# Lindelof Exercise 2

The preceding post is an exercise showing that the product of countably many $\sigma$-compact spaces is a Lindelof space. The result is an example of a situation where the Lindelof property is countably productive if each factor is a “nice” Lindelof space. In this case, “nice” means $\sigma$-compact. This post gives several exercises surrounding the notion of $\sigma$-compactness.

Exercise 2.A

According to the preceding exercise, the product of countably many $\sigma$-compact spaces is a Lindelof space. Give an example showing that the result cannot be extended to the product of uncountably many $\sigma$-compact spaces. More specifically, give an example of a product of uncountably many $\sigma$-compact spaces such that the product space is not Lindelof.

Exercise 2.B

Any $\sigma$-compact space is Lindelof. Since $\mathbb{R}=\bigcup_{n=1}^\infty [-n,n]$, the real line with the usual Euclidean topology is $\sigma$-compact. This exercise is to find an example of “Lindelof does not imply $\sigma$-compact.” Find one such example among the subspaces of the real line. Note that as a subspace of the real line, the example would be a separable metric space, hence would be a Lindelof space.

Exercise 2.C

This exercise is also to look for an example of a space that is Lindelof and not $\sigma$-compact. The example sought is a non-metric one, preferably a space whose underlying set is the real line and whose topology is finer than the Euclidean topology.

Exercise 2.D

Show that the product of two Lindelof spaces is a Lindelof space whenever one of the factors is a $\sigma$-compact space.

Exercise 2.E

Prove that the product of finitely many $\sigma$-compact spaces is a $\sigma$-compact space. Give an example of a space showing that the product of countably and infinitely many $\sigma$-compact spaces does not have to be $\sigma$-compact. For example, show that $\mathbb{R}^\omega$, the product of countably many copies of the real line, is not $\sigma$-compact.

The Lindelof property and $\sigma$-compactness are basic topological notions. The above exercises are natural questions based on these two basic notions. One immediate purpose of these exercises is that they provide further interaction with the two basic notions. More importantly, working on these exercise give exposure to mathematics that is seemingly unrelated to the two basic notions. For example, finding $\sigma$-compactness on subspaces of the real line and subspaces of compact spaces naturally uses a Baire category argument, which is a deep and rich topic that finds uses in multiple areas of mathematics. For this reason, these exercises present excellent learning opportunities not only in topology but also in other useful mathematical topics.

If preferred, the exercises can be attacked head on. The exercises are also intended to be a guided tour. Hints are also provided below. Two sets of hints are given – Hints (blue dividers) and Further Hints (maroon dividers). The proofs of certain key facts are also given (orange dividers). Concluding remarks are given at the end of the post.

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Hints for Exercise 2.A

Prove that the Lindelof property is hereditary with respect to closed subspaces. That is, if $X$ is a Lindelof space, then every closed subspace of $X$ is also Lindelof.

Prove that if $X$ is a Lindelof space, then every closed and discrete subset of $X$ is countable (every space that has this property is said to have countable extent).

Show that the product of uncountably many copies of the real line does not have countable extent. Specifically, focus on either one of the following two examples.

• Show that the product space $\mathbb{R}^c$ has a closed and discrete subspace of cardinality continuum where $c$ is cardinality of continuum. Hence $\mathbb{R}^c$ is not Lindelof.
• Show that the product space $\mathbb{R}^{\omega_1}$ has a closed and discrete subspace of cardinality $\omega_1$ where $\omega_1$ is the first uncountable ordinal. Hence $\mathbb{R}^{\omega_1}$ is not Lindelof.

Hints for Exercise 2.B

Let $\mathbb{P}$ be the set of all irrational numbers. Show that $\mathbb{P}$ as a subspace of the real line is not $\sigma$-compact.

Hints for Exercise 2.C

Let $S$ be the real line with the topology generated by the half open and half closed intervals of the form $[a,b)=\{ x \in \mathbb{R}: a \le x < b \}$. The real line with this topology is called the Sorgenfrey line. Show that $S$ is Lindelof and is not $\sigma$-compact.

Hints for Exercise 2.D

It is helpful to first prove: the product of two Lindelof space is Lindelof if one of the factors is a compact space. The Tube lemma is helpful.

Tube Lemma
Let $X$ be a space. Let $Y$ be a compact space. Suppose that $U$ is an open subset of $X \times Y$ and suppose that $\{ x \} \times Y \subset U$ where $x \in X$. Then there exists an open subset $V$ of $X$ such that $\{ x \} \times Y \subset V \times Y \subset U$.

Hints for Exercise 2.E

Since the real line $\mathbb{R}$ is homeomorphic to the open interval $(0,1)$, $\mathbb{R}^\omega$ is homeomorphic to $(0,1)^\omega$. Show that $(0,1)^\omega$ is not $\sigma$-compact.

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Further Hints for Exercise 2.A

The hints here focus on the example $\mathbb{R}^c$.

Let $I=[0,1]$. Let $\omega$ be the first infinite ordinal. For convenience, consider $\omega$ the set $\{ 0,1,2,3,\cdots \}$, the set of all non-negative integers. Since $\omega^I$ is a closed subset of $\mathbb{R}^I$, any closed and discrete subset of $\omega^I$ is a closed and discrete subset of $\mathbb{R}^I$. The task at hand is to find a closed and discrete subset of $Y=\omega^I$. To this end, we define $W=\{W_x: x \in I \}$ after setting up background information.

For each $t \in I$, choose a sequence $O_{t,1},O_{t,2},O_{t,3},\cdots$ of open intervals (in the usual topology of $I$) such that

• $\{ t \}=\bigcap_{j=1}^\infty O_{t,j}$,
• $\overline{O_{t,j+1}} \subset O_{t,j}$ for each $j$ (the closure is in the usual topology of $I$).

Note. For each $t \in I-\{0,1 \}$, the open intervals $O_{t,j}$ are of the form $(a,b)$. For $t=0$, the open intervals $O_{t,j}$ are of the form $[0,b)$. For $t=1$, the open intervals $O_{t,j}$ are of the form $(a,1]$.

For each $t \in I$, define the map $f_t: I \rightarrow \omega$ as follows:

$f_t(x) = \begin{cases} 0 & \ \ \ \mbox{if } x=t \\ 1 & \ \ \ \mbox{if } x \in I-O_{t,1} \\ 2 & \ \ \ \mbox{if } x \in I-O_{t,2} \text{ and } x \in O_{t,1} \\ 3 & \ \ \ \mbox{if } x \in I-O_{t,3} \text{ and } x \in O_{t,2} \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \\ j & \ \ \ \mbox{if } x \in I-O_{t,j} \text{ and } x \in O_{t,j-1} \\ \vdots & \ \ \ \ \ \ \ \ \ \ \vdots \end{cases}$

We are now ready to define $W=\{W_x: x \in I \}$. For each $x \in I$, $W_x$ is the mapping $W_x:I \rightarrow \omega$ defined by $W_x(t)=f_t(x)$ for each $t \in I$.

Show the following:

• The set $W=\{W_x: x \in I \}$ has cardinality continuum.
• The set $W$ is a discrete space.
• The set $W$ is a closed subspace of $Y$.

Further Hints for Exercise 2.B

A subset $A$ of the real line $\mathbb{R}$ is nowhere dense in $\mathbb{R}$ if for any nonempty open subset $U$ of $\mathbb{R}$, there is a nonempty open subset $V$ of $U$ such that $V \cap A=\varnothing$. If we replace open sets by open intervals, we have the same notion.

Show that the real line $\mathbb{R}$ with the usual Euclidean topology cannot be the union of countably many closed and nowhere dense sets.

Further Hints for Exercise 2.C

Prove that if $X$ and $Y$ are $\sigma$-compact, then the product $X \times Y$ is $\sigma$-compact, hence Lindelof.

Prove that $S$, the Sorgenfrey line, is Lindelof while its square $S \times S$ is not Lindelof.

Further Hints for Exercise 2.D

As suggested in the hints given earlier, prove that $X \times Y$ is Lindelof if $X$ is Lindelof and $Y$ is compact. As suggested, the Tube lemma is a useful tool.

Further Hints for Exercise 2.E

The product space $(0,1)^\omega$ is a subspace of the product space $[0,1]^\omega$. Since $[0,1]^\omega$ is compact, we can fall back on a Baire category theorem argument to show why $(0,1)^\omega$ cannot be $\sigma$-compact. To this end, we consider the notion of Baire space. A space $X$ is said to be a Baire space if for each countable family $\{ U_1,U_2,U_3,\cdots \}$ of open and dense subsets of $X$, the intersection $\bigcap_{i=1}^\infty U_i$ is a dense subset of $X$. Prove the following results.

Fact E.1
Let $X$ be a compact Hausdorff space. Let $O_1,O_2,O_3,\cdots$ be a sequence of non-empty open subsets of $X$ such that $\overline{O_{n+1}} \subset O_n$ for each $n$. Then the intersection $\bigcap_{i=1}^\infty O_i$ is non-empty.

Fact E.2
Any compact Hausdorff space is Baire space.

Fact E.3
Let $X$ be a Baire space. Let $Y$ be a dense $G_\delta$-subset of $X$ such that $X-Y$ is a dense subset of $X$. Then $Y$ is not a $\sigma$-compact space.

Since $X=[0,1]^\omega$ is compact, it follows from Fact E.2 that the product space $X=[0,1]^\omega$ is a Baire space.

Fact E.4
Let $X=[0,1]^\omega$ and $Y=(0,1)^\omega$. The product space $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X=[0,1]^\omega$. Furthermore, $X-Y$ is a dense subset of $X$.

It follows from the above facts that the product space $(0,1)^\omega$ cannot be a $\sigma$-compact space.

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Proofs of Key Steps for Exercise 2.A

The proof here focuses on the example $\mathbb{R}^c$.

To see that $W=\{W_x: x \in I \}$ has the same cardinality as that of $I$, show that $W_x \ne W_y$ for $x \ne y$. This follows from the definition of the mapping $W_x$.

To see that $W$ is discrete, for each $x \in I$, consider the open set $U_x=\{ b \in Y: b(x)=0 \}$. Note that $W_x \in U_x$. Further note that $W_y \notin U_x$ for all $y \ne x$.

To see that $W$ is a closed subset of $Y$, let $k: I \rightarrow \omega$ such that $k \notin W$. Consider two cases.

Case 1. $k(r) \ne 0$ for all $r \in I$.
Note that $\{ O_{t,k(t)}: t \in I \}$ is an open cover of $I$ (in the usual topology). There exists a finite $H \subset I$ such that $\{ O_{h,k(h)}: h \in H \}$ is a cover of $I$. Consider the open set $G=\{ b \in Y: \forall \ h \in H, \ b(h)=k(h) \}$. Define the set $F$ as follows:

$F=\{ c \in I: W_c \in G \}$

The set $F$ can be further described as follows:

\displaystyle \begin{aligned} F&=\{ c \in I: W_c \in G \} \\&=\{ c \in I: \forall \ h \in H, \ W_c(h)=f_h(c)=k(h) \ne 0 \} \\&=\{ c \in I: \forall \ h \in H, \ c \in I-O_{h,k(h)} \} \\&=\bigcap_{h \in H} (I-O_{h,k(h)}) \\&=I-\bigcup_{h \in H} O_{h,k(h)}=I-I =\varnothing \end{aligned}

The last step is $\varnothing$ because $\{ O_{h,k(h)}: h \in H \}$ is a cover of $I$. The fact that $F=\varnothing$ means that $G$ is an open subset of $Y$ containing the point $k$ such that $G$ contains no point of $W$.

