Perfect preimages of Lindelof spaces

Posted on April 7, 2023 by Dan Ma

Let $f:X \rightarrow Y$ be a mapping from a topological space $X$ onto a topological space $Y$ . If $f$ is a perfect map and $Y$ is a Lindelof space, then so is $X$ . If $f$ is a closed map and $Y$ is a paracompact space, then so is $X$ . In other words, the pre-image of a Lindelof space under a perfect map is always a Lindelof space. Likewise, the pre-image of a paracompact space under a closed map is always a paracompact space. After proving these two facts, we show that for any compact space $Y$ , the product $X \times Y$ is Lindelof (paracompact) for any Lindelof (paracompact) space $X$ . All spaces under consideration are Hausdorff.

Another way to state the above two facts is that Lindelofness is an inverse invariance under perfect maps and that paracompactness is an inverse invariance of the closed maps. In general, a topological property is an inverse invariance of a class of mappings $\mathcal{M}$ if the following holds: for any mapping $f:X \rightarrow Y$ belonging to $\mathcal{M}$ , if $Y$ has the property, then so does $X$ . In contrast, a topological property is an invariance of a class of mappings $\mathcal{M}$ if for any mapping $f:X \rightarrow Y$ belonging to $\mathcal{M}$ , if $X$ has the property, then so does $Y$ .

All mappings under consideration are continuous maps. A mapping $f:X \rightarrow Y$ , where $f(X)=Y$ , is a closed map if for any closed subset $A$ of $X$ , $f(A)$ is closed in $Y$ . A mapping $f:X \rightarrow Y$ , where $f(X)=Y$ , is a perfect map if $f$ is a closed map and that the point inverse $f^{-1}(y)$ is compact for each $y \in Y$ .

Perfect mappings and closed mappings are objects with strong properties. Such a map places a restriction on what topological properties the “domain” space or the “range” space can have. The theorems below indicate that it is not possible to map a non-Lindelof space onto a Lindelof space using a perfect map and that it is not possible to map a non-paracompact space onto a paracompact space using a closed map. On the other hand, it is not possible to map a separable metric space onto a separable but non-metric space using a perfect map (see here). We prove the following theorems.

Theorem 1…. Lindelofness (or the Lindelof property) is an inverse invariant of the perfect maps.

Theorem 2 ….Paracompactness is an inverse invariant of the closed maps.

Lemma 3 …. Let $f:X \longrightarrow Y$ be a closed map with $f(X)=Y$ . Let $V$ be an open subset of $X$ . Define $f_*(V)=\{ y \in Y: f^{-1}(y) \subset V \}$ . Then the set $f_*(V)$ is open in $Y$ and that $f_*(V) \subset f(V)$ .

For the proof of Lemma 3, see Lemma 2 here.

Proof of Theorem 1
Let $f:X \longrightarrow Y$ be a perfect map with $f(X)=Y$ . Suppose $Y$ is Lindelof. Let $\mathcal{U}$ be an open cover of $X$ . Without loss of generality, we can assume that $\mathcal{U}$ is closed under finite unions. For each $U \in \mathcal{U}$ , define $f_*(U)=\{ y \in Y: f^{-1}(y) \subset U \}$ . By Lemma 3, each $f_*(U)$ is an open subset of $Y$ . We claim that $\mathcal{V}=\{ f_*(U): U \in \mathcal{U} \}$ is an open cover of $Y$ . To this end, let $y \in Y$ . Since $f$ is a perfect map, the point inverse $f^{-1}(y)$ is compact. As a result, we can find a finite $\mathcal{F} \subset \mathcal{U}$ such that $f^{-1}(y) \subset \bigcup \mathcal{F}=W$ . Since $\mathcal{U}$ is closed under finite unions, $W \in \mathcal{U}$ . It follows that $y \in f_*(W)$ . Since $Y$ is Lindelof, there exists a countable $\{W_0,W_1,W_2,\cdots \} \subset \mathcal{V}$ such that $Y = \bigcup_{n=0}^\infty W_n$ . For each $n$ , $W_n=f_*(U_n)$ for some $U_n \in \mathcal{U}$ . We claim that $\{U_0,U_1,U_2,\cdots \}$ is a cover of $X$ . To this end, let $x \in X$ . Then for some $n$ , $y=f(x) \in W_n=f_*(U_n)$ . This implies that $x \in f^{-1}(y) \subset U_n$ . Thus, the open cover $\mathcal{U}$ has a countable subcover. This concludes the proof of Theorem 1. $\square$

Proof of Theorem 2
Let $f:X \longrightarrow Y$ be a perfect map with $f(X)=Y$ . Suppose $Y$ is paracompact. Let $\mathcal{U}$ be an open cover of $X$ . For each $U \in \mathcal{U}$ , define $f_*(U)$ as in Lemma 3. By Lemma 3, each $f_*(U)$ is an open subset of $Y$ . Let $\mathcal{V}=\{ f_*(U): U \in \mathcal{U} \}$ . As shown in the proof of Theorem 1, $\mathcal{V}$ is an open cover of $Y$ . Since $Y$ is paracompact, there exists a locally finite open refinement $\mathcal{W}$ of $\mathcal{V}$ . Let $\mathcal{W}_0=\{ f^{-1}(W): W \in \mathcal{W} \}$ .

