open refinement.

Proposition 2
Every $F_\sigma$-subset of a paracompact space is paracompact.

Proposition 3
Any paracompact space with a dense Lindelof subspace is Lindelof.

Proof of Proposition 3
Let $L$ be a paracompact space. Let $M \subset L$ be a dense Lindelof subspace. Let $\mathcal{U}$ be an open cover of $L$. Since we are working with a regular space, let $\mathcal{V}$ be an open cover of $L$ such that $\left\{\overline{V}: V \in \mathcal{V} \right\}$ refines $\mathcal{U}$. Let $\mathcal{W}$ be a locally finite open refinement of $\mathcal{V}$. Choose $\left\{W_1,W_2,W_3,\cdots \right\} \subset \mathcal{W}$ such that it is a cover of $M$. Since $M \subset \bigcup \limits_{i=1}^\infty W_i$, $\overline{\bigcup \limits_{i=1}^\infty W_i}=L$.

Since the sets $W_i$ come from a locally finite collection, they are closure preserving. Hence we have:

$\overline{\bigcup \limits_{i=1}^\infty W_i}=\bigcup \limits_{i=1}^\infty \overline{W_i}=L$

For each $i$, choose some $U_i \in \mathcal{U}$ such that $\overline{W_i} \subset U_i$. Then $\left\{U_1,U_2,U_3,\cdots \right\}$ is a countable subcollection of $\mathcal{U}$ covering the space $L$. $\blacksquare$

A space is said to be a perfectly normal if it is a normal space with the additional property that every closed subset is a $G_\delta$-set in the space (equivalently every open subset is an $F_\sigma$-set). We need two basic results about hereditarily Lindelof spaces. A space is Lindelof if every open cover of that space has a countable subcover. A space is hereditarily Lindelof if every subspace of that space is Lindelof. Proposition 4 below, stated without proof, shows that to prove a space is hereditarily Lindelof, we only need to show that every open subspace is Lindelof.

Proposition 4
Let $L$ be a space. Then $L$ is hereditarily Lindelof if and only if every open subspace of $L$ is Lindelof.

Proposition 5
Let $L$ be a Lindelof space. Then $L$ is hereditarily Lindelof if and only if $L$ is perfectly normal.

Proof of Proposition 5
$\Rightarrow$ Suppose $L$ is hereditarily Lindelof. It is well known that regular Lindelof space is normal. Thus $L$ is normal. It remains to show that every open subset of $L$ is $F_\sigma$. Let $U \subset L$ be an non-empty open set. For each $x \in U$, let $V_x$ be open such that $x \in V_x$ and $\overline{V_x} \subset U$ (the space is assumed to be regular). By assumption, the open set $U$ is Lindelof. The open sets $V_x$ form an open cover of $U$. Thus $U$ is the union of countably many $\overline{V}_x$.

$\Leftarrow$ Suppose $L$ is perfectly normal. To show that $L$ is hereditarily Lindelof, it suffices to show that every open subset of $L$ is Lindelof (by Proposition 4). Let $U \subset L$ be non-empty open. By assumption, $U=\bigcup \limits_{i=1}^\infty F_i$ where each $F_i$ is a closed set in $L$. Since the Lindelof property is hereditary with respect to closed subsets, $U$ is Lindelof. $\blacksquare$

Another important piece of information that we need is the following metrization theorem. It shows that being a metrizable space is equivalent to have a base that is $\sigma$-locally finite. In proving Result 3, we will assume that the metric factor has such a base. This is a classic metrization theorem (see [1] or [2] or any other standard topology text).

Theorem 6
Let $X$ be a space. Then $X$ is metrizable if and only if $X$ has a $\sigma$-locally finite base.

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Result 3

Result 3 is the statement that:

If $X$ is paracompact and perfectly normal and Y is a metric space then $X \times Y$ is paracompact and perfectly normal.

Result 3 follows from the following two lemmas.

Lemma 7
If the following two conditions hold:

• every open subset of $X$ is an $F_\sigma$-set in $X$,
• $Y$ is a metric space,

then every open subset of $X \times Y$ is an $F_\sigma$-set in $X \times Y$.

Proof of Lemma 7
Let $U$ be a open subset of $X \times Y$. If $U=\varnothing$, then $U$ is certainly the union of countably many closed sets. So assume $U \ne \varnothing$. Let $\mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i$ be a base for $Y$ such that each $\mathcal{B}_i$ is locally finite in $Y$ (by Theorem 6, such a base exists since $Y$ is metrizable).