Case 2. $k(r) = 0$ for some $r \in I$.
Since $k \notin W$, $k \ne W_x$ for all $x \in I$. In particular, $k \ne W_r$. This means that $k(t) \ne W_r(t)$ for some $t \in I$. Define the open set $G$ as follows:

$G=\{ b \in Y: b(r)=0 \text{ and } b(t)=k(t) \}$

Clearly $k \in G$. Observe that $W_r \notin G$ since $W_r(t) \ne k(t)$. For each $p \in I-\{ r \}$, $W_p \notin G$ since $W_p(r) \ne 0$. Thus $G$ is an open set containing $k$ such that $G \cap W=\varnothing$.

Both cases show that $W$ is a closed subset of $Y=\omega^I$.

Proofs of Key Steps for Exercise 2.B

Suppose that $\mathbb{P}$, the set of all irrational numbers, is $\sigma$-compact. That is, $\mathbb{P}=A_1 \cup A_2 \cup A_3 \cup \cdots$ where each $A_i$ is a compact space as a subspace of $\mathbb{P}$. Any compact subspace of $\mathbb{P}$ is also a compact subspace of $\mathbb{R}$. As a result, each $A_i$ is a closed subset of $\mathbb{R}$. Furthermore, prove the following:

Each $A_i$ is a nowhere dense subset of $\mathbb{R}$.

Each singleton set $\{ r \}$ where $r$ is any rational number is also a closed and nowhere dense subset of $\mathbb{R}$. This means that the real line is the union of countably many closed and nowhere dense subsets, contracting the hints given earlier. Thus $\mathbb{P}$ cannot be $\sigma$-compact.

Proofs of Key Steps for Exercise 2.C

The Sorgenfrey line $S$ is a Lindelof space whose square $S \times S$ is not normal. This is a famous example of a Lindelof space whose square is not Lindelof (not even normal). For reference, a proof is found here. An alternative proof of the non-normality of $S \times S$ uses the Baire category theorem and is found here.

If the Sorgenfrey line is $\sigma$-compact, then $S \times S$ would be $\sigma$-compact and hence Lindelof. Thus $S$ cannot be $\sigma$-compact.

Proofs of Key Steps for Exercise 2.D

Suppose that $X$ is Lindelof and that $Y$ is compact. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x \subset \mathcal{U}$ be finite such that $\mathcal{U}_x$ is a cover of $\{ x \} \times Y$. Putting it another way, $\{ x \} \times Y \subset \cup \mathcal{U}_x$. By the Tube lemma, for each $x \in X$, there is an open $O_x$ such that $\{ x \} \times Y \subset O_x \times Y \subset \cup \mathcal{U}_x$. Since $X$ is Lindelof, there exists a countable set $\{ x_1,x_2,x_3,\cdots \} \subset X$ such that $\{ O_{x_1},O_{x_2},O_{x_3},\cdots \}$ is a cover of $X$. Then $\mathcal{U}_{x_1} \cup \mathcal{U}_{x_2} \cup \mathcal{U}_{x_3} \cup \cdots$ is a countable subcover of $\mathcal{U}$. This completes the proof that $X \times Y$ is Lindelof when $X$ is Lindelof and $Y$ is compact.

To complete the exercise, observe that if $X$ is Lindelof and $Y$ is $\sigma$-compact, then $X \times Y$ is the union of countably many Lindelof subspaces.

Proofs of Key Steps for Exercise 2.E

Proof of Fact E.1
Let $X$ be a compact Hausdorff space. Let $O_1,O_2,O_3,\cdots$ be a sequence of non-empty open subsets of $X$ such that \$latex $\overline{O_{n+1}} \subset O_n$ for each $n$. Show that the intersection $\bigcap_{i=1}^\infty O_i$ is non-empty.

Suppose that $\bigcap_{i=1}^\infty O_i=\varnothing$. Choose $x_1 \in O_1$. There must exist some $n_1$ such that $x_1 \notin O_{n_1}$. Choose $x_2 \in O_{n_1}$. There must exist some $n_2>n_1$ such that $x_2 \notin O_{n_2}$. Continue in this manner we can choose inductively an infinite set $A=\{ x_1,x_2,x_3,\cdots \} \subset X$ such that $x_i \ne x_j$ for $i \ne j$. Since $X$ is compact, the infinite set $A$ has a limit point $p$. This means that every open set containing $p$ contains some $x_j$ (in fact for infinitely many $j$). The point $p$ cannot be in the intersection $\bigcap_{i=1}^\infty O_i$. Thus for some $n$, $p \notin O_n$. Thus $p \notin \overline{O_{n+1}}$. We can choose an open set $U$ such that $p \in U$ and $U \cap \overline{O_{n+1}}=\varnothing$. However, $U$ must contain some point $x_j$ where $j>n+1$. This is a contradiction since $O_j \subset \overline{O_{n+1}}$ for all $j>n+1$. Thus Fact E.1 is established.

Proof of Fact E.2
Let $X$ be a compact space. Let $U_1,U_2,U_3,\cdots$ be open subsets of $X$ such that each $U_i$ is also a dense subset of $X$. Let $V$ a non-empty open subset of $X$. We wish to show that $V$ contains a point that belongs to each $U_i$. Since $U_1$ is dense in $X$, $O_1=V \cap U_1$ is non-empty. Since $U_2$ is dense in $X$, choose non-empty open $O_2$ such that $\overline{O_2} \subset O_1$ and $O_2 \subset U_2$. Since $U_3$ is dense in $X$, choose non-empty open $O_3$ such that $\overline{O_3} \subset O_2$ and $O_3 \subset U_3$. Continue inductively in this manner and we have a sequence of open sets $O_1,O_2,O_3,\cdots$ just like in Fact E.1. Then the intersection of the open sets $O_n$ is non-empty. Points in the intersection are in $V$ and in all the $U_n$. This completes the proof of Fact E.2.

Proof of Fact E.3
Let $X$ be a Baire space. Let $Y$ be a dense $G_\delta$-subset of $X$ such that $X-Y$ is a dense subset of $X$. Show that $Y$ is not a $\sigma$-compact space.

Suppose $Y$ is $\sigma$-compact. Let $Y=\bigcup_{n=1}^\infty B_n$ where each $B_n$ is compact. Each $B_n$ is obviously a closed subset of $X$. We claim that each $B_n$ is a closed nowhere dense subset of $X$. To see this, let $U$ be a non-empty open subset of $X$. Since $X-Y$ is dense in $X$, $U$ contains a point $p$ where $p \notin Y$. Since $p \notin B_n$, there exists a non-empty open $V \subset U$ such that $V \cap B_n=\varnothing$. This shows that each $B_n$ is a nowhere dense subset of $X$.

Since $Y$ is a dense $G_\delta$-subset of $X$, $Y=\bigcap_{n=1}^\infty O_n$ where each $O_n$ is an open and dense subset of $X$. Then each $A_n=X-O_n$ is a closed nowhere dense subset of $X$. This means that $X$ is the union of countably many closed and nowhere dense subsets of $X$. More specifically, we have the following.

(1)………$X= \biggl( \bigcup_{n=1}^\infty A_n \biggr) \cup \biggl( \bigcup_{n=1}^\infty B_n \biggr)$

Statement (1) contradicts the fact that $X$ is a Baire space. Note that all $X-A_n$ and $X-B_n$ are open and dense subsets of $X$. Further note that the intersection of all these countably many open and dense subsets of $X$ is empty according to (1). Thus $Y$ cannot not a $\sigma$-compact space.

Proof of Fact E.4
The space $X=[0,1]^\omega$ is compact since it is a product of compact spaces. To see that $Y=(0,1)^\omega$ is a dense $G_\delta$-subset of $X$, note that $Y=\bigcap_{n=1}^\infty U_n$ where for each integer $n \ge 1$

(2)………$U_n=(0,1) \times \cdots \times (0,1) \times [0,1] \times [0,1] \times \cdots$

Note that the first $n$ factors of $U_n$ are the open interval $(0,1)$ and the remaining factors are the closed interval $[0,1]$. It is also clear that $X-Y$ is a dense subset of $X$. This completes the proof of Fact E.4.

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Concluding Remarks

Exercise 2.A
The exercise is to show that the product of uncountably many $\sigma$-compact spaces does not need to be Lindelof. The approach suggested in the hints is to show that $\mathbb{R}^{c}$ has uncountable extent where $c$ is continuum. Having uncountable extent (i.e. having an uncountable subset that is both closed and discrete) implies the space is not Lindelof. The uncountable extent of the product space $\mathbb{R}^{\omega_1}$ is discussed in this post.

For $\mathbb{R}^{c}$ and $\mathbb{R}^{\omega_1}$, there is another way to show non-Lindelof. For example, both product spaces are not normal. As a result, both product spaces cannot be Lindelof. Note that every regular Lindelof space is normal. Both product spaces contain the product $\omega^{\omega_1}$ as a closed subspace. The non-normality of $\omega^{\omega_1}$ is discussed here.

Exercise 2.B
The hints given above is to show that the set of all irrational numbers, $\mathbb{P}$, is not $\sigma$-compact (as a subspace of the real line). The same argument showing that $\mathbb{P}$ is not $\sigma$-compact can be generalized. Note that the complement of $\mathbb{P}$ is $\mathbb{Q}$, the set of all rational numbers (a countable set). In this case, $\mathbb{Q}$ is a dense subset of the real line and is the union of countably many singleton sets. Each singleton set is a closed and nowhere dense subset of the real line. In general, we can let $B$, the complement of a set $A$, be dense in the real line and be the union of countably many closed nowhere dense subsets of the real line (not necessarily singleton sets). The same argument will show that $A$ cannot be a $\sigma$-compact space. This argument is captured in Fact E.3 in Exercise 2.E. Thus both Exercise 2.B and Exercise 2.E use a Baire category argument.

Exercise 2.E
Like Exercise 2.B, this exercise is also to show a certain space is not $\sigma$-compact. In this case, the suggested space is $\mathbb{R}^{\omega}$, the product of countably many copies of the real line. The hints given use a Baire category argument, as outlined in Fact E.1 through Fact E.4. The product space $\mathbb{R}^{\omega}$ is embedded in the compact space $[0,1]^{\omega}$, which is a Baire space. As mentioned earlier, Fact E.3 is essentially the same argument used for Exercise 2.B.

Using the same Baire category argument, it can be shown that $\omega^{\omega}$, the product of countably many copies of the countably infinite discrete space, is not $\sigma$-compact. The space $\omega$ of the non-negative integers, as a subspace of the real line, is certainly $\sigma$-compact. Using the same Baire category argument, we can see that the product of countably many copies of this discrete space is not $\sigma$-compact. With the product space $\omega^{\omega}$, there is a connection with Exercise 2.B. The product $\omega^{\omega}$ is homeomorphic to $\mathbb{P}$. The idea of the homeomorphism is discussed here. Thus the non-$\sigma$-compactness of $\omega^{\omega}$ can be achieved by mapping it to the irrationals. Of course, the same Baire category argument runs through both exercises.