We show three facts about $\mathcal{W}_0$ . (1) It is an open cover of $X$ . (2) It is a locally finite collection in $X$ . (3) It is a refinement of $\mathcal{U}$ . To see (1), note that $\mathcal{W}$ is an open cover of $Y$ . As a result, $\mathcal{W}_0$ is an open cover of $X$ . To see (2), let $x \in X$ . We find an open $O \subset X$ such that $x \in O$ and such that $O$ intersects only finitely many elements of $\mathcal{W}_0$ . Since $\mathcal{W}$ is locally finite in $Y$ , there exists an open $B \subset Y$ such that $y=f(x) \in B$ and such that $B$ intersects only finitely many elements of $\mathcal{W}$ , say, $W_0,W_1,\cdots,W_n$ . Let $O=f^{-1}(B)$ . Clearly, $x \in O$ . It can be verified that the only elements of $\mathcal{W}_0$ having non-empty intersections with $O$ are $f^{-1}(W_0)$ , $f^{-1}(W_1),\cdots$ , $f^{-1}(W_n)$ . To see (3), let $f^{-1}(W) \in \mathcal{W}_0$ where $W \in \mathcal{W}$ . Then $W \subset V=f_*(U)$ for some $V \in \mathcal{V}$ and some $U \in \mathcal{U}$ . We claim that $f^{-1}(W) \subset U$ . Let $x \in f^{-1}(W)$ . Then $y=f(x) \in W \subset V=f_*(U)$ . This implies that $x \in f^{-1}(y) \subset U$ . It follows that $\mathcal{W}_0$ is a locally finite open refinement of the open cover $\mathcal{U}$ . This completes the proof of Theorem 2. $\square$

Productively Paracompact Spaces

A space $X$ is productively paramcompact if $X \times Y$ is paracompact for every paracompact space $Y$ . The definition for productively Lindelof can be stated in a similar way. For some reason, the term “productively paracompact” is not used in the literature but is a topic that had been extensively studied. It is also a topic found in this site. The following four classes of spaces are productively paracompact (see here and here).

Compact spaces
$\sigma$ -compact spaces
Locally compact spaces
$\sigma$ -locally compact spaces

The proof for compact spaces being productively paracompact given here uses the Tube Lemma (see here). As applications of Theorem 1 and Theorem 2, we use the two theorems to show that compact spaces are both productively Lindelof and productive paracompact.

Theorem 4…. Let $Y$ any compact space. Then $X \times Y$ is Lindelof for every Lindelof space $X$ .

Theorem 5 ….Let $Y$ any compact space. Then $X \times Y$ is paracompact for every paracompact space $X$ .

Theorems 4 and 5 are corollaries to the Kuratowski theorem (see here) and Theorems 1 and 2 above. Suppose $Y$ is compact. Then the projection map from $X \times Y$ onto $X$ is a closed map. The paracompactness of $X \times Y$ follows whenever $X$ is paracompact. The projection map is also perfect since the point inverses are compact due to the compactness of the factor $Y$ . Then the Lindelofness of $X \times Y$ follows whenever $X$ is Lindelof.

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Making sense of the spaces with small diagonal

Posted on January 26, 2021 by Dan Ma

This is a small attempt toward making sense of the spaces with small diagonal. What makes the property of small diagonal interesting is the longstanding open problem that is discussed below. We use examples to help make sense of the definition of small diagonal. Then we discuss briefly the open problem regarding small diagonal. We discuss the following three examples of compact spaces:

$\omega_1 + 1$ with the ordered topology
The one-point compactification of a discrete space of cardinality $\omega_1$
The double arrow space

By a space we mean a topological space that is Tychonoff, i.e. Hausdorff and completely regular (defined here). Let $X$ be a space. The subset $\Delta=\{ (x,x): x \in X \}$ of the square $X^2=X \times X$ is called the diagonal of the space $X$ .

We want to focus on two diagonal properties. The space $X$ is said to have a $G_\delta$ -diagonal if $\Delta$ is a $G_\delta$ -set in $X^2$ , i.e. $\Delta$ is the intersection of countably many open subsets of $X^2$ . The space $X$ is said to have a small diagonal if for each uncountable subset $A$ of $X^2 \backslash \Delta$ , there is an open subset $O$ of $X^2$ such that $\Delta \subset O$ and $O$ misses uncountably many points of $A$ .