Consider all non-empty $B \in \mathcal{B}$ such that we can choose nonempty open set $W_B \subset X$ with $W_B \times \overline{B} \subset U$. Since $U$ is non-empty open, such pairs $(B, W_B)$ exist. Let $\mathcal{B}^*$ be the collection of all non-empty $B \in \mathcal{B}$ for which there is a matching non-empty $W_B$. For each $i$, let $\mathcal{B}_i^*=\mathcal{B}^* \cap \mathcal{B}_i$. Of course, each $\mathcal{B}_i^*$ is still locally finite.

Since every open subset of $X$ is an $F_\sigma$-set in $X$, for each $W_B$, we can write $W_B$ as

$W_B=\bigcup \limits_{j=1}^\infty W_{B,j}$

where each $W_{B,i}$ is closed in $X$.

For each $i=1,2,3,\cdots$ and each $j=1,2,3,\cdots$, consider the following collection:

$\mathcal{V}_{i,j}=\left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}$

Each element of $\mathcal{V}_{i,j}$ is a closed set in $X \times Y$. Since $\mathcal{B}_i^*$ is a locally finite collection in $Y$, $\mathcal{V}_{i,j}$ is a locally finite collection in $X \times Y$. Define $V_{i,j}=\bigcup \mathcal{V}_{i,j}$. The set $V_{i,j}$ is a union of closed sets. In general, the union of closed sets needs not be closed. However, $V_{i,j}$ is still a closed set in $X \times Y$ since $\mathcal{V}_{i,j}$ is a locally finite collection of closed sets. This is because a locally finite collection of sets is closure preserving. Note the following:

$\overline{V_{i,j}}=\overline{\bigcup \mathcal{V}_{i,j}}=\overline{\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}}=\bigcup \left\{\overline{W_{B,j} \times \overline{B}}: B \in \mathcal{B}_i^* \right\}$

$=\bigcup \left\{W_{B,j} \times \overline{B}: B \in \mathcal{B}_i^* \right\}=V_{i,j}$

Finally, we have $U=\bigcup \limits_{i=1}^\infty \bigcup \limits_{j=1}^\infty V_{i,j}$, which is the union of countably many closed sets. $\blacksquare$

Lemma 8
If $X$ is a paracompact space satisfying the following two conditions:

• every open subset of $X$ is an $F_\sigma$-set in $X$,
• $Y$ is a metric space,

then $X \times Y$ is paracompact.

Proof of Lemma 8
As in the proof of the above lemma, let $\mathcal{B}=\bigcup \limits_{i=1}^\infty \mathcal{B}_i$ be a base for $Y$ such that each $\mathcal{B}_i$ is locally finite in $Y$. Let $\mathcal{U}$ be an open cover of $X \times Y$. Assume that elements of $\mathcal{U}$ are of the form $A \times B$ where $A$ is open in $X$ and $B \in \mathcal{B}$.

For each $B \in \mathcal{B}$, consider the following two items:

$\mathcal{W}_B=\left\{A: A \times B \in \mathcal{U} \right\}$

$W_B=\bigcup \mathcal{W}_B$

To simplify matter, we only consider $B \in \mathcal{B}$ such that $\mathcal{W}_B \ne \varnothing$. Each $W_B$ is open in $X$ and hence by assumption an $F_\sigma$-set in $X$. Thus by Proposition 2, each $W_B$ is paracompact. Note that $\mathcal{W}_B$ is an open cover of $W_B$. Let $\mathcal{H}_B$ be a locally finite open refinement of $\mathcal{W}_B$. Consider the following two items:

For each $j=1,2,3,\cdots$, let $\mathcal{V}_j=\left\{A \times B: A \in \mathcal{H}_B \text{ and } B \in \mathcal{B}_j \right\}$

$\mathcal{V}=\bigcup \limits_{j=1}^\infty \mathcal{V}_j$

We observe that $\mathcal{V}$ is an open cover of $X \times Y$ and that $\mathcal{V}$ refines $\mathcal{U}$. Furthermore each $\mathcal{V}_j$ is a locally finite collection. The open cover $\mathcal{U}$ we start with has a $\sigma$-locally finite open refinement. Thus $X \times Y$ is paracompact. $\blacksquare$

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Result 4

Result 4 is the statement that:

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

Proof of Result 4
Suppose $X$ is hereditarily Lindelof and that $Y$ is a separable metric space. It is well known that regular Lindelof spaces are paracompact. Thus $X$ is paracompact. By Proposition 5, $X$ is perfectly normal. By Result 3, $X \times Y$ is paracompact and perfectly normal.