Exercise 2.C
Even the non-$\sigma$-compactness of the Sorgenfrey line $S$ can be achieved by a Baire category argument. The non-normality of the Sorgenfrey plane $S \times S$ can be achieved by Jones’ lemma argument or by the fact that $\mathbb{P}$ is not a first category set. Links to both arguments are given in the Proof section above.

See here for another introduction to the Baire category theorem.

The Tube lemma is discussed here.

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# Lindelof Exercise 1

A space $X$ is called a $\sigma$-compact space if it is the union of countably many compact subspaces. Clearly, any $\sigma$-compact space is Lindelof. It is well known that the product of Lindelof spaces does not need to be Lindelof. The most well known example is perhaps the square of the Sorgenfrey line. In certain cases, the Lindelof property can be productive. For example, the product of countably many $\sigma$-compact spaces is a Lindelof space. The discussion here centers on the following theorem.

Theorem 1
Let $X_1,X_2,X_3,\cdots$ be $\sigma$-compact spaces. Then the product space $\prod_{i=1}^\infty X_i$ is Lindelof.

Theorem 1 is Exercise 3.8G in page 195 of General Topology by Engelking [1]. The reference for Exercise 3.8G is [2]. But the theorem is not found in [2] (it is not stated directly and it does not seem to be an obvious corollary of a theorem discussed in that paper). However, a hint is provided in Engelking for Exercise 3.8G. In this post, we discuss Theorem 1 as an exercise by giving expanded hint. Solutions to some of the key steps in the expanded hint are given at the end of the post.

Expanded Hint

It is helpful to first prove the following theorem.

Theorem 2
For each integer $i \ge 1$, let $C_{i,1},C_{i,2},\cdots$ be compact spaces and let $C_i$ be the topological sum:

$C_i=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots=\oplus_{j=1}^\infty C_{i,j}$

Then the product $\prod_{i=1}^\infty C_i$ is Lindelof.

Note that in the topological sum $C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots$, the spaces $C_{i,1},C_{i,2},C_{i,3},\cdots$ are considered pairwise disjoint. The open sets in the sum are simply unions of the open sets in the individual spaces. Another way to view this topology: each of the $C_{i,j}$ is both closed and open in the topological sum. Theorem 2 is essentially saying that the product of countably many $\sigma$-compact spaces is Lindelof if each $\sigma$-compact space is the union of countably many disjoint compact spaces. The hint for Exercise 3.8G can be applied much more naturally on Theorem 2 than on Theorem 1. The following is Exercise 3.8F (a), which is the hint for Exercise 3.8G.

Lemma 3
Let $Z$ be a compact space. Let $X$ be a subspace of $Z$. Suppose that there exist $F_1,F_2,F_3,\cdots$, closed subsets of $Z$, such that for all $x$ and $y$ where $x \in X$ and $y \in Z-X$, there exists $F_i$ such that $x \in F_i$ and $y \notin F_i$. Then $X$ is a Lindelof space.

The following theorem connects the hint (Lemma 3) with Theorem 2.

Theorem 4
For each integer $i \ge 1$, let $Z_i$ be the one-point compactification of $C_i$ in Theorem 2. Then the product $Z=\prod_{i=1}^\infty Z_i$ is a compact space. Furthermore, $X=\prod_{i=1}^\infty C_i$ is a subspace of $Z$. Prove that $Z$ and $X$ satisfy Lemma 3.

Each $C_i$ in Theorem 2 is a locally compact space. To define the one-point compactifications, for each $i$, choose $p_i \notin C_i$. Make sure that $p_i \ne p_j$ for $i \ne j$. Then $Z_i$ is simply

$Z_i=C_i \cup \{ p_i \}=C_{i,1} \oplus C_{i,2} \oplus C_{i,3} \oplus \cdots \cup \{ p_i \}$

with the topology defined as follows:

• Open subsets of $C_i$ continue to be open in $Z_i$.
• An open set containing $p_i$ is of the form $\{ p_i \} \cup (C_i - \overline{D})$ where $D$ is open in $C_i$ and $D$ is contained in the union of finitely many $C_{i,j}$.

For convenience, each point $p_i$ is called a point at infinity.

Note that Theorem 2 follows from Lemma 3 and Theorem 4. In order to establish Theorem 1 from Theorem 2, observe that the Lindelof property is preserved by any continuous mapping and that there is a natural continuous map from the product space in Theorem 2 to the product space in Theorem 1.

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Proofs of Key Steps

Proof of Lemma 3
Let $Z$, $X$ and $F_1,F_2,F_3,\cdots$ be as described in the statement for Lemma 3. Let $\mathcal{U}$ be a collection of open subsets of $Z$ such that $\mathcal{U}$ covers $X$. We would like to show that a countable subcollection of $\mathcal{U}$ is also a cover of $X$. Let $O=\cup \mathcal{U}$. If $Z-O=\varnothing$, then $\mathcal{U}$ is an open cover of $Z$ and there is a finite subset of $\mathcal{U}$ that is a cover of $Z$ and thus a cover of $X$. Thus we can assume that $Z-O \ne \varnothing$.

Let $F=\{ F_1,F_2,F_3,\cdots \}$. Let $K=Z-O$, which is compact. We make the following claim.

Claim. Let $Y$ be the union of all possible $\cap G$ where $G \subset F$ is finite and $\cap G \subset O$. Then $X \subset Y \subset O$.

To establish the claim, let $x \in X$. For each $y \in K=Z-O$, there exists $F_{n(y)}$ such that $x \in F_{n(y)}$ and $y \notin F_{n(y)}$. This means that $\{ Z-F_{n(y)}: y \in K \}$ is an open cover of $K$. By the compactness of $K$, there are finitely many $F_{n(y_1)}, \cdots, F_{n(y_k)}$ such that $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)}$ misses $K$, or equivalently $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset O$. Note that $x \in F_{n(y_1)} \cap \cdots \cap F_{n(y_k)}$. Further note that $F_{n(y_1)} \cap \cdots \cap F_{n(y_k)} \subset Y$. This establishes the claim that $X \subset Y$. The claim that $Y \subset O$ is clear from the definition of $Y$.

Each set $F_i$ is compact since it is closed in $Z$. The intersection of finitely many $F_i$ is also compact. Thus the $\cap G$ in the definition of $Y$ in the above claim is compact. There can be only countably many $\cap G$ in the definition of $Y$. Thus $Y$ is a $\sigma$-compact space that is covered by the open cover $\mathcal{U}$. Choose a countable $\mathcal{V} \subset \mathcal{U}$ such that $\mathcal{V}$ covers $Y$. Then $\mathcal{V}$ is a cover of $X$ too. This completes the proof that $X$ is Lindelof.

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Proof of Theorem 4
Recall that $Z=\prod_{i=1}^\infty Z_i$ and that $X=\prod_{i=1}^\infty C_i$. Each $Z_i$ is the one-point compactification of $C_i$, which is the topological sum of the disjoint compact spaces $C_{i,1},C_{i,2},\cdots$.

For integers $i,j \ge 1$, define $K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}$. For integers $n,j \ge 1$, define the product $F_{n,j}$ as follows:

$F_{n,j}=K_{1,j} \times \cdots \times K_{n,j} \times Z_{n+1} \times Z_{n+2} \times \cdots$

Since $F_{n,j}$ is a product of compact spaces, $F_{n,j}$ is compact and thus closed in $Z$. There are only countably many $F_{n,j}$.

We claim that the countably many $F_{n,j}$ have the property indicated in Lemma 3. To this end, let $f \in X=\prod_{i=1}^\infty C_i$ and $g \in Z-X$. There exists an integer $n \ge 1$ such that $g(n) \notin C_{n}$. This means that $g(n) \notin C_{n,j}$ for all $j$, i.e. $g(n)=p_n$ (so $g(n)$ must be the point at infinity). Choose $j \ge 1$ large enough such that

$f(i) \in K_{i,j}=C_{i,1} \oplus C_{i,2} \oplus \cdots \oplus C_{i,j}$

for all $i \le n$. It follows that $f \in F_{n,j}$ and $g \notin F_{n,j}$. Thus the sequence of closed sets $F_{n,j}$ satisfies Lemma 3. By Lemma 3, $X=\prod_{i=1}^\infty C_i$ is Lindelof.

Reference

1. Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.
2. Hager A. W., Approximation of real continuous functions on Lindelof spaces, Proc. Amer. Math. Soc., 22, 156-163, 1969.

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# Every space is star discrete

The statement in the title is a folklore fact, though the term star discrete is usually not used whenever this well known fact is invoked in the literature. We present a proof to this well known fact. We also discuss some related concepts.

All spaces are assumed to be Hausdorff and regular.

First, let’s define the star notation. Let $X$ be a space. Let $\mathcal{U}$ be a collection of subsets of $X$. Let $A \subset X$. Define $\text{St}(A,\mathcal{U})$ to be the set $\bigcup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}$. In other words, the set $\text{St}(A,\mathcal{U})$ is simply the union of all elements of $\mathcal{U}$ that contains points of the set $A$. The set $\text{St}(A,\mathcal{U})$ is also called the star of the set $A$ with respect to the collection $\mathcal{U}$. If $A=\{ x \}$, we use the notation $\text{St}(x,\mathcal{U})$ instead of $\text{St}( \{ x \},\mathcal{U})$. The following is the well known result in question.

Lemma 1
Let $X$ be a space. For any open cover $\mathcal{U}$ of $X$, there exists a discrete subspace $A$ of $X$ such that $X=\text{St}(A,\mathcal{U})$. Furthermore, the set $A$ can be chosen in such a way that it is also a closed subset of the space $X$.

Any space that satisfies the condition in Lemma 1 is said to be a star discrete space. The proof shown below will work for any topological space. Hence every space is star discrete. We come across three references in which the lemma is stated or is used – Lemma IV.2.20 in page 135 of [3], page 137 of [2] and [1]. The first two references do not use the term star discrete. Star discrete is mentioned in [1] since that paper focuses on star properties. This property that is present in every topological space is at heart a covering property. Here’s a rewording of the lemma that makes it look like a covering property.

Lemma 1a
Let $X$ be a space. For any open cover $\mathcal{U}$ of $X$, there exists a discrete subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. Furthermore, the set $A$ can be chosen in such a way that it is also a closed subset of the space $X$.

Lemma 1a is clearly identical to Lemma 1. However, Lemma 1a makes it extra clear that this is a covering property. For every open cover of a space, instead of finding a sub cover or an open refinement, we find a discrete subspace so that the stars of the points of the discrete subspace with respect to the given open cover also cover the space.

Lemma 1a naturally leads to other star covering properties. For example, a space $X$ is said to be a star countable space if for any open cover $\mathcal{U}$ of $X$, there exists a countable subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. A space $X$ is said to be a star Lindelof space if for any open cover $\mathcal{U}$ of $X$, there exists a Lindelof subspace $A$ of $X$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$. In general, for any topological property $\mathcal{P}$, a space $X$ is a star $\mathcal{P}$ space if for any open cover $\mathcal{U}$ of $X$, there exists a subspace $A$ of $X$ with property $\mathcal{P}$ such that $\{ \text{St}(x,\mathcal{U}): x \in A \}$ is a cover of $X$.

