# A little corner in the world of set-theoretic topology

This post puts a spot light on a little corner in the world of set-theoretic topology. There lies in this corner a simple topological statement that opens a door to the esoteric world of independence results. In this post, we give a proof of this basic fact and discuss its ramifications. This basic result is an excellent entry point to the study of S and L spaces.

The following paragraph is found in the paper called Gently killing S-spaces by Todd Eisworth, Peter Nyikos and Saharon Shelah [1]. The basic fact in question is highlighted in blue.

A simultaneous generalization of hereditarily separable and hereditarily Lindelof spaces is the class of spaces of countable spread – those spaces in which every discrete subspace is countable. One of the basic facts in this little corner of set-theoretic topology is that if a regular space of countable spread is not hereditarily separable, it contains an L-space, and if it is not hereditarily Lindelof, it contains an S-space. [1]

The same basic fact is also mentioned in the paper called The spread of regular spaces by Judith Roitman [2].

It is also well known that a regular space of countable spread which is not hereditarily separable contains an L-space and a regular space of countable spread which is not hereditarily Lindelof contains an S-space. Thus an absolute example of a space satisfying (Statement) A would contain a proof of the existence of S and L space – a consummation which some may devoutly wish, but which this paper does not attempt. [2]

Statement A in [2] is: There exists a 0-dimensional Hausdorff space of countable spread that is not the union of a hereditarily separable and a hereditarily Lindelof space. Statement A would mean the existence of a regular space of countable spread that is not hereditarily separable and that is also not hereditarily Lindelof. By the well known fact just mentioned, statement A would imply the existence of a space that is simultaneously an S-space and an L-space!

Let’s unpack the preceding section. First some basic definitions. A space $X$ is of countable spread (has countable spread) if every discrete subspace of $X$ is countable. A space $X$ is hereditarily separable if every subspace of $X$ is separable. A space $X$ is hereditarily Lindelof if every subspace of $X$ is Lindelof. A space is an S-space if it is hereditarily separable but not Lindelof. A space is an L-space if it is hereditarily Lindelof but not separable. See [3] for a basic discussion of S and L spaces.

Hereditarily separable but not Lindelof spaces as well as hereditarily Lindelof but not separable spaces can be easily defined in ZFC [3]. However, such examples are not regular. For the notions of S and L-spaces to be interesting, the definitions must include regularity. Thus in the discussion that follows, all spaces are assumed to be Hausdorff and regular.

One amazing aspect about set-theoretic topology is that one sometimes does not have to stray far from basic topological notions to encounter pathological objects such as S-spaces and L-spaces. The definition of a topological space is of course a basic definition. Separable spaces and Lindelof spaces are basic notions that are not far from the definition of topological spaces. The same can be said about hereditarily separable and hereditarily Lindelof spaces. Out of these basic ingredients come the notion of S-spaces and L-spaces, the existence of which is one of the key motivating questions in set-theoretic topology in the twentieth century. The study of S and L-spaces is a body of mathematics that had been developed for nearly a century. It is a fruitful area of research at the boundary of topology and axiomatic set theory.

The existence of an S-space is independent of ZFC (as a result of the work by Todorcevic in early 1980s). This means that there is a model of set theory in which an S-space exists and there is also a model of set theory in which S-spaces cannot exist. One half of the basic result mentioned in the preceding section is intimately tied to the existence of S-spaces and thus has interesting set-theoretic implications. The other half of the basic result involves the existence of L-spaces, which are shown to exist without using extra set theory axioms beyond ZFC by Justin Moore in 2005, which went against the common expectation that the existence of L-spaces would be independent of ZFC as well.

Let’s examine the basic notions in a little more details. The following diagram shows the properties surrounding the notion of countable spread.

Diagram 1 – Properties surrounding countable spread

The implications (the arrows) in Diagram 1 can be verified easily. Central to the discussion at hand, both hereditarily separable and hereditarily Lindelof imply countable spread. The best way to see this is that if a space has an uncountable discrete subspace, that subspace is simultaneously a non-separable subspace and a non-Lindelof subspace. A natural question is whether these implications can be reversed. Another question is whether the properties in Diagram 1 can be related in other ways. The following diagram attempts to ask these questions.

Diagram 2 – Reverse implications surrounding countable spread

Not shown in Diagram 2 are these four facts: separable $\not \rightarrow$ hereditarily separable, Lindelof $\not \rightarrow$ hereditarily Lindelof, separable $\not \rightarrow$ countable spread and Lindelof $\not \rightarrow$ countable spread. The examples supporting these facts are not set-theoretic in nature and are not discussed here.

