Countably paracompact spaces

This post is a basic discussion on countably paracompact space. A space is a paracompact space if every open cover has a locally finite open refinement. The definition can be tweaked by saying that only open covers of size not more than a certain cardinal number \tau can have a locally finite open refinement (any space with this property is called a \tau-paracompact space). The focus here is that the open covers of interest are countable in size. Specifically, a space is a countably paracompact space if every countable open cover has a locally finite open refinement. Even though the property appears to be weaker than paracompact spaces, the notion of countably paracompactness is important in general topology. This post discusses basic properties of such spaces. All spaces under consideration are Hausdorff.

Basic discussion of paracompact spaces and their Cartesian products are discussed in these two posts (here and here).

A related notion is that of metacompactness. A space is a metacompact space if every open cover has a point-finite open refinement. For a given open cover, any locally finite refinement is a point-finite refinement. Thus paracompactness implies metacompactness. The countable version of metacompactness is also interesting. A space is countably metacompact if every countable open cover has a point-finite open refinement. In fact, for any normal space, the space is countably paracompact if and only of it is countably metacompact (see Corollary 2 below).

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Normal Countably Paracompact Spaces

A good place to begin is to look at countably paracompactness along with normality. In 1951, C. H. Dowker characterized countably paracompactness in the class of normal spaces.

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 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. 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.

Dowker’s Theorem is proved in this previous post. Condition 2 in the above formulation of the Dowker’s theorem is not in the Dowker’s theorem in the previous post. In the proof for 1 \rightarrow 2 in the previous post is essentially 1 \rightarrow 2 \rightarrow 3 for Theorem 1 above. As a result, we have the following.

Corollary 2
Let X be a normal space. Then X is countably paracompact if and only of X is countably metacompact.

Theorem 1 indicates that normal countably paracompact spaces are important for the discussion of normality in product spaces. As a result of this theorem, we know that normal countably paracompact spaces are productively normal with compact metric spaces. The Cartesian product of normal spaces with compact spaces can be non-normal (an example is found here). When the normal factor is countably paracompact and the compact factor is upgraded to a metric space, the product is always normal. The connection with normality in products is further demonstrated by the following corollary of Theorem 1.

Corollary 3
Let X be a normal space. Let Y be a non-discrete metric space. If X \times Y is normal, then X is countably paracompact.

Since Y is non-discrete, there is a non-trivial convergent sequence (i.e. the sequence represents infinitely many points). Then the sequence along with the limit point is a compact metric subspace of Y. Let’s call this subspace S. Then X \times S is a closed subspace of the normal X \times Y. As a result, X \times S is normal. By Theorem 1, X is countably paracompact.

C. H. Dowker in 1951 raised the question: is every normal space countably compact? Put it in another way, is the product of a normal space and the unit interval always a normal space? As a result of Theorem 1, any normal space that is not countably paracompact is called a Dowker space. The search for a Dowker space took about 20 years. In 1955, M. E. Rudin showed that a Dowker space can be constructed from assuming a Souslin line. In the mid 1960s, the existence of a Souslin line was shown to be independent of the usual axioms of set theorey (ZFC). Thus the existence of a Dowker space was known to be consistent with ZFC. In 1971, Rudin constructed a Dowker space in ZFC. Rudin’s Dowker space has large cardinality and is pathological in many ways. Zoltan Balogh constructed a small Dowker space (cardinality continuum) in 1996. Various Dowker space with nicer properties have also been constructed using extra set theory axioms. The first ZFC Dowker space constructed by Rudin is found in [2]. An in-depth discussion of Dowker spaces is found in [3]. Other references on Dowker spaces is found in [4].

Since Dowker spaces are rare and are difficult to come by, we can employ a “probabilistic” argument. For example, any “concrete” normal space (i.e. normality can be shown without using extra set theory axioms) is likely to be countably paracompact. Thus any space that is normal and not paracompact is likely countably paracompact (if the fact of being normal and not paracompact is established in ZFC). Indeed, any well known ZFC example of normal and not paracompact must be countably paracompact. In the long search for Dowker spaces, researchers must have checked all the well known examples! This probability thinking is not meant to be a proof that a given normal space is countably paracompact. It is just a way to suggest a possible answer. In fact, a good exercise is to pick a normal and non-paracompact space and show that it is countably paracompact.