It follows that every Lindelof space is a star countable space. It is also clear that every star countable space is a star Lindelof space.

Lemma 1 or Lemma 1a, at first glance, may seem like a surprising result. However, one can argue that it is not a strong result at all since the property is possessed by every space. Indeed, the lemma has nothing to say about the size of the discrete set. It only says that there exists a star cover based on a discrete set for a given open cover. To derive more information about the given space, we may need to work with more information on the space in question.

Consider spaces such that every discrete subspace is countable (such a space is said to have countable spread or a space of countable spread). Also consider spaces such that every closed and discrete subspace is countable (such a space is said to have countable extent or a space of countable extent). Any space that has countable spread is also a space that has countable extent for the simple reason that if every discrete subspace is countable, then every closed and discrete subspace is countable.

Then it follows from Lemma 1 that any space $X$ that has countable extent is star countable. Any star countable space is obviously a star Lindelof space. The following diagram displays these relationships.

According to the diagram, the star countable and star Lindelof are both downstream from the countable spread property and the Lindelof property. The star properties being downstream from the Lindelof property is not surprising. What is interesting is that if a space has countable spread, then it is star countable and hence star Lindelof.

Do “countable spread” and “Lindelof” relate to each other? Lindelof spaces do not have to have countable spread. The simplest example is the one-point compactification of an uncountable discrete space. More specifically, let $X$ be an uncountable discrete space. Let $p$ be a point not in $X$. Then $Y=X \cup \{ p \}$ is a compact space (hence Lindelof) where $X$ is discrete and an open neighborhood of $p$ is of the form $\{ p \} \cup U$ where $X-U$ is a finite subset of $X$. The space $Y$ is not of countable spread since $X$ is an uncountable discrete subspace.

Does “countable spread” imply “Lindelof”? Is there a non-Lindelof space that has countable spread? It turns out that the answers are independent of ZFC. The next post has more details.

We now give a proof to Lemma 1. Suppose that $X$ is an infinite space (if it is finite, the lemma is true since the space is Hausdorff). Let $\kappa=\lvert X \lvert$. Let $\kappa^+$ be the next cardinal greater than $\kappa$. Let $\mathcal{U}$ be an open cover of the space $X$. Choose $x_0 \in X$. We choose a sequence of points $x_0,x_1,\cdots,x_\alpha,\cdots$ inductively. If $\text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}) \ne X$, we can choose a point $x_\alpha \in X$ such that $x_\alpha \notin \text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U})$.

We claim that the induction process must stop at some $\alpha<\kappa^+$. In other words, at some $\alpha<\kappa^+$, the star of the previous points must be the entire space and we run out of points to choose. Otherwise, we would have obtained a subset of $X$ with cardinality $\kappa^+$, a contradiction. Choose the least $\alpha<\kappa^+$ such that $\text{St}(\{x_\beta: \beta<\alpha \},\mathcal{U}) = X$. Let $A=\{x_\beta: \beta<\alpha \}$.

Then it can be verified that the set $A$ is a discrete subspace of $X$ and that $A$ is a closed subset of $X$. Note that $x_\beta \in \text{St}(x_\beta, \mathcal{U})$ while $x_\gamma \notin \text{St}(x_\beta, \mathcal{U})$ for all $\gamma \ne \beta$. This follows from the way the points are chosen in the induction process. On the other hand, for any $x \in X-A$, $x \in \text{St}(x_\beta, \mathcal{U})$ for some $\beta<\alpha$. As discussed, the open set $\text{St}(x_\beta, \mathcal{U})$ contains only one point of $A$, namely $x_\beta$.

Reference

1. Alas O., Jumqueira L., van Mill J., Tkachuk V., Wilson R.On the extent of star countable spaces, Cent. Eur. J. Math., 9 (3), 603-615, 2011.
2. Alster, K., Pol, R.,On function spaces of compact subspaces of $\Sigma$-products of the real line, Fund. Math., 107, 35-46, 1980.
3. Arkhangelskii, A. V.,Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright$ 2018 – Dan Ma

# Equivalent conditions for hereditarily Lindelof spaces

A topological space $X$ is Lindelof if every open cover $X$ has a countable subcollection that also is a cover of $X$. A topological space $X$ is hereditarily Lindelof if every subspace of $X$, with respect to the subspace topology, is a Lindelof space. In this post, we prove a theorem that gives two equivalent conditions for the hereditarily Lindelof property. We consider the following theorem.

Theorem 1
Let $X$ be a topological space. The following conditions are equivalent.

1. The space $X$ is a hereditarily Lindelof space.
2. Every open subspace of $X$ is Lindelof.
3. For every uncountable subspace $Y$ of $X$, there exists a point $y \in Y$ such that every open subset of $X$ containing $y$ contains uncountably many points of $Y$.

This is an excellent exercise for the hereditarily Lindelof property and for transfinite induction (for one of the directions). The equivalence $1 \longleftrightarrow 3$ is the exercise 3.12.7(d) on page 224 of [1]. The equivalence of the 3 conditions of Theorem 1 is mentioned on page 182 (chapter d-8) of [2].

Proof of Theorem 1
The direction $1 \longrightarrow 2$ is immediate. The direction $2 \longrightarrow 3$ is straightforward.

$3 \longrightarrow 1$
We show $\text{not } 1 \longrightarrow \text{not } 3$. Suppose $T$ is a non-Lindelof subspace of $X$. Let $\mathcal{U}$ be an open cover of $T$ such that no countable subcollection of $\mathcal{U}$ can cover $T$. By a transfinite inductive process, choose a set of points $\left\{t_\alpha \in T: \alpha < \omega_1 \right\}$ and a collection of open sets $\left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}$ such that for each $\alpha < \omega_1$, $t_\alpha \in U_\alpha$ and $t_\alpha \notin \cup \left\{U_\beta: \beta<\alpha \right\}$. The inductive process is possible since no countable subcollection of $\mathcal{U}$ can cover $T$. Now let $Y=\left\{t_\alpha: \alpha<\omega_1 \right\}$. Note that each $U_\alpha$ can at most contain countably many points of $Y$, namely the points in $\left\{t_\beta: \beta \le \alpha \right\}$.

For each $\alpha$, let $V_\alpha$ be an open subset of $X$ such that $U_\alpha=V_\alpha \cap Y$. We can now conclude: for every point $t_\alpha$ of $Y$, there exists an open set $V_\alpha$ containing $t_\alpha$ such that $V_\alpha$ contains only countably many points of $Y$. This is the negation of condition 3. $\blacksquare$

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Remarks

Condition 3 indicates that every uncountable set has a certain special type of limit points. Let $p \in X$. We say $p$ is a limit point of the set $Y \subset X$ if every open set containing $p$ contains a point of $Y$ different from $p$. Being a limit point of $Y$, we only know that each open set containing $p$ contain infinitely many points of $Y$ (assuming a $T_1$ space). Thus the limit points indicated in condition 3 are a special type of limit points. According to the terminology of [1], if $p$ is a limit point of $Y$ satisfying condition 3, then $p$ is said to be a condensation point of $Y$. According to Theorem 1, existence of condensation point in every uncountable set is a strong topological property (being equivalent to the hereditarily property). It is easy to see that of condition 3 holds, all but countably many points of any uncountable set $Y$ is a condensation point of $Y$.

In some situations, we may not need the full strength of condition 3. In such situations, the following corollary may be sufficient.

Corollary 2
If the space $X$ is hereditarily Lindelof, then every uncountable subspace $Y$ of $X$ contains one of its limit points.

As noted earlier, if every uncountable set contains one of its limits, then all but countably many points of any uncountable set are limit points. To contrast the hereditarily Lindelof property with the Lindelof property, consider the following theorem.

Theorem 3
If the space $X$ is Lindelof, then every uncountable subspace $Y$ of $X$ has a limit point.

The condition “every uncountable subspace $Y$ of $X$ has a limit point” has another name. When a space satisfies this condition, it is said to have countable extent. The ideas in Corollary 2 and Theorem 3 are also discussed in this previous post.

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

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

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

# (Lower case) sigma-products of separable metric spaces are Lindelof

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$. Fix a point $b \in \prod_{\alpha \in A} X_\alpha$, called the base point. The $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ is the following subspace of the product space $X$:

$\Sigma_{\alpha \in A} X_\alpha=\left\{ x \in X: x_\alpha \ne b_\alpha \text{ for at most countably many } \alpha \in A \right\}$

In other words, the space $\Sigma_{\alpha \in A} X_\alpha$ is the subspace of the product space $X=\prod_{\alpha \in A} X_\alpha$ consisting of all points that deviate from the base point on at most countably many coordinates $\alpha \in A$. We also consider the following subspace of $\Sigma_{\alpha \in A} X_\alpha$.

$\sigma=\left\{ x \in \Sigma_{\alpha \in A} X_\alpha: x_\alpha \ne b_\alpha \text{ for at most finitely many } \alpha \in A \right\}$

For convenience , we call $\Sigma_{\alpha \in A} X_\alpha$ the (upper case) Sigma-product (or $\Sigma$-product) of the spaces $X_\alpha$ and we call the space $\sigma$ the (lower case) sigma-product (or $\sigma$-product). Clearly, the space $\sigma$ is a dense subspace of $\Sigma_{\alpha \in A} X_\alpha$. In a previous post, we show that the upper case Sigma-product of separable metric spaces is collectionwise normal. In this post, we show that the (lower case) sigma-product of separable metric spaces is Lindelof. Thus when each factor $X_\alpha$ is a separable metric space with at least two points, the $\Sigma$-product, though not Lindelof, has a dense Lindelof subspace. The (upper case) $\Sigma$-product of separable metric spaces is a handy example of a non-Lindelof space that contains a dense Lindelof subspace.

Naturally, the lower case sigma-product can be further broken down into countably many subspaces. For each integer $n=0,1,2,3,\cdots$, we define $\sigma_n$ as follows:

$\sigma_n=\left\{ x \in \sigma: x_\alpha \ne b_\alpha \text{ for at most } n \text{ many } \alpha \in A \right\}$

Clearly, $\sigma=\bigcup_{n=0}^\infty \sigma_n$. We prove the following theorem. The fact that $\sigma$ is Lindelof will follow as a corollary. Understanding the following proof for Theorem 1 is a matter of keeping straight the notations involving standard basic open sets in the product space $X=\prod_{\alpha \in A} X_\alpha$. We say $V$ is a standard basic open subset of the product space $X$ if $V$ is of the form $V=\prod_{\alpha \in A} V_\alpha$ such that each $V_\alpha$ is an open subset of the factor space $X_\alpha$ and $V_\alpha=X_\alpha$ for all but finitely many $\alpha \in A$. The finite set $F$ of all $\alpha \in A$ such that $V_\alpha \ne X_\alpha$ is called the support of the open set $V$.