Let’s focus on each question mark in Diagram 2. The two horizontal arrows with question marks at the top are about S-space and L-space. If $X$ is hereditarily separable, then is $X$ hereditarily Lindelof? A “no” answer would mean there is an S-space. A “yes” answer would mean there exists no S-space. So the top arrow from left to right is independent of ZFC. Since an L-space can be constructed within ZFC, the question mark in the top arrow in Diagram 2 from right to left has a “no” answer.

Now focus on the arrows emanating from countable spread in Diagram 2. These arrows are about the basic fact discussed earlier. From Diagram 1, we know that hereditarily separable implies countable spread. Can the implication be reversed? Any L-space would be an example showing that the implication cannot be reversed. Note that any L-space is of countable spread and is not separable and hence not hereditarily separable. Since L-space exists in ZFC, the question mark in the arrow from countable spread to hereditarily separable has a “no” answer. The same is true for the question mark in the arrow from countable spread to separable

We know that hereditarily Lindelof implies countable spread. Can the implication be reversed? According to the basic fact mentioned earlier, if the implication cannot be reversed, there exists an S-space. Thus if S-space does not exist, the implication can be reversed. Any S-space is an example showing that the implication cannot be reversed. Thus the question in the arrow from countable spread to hereditarily Lindelof cannot be answered without assuming axioms beyond ZFC. The same is true for the question mark for the arrow from countable spread to Lindelf.

Diagram 2 is set-theoretic in nature. The diagram is remarkable in that the properties in the diagram are basic notions that are only brief steps away from the definition of a topological space. Thus the basic highlighted here is a quick route to the world of independence results.

We now give a proof of the basic result, which is stated in the following theorem.

Theorem 1
Let $X$ is regular and Hausdorff space. Then the following is true.

• If $X$ is of countable spread and is not a hereditarily separable space, then $X$ contains an L-space.
• If $X$ is of countable spread and is not a hereditarily Lindelof space, then $X$ contains an S-space.

To that end, we use the concepts of right separated space and left separated space. Recall that an initial segment of a well-ordered set $(X,<)$ is a set of the form $\{y \in X: y where $x \in X$. A space $X$ is a right separated space if $X$ can be well-ordered in such a way that every initial segment is open. A right separated space is in type $\kappa$ if the well-ordering is of type $\kappa$. A space $X$ is a left separated space if $X$ can be well-ordered in such a way that every initial segment is closed. A left separated space is in type $\kappa$ if the well-ordering is of type $\kappa$. The following results are used in proving Theorem 1.

Theorem A
Let $X$ is regular and Hausdorff space. Then the following is true.

• The space $X$ is hereditarily separable space if and only if $X$ has no uncountable left separated subspace.
• The space $X$ is hereditarily Lindelof space if and only if $X$ has no uncountable right separated subspace.

Proof of Theorem A
$\Longrightarrow$ of the first bullet point.
Suppose $Y \subset X$ is an uncountable left separated subspace. Suppose that the well-ordering of $Y$ is of type $\kappa$ where $\kappa>\omega$. Further suppose that $Y=\{ x_\alpha: \alpha<\kappa \}$ such that for each $\alpha<\kappa$, $C_\alpha=\{ x_\beta: \beta<\alpha \}$ is a closed subset of $Y$. Since $\kappa$ is uncountable, the well-ordering has an initial segment of type $\omega_1$. So we might as well assume $\kappa=\omega_1$. Note that for any countable $A \subset Y$, $A \subset C_\alpha$ for some $\alpha<\omega_1$. It follows that $Y$ is not separable. This means that $X$ is not hereditarily separable.

$\Longleftarrow$ of the first bullet point.
Suppose that $X$ is not hereditarily separable. Let $Y \subset X$ be a subspace that is not separable. We now inductively derive an uncountable left separated subspace of $Y$. Choose $y_0 \in Y$. For each $\alpha<\omega_1$, let $A_\alpha=\{ y_\beta \in Y: \beta <\alpha \}$. The set $A_\alpha$ is the set of all the points of $Y$ chosen before the step at $\alpha<\omega_1$. Since $A_\alpha$ is countable, its closure in $Y$ is not the entire space $Y$. Choose $y_\alpha \in Y-\overline{A_\alpha}=O_\alpha$.