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Some Examples

The following lists out a few classes of spaces that are always countably paracompact.

  • Metric spaces are countably paracompact.
  • Paracompact spaces are countably paracompact.
  • Compact spaces are countably paracompact.
  • Countably compact spaces are countably paracompact.
  • Perfectly normal spaces are countably paracompact.
  • Normal Moore spaces are countably paracompact.
  • Linearly ordered spaces are countably paracompact.
  • Shrinking spaces are countably paracompact.

The first four bullet points are clear. Metric spaces are paracompact. It is clear from definition that paracompact spaces, compact and countably compact spaces are countably paracompact. One way to show perfect normal spaces are countably paracompact is to show that they satisfy condition 6 in Theorem 1 (shown here). Any Moore space is perfect (closed sets are G_\delta). Thus normal Moore space are perfectly normal and hence countably paracompact. The proof of the countably paracompactness of linearly ordered spaces can be found in [1]. See Theorem 5 and Corollary 6 below for the proof of the last bullet point.

As suggested by the probability thinking in the last section, we now look at examples of countably paracompact spaces among spaces that are “normal and not paracompact”. The first uncountable ordinal \omega_1 is normal and not paracompact. But it is countably compact and is thus countably paracompact.

Example 1
Any \Sigma-product of uncountably many metric spaces is normal and countably paracompact.

For each \alpha<\omega_1, let X_\alpha be a metric space that has at least two points. Assume that each X_\alpha has a point that is labeled 0. Consider the following subspace of the product space \prod_{\alpha<\omega_1} X_\alpha.

    \displaystyle \Sigma_{\alpha<\omega_1} X_\alpha =\left\{f \in \prod_{\alpha<\omega_1} X_\alpha: \ f(\alpha) \ne 0 \text{ for at most countably many } \alpha \right\}

The space \Sigma_{\alpha<\omega_1} X_\alpha is said to be the \Sigma-product of the spaces X_\alpha. It is well known that the \Sigma-product of metric spaces is normal, in fact collectionwise normal (this previous post has a proof that \Sigma-product of separable metric spaces is collectionwise normal). On the other hand, any \Sigma-product always contains \omega_1 as a closed subset as long as there are uncountably many factors and each factor has at least two points (see the lemma in this previous post). Thus any such \Sigma-product, including the one being discussed, cannot be paracompact.

Next we show that T=(\Sigma_{\alpha<\omega_1} X_\alpha) \times [0,1] is normal. The space T can be reformulated as a \Sigma-product of metric spaces and is thus normal. Note that T=\Sigma_{\alpha<\omega_1} Y_\alpha where Y_0=[0,1], for any n with 1 \le n<\omega, Y_n=X_{n-1} and for any \alpha with \alpha>\omega, Y_\alpha=X_\alpha. Thus T is normal since it is the \Sigma-product of metric spaces. By Theorem 1, the space \Sigma_{\alpha<\omega_1} X_\alpha is countably paracompact. \square

Example 2
Let \tau be any uncountable cardinal number. Let D_\tau be the discrete space of cardinality \tau. Let L_\tau be the one-point Lindelofication of D_\tau. This means that L_\tau=D_\tau \cup \left\{\infty \right\} where \infty is a point not in D_\tau. In the topology for L_\tau, points in D_\tau are isolated as before and open neighborhoods at \infty are of the form L_\tau - C where C is any countable subset of D_\tau. Now consider C_p(L_\tau), the space of real-valued continuous functions defined on L_\tau endowed with the pointwise convergence topology. The space C_p(L_\tau) is normal and not Lindelof, hence not paracompact (discussed here). The space C_p(L_\tau) is also homeomorphic to a \Sigma-product of \tau many copies of the real lines. By the same discussion in Example 1, C_p(L_\tau) is countably paracompact. For the purpose at hand, Example 2 is similar to Example 1. \square

Example 3
Consider R. H. Bing’s example G, which is a classic example of a normal and not collectionwise normal space. It is also countably paracompact. This previous post shows that Bing’s Example G is countably metacompact. By Corollary 2, it is countably paracompact. \square

Based on the “probabilistic” reasoning discussed at the end of the last section (based on the idea that Dowker spaces are rare), “normal countably paracompact and not paracompact” should be in plentiful supply. The above three examples are a small demonstration of this phenomenon.