Theorem 1
Let $\sigma$ be the $\sigma$-product of the separable metrizable spaces $\left\{X_\alpha: \alpha \in A \right\}$. For each $n$, let $\sigma_n$ be defined as above. The product space $\sigma_n \times Y$ is Lindelof for each non-negative integer $n$ and for all separable metric space $Y$.

Proof of Theorem 1
We prove by induction on $n$. Note that $\sigma_0=\left\{b \right\}$, the base point. Clearly $\sigma_0 \times Y$ is Lindelof for all separable metric space $Y$. Suppose the theorem hold for the integer $n$. We show that $\sigma_{n+1} \times Y$ for all separable metric space $Y$. To this end, let $\mathcal{U}$ be an open cover of $\sigma_{n+1} \times Y$ where $Y$ is a separable metric space. Without loss of generality, we assume that each element of $\mathcal{U}$ is of the form $V \times W$ where $V=\prod_{\alpha \in A} V_\alpha$ is a standard basic open subset of the product space $X=\prod_{\alpha \in A} X_\alpha$ and $W$ is an open subset of $Y$.

Let $\mathcal{U}_0=\left\{U_1,U_2,U_3,\cdots \right\}$ be a countable subcollection of $\mathcal{U}$ such that $\mathcal{U}_0$ covers $\left\{b \right\} \times Y$. For each $j$, let $U_j=V_j \times W_j$ where $V_j=\prod_{\alpha \in A} V_{j,\alpha}$ is a standard basic open subset of the product space $X$ with $b \in V_j$ and $W_j$ is an open subset of $Y$. For each $j$, let $F_j$ be the support of $V_j$. Note that $\alpha \in F_j$ if and only if $V_{j,\alpha} \ne X_\alpha$. Also for each $\alpha \in F_j$, $b_\alpha \in V_{j,\alpha}$. Furthermore, for each $\alpha \in F_j$, let $V^c_{j,\alpha}=X_\alpha- V_{j,\alpha}$. With all these notations in mind, we define the following open set for each $\beta \in F_j$:

$H_{j,\beta}= \biggl( V^c_{j,\beta} \times \prod_{\alpha \in A, \alpha \ne \beta} X_\alpha \biggr) \times W_j=\biggl( V^c_{j,\beta} \times T_\beta \biggr) \times W_j$

Observe that for each point $y \in \sigma_{n+1}$ such that $y \in V^c_{j,\beta} \times T_\beta$, the point $y$ already deviates from the base point $b$ on one coordinate, namely $\beta$. Thus on the coordinates other than $\beta$, the point $y$ can only deviates from $b$ on at most $n$ many coordinates. Thus $\sigma_{n+1} \cap (V^c_{j,\beta} \times T_\beta)$ is homeomorphic to $V^c_{j,\beta} \times \sigma_n$. Note that $V^c_{j,\beta} \times W_j$ is a separable metric space. By inductive hypothesis, $V^c_{j,\beta} \times \sigma_n \times W_j$ is Lindelof. Thus there are countably many open sets in the open cover $\mathcal{U}$ that covers points of $H_{j,\beta} \cap (\sigma_{n+1} \times W_j)$.

Note that

$\sigma_{n+1} \times Y=\biggl( \bigcup_{j=1}^\infty U_j \cap \sigma_{n+1} \biggr) \cup \biggl( \bigcup \left\{H_{j,\beta} \cap (\sigma_{n+1} \times W_j): j=1,2,3,\cdots, \beta \in F_j \right\} \biggr)$

To see that the left-side is a subset of the right-side, let $t=(x,y) \in \sigma_{n+1} \times Y$. If $t \in U_j$ for some $j$, we are done. Suppose $t \notin U_j$ for all $j$. Observe that $y \in W_j$ for some $j$. Since $t=(x,y) \notin U_j$, $x_\beta \notin V_{j,\beta}$ for some $\beta \in F_j$. Then $t=(x,y) \in H_{j,\beta}$. It is now clear that $t=(x,y) \in H_{j,\beta} \cap (\sigma_{n+1} \times W_j)$. Thus the above set equality is established. Thus one part of $\sigma_{n+1} \times Y$ is covered by countably many open sets in $\mathcal{U}$ while the other part is the union of countably many Lindelof subspaces. It follows that a countable subcollection of $\mathcal{U}$ covers $\sigma_{n+1} \times Y$. $\blacksquare$

Corollary 2
It follows from Theorem 1 that

• If each factor space $X_\alpha$ is a separable metric space, then each $\sigma_n$ is a Lindelof space and that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a Lindelof space.
• If each factor space $X_\alpha$ is a compact separable metric space, then each $\sigma_n$ is a compact space and that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a $\sigma$-compact space.

Proof of Corollary 2
The first bullet point is a clear corollary of Theorem 1. A previous post shows that $\Sigma$-product of compact spaces is countably compact. Thus $\Sigma_{\alpha \in A} X_\alpha$ is a countably compact space if each $X_\alpha$ is compact. Note that each $\sigma_n$ is a closed subset of $\Sigma_{\alpha \in A} X_\alpha$ and is thus countably compact. Being a Lindelof space, each $\sigma_n$ is compact. It follows that $\sigma=\bigcup_{n=0}^\infty \sigma_n$ is a $\sigma$-compact space. $\blacksquare$

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A non-Lindelof space with a dense Lindelof subspace

Now we put everything together to obtain the example described at the beginning. For each $\alpha \in A$, let $X_\alpha$ be a separable metric space with at least two points. Then the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is collectionwise normal (see this previous post). According to the lemma in this previous post, the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ contains a closed copy of $\omega_1$. Thus the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is not Lindelof. It is clear that the $\sigma$-product is a dense subspace of $\Sigma_{\alpha \in A} X_\alpha$. By Corollary 2, the $\sigma$-product is a Lindelof subspace of $\Sigma_{\alpha \in A} X_\alpha$.

Using specific factor spaces, if each $X_\alpha=\mathbb{R}$ with the usual topology, then $\Sigma_{\alpha<\omega_1} X_\alpha$ is a non-Lindelof space with a dense Lindelof subspace. On the other hand, if each $X_\alpha=[0,1]$ with the usual topology, then $\Sigma_{\alpha<\omega_1} X_\alpha$ is a non-Lindelof space with a dense $\sigma$-compact subspace. Another example of a non-Lindelof space with a dense Lindelof subspace is given In this previous post (see Example 1).

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

# Weakly Lindelof spaces

The weakly Lindelof property is a natural weakening of the familiar Lindelof property. In this post, we discuss some of the basic properties of weakly Lindelof spaces.

We consider topological spaces that are at least $T_1$ (i.e. finite sets are closed) and regular. A space $X$ is said to be Lindelof if for any open cover $\mathcal{U}$ of $X$, there is a countable $\mathcal{V} \subset \mathcal{U}$ such that $X=\bigcup \mathcal{V}$. A natural weakening of the Lindelof property is that we only require the countable $\mathcal{V}$ to cover a dense subset of the space $X$. Specifically, a space $X$ is said to be a weakly Lindelof space if for any open cover $\mathcal{U}$ of $X$, there is a countable $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{V}$ is dense in $X$.

The notion of weakly Lindelof has a brief mention in the Encyclopedia of General Topology (see page 183 in [4]), pointing out a connection to Banach space theory. Furthermore, assuming CH, the weakly Lindelof subspaces of $\beta \mathbb{N}$ are precisely those subspaces which are $C^*$-embedded into $\beta \mathbb{N}$. In this post, we focus on the basic properties.

Clearly separable spaces and Lindelof spaces are weakly Lindelof. Another obvious property that implies weakly Lindelof is the existence of a dense Lindelof subspace. It is slightly less obvious that the countable chain condition implies the weakly Lindelof property. We have the following implications.

All the affirmative implications in the above diagram cannot be reversed (see Examples 1, 2 and 3 below).

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Some Cardinal Functions

Some of the properties discussed below can be described by cardinal functions, e.g., Lindelof number and weak Lindelof numbers. So we describe these before going into the basic properties. Let $X$ be a space. The Lindelof number of the space $X$, denoted by $L(X)$, is the least cardinal number $\mathcal{K}$ such that every open cover $\mathcal{U}$ of $X$ has a subcollection $\mathcal{V} \subset \mathcal{U}$ with $\lvert \mathcal{V} \lvert \le \mathcal{K}$ such that $\mathcal{V}$ is a cover of $X$. When $L(X)=\omega$, we say that the space is Lindelof.

The weak Lindelof number of the space $X$, denoted by $wL(X)$, is the least cardinal number $\mathcal{K}$ such that every open cover $\mathcal{U}$ of $X$ has a subcollection $\mathcal{V} \subset \mathcal{U}$ with $\lvert \mathcal{V} \lvert \le \mathcal{K}$ such that $X=\overline{\bigcup \mathcal{V}}$. When $wL(X)=\omega$, we say that the space is weakly Lindelof.

The character at $x \in X$, denoted by $\chi(x,X)$, is the least cardinal number of a local base at the point $x \in X$. The character of the space $X$, denoted by $\chi(X)$, is the supremum of all the cardinal numbers $\chi(x,X)$ over all $x \in X$. When $\chi(X)=\omega$, we say that $X$ is first countable.

The cellularity of the space $X$, denoted by $c(X)$, is the least infinite cardinal number $\mathcal{K}$ such that every collection of pairwise disjoint non-empty open subsets of $X$ has cardinality $\le \mathcal{K}$. When $c(X)=\omega$, we say that $X$ has the countable chain condition.

The extent of the space $X$, denoted by $e(X)$, is the least infinite cardinal number $\mathcal{K}$ such that if $A$ is a closed and discrete subset of $X$, then $\lvert A \lvert \le \mathcal{K}$. If $e(X)=\omega$, then $X$ is said to have countable extent (there are no uncountable closed and discrete subset). It is well known that Lindelof spaces have countable extent. The Lindelof number and the extent is related by the inequality: $e(X) \le L(X)$.

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Basic Properties

Weakly Lindelof spaces behave differently from Lindelof spaces in some ways. For example, closed subsets of a weakly Lindelof space do not have to be weakly Lindelof. In other ways, weakly Lindelof spaces and Lindelof spaces behave similarly. For example, the product of weakly Lindelof spaces needs not be weakly Lindelof and that every continuous image of a weakly Lindelof space is weakly Lindelof. Any Lindelof, Hausdorff and first countable space has cardinality no more than continuum. There is a similar theorem for weakly Lindelof spaces. Despite all these similarities with Lindelof spaces, the weak Lindelof property is a very weak property. It is well known that every Lindelof space has countable extent. There is no bound on the extent of weakly Lindelof spaces. The extent of a weakly Lindelof space can be arbitrarily large (see Example 4 below).

We discuss the following properties of weakly Lindelof spaces.

1. Any space with the countable chain condition is weakly Lindelof.
2. Any paracompact weakly Lindelof space is Lindelof.
3. Every continuous image of a weakly Lindelof space is weakly Lindelof.
4. The product of a compact space and a weakly lindelof space is weakly Lindelof.
5. The product of two Lindelof spaces needs not be weakly Lindelof.
6. Any normal first countable weakly Lindelof space has cardinality $\le 2^\omega$.
7. For any infinite cardinal $\mathcal{K}$, there exists a weakly Lindelof space $X$ such that $e(X) \ge \mathcal{K}$, i.e., the extent is at least $\mathcal{K}$. See Example 4 below.