Let $Y_L=\{ y_\alpha: \alpha<\omega_1 \}$. We claim that $Y_L$ is a left separated space. To this end, we need to show that each initial segment $A_\alpha$ is a closed subset of $Y_L$. Note that for each $\gamma \ge \alpha$, $O_\gamma=Y-\overline{A_\gamma}$ is an open subset of $Y$ with $y_\gamma \in O_\gamma$ such that $O_\gamma \cap \overline{A_\gamma}=\varnothing$ and thus $O_\gamma \cap \overline{A_\alpha}=\varnothing$ (closure in $Y$). Then $U_\gamma=O_\gamma \cap Y_L$ is an open subset of $Y_L$ containing $y_\gamma$ such that $U_\gamma \cap A_\alpha=\varnothing$. It follows that $Y-A_\alpha$ is open in $Y_L$ and that $A_\alpha$ is a closed subset of $Y_L$.

$\Longrightarrow$ of the second bullet point.
Suppose $Y \subset X$ is an uncountable right separated subspace. Suppose that the well-ordering of $Y$ is of type $\kappa$ where $\kappa>\omega$. Further suppose that $Y=\{ x_\alpha: \alpha<\kappa \}$ such that for each $\alpha<\kappa$, $U_\alpha=\{ x_\beta: \beta<\alpha \}$ is an open subset of $Y$.

Since $\kappa$ is uncountable, the well-ordering has an initial segment of type $\omega_1$. So we might as well assume $\kappa=\omega_1$. Note that $\{ U_\alpha: \alpha<\omega_1 \}$ is an open cover of $Y$ that has no countable subcover. It follows that $Y$ is not Lindelof. This means that $X$ is not hereditarily Lindelof.

$\Longleftarrow$ of the second bullet point.
Suppose that $X$ is not hereditarily Lindelof. Let $Y \subset X$ be a subspace that is not Lindelof. Let $\mathcal{U}$ be an open cover of $Y$ that has no countable subcover. We now inductively derive a right separated subspace of $Y$ of type $\omega_1$.

Choose $U_0 \in \mathcal{U}$ and choose $y_0 \in U_0$. Choose $y_1 \in Y-U_0$ and choose $U_1 \in \mathcal{U}$ such that $y_1 \in U_1$. Let $\alpha<\omega_1$. Suppose that points $y_\beta$ and open sets $U_\beta$, $\beta<\alpha$, have been chosen such that $y_\beta \in Y-\bigcup_{\delta<\beta} U_\delta$ and $y_\beta \in U_\beta$. The countably many chosen open sets $U_\beta$, $\beta<\alpha$, cannot cover $Y$. Choose $y_\alpha \in Y-\bigcup_{\beta<\alpha} U_\beta$. Choose $U_\alpha \in \mathcal{U}$ such that $y_\alpha \in U_\alpha$.

Let $Y_R=\{ y_\alpha: \alpha<\omega_1 \}$. It follows that $Y_R$ is a right separated space. Note that for each $\alpha<\omega_1$, $\{ y_\beta: \beta<\alpha \} \subset \bigcup_{\beta<\alpha} U_\beta$ and the open set $\bigcup_{\beta<\alpha} U_\beta$ does not contain $y_\gamma$ for any $\gamma \ge \alpha$. This means that the initial segment $\{ y_\beta: \beta<\alpha \}$ is open in $Y_L$. $\square$

Lemma B
Let $X$ be a space that is a right separated space and also a left separated space based on the same well ordering. Then $X$ is a discrete space.

Proof of Lemma B
Let $X=\{ w_\alpha: \alpha<\kappa \}$ such that the well-ordering is given by the ordinals in the subscripts, i.e. $w_\beta if and only if $\beta<\gamma$. Suppose that $X$ with this well-ordering is both a right separated space and a left separated space. We claim that every point is a discrete point, i.e. $\{ x_\alpha \}$ is open for any $\alpha<\kappa$.

To see this, fix $\alpha<\kappa$. The initial segment $A_\alpha=\{ w_\beta: \beta<\alpha \}$ is closed in $X$ since $X$ is a left separated space. On the other hand, the initial segment $\{ w_\beta: \beta < \alpha+1 \}$ is open in $X$ since $X$ is a right separated space. Then $B_{\alpha}=\{ w_\beta: \beta \ge \alpha+1 \}$ is closed in $X$. It follows that $\{ x_\alpha \}$ must be open since $X=A_\alpha \cup B_\alpha \cup \{ w_\alpha \}$. $\square$

Theorem C
Let $X$ is regular and Hausdorff space. Then the following is true.