Existence of Dowker spaces shows that normality by itself does not imply countably paracompactness. On the other hand, paracompact implies countably paracompact. Is there some intermediate property that always implies countably paracompactness? We point that even though collectionwise normality is intermediate between paracompactness and normality, it is not sufficiently strong to imply countably paracompactness. In fact, the Dowker space constructed by Rudin in 1971 is collectionwise normal.

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More on Countably Paracompactness

Without assuming normality, the following is a characterization of countably paracompact spaces.

Theorem 4
Let X be a topological space. Then the space X is countably paracompact if and only of the following condition holds.

  • For any decreasing sequence \left\{A_n: n=1,2,3,\cdots \right\} of closed subsets of X such that \cap_n A_n=\varnothing, there exists a decreasing sequence \left\{B_n: n=1,2,3,\cdots \right\} of open subsets of X such that A_n \subset B_n for each n and \cap_n \overline{B_n}=\varnothing.

Proof of Theorem 4
Suppose that X is countably paracompact. Suppose that \left\{A_n: n=1,2,3,\cdots \right\} is a decreasing sequence of closed subsets of X as in the condition in the theorem. Then \mathcal{U}=\left\{X-A_n: n=1,2,3,\cdots \right\} is an open cover of X. Let \mathcal{V} be a locally finite open refinement of \mathcal{U}. For each n=1,2,3,\cdots, define the following:

    B_n=\cup \left\{V \in \mathcal{V}: V \cap A_n \ne \varnothing  \right\}

It is clear that A_n \subset B_n for each n. The open sets B_n are decreasing, i.e. B_1 \supset B_2 \supset \cdots since the closed sets A_n are decreasing. To show that \cap_n \overline{B_n}=\varnothing, let x \in X. The goal is to find B_j such that x \notin \overline{B_j}. Once B_j is found, we will obtain an open set V such that x \in V and V contains no points of B_j.

Since \mathcal{V} is locally finite, there exists an open set V such that x \in V and V meets only finitely many sets in \mathcal{V}. Suppose that these finitely many open sets in \mathcal{V} are V_1,V_2,\cdots,V_m. Observe that for each i=1,2,\cdots,m, there is some j(i) such that V_i \cap A_{j(i)}=\varnothing (i.e. V_i \subset X-A_{j(i)}). This follows from the fact that \mathcal{V} is a refinement \mathcal{U}. Let j be the maximum of all j(i) where i=1,2,\cdots,m. Then V_i \cap A_{j}=\varnothing for all i=1,2,\cdots,m. It follows that the open set V contains no points of B_j. Thus x \notin \overline{B_j}.

For the other direction, suppose that the space X satisfies the condition given in the theorem. Let \mathcal{U}=\left\{U_n: n=1,2,3,\cdots \right\} be an open cover of X. For each n, define A_n as follows:

    A_n=X-U_1 \cup U_2 \cup \cdots \cup U_n

Then the closed sets A_n form a decreasing sequence of closed sets with empty intersection. Let B_n be decreasing open sets such that \bigcap_{i=1}^\infty \overline{B_i}=\varnothing and A_n \subset B_n for each n. Let C_n=X-B_n for each n. Then C_n \subset \cup_{j=1}^n U_j. Define V_1=U_1. For each n \ge 2, define V_n=U_n-\bigcup_{j=1}^{n-1}C_{j}. Clearly each V_n is open and V_n \subset U_n. It is straightforward to verify that \mathcal{V}=\left\{V_n: n=1,2,3,\cdots \right\} is a cover of X.

We claim that \mathcal{V} is locally finite in X. Let x \in X. Choose the least n such that x \notin \overline{B_n}. Choose an open set O such that x \in O and O \cap \overline{B_n}=\varnothing. Then O \cap B_n=\varnothing and O \subset C_n. This means that O \cap V_k=\varnothing for all k \ge n+1. Thus the open cover \mathcal{V} is a locally finite refinement of \mathcal{U}. \square

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We present another characterization of countably paracompact spaces that involves the notion of shrinkable open covers. An open cover \mathcal{U} of a space X is said to be shrinkable if there exists an open cover \mathcal{V}=\left\{V(U): U \in \mathcal{U} \right\} of the space X such that for each U \in \mathcal{U}, \overline{V(U)} \subset U. If \mathcal{U} is shrinkable by \mathcal{V}, then we also say that \mathcal{V} is a shrinking of \mathcal{U}. Note that Theorem 1 involves a shrinking. Condition 3 in Theorem 1 (Dowker’s Theorem) can rephrased as: every countable open cover of X has a shrinking. This for any normal countably paracompact space, every countable open cover has a shrinking (or is shrinkable).