Let $D$ be a countable dense subset of $Y$. We can think of $D$ as a $\sigma$-compact space. The product of any Lindelof space with a $\sigma$-compact space is Lindelof (see Corollary 3 in the post “The Tube Lemma”). Thus $X \times D$ is Lindelof. Furthermore $X \times D$ is a dense Lindelof subspace of $X \times Y$. By Proposition 3, $X \times Y$ is Lindelof. By Proposition 5, $X \times Y$ is hereditarily Lindelof. $\blacksquare$

Remark
In the previous post “Bernstein Sets and the Michael Line”, a non-normal product space where one factor is Lindelof and the other factor is a separable metric space is presented. That Lindelof space is not hereditarily Lindelof (it has uncountably many isolated points). Note that by Result 4, for any such non-normal product space, the Lindelof factor cannot be hereditarily Lindelof.

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Reference

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

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$\copyright \ \ 2012$

# Cartesian Products of Two Paracompact Spaces

In some previous posts we discuss examples surrounding the Michael line showing that the product of a paracompact space and a complete metric space needs not be normal (see “Michael Line Basics”) and that the product of a Lindelof space and a separable metric space need not be normal (see “Bernstein Sets and the Michael Line”). These examples are classic counterexamples demonstrating that both paracompactness and Lindelofness are not preserved by taking two-factor cartesian products even when one of the factors is nice (complete metric space in the first example and separable metric space in the second example). We now show some positive results. Of course, these results require additional conditions on one or both of the factors. We prove the following results.

Result 1

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

Result 2

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Result 3

If $X$ is paracompact and perfectly normal and $Y$ is metrizable, then $X \times Y$ is paracompact and perfectly normal.

Result 4

If $X$ is hereditarily Lindelof and $Y$ is a separable metric space, then $X \times Y$ is hereditarily Lindelof.

With Results 1 and 2, compact spaces and $\sigma$-compact spaces can be called productively paracompact since the product of each of these spaces with any paracompact space is paracompact. We prove Result 1 and Result 2 below.

Result 3 and Result 4 are proved in another post Cartesian Products of Two Paracompact Spaces – Continued.

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Paracompact Spaces

First, recall some definitions. All spaces are at least regular (to us regular implies Hausdorff). Let $X$ be a space. A collection $\mathcal{A}$ of subsets of $X$ is said to be a cover of $X$ if $X=\bigcup \mathcal{A}$ (in words every point of the space belongs to one set in the collection). Furthermore, $\mathcal{A}$ is an open cover of $X$ is it is a cover of $X$ consisting of open subsets of $X$.

Let $\mathcal{A}$ and $\mathcal{B}$ be covers of the space $X$. The cover $\mathcal{B}$ is said to be a refinement of $\mathcal{A}$ ($\mathcal{B}$ is said to refine $\mathcal{A}$) if for every $B \in \mathcal{B}$, there is some $A \in \mathcal{A}$ such that $B \subset A$. The cover $\mathcal{B}$ is said to be an open refinement of $\mathcal{A}$ if $\mathcal{B}$ refines $\mathcal{A}$ and $\mathcal{B}$ is an open cover.

A collection $\mathcal{A}$ of subsets of $X$ is said to be a locally finite collection if for each point $x \in X$, there is a non-empty open subset $V$ of $X$ such that $x \in V$ and $V$ has non-empty intersection with at most finitely many sets in $\mathcal{A}$. An open cover $\mathcal{A}$ of $X$ is said to have a locally finite open refinement if there exists an open cover $\mathcal{C}$ of $X$ such that $\mathcal{C}$ refines $\mathcal{A}$ and $\mathcal{C}$ is a locally finite collection. We have the following definition.

Definition

The space $X$ is said to be paracompact if every open cover of $X$ has a locally finite open refinement.

A collection $\mathcal{U}$ of subsets of the space $X$ is said to be a $\sigma$-locally finite collection if $\mathcal{U}=\bigcup \limits_{i=1}^\infty \mathcal{U}_i$ such that each $\mathcal{U}_i$ is a locally finite collection of subsets of $X$. Consider the property that every open cover of $X$ has a $\sigma$-locally finite open refinement. This on the surface is a stronger property than paracompactness. However, Theorem 1 below shows that it is actually equivalent to paracompactness. The proof of Theorem 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [2] (Theorem 20.7 in page 146).

Theorem 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover $\mathcal{U}$ of $X$ has a $\sigma$-locally finite open refinement.

Theorem 2 below is another characterization of paracompactness that is useful. For a proof of Theorem 2, see “Finite and Countable Products of the Michael Line”.

Theorem 2
Let $X$ be a regular space. Then $X$ is paracompact if and only if the following holds:

For each open cover $\left\{U_t: t \in T \right\}$ of $X$, there exists a locally finite open cover $\left\{V_t: t \in T \right\}$ such that $\overline{V_t} \subset U_t$ for each $t \in T$.

Theorem 3 below shows that paracompactness is hereditary with respect to $F_\sigma$-subsets.