• Suppose the space $X$ is right separated space of type $\omega_1$. Then if $X$ has no uncountable discrete subspaces, then $X$ is an S-space or $X$ contains an S-space.
• Suppose the space $X$ is left separated space of type $\omega_1$. Then if $X$ has no uncountable discrete subspaces, then $X$ is an L-space or $X$ contains an L-space.

Proof of Theorem C
For the first bullet point, suppose the space $X$ is right separated space of type $\omega_1$. Then by Theorem A, $X$ is not hereditarily Lindelof. If $X$ is hereditarily separable, then $X$ is an S-space (if $X$ is not Lindelof) or $X$ contains an S-space (a non-Lindelof subspace of $X$). Suppose $X$ is not hereditarily separable. By Theorem A, $X$ has an uncountable left separated subspace of type $\omega_1$.

Let $X=\{ x_\alpha: \alpha<\omega_1 \}$ such that the well-ordering represented by the ordinals in the subscripts is a right separated space. Let $<_R$ be the symbol for the right separated well-ordering, i.e. $x_\beta <_R \ x_\delta$ if and only if $\beta<\delta$. As indicated in the preceding paragraph, $X$ has an uncountable left separated subspace. Let $Y=\{ y_\alpha \in X: \alpha<\omega_1 \}$ be this left separated subspace. Let $<_L$ be the symbol for the left separated well-ordering. The well-ordering $<_R$ may be different from the well-ordering $<_L$. However, we can obtain an uncountable subset of $Y$ such that the two well-orderings coincide on this subset.

To start, pick any $y_\gamma$ in $Y$ and relabel it $t_0$. The final segment $\{y_\beta \in Y: t_0 <_L \ y_\beta \}$ must intersect the final segment $\{x_\beta \in X: t_0 <_R \ x_\beta \}$ in uncountably many points. Choose the least such point (according to $<_R$) and call it $t_1$. It is clear how $t_{\delta+1}$ is chosen if $t_\delta$ has been chosen.

Suppose $\alpha<\omega_1$ is a limit ordinal and that $t_\beta$ has been chosen for all $\beta<\alpha$. Then the set $\{y_\tau: \forall \ \beta<\alpha, t_\beta <_L \ y_\tau \}$ and the set $\{x_\tau: \forall \ \beta<\alpha, t_\beta <_R \ x_\tau \}$ must intersect in uncountably many points. Choose the least such point and call it $t_\alpha$ (according to $<_R$). As a result, we have obtained $T=\{ t_\alpha: \alpha<\omega_1 \}$. It follows that T with the well-ordering represented by the ordinals in the subscript is a subset of $(X,<_R)$ and a subset of $(Y,<_L)$. Thus $T$ is both right separated and left separated.

By Lemma B, $T$ is a discrete subspace of $X$. However, $X$ is assumed to have no uncountable discrete subspace. Thus if $X$ has no uncountable discrete subspace, then $X$ must be hereditarily separable and as a result, must be an S-space or must contain an S-space.

The proof for the second bullet point is analogous to that of the first bullet point. $\square$

We are now ready to prove Theorem 1.

Proof of Theorem 1
Suppose that $X$ is of countable spread and that $X$ is not hereditarily separable. By Theorem A, $X$ has an uncountable left separated subspace $Y$ (assume it is of type $\omega_1$). The property of countable spread is hereditary. So $Y$ is of countable spread. By Theorem C, $Y$ is an L-space or $Y$ contains an L-space. In either way, $X$ contains an L-space.

Suppose that $X$ is of countable spread and that $X$ is not hereditarily Lindelof. By Theorem A, $X$ has an uncountable right separated subspace $Y$ (assume it is of type $\omega_1$). By Theorem C, $Y$ is an S-space or $Y$ contains an S-space. In either way, $X$ contains an S-space.

Reference

1. Eisworth T., Nyikos P., Shelah S., Gently killing S-spaces, Israel Journal of Mathmatics, 136, 189-220, 2003.
2. Roitman J., The spread of regular spaces, General Topology and Its Applications, 8, 85-91, 1978.
3. Roitman, J., Basic S and L, Handbook of Set-Theoretic Topology, (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 295-326, 1984.
4. Tatch-Moore J., A solution to the L space problem, Journal of the American Mathematical Society, 19, 717-736, 2006.

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$\copyright$ 2018 – Dan Ma

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