A space X is a shrinking space if every open cover of X is shrinkable. Every shrinking space is a normal space. This follows from this lemma: A space X is normal if and only if every point-finite open cover of X is shrinkable (see here for a proof). With this lemma, it follows that every shrinking space is normal. The converse is not true. To see this we first show that any shrinking space is countably paracompact. Since any Dowker space is a normal space that is not countably paracompact, any Dowker space is an example of a normal space that is not a shrinking space. To show that any shrinking space is countably paracompact, we first prove the following characterization of countably paracompactness.

Theorem 5
Let X be a space. Then X is countably paracompact if and only of every countable increasing open cover of X is shrinkable.

Proof of Theorem 5
Suppose that X is countably paracompact. Let \mathcal{U}=\left\{U_1,U_2,U_3,\cdots \right\} be an increasing open cover of X. Then there exists a locally open refinement \mathcal{V}_0 of \mathcal{U}. For each n, define V_n=\cup \left\{O \in \mathcal{V}_0: O \subset U_n \right\}. Then \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} is also a locally finite refinement of \mathcal{U}. For each n, define

    G_n=\cup \left\{O \subset X: O \text{ is open and } \forall \ m > n, O \cap V_m=\varnothing \right\}

Let \mathcal{G}=\left\{G_n: n=1,2,3,\cdots \right\}. It follows that G_n \subset G_m if n<m. Then \mathcal{G} is an increasing open cover of X. Observe that for each n, \overline{G_n} \cap V_m=\varnothing for all m > n. Then we have the following:

    \displaystyle \begin{aligned} \overline{G_n}&\subset X-\cup \left\{V_m: m > n \right\} \\&\subset \cup \left\{V_k: k=1,2,\cdots,n \right\} \\&\subset \cup \left\{U_k: k=1,2,\cdots,n \right\}=U_n  \end{aligned}

We have just established that \mathcal{G} is a shrinking of \mathcal{U}, or that \mathcal{U} is shrinkable.

For the other direction, to show that X is countably paracompact, we show that the condition in Theorem 4 is satisfied. Let \left\{A_1,A_2,A_3,\cdots \right\} be a decreasing sequence of closed subsets of X with empty intersection. Then \mathcal{U}=\left\{U_1,U_2,U_3,\cdots \right\} be an open cover of X where U_n=X-A_n for each n. By assumption, \mathcal{U} is shrinkable. Let \mathcal{V}=\left\{V_1,V_2,V_3,\cdots \right\} be a shrinking. We can assume that \mathcal{V} is an increasing sequence of open sets.

For each n, let B_n=X-\overline{V_n}. We claim that \left\{B_1,B_2,B_3,\cdots \right\} is a decreasing sequence of open sets that expand the closed sets A_n and that \bigcap_{n=1}^\infty \overline{B_n}=\varnothing. The expansion part follows from the following:

    A_n=X-U_n \subset X-\overline{V_n}=B_n

The part about decreasing follows from:

    B_{n+1}=X-\overline{V_{n+1}} \subset X-\overline{V_n}=B_n

We show that \bigcap_{n=1}^\infty \overline{B_n}=\varnothing. To this end, let x \in X. Then x \in V_n for some n. We claim that x \notin \overline{B_n}. Suppose x \in \overline{B_n}. Since V_n is an open set containing x, V_n must contain a point of B_n, say y. Since y \in B_n, y \notin \overline{V_n}. This in turns means that y \notin V_n, a contradiction. Thus we have x \notin \overline{B_n} as claimed. We have established that every point of X is not in \overline{B_n} for some n. Thus the intersection of all the \overline{B_n} must be empty. We have established the condition in Theorem 4 is satisfied. Thus X is countably paracompact. \square

Corollary 6
If X is a shrinking space, then X is countably paracompact.