Proof of 1
A space $X$ has the countable chain condition (has the CCC or is CCC for short) if there exists no uncountable collection of non-empty pairwise disjoint open subsets of $X$. “CCC $\Longrightarrow$ weakly Lindelof” follows from the following theorem (proved in this previous post).

Theorem
A space $X$ has the CCC if and only if for every collection $\mathcal{U}$ of non-empty open subsets of $X$, there is a countable $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}$.

To finish off, let $\mathcal{U}$ be an open cover of $X$. By the theorem, there exists a countable $\mathcal{V} \subset \mathcal{U}$ such that $\bigcup \mathcal{U} \subset \overline{\bigcup \mathcal{V}}$. This means that $X=\overline{\bigcup \mathcal{V}}$. $\blacksquare$

Even though CCC implies weakly Lindelof, CCC does not imply the stronger property of having a dense Lindelof subspace (see Example 3 below).

The proof of 1 can be generalized to show that $wL(X) \le c(X)$ for any space $X$. However, the inequality cannot be made an equality. In fact, the inequality $wL(X) \le c(X)$ can be made as wide as one wishes. Specifically, we can keep $wL(X)=\omega$ while making $c(X)$ as large as one wishes (see Example 2 below). Thus the notions of countable chain condition and the weakly Lindelof property are far apart.

Proof of 2
Let $\mathcal{U}$ be an open cover of a paracompact weakly Lindelof space $X$. Using the regularity of the space, there is an open refinement $\mathcal{V}$ of $\mathcal{U}$ for each $V \in \mathcal{V}$, $\overline{V} \subset U$ for some $U \in \mathcal{U}$. Using the paracompactness, let $\mathcal{W}$ be a locally finite open refinement of $\mathcal{V}$. Using the weakly Lindelof property, choose a countable $\mathcal{C} \subset \mathcal{W}$ such that $X=\overline{\bigcup \mathcal{C}}$. With the collection $\mathcal{C}$ being locally finite, we have $X=\overline{\bigcup \mathcal{C}}=\bigcup \left\{\overline{C}: C \in \mathcal{C} \right\}$. Thus every point of $X$ belongs to some $\overline{C}$ for some $C \in \mathcal{C}$. Tracing from $\mathcal{C}$ to $\mathcal{W}$, to $\mathcal{V}$ and then to $\mathcal{U}$, we see that for every $C \in \mathcal{C}$, $\overline{C} \subset U$ for some $U \in \mathcal{U}$. It follows that a countable subcollection of $\mathcal{U}$ is a cover of $X$. This completes the proof of bullet point 2.

This result implies that in any metrizable space, the weakly Lindelof number coincides with the Lindelof number. So in metrizable spaces, the weak Lindelof number is just as good as an indicator of weight as the other cardinal functions such as density and Lindelof number.

Among CCC spaces, paracompactness and the Lindelof property coincide. This result shows that among weakly Lindelof spaces, paracompactness and the Lindelof property also coincide. $\blacksquare$

The proof of 3 is straightforward. It is very similar to the proof that continuous image of a Lindelof space is Lindelof.

Proof of 4
The proof that the product of a compact space and a weakly Lindelof space is weakly Lindelof makes use of the tube lemma, as in the proof that the product of a compact space and a Lindelof space is Lindelof.

Let $X$ be weakly Lindelof. Let $Y$ be compact. Let $\mathcal{U}$ be an open cover of $X \times Y$. For each $x \in X$, let $\mathcal{U}_x \subset \mathcal{U}$ be finite such that $\mathcal{U}_x$ is a cover of $\left\{ x \right\} \times Y$. By the tube lemma, there exists some open set $O_x \subset X$ such that $\left\{ x \right\} \times Y \subset O_x \times Y \subset \bigcup \mathcal{U}_x$.

Since $X$ is weakly Lindelof, there exists a countable $A \subset X$ such that $X=\overline{\bigcup \limits_{x \in A} O_x}$. Let $\mathcal{U}_A=\bigcup \limits_{x \in A} \mathcal{U}_x$. It is clear that $\mathcal{U}_A$ is a countable subcollection of $\mathcal{U}$. Note that the set $\bigcup \limits_{x \in A} (O_x \times Y)$ is dense in $X \times Y$. Thus the set $\bigcup \bigcup \limits_{x \in A} \mathcal{U}_x$ is dense in $X \times Y$ too. Thus $X \times Y=\overline{\bigcup \bigcup \limits_{x \in A} \mathcal{U}_x}$. This completes the proof that $X \times Y$ is weakly Lindelof. $\blacksquare$

Proof of 5
An example of two Lindelof spaces whose product is not weakly Lindelof is provided in [3]. $\blacksquare$

Discussion of 6
Any Lindelof first countable Hausdorff space has cardinality no more than continuum (discussed in this previous post). This fact is a specific case of the general theorem that

$\lvert X \lvert \le 2^{\chi(X) \cdot L(X)}$

for any Hausdorff space $X$. Hence, the cardinality of any first countable Lindelof space is bounded by $2^\omega$. It is interesting that there is an analogous result for weakly Lindelof space. In [2], the following inequality was proved:

$\lvert X \lvert \le 2^{\chi(X) \cdot wL(X)}$

for any normal space (Theorem 2.1 in [2]). Thus the cardinality of any normal weakly Lindelof space is bounded by $2^\omega$.

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Examples

Example 1 and Example 2 below use Lindelof or compact spaces that do not have the CCC as starting point. Here’s several examples of Lindelof non-CCC spaces:

• One-point Lindelofication of an uncountable set. The space is denoted by $L(\mathcal{K})$ and is the set $\left\{p \right\} \cup D(\mathcal{K})$ where $D(\mathcal{K})$ is the discrete space of cardinality $\mathcal{K}$ and $p$ is a point not in $D(\mathcal{K})$. The open neighborhoods at $p$ have the form $\left\{p \right\} \cup (D(\mathcal{K})-C)$ where $C \subset D(\mathcal{K})$ is countable.
• The space $\omega_1+1$ with the order topology. Note that $\omega_1+1$ is the immediate successor of $\omega_1$, the first uncountable ordinal. See here.
• The unit square $[0,1] \times [0,1]$ with the lexicographic order. See here.
• The Alexandroff Double Circle. See here.

In the above four spaces, the first one is Lindelof and the other three are compact. All four do not have the countable chain condition.

Example 1
A non-Lindelof space $X_1$ that has a dense Lindelof subspace. As a bonus, this space does not have the CCC.

The idea is to start with a space that has a countable dense set of isolated points and an uncountable closed and discrete subset. One such space is a so called psi-space, a space defined using an uncountable almost disjoint family of subsets of $\omega$. Then replace each of the countably many isolated points with a copy of one of the above examples of a Lindelof space without the CCC.

Let $\omega$ the first infinite ordinal (or the set of all nonnegative integers). Let $\mathcal{A}$ be an uncountable almost disjoint family of subsets of $\omega$ (for the purpose of this example, it does not have to be an maximal almost disjoint family). Let $\Psi(\mathcal{A})=\mathcal{A} \cup \omega$, where each $n \in \omega$ is isolated and each $A \in \mathcal{A}$ has open neighborhoods of the form $\left\{A \right\} \cup (A-F)$ where $F \subset \omega$ is finite. For a more detailed discussion about Psi-space, see this previous post.

Let $Y$ be any one of the above Lindelof space that is not CCC. For each $n \in \omega$, let $Y_n=Y \times \left\{n \right\}$. So the $Y_n$ are distinct copies of the space $Y$. The underlying set of this example is the following set:

$X_1=\mathcal{A} \cup \bigcup \limits_{n \in \omega} Y_n$

The topology on $X_1$ is defined in such a way that each $Y_n$ is considered a copy of the space $Y$ and each $A \in \mathcal{A}$ has open neighborhoods of the form:

$\left\{A \right\} \cup \bigcup \limits_{n \in A-F} Y_n$

where $F \subset \omega$ is finite. The union of all $Y_n$ is a dense Lindelof subspace of $X_1$. The set $\mathcal{A}$ is an uncountable closed and discrete subset of $X_1$. Thus $X_1$ is not Lindelof. Each $Y_n$ has uncountably many disjoint open sets. Thus $X_1$ does not have the CCC. This example shows that the existence of a dense Lindelof subspace implies neither the CCC nor the Lindelof property.

Example 2
A weakly Lindelof non-CCC space $X_2$.

Let $X$ be any one of the above three non-CCC compact spaces. Let $Y$ be any space with the CCC, hence is weakly Lindelof. Let $X_2=X \times Y$. Then $X \times Y$ is weakly Lindelof. It is also clear that $X \times Y$ does not have the CCC. This example shows that the weakly Lindelof property does not imply the countable chain condition.

This example shows that $\omega=wL(X_2). In fact, it is possible to make $c(X_2)$ as large as possible. In the definition of $X \times Y$ in this example, let $X$ be the one-point Lindelofication $L(\mathcal{K})$ and $Y$ be any CCC space. Then $c(L(\mathcal{K}))$ can be made as large as possible. Hence $c(X \times Y)$ can be made as large as possible.

Example 3
A CCC space $X_3$ that has no dense Lindelof subspace.

This example is found in a paper of Arhangel’skii (Theorem 1.1 in [1]). Let $C(\omega_1+1)$ be the set of all continuous real-valued functions defined on $\omega_1+1$. The set $C(\omega_1+1)$ endowed with the pointwise convergence topology is typically denoted by $C_p(\omega_1+1)$. The space we want to use is $X_3=C_p(\omega_1+1)$.

The space $C_p(\omega_1+1)$ is a dense subspace of the product space $\mathbb{R}^{\omega_1}$. Thus $C_p(\omega_1+1)$ has the CCC. In [1], it is shown that $C_p(\omega_1+1)$ does not contain a dense normal subspace. Hence it does not contain a dense Lindelof subspace. The proof that $C_p(\omega_1+1)$ does not contain a dense normal subspace is a deep and non-trivial result.

The example $X_3=C_p(\omega_1+1)$ shows that even though CCC implies the weakly Lindelof property, it cannot give the stronger property of the existence of a dense Lindelof subspace. It is also an example showing that the implication “existence of a dense Lindelof subspace $\Longrightarrow$ weakly Lindelof” cannot be reversed.

Example 4
An weakly Lindelof space $X_4$ such that the extent can be made arbitrarily large.

Let $\mathcal{K}$ be any uncountable cardinal. Let $W$ be a discrete space of cardinality $\mathcal{K}$. Let $\beta W$ be the Stone-Cech compactification of $W$. Consider the ordinal $S=\omega+1$ with the order topology (can just think of it as a sequence of isolated points converging to the limit $\omega$). The space $X_4$ is defined as follows:

$X_4=\beta W \times S-(\beta W-W) \times \left\{\omega \right\}$

Note that $\beta W \times \omega$ is a $\sigma$-compact dense subspace of $X_4$. Hence $X_4$ is weakly Lindelof. On the other hand, the set $W \times \left\{\omega \right\}$ is a closed and discrete subset of $X_4$. Since the cardinality of $W$ can be made arbitrarily large, the extent of $X_4$ can be made arbitrarily large. Thus there is no upper bound on the extent of weakly Lindelof spaces (unlike Lindelof spaces).