Theorem 3
Every $F_\sigma$-subset of a paracompact space is paracompact.

Proof of Theorem 3
Let $X$ be paracompact. Let $Y \subset X$ such that $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is a closed subset of $X$. Let $\mathcal{U}$ be an open cover of $Y$. For each $U \in \mathcal{U}$, let $U^*$ be open in $X$ such that $U^* \cap Y=U$.

For each $i$, let $\mathcal{U}_i^*$ be the set of all $U^*$ such that $U \cap Y_i \ne \varnothing$. Let $\mathcal{V}_i^*$ be a locally finite refinement of $\mathcal{U}_i^* \cup \left\{X-Y_i \right\}$. Let $\mathcal{V}_i$ be the following:

$\mathcal{V}_i=\left\{V \cap Y: V \in \mathcal{V}_i^* \text{ and } V \cap Y_i \ne \varnothing \right\}$

It is clear that each $\mathcal{V}_i$ is a locally finite collection of open set in $Y$ covering $Y_i$. All the $\mathcal{V}_i$ together form a refinement of $\mathcal{U}$. Thus $\mathcal{V}=\bigcup \limits_{i=1}^\infty \mathcal{V}_i$ is a $\sigma$-locally finite open refinement of $\mathcal{U}$. By Theorem 1, the $F_\sigma$-set $Y$ is paracompact. $\blacksquare$
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Result 1

Result 1 is the statement that:

If $X$ is paracompact and $Y$ is compact, then $X \times Y$ is paracompact.

To prove Result 1, we use the Tube lemma (for a proof, see “The Tube Lemma”).

The Tube Lemma
Let $X$ be any space and $Y$ be compact. For each $x \in X$ and for each open set $U \subset X \times Y$ such that $\left\{x \right\} \times Y \subset U$, there is an open set $O \subset X$ such that $\left\{x \right\} \times Y \subset O \times Y \subset U$.

Proof of Result 1
Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, choose a finite $\mathcal{U}_x \subset \mathcal{U}$ such that $\mathcal{U}_x$ is a cover of $\left\{x \right\} \times Y$. By the Tube Lemma, for each $x \in X$, there is an open set $O_x \subset X$ such that $\left\{x \right\} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is paracompact, by Theorem 2, let $\left\{W_x: x \in X \right\}$ be a locally finite open refinement of $\left\{O_x: x \in X \right\}$ such that $W_x \subset O_x$ for each $x \in X$.

Let $\mathcal{W}=\left\{(W_x \times Y) \cap U: x \in X, U \in \mathcal{U}_x \right\}$. We claim that $\mathcal{W}$ is a locally finite open refinement of $\mathcal{U}$. First, this is an open cover of $X \times Y$. To see this, let $(a,b) \in X \times Y$. Then $a \in W_x$ for some $x \in X$. Furthermore, $a \in O_x$ and $(a,b) \in \cup \mathcal{U}_x$. Thus, $(a,b) \in (W_x \times Y) \cap U$ for some $U \in \mathcal{U}_x$. Secondly, it is clear that $\mathcal{W}$ is a refinement of the original cover $\mathcal{U}$.

It remains to show that $\mathcal{W}$ is locally finite. To see this, let $(a,b) \in X \times Y$. Then there is an open $V$ in $X$ such that $x \in V$ and $V$ can meets only finitely many $W_x$. Then $V \times Y$ can meet only finitely many sets in $\mathcal{W}$. $\blacksquare$

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Result 2

Result 2 is the statement that:

If $X$ is paracompact and $Y$ is $\sigma$-compact, then $X \times Y$ is paracompact.

Proof of Result 2
Note that the $\sigma$-compact space $Y$ is Lindelof. Since regular Lindelof are normal, $Y$ is normal and is thus completely regular. So we can embed $Y$ into a compact space $K$. For example, we can let $K=\beta Y$, which is the Stone-Cech compactification of $Y$ (see “Embedding Completely Regular Spaces into a Cube”). For our purpose here, any compact space containing $Y$ will do. By Result 1, $X \times K$ is paracompact. Note that $X \times Y$ can be regarded as a subspace of $X \times K$.

Let $Y=\bigcup \limits_{i=1}^\infty Y_i$ where each $Y_i$ is compact in $Y$. Note that $X \times Y=\bigcup \limits_{i=1}^\infty X \times Y_i$ and each $X \times Y_i$ is a closed subset of $X \times K$. Thus the product $X \times Y$ is an $F_\sigma$-subset of $X \times K$. According to Theorem 3, $F_\sigma$-subsets of any paracompact space is paracompact space. Thus $X \times Y$ is paracompact. $\blacksquare$

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Reference

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

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$\copyright \ \ 2012$