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Reference

  1. Ball, B. J., Countable Paracompactness in Linearly Ordered Spaces, Proc. Amer. Math. Soc., 5, 190-192, 1954. (link)
  2. Rudin, M. E., A Normal Space X for which X \times I is not Normal, Fund. Math., 73, 179-486, 1971. (link)
  3. Rudin, M. E., Dowker Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 761-780.
  4. Wikipedia Entry on Dowker Spaces (link)

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\copyright \ 2016 \text{ by 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}

Cp(X) where X is a separable metric space

Let \tau be an uncountable cardinal. Let \prod_{\alpha < \tau} \mathbb{R}=\mathbb{R}^{\tau} be the Cartesian product of \tau many copies of the real line. This product space is not normal since it contains \prod_{\alpha \in \omega_1} \omega=\omega^{\omega_1} as a closed subspace. However, there are dense subspaces of \mathbb{R}^{\tau} are normal. For example, the \Sigma-product of \tau copies of the real line is normal, i.e., the subspace of \mathbb{R}^{\tau} consisting of points which have at most countably many non-zero coordinates (see this post). In this post, we look for more normal spaces among the subspaces of \mathbb{R}^{\tau} that are function spaces. In particular, we look at spaces of continuous real-valued functions defined on a separable metrizable space, i.e., the function space C_p(X) where X is a separable metrizable space.

For definitions of basic open sets and other background information on the function space C_p(X), see this previous post.

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C_p(X) when X is a separable metric space

In the remainder of the post, X denotes a separable metrizable space. Then, C_p(X) is more than normal. The function space C_p(X) has the following properties:

  • normal,
  • Lindelof (hence paracompact and collectionwise normal),
  • hereditarily Lindelof (hence hereditarily normal),
  • hereditarily separable,
  • perfectly normal.

All such properties stem from the fact that C_p(X) has a countable network whenever X is a separable metrizable space.

Let L be a topological space. A collection \mathcal{N} of subsets of L is said to be a network for L if for each x \in L and for each open O \subset L with x \in O, there exists some A \in \mathcal{N} such that x \in A \subset O. A countable network is a network that has only countably many elements. The property of having a countable network is a very strong property, e.g., having all the properties listed above. For a basic discussion of this property, see this previous post and this previous post.

To define a countable network for C_p(X), let \mathcal{B} be a countable base for the domain space X. For each B \subset \mathcal{B} and for any open interval (a,b) in the real line with rational endpoints, consider the following set:

    [B,(a,b)]=\left\{f \in C(X): f(B) \subset (a,b) \right\}

There are only countably many sets of the form [B,(a,b)]. Let \mathcal{N} be the collection of sets, each of which is the intersection of finitely many sets of the form [B,(a,b)]. Then \mathcal{N} is a network for the function space C_p(X). To see this, let f \in O where O=\bigcap_{x \in F} [x,O_x] is a basic open set in C_p(X) where F \subset X is finite and each O_x is an open interval with rational endpoints. For each point x \in F, choose B_x \in \mathcal{B} with x \in B_x such that f(B_x) \subset O_x. Clearly f \in \bigcap_{x \in F} \ [B_x,O_x]. It follows that \bigcap_{x \in F} \ [B_x,O_x] \subset O.

Examples include C_p(\mathbb{R}), C_p([0,1]) and C_p(\mathbb{R}^\omega). All three can be considered subspaces of the product space \mathbb{R}^c where c is the cardinality of the continuum. This is true for any separable metrizable X. Note that any separable metrizable X can be embedded in the product space \mathbb{R}^\omega. The product space \mathbb{R}^\omega has cardinality c. Thus the cardinality of any separable metrizable space X is at most continuum. So C_p(X) is the subspace of a product space of \le continuum many copies of the real lines, hence can be regarded as a subspace of \mathbb{R}^c.

A space L has countable extent if every closed and discrete subset of L is countable. The \Sigma-product \Sigma_{\alpha \in A} X_\alpha of the separable metric spaces \left\{X_\alpha: \alpha \in A \right\} is a dense and normal subspace of the product space \prod_{\alpha \in A} X_\alpha. The normal space \Sigma_{\alpha \in A} X_\alpha has countable extent (hence collectionwise normal). The examples of C_p(X) discussed here are Lindelof and hence have countable extent. Many, though not all, dense normal subspaces of products of separable metric spaces have countable extent. For a dense normal subspace of a product of separable metric spaces, one interesting problem is to find out whether it has countable extent.

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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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A Theorem about CCC Spaces

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 <n. 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 A, 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}=\left\{V_x: x \in X \right\} such that V_x \subset O_x for each x \in X.

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

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

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