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Reference

1. Arhangel’skii A. V., Normality and Dense Subspaces, Proc. Amer. Math. Soc., 48, no. 2, 283-291, 2001.
2. Bell M., Ginsburg J., Woods G., Cardinal Inequalities for Topological Spaces Involving the Weak Lindelof Number, Pacific J. Math., 79, no. 1, 37-45, 1978.
3. Hajnal A., Juhasz I., On the Products of Weakly Lindelof Spaces, Proc. Amer. Math. Soc., 130, no. 1, 454-456, 1975.
4. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

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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.
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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 Bingâ€™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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$\copyright \ 2014 \text{ by Dan Ma}$

# Pixley-Roy hyperspaces

In this post, we introduce a class of hyperspaces called Pixley-Roy spaces. This is a well-known and well studied set of topological spaces. Our goal here is not to be comprehensive but rather to present some selected basic results to give a sense of what Pixley-Roy spaces are like.

A hyperspace refers to a space in which the points are subsets of a given “ground” space. There are more than one way to define a hyperspace. Pixley-Roy spaces were first described by Carl Pixley and Prabir Roy in 1969 (see [5]). In such a space, the points are the non-empty finite subsets of a given ground space. More precisely, let $X$ be a $T_1$ space (i.e. finite sets are closed). Let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$. For each $F \in \mathcal{F}[X]$ and for each open subset $U$ of $X$ with $F \subset U$, we define:

$[F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}$

The sets $[F,U]$ over all possible $F$ and $U$ form a base for a topology on $\mathcal{F}[X]$. This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set $\mathcal{F}[X]$ with this topology is called a Pixley-Roy space.

The hyperspace as defined above was first defined by Pixley and Roy on the real line (see [5]) and was later generalized by van Douwen (see [7]). These spaces are easy to define and is useful for constructing various kinds of counterexamples. Pixley-Roy played an important part in answering the normal Moore space conjecture. Pixley-Roy spaces have also been studied in their own right. Over the years, many authors have investigated when the Pixley-Roy spaces are metrizable, normal, collectionwise Hausdorff, CCC and homogeneous. For a small sample of such investigations, see the references listed at the end of the post. Our goal here is not to discuss the results in these references. Instead, we discuss some basic properties of Pixley-Roy to solidify the definition as well as to give a sense of what these spaces are like. Good survey articles of Pixley-Roy are [3] and [7].

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Basic Discussion

In this section, we focus on properties that are always possessed by a Pixley-Roy space given that the ground space is at least $T_1$. Let $X$ be a $T_1$ space. We discuss the following points:

1. The topology defined above is a legitimate one, i.e., the sets $[F,U]$ indeed form a base for a topology on $\mathcal{F}[X]$.
2. $\mathcal{F}[X]$ is a Hausdorff space.
3. $\mathcal{F}[X]$ is a zero-dimensional space.
4. $\mathcal{F}[X]$ is a completely regular space.
5. $\mathcal{F}[X]$ is a hereditarily metacompact space.

Let $\mathcal{B}=\left\{[F,U]: F \in \mathcal{F}[X] \text{ and } U \text{ is open in } X \right\}$. Note that every finite set $F$ belongs to at least one set in $\mathcal{B}$, namely $[F,X]$. So $\mathcal{B}$ is a cover of $\mathcal{F}[X]$. For $A \in [F_1,U_1] \cap [F_2,U_2]$, we have $A \in [A,U_1 \cap U_2] \subset [F_1,U_1] \cap [F_2,U_2]$. So $\mathcal{B}$ is indeed a base for a topology on $\mathcal{F}[X]$.

To show $\mathcal{F}[X]$ is Hausdorff, let $A$ and $B$ be finite subsets of $X$ where $A \ne B$. Then one of the two sets has a point that is not in the other one. Assume we have $x \in A-B$. Since $X$ is $T_1$, we can find open sets $U, V \subset X$ such that $x \in U$, $x \notin V$ and $A \cup B-\left\{ x \right\} \subset V$. Then $[A,U \cup V]$ and $[B,V]$ are disjoint open sets containing $A$ and $B$ respectively.

To see that $\mathcal{F}[X]$ is a zero-dimensional space, we show that $\mathcal{B}$ is a base consisting of closed and open sets. To see that $[F,U]$ is closed, let $C \notin [F,U]$. Either $F \not \subset C$ or $C \not \subset U$. In either case, we can choose open $V \subset X$ with $C \subset V$ such that $[C,V] \cap [F,U]=\varnothing$.

The fact that $\mathcal{F}[X]$ is completely regular follows from the fact that it is zero-dimensional.

To show that $\mathcal{F}[X]$ is metacompact, let $\mathcal{G}$ be an open cover of $\mathcal{F}[X]$. For each $F \in \mathcal{F}[X]$, choose $G_F \in \mathcal{G}$ such that $F \in G_F$ and let $V_F=[F,X] \cap G_F$. Then $\mathcal{V}=\left\{V_F: F \in \mathcal{F}[X] \right\}$ is a point-finite open refinement of $\mathcal{G}$. For each $A \in \mathcal{F}[X]$, $A$ can only possibly belong to $V_F$ for the finitely many $F \subset A$.

A similar argument show that $\mathcal{F}[X]$ is hereditarily metacompact. Let $Y \subset \mathcal{F}[X]$. Let $\mathcal{H}$ be an open cover of $Y$. For each $F \in Y$, choose $H_F \in \mathcal{H}$ such that $F \in H_F$ and let $W_F=([F,X] \cap Y) \cap H_F$. Then $\mathcal{W}=\left\{W_F: F \in Y \right\}$ is a point-finite open refinement of $\mathcal{H}$. For each $A \in Y$, $A$ can only possibly belong to $W_F$ for the finitely many $F \subset A$ such that $F \in Y$.

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More Basic Results

We now discuss various basic topological properties of $\mathcal{F}[X]$. We first note that $\mathcal{F}[X]$ is a discrete space if and only if the ground space $X$ is discrete. Though we do not need to make this explicit, it makes sense to focus on non-discrete spaces $X$ when we look at topological properties of $\mathcal{F}[X]$. We discuss the following points:

1. If $X$ is uncountable, then $\mathcal{F}[X]$ is not separable.
2. If $X$ is uncountable, then every uncountable subspace of $\mathcal{F}[X]$ is not separable.
3. If $\mathcal{F}[X]$ is Lindelof, then $X$ is countable.
4. If $\mathcal{F}[X]$ is Baire space, then $X$ is discrete.
5. If $\mathcal{F}[X]$ has the CCC, then $X$ has the CCC.
6. If $\mathcal{F}[X]$ has the CCC, then $X$ has no uncountable discrete subspaces,i.e., $X$ has countable spread, which of course implies CCC.
7. If $\mathcal{F}[X]$ has the CCC, then $X$ is hereditarily Lindelof.
8. If $\mathcal{F}[X]$ has the CCC, then $X$ is hereditarily separable.
9. If $X$ has a countable network, then $\mathcal{F}[X]$ has the CCC.
10. The Pixley-Roy space of the Sorgenfrey line does not have the CCC.
11. If $X$ is a first countable space, then $\mathcal{F}[X]$ is a Moore space.

Bullet points 6 to 9 refer to properties that are never possessed by Pixley-Roy spaces except in trivial cases. Bullet points 6 to 8 indicate that $\mathcal{F}[X]$ can never be separable and Lindelof as long as the ground space $X$ is uncountable. Note that $\mathcal{F}[X]$ is discrete if and only if $X$ is discrete. Bullet point 9 indicates that any non-discrete $\mathcal{F}[X]$ can never be a Baire space. Bullet points 10 to 13 give some necessary conditions for $\mathcal{F}[X]$ to be CCC. Bullet 14 gives a sufficient condition for $\mathcal{F}[X]$ to have the CCC. Bullet 15 indicates that the hereditary separability and the hereditary Lindelof property are not sufficient conditions for the CCC of Pixley-Roy space (though they are necessary conditions). Bullet 16 indicates that the first countability of the ground space is a strong condition, making $\mathcal{F}[X]$ a Moore space.

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To see bullet point 6, let $X$ be an uncountable space. Let $\left\{F_1,F_2,F_3,\cdots \right\}$ be any countable subset of $\mathcal{F}[X]$. Choose a point $x \in X$ that is not in any $F_n$. Then none of the sets $F_i$ belongs to the basic open set $[\left\{x \right\} ,X]$. Thus $\mathcal{F}[X]$ can never be separable if $X$ is uncountable.

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To see bullet point 7, let $Y \subset \mathcal{F}[X]$ be uncountable. Let $W=\cup \left\{F: F \in Y \right\}$. Let $\left\{F_1,F_2,F_3,\cdots \right\}$ be any countable subset of $Y$. We can choose a point $x \in W$ that is not in any $F_n$. Choose some $A \in Y$ such that $x \in A$. Then none of the sets $F_n$ belongs to the open set $[A ,X] \cap Y$. So not only $\mathcal{F}[X]$ is not separable, no uncountable subset of $\mathcal{F}[X]$ is separable if $X$ is uncountable.

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To see bullet point 8, note that $\mathcal{F}[X]$ has no countable open cover consisting of basic open sets, assuming that $X$ is uncountable. Consider the open collection $\left\{[F_1,U_1],[F_2,U_2],[F_3,U_3],\cdots \right\}$. Choose $x \in X$ that is not in any of the sets $F_n$. Then $\left\{ x \right\}$ cannot belong to $[F_n,U_n]$ for any $n$. Thus $\mathcal{F}[X]$ can never be Lindelof if $X$ is uncountable.

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For an elementary discussion on Baire spaces, see this previous post.

To see bullet point 9, let $X$ be a non-discrete space. To show $\mathcal{F}[X]$ is not Baire, we produce an open subset that is of first category (i.e. the union of countably many closed nowhere dense sets). Let $x \in X$ a limit point (i.e. an non-isolated point). We claim that the basic open set $V=[\left\{ x \right\},X]$ is a desired open set. Note that $V=\bigcup \limits_{n=1}^\infty H_n$ where

$H_n=\left\{F \in \mathcal{F}[X]: x \in F \text{ and } \lvert F \lvert \le n \right\}$

We show that each $H_n$ is closed and nowhere dense in the open subspace $V$. To see that it is closed, let $A \notin H_n$ with $x \in A$. We have $\lvert A \lvert>n$. Then $[A,X]$ is open and every point of $[A,X]$ has more than $n$ points of the space $X$. To see that $H_n$ is nowhere dense in $V$, let $[B,U]$ be open with $[B,U] \subset V$. It is clear that $x \in B \subset U$ where $U$ is open in the ground space $X$. Since the point $x$ is not an isolated point in the space $X$, $U$ contains infinitely many points of $X$. So choose an finite set $C$ with at least $2 \times n$ points such that $B \subset C \subset U$. For the the open set $[C,U]$, we have $[C,U] \subset [B,U]$ and $[C,U]$ contains no point of $H_n$. With the open set $V$ being a union of countably many closed and nowhere dense sets in $V$, the open set $V$ is not of second category. We complete the proof that $\mathcal{F}[X]$ is not a Baire space.

