A little corner in the world of set-theoretic topology

Posted on October 15, 2018 by Dan Ma

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 $\{y \in X: y<x \}$ 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. $w_\beta<w_\gamma$ 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

Eisworth T., Nyikos P., Shelah S., Gently killing S-spaces, Israel Journal of Mathmatics, 136, 189-220, 2003.
Roitman J., The spread of regular spaces, General Topology and Its Applications, 8, 85-91, 1978.
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.
Tatch-Moore J., A solution to the L space problem, Journal of the American Mathematical Society, 19, 717-736, 2006.

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Every space is star discrete

Posted on October 2, 2018 by Dan Ma

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.

Countable spread and Lindelof property

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

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.
Alster, K., Pol, R.,On function spaces of compact subspaces of $\Sigma$ -products of the real line, Fund. Math., 107, 35-46, 1980.
Arkhangelskii, A. V.,Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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Every space is star discrete