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To see bullet point 10, let $\mathcal{O}$ be an uncountable and pairwise disjoint collection of open subsets of $X$. For each $O \in \mathcal{O}$, choose a point $x_O \in O$. Then $\left\{[\left\{ x_O \right\},O]: O \in \mathcal{O} \right\}$ is an uncountable and pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ is CCC then $X$ must have the CCC.

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To see bullet point 11, let $Y \subset X$ be uncountable such that $Y$ as a space is discrete. This means that for each $y \in Y$, there exists an open $O_y \subset X$ such that $y \in O_y$ and $O_y$ contains no point of $Y$ other than $y$. Then $\left\{[\left\{y \right\},O_y]: y \in Y \right\}$ is an uncountable and pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ has the CCC, then the ground space $X$ has no uncountable discrete subspace (such a space is said to have countable spread).

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To see bullet point 12, let $Y \subset X$ be uncountable such that $Y$ is not Lindelof. Then there exists an open cover $\mathcal{U}$ of $Y$ such that no countable subcollection of $\mathcal{U}$ can cover $Y$. We can assume that sets in $\mathcal{U}$ are open subsets of $X$. Also by considering a subcollection of $\mathcal{U}$ if necessary, we can assume that cardinality of $\mathcal{U}$ is $\aleph_1$ or $\omega_1$. Now by doing a transfinite induction we can choose the following sequence of points and the following sequence of open sets:

$\left\{x_\alpha \in Y: \alpha < \omega_1 \right\}$

$\left\{U_\alpha \in \mathcal{U}: \alpha < \omega_1 \right\}$

such that $x_\beta \ne x_\gamma$ if $\beta \ne \gamma$, $x_\alpha \in U_\alpha$ and $x_\alpha \notin \bigcup \limits_{\beta < \alpha} U_\beta$ for each $\alpha < \omega_1$. At each step $\alpha$, all the previously chosen open sets cannot cover $Y$. So we can always choose another point $x_\alpha$ of $Y$ and then choose an open set in $\mathcal{U}$ that contains $x_\alpha$.

Then $\left\{[\left\{x_\alpha \right\},U_\alpha]: \alpha < \omega_1 \right\}$ is a pairwise disjoint collection of open subsets of $\mathcal{F}[X]$. Thus if $\mathcal{F}[X]$ has the CCC, then $X$ must be hereditarily Lindelof.

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To see bullet point 13, let $Y \subset X$. Consider open sets $[A,U]$ where $A$ ranges over all finite subsets of $Y$ and $U$ ranges over all open subsets of $X$ with $A \subset U$. Let $\mathcal{G}$ be a collection of such $[A,U]$ such that $\mathcal{G}$ is pairwise disjoint and $\mathcal{G}$ is maximal (i.e. by adding one more open set, the collection will no longer be pairwise disjoint). We can apply a Zorn lemma argument to obtain such a maximal collection. Let $D$ be the following subset of $Y$.

$D=\bigcup \left\{A: [A,U] \in \mathcal{G} \text{ for some open } U \right\}$

We claim that the set $D$ is dense in $Y$. Suppose that there is some open set $W \subset X$ such that $W \cap Y \ne \varnothing$ and $W \cap D=\varnothing$. Let $y \in W \cap Y$. Then $[\left\{y \right\},W] \cap [A,U]=\varnothing$ for all $[A,U] \in \mathcal{G}$. So adding $[\left\{y \right\},W]$ to $\mathcal{G}$, we still get a pairwise disjoint collection of open sets, contradicting that $\mathcal{G}$ is maximal. So $D$ is dense in $Y$.

If $\mathcal{F}[X]$ has the CCC, then $\mathcal{G}$ is countable and $D$ is a countable dense subset of $Y$. Thus if $\mathcal{F}[X]$ has the CCC, the ground space $X$ is hereditarily separable.

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A collection $\mathcal{N}$ of subsets of a space $Y$ is said to be a network for the space $Y$ if any non-empty open subset of $Y$ is the union of elements of $\mathcal{N}$, equivalently, for each $y \in Y$ and for each open $U \subset Y$ with $y \in U$, there is some $A \in \mathcal{N}$ with $x \in A \subset U$. Note that a network works like a base but the elements of a network do not have to be open. The concept of network and spaces with countable network are discussed in these previous posts Network Weight of Topological Spaces â€“ I and Network Weight of Topological Spaces â€“ II.

To see bullet point 14, let $\mathcal{N}$ be a network for the ground space $X$ such that $\mathcal{N}$ is also countable. Assume that $\mathcal{N}$ is closed under finite unions (for example, adding all the finite unions if necessary). Let $\left\{[A_\alpha,U_\alpha]: \alpha < \omega_1 \right\}$ be a collection of basic open sets in $\mathcal{F}[X]$. Then for each $\alpha$, find $B_\alpha \in \mathcal{N}$ such that $A_\alpha \subset B_\alpha \subset U_\alpha$. Since $\mathcal{N}$ is countable, there is some $B \in \mathcal{N}$ such that $M=\left\{\alpha< \omega_1: B=B_\alpha \right\}$ is uncountable. It follows that for any finite $E \subset M$, $\bigcap \limits_{\alpha \in E} [A_\alpha,U_\alpha] \ne \varnothing$.

Thus if the ground space $X$ has a countable network, then $\mathcal{F}[X]$ has the CCC.

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The implications in bullet points 12 and 13 cannot be reversed. Hereditarily Lindelof property and hereditarily separability are not sufficient conditions for $\mathcal{F}[X]$ to have the CCC. See [4] for a study of the CCC property of the Pixley-Roy spaces.

To see bullet point 15, let $S$ be the Sorgenfrey line, i.e. the real line $\mathbb{R}$ with the topology generated by the half closed intervals of the form $[a,b)$. For each $x \in S$, let $U_x=[x,x+1)$. Then $\left\{[ \left\{ x \right\},U_x]: x \in S \right\}$ is a collection of pairwise disjoint open sets in $\mathcal{F}[S]$.

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A Moore space is a space with a development. For the definition, see this previous post.

To see bullet point 16, for each $x \in X$, let $\left\{B_n(x): n=1,2,3,\cdots \right\}$ be a decreasing local base at $x$. We define a development for the space $\mathcal{F}[X]$.

For each finite $F \subset X$ and for each $n$, let $B_n(F)=\bigcup \limits_{x \in F} B_n(x)$. Clearly, the sets $B_n(F)$ form a decreasing local base at the finite set $F$. For each $n$, let $\mathcal{H}_n$ be the following collection:

$\mathcal{H}_n=\left\{[F,B_n(F)]: F \in \mathcal{F}[X] \right\}$

We claim that $\left\{\mathcal{H}_n: n=1,2,3,\cdots \right\}$ is a development for $\mathcal{F}[X]$. To this end, let $V$ be open in $\mathcal{F}[X]$ with $F \in V$. If we make $n$ large enough, we have $[F,B_n(F)] \subset V$.

For each non-empty proper $G \subset F$, choose an integer $f(G)$ such that $[F,B_{f(G)}(F)] \subset V$ and $F \not \subset B_{f(G)}(G)$. Let $m$ be defined by:

$m=\text{max} \left\{f(G): G \ne \varnothing \text{ and } G \subset F \text{ and } G \text{ is proper} \right\}$

We have $F \not \subset B_{m}(G)$ for all non-empty proper $G \subset F$. Thus $F \notin [G,B_m(G)]$ for all non-empty proper $G \subset F$. But in $\mathcal{H}_m$, the only sets that contain $F$ are $[F,B_m(F)]$ and $[G,B_m(G)]$ for all non-empty proper $G \subset F$. So $[F,B_m(F)]$ is the only set in $\mathcal{H}_m$ that contains $F$, and clearly $[F,B_m(F)] \subset V$.

We have shown that for each open $V$ in $\mathcal{F}[X]$ with $F \in V$, there exists an $m$ such that any open set in $\mathcal{H}_m$ that contains $F$ must be a subset of $V$. This shows that the $\mathcal{H}_n$ defined above form a development for $\mathcal{F}[X]$.

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Examples

In the original construction of Pixley and Roy, the example was $\mathcal{F}[\mathbb{R}]$. Based on the above discussion, $\mathcal{F}[\mathbb{R}]$ is a non-separable CCC Moore space. Because the density (greater than $\omega$ for not separable) and the cellularity ($=\omega$ for CCC) do not agree, $\mathcal{F}[\mathbb{R}]$ is not metrizable. In fact, it does not even have a dense metrizable subspace. Note that countable subspaces of $\mathcal{F}[\mathbb{R}]$ are metrizable but are not dense. Any uncountable dense subspace of $\mathcal{F}[\mathbb{R}]$ is not separable but has the CCC. Not only $\mathcal{F}[\mathbb{R}]$ is not metrizable, it is not normal. The problem of finding $X \subset \mathbb{R}$ for which $\mathcal{F}[X]$ is normal requires extra set-theoretic axioms beyond ZFC (see [6]). In fact, Pixley-Roy spaces played a large role in the normal Moore space conjecture. Assuming some extra set theory beyond ZFC, there is a subset $M \subset \mathbb{R}$ such that $\mathcal{F}[M]$ is a CCC metacompact normal Moore space that is not metrizable (see Example I in [8]).

On the other hand, Pixley-Roy space of the Sorgenfrey line and the Pixley-Roy space of $\omega_1$ (the first uncountable ordinal with the order topology) are metrizable (see [3]).

The Sorgenfrey line and the first uncountable ordinal are classic examples of topological spaces that demonstrate that topological spaces in general are not as well behaved like metrizable spaces. Yet their Pixley-Roy spaces are nice. The real line and other separable metric spaces are nice spaces that behave well. Yet their Pixley-Roy spaces are very much unlike the ground spaces. This inverse relation between the ground space and the Pixley-Roy space was noted by van Douwen (see [3] and [7]) and is one reason that Pixley-Roy hyperspaces are a good source of counterexamples.

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Reference

1. Bennett, H. R., Fleissner, W. G., Lutzer, D. J., Metrizability of certain Pixley-Roy spaces, Fund. Math. 110, 51-61, 1980.
2. Daniels, P, Pixley-Roy Spaces Over Subsets of the Reals, Topology Appl. 29, 93-106, 1988.
3. Lutzer, D. J., Pixley-Roy topology, Topology Proc. 3, 139-158, 1978.
4. Hajnal, A., Juahasz, I., When is a Pixley-Roy Hyperspace CCC?, Topology Appl. 13, 33-41, 1982.
5. Pixley, C., Roy, P., Uncompletable Moore spaces, Proc. Auburn Univ. Conf. Auburn, AL, 1969.
6. Przymusinski, T., Normality and paracompactness of Pixley-Roy hyperspaces, Fund. Math. 113, 291-297, 1981.
7. van Douwen, E. K., The Pixley-Roy topology on spaces of subsets, Set-theoretic Topology, Academic Press, New York, 111-134, 1977.
8. Tall, F. D., Normality versus Collectionwise Normality, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 685-732, 1984.
9. Tanaka, H, Normality and hereditary countable paracompactness of Pixley-Roy hyperspaces, Fund. Math. 126, 201-208, 1986.

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