How do these two diagonal properties relate? Any space that has a $G_\delta$ -diagonal also has a small diagonal. This fact can be worked out quite easily based on the definitions. The opposite direction is a totally different matter. In fact, the question of whether having a small diagonal implies having a $G_\delta$ -diagonal is related to a longstanding open question.

Let’s focus on compact spaces. A classic metrization theorem for compact spaces states that any compact space with a $G_\delta$ -diagonal is metrizable (see here). A natural question is: if a compact space has a small diagonal, must it be metrizable? Indeed this is a well known open problem, a longstanding problem that has not completely resolved completely. Before discussion the open problem, we look at the three examples indicated above.

The three examples are all non-metrizable compact spaces and thus do not have a $G_\delta$ -diagonal. We show that they also not have a small diagonal. To appreciate the definition of small diagonal, it is helpful to look spaces that do not have a small diagonal.

Example 1

Consider $X=\omega_1+1$ with the order topology. Using interval notion, $X=[0,\omega_1]$ . This is the space of all countable ordinals $\alpha<\omega_1$ plus the point $\omega_1$ at the end. This is a compact space that is not metrizable. For example, any compact metrizable space would be separable. The last point $\omega_1$ cannot be in the closure of any countable subset. According to the classic theorem mentioned above, $X=[0,\omega_1]$ cannot have a $G_\delta$ -diagonal. We show that it does have a small diagonal too.

Let $A=\{ (\gamma, \gamma+1): \gamma< \omega_1 \}$ . Note that any open set of the form $(\alpha, \omega_1] \times (\alpha, \omega_1]$ contains the point $(\omega_1, \omega_1)$ and contains all but countably many points of $A$ . As a result, any open set containing the diagonal $\Delta$ contains all but countably many points of $A$ . This violates the definition of having a small diagonal. Thus the space $X=[0,\omega_1]$ does not have a small diagonal.

More on the Definition

Example 1 suggests a different angle in looking at the definition. The set $A=\{ (\gamma, \gamma+1): \gamma< \omega_1 \}$ is a $\omega_1$ -length sequence convergent to the diagonal, meaning that any open set containing the diagonal contains a tail of the sequence (in this case containing all but countably many elements in the sequence). The convergent sequence view point is the way Husek defined small diagonal [7].

In [7], $X$ is said to have an $\omega_1$ -accessible diagonal if there is an $\omega_1$ -length sequence $\{ (x_\alpha, y_\alpha) \in X^2 \backslash \Delta: \alpha < \omega_1 \}$ that converges to the diagonal $\Delta$ , meaning that any open set containing $\Delta$ contains all but countably many terms in the sequence. This is exactly what is occurring in the space $X=[0,\omega_1]$ . The set $A=\{ (\gamma, \gamma+1): \gamma< \omega_1 \}$ is precisely a convergent sequence converging to the diagonal. The space $X=[0,\omega_1]$ has an $\omega_1$ -accessible diagonal.

Seen in this light, spaces with a small diagonal are the the spaces that do not have an $\omega_1$ -accessible diagonal (or spaces that have an $\omega_1$ -inaccessible diagonal). Though the definition of Husek is more descriptive, the term small diagonal, suggested by E. van Douwen, has become more popular. The small diagonal defined above is more positive sounding. For example, we say the space $X$ has a small diagonal. In Husek [7], the spaces of interest would be the spaces without an $\omega_1$ -accessible diagonal. This definition is a negative one (defined by the lack of certain thing) and takes more syllables to express. Personally speaking, we prefer the term small diagonal though we understand that $\omega_1$ -accessible diagonal is more descriptive.

A slightly different (but equivalent) way of stating the definition of a space with a small diagonal: a space $X$ has a small diagonal if for any uncountable $A \subset X^2 \backslash \Delta$ , there is an uncountable $B \subset A$ such that $\Delta \cap \overline{B}=\varnothing$ . In the literature, the definition of a space with a small diagonal is usually either this one or the one given at the beginning of this post.

Example 2

Now consider the space $X=\omega_1 \cup \{ \infty \}$ where points in $\omega_1$ are isolated and open neighborhoods of the point $\infty$ are of the form $\{ \infty \} \cup (\omega_1 \backslash F)$ with $F$ being any finite subset of $\omega_1$ . This is usually called the one-point compactification of a discrete space (in this case of size $\omega_1$ ). This space is compact. It is also non-metrizable since it has an uncountable discrete subset. Thus it cannot have a $G_\delta$ -diagonal. The weight of $X$ here is $\omega_1$ . Any compact space whose weight is $\omega_1$ cannot have a small diagonal. This is Fact 1 below.

Some Basic Results

Fact 1
Let $X$ any compact space with $w(X)=\omega_1$ . Then $X$ does not have a small diagonal.

Proof of Fact 1
Let $\mathcal{B}=\{ B_\alpha: \alpha < \omega_1 \}$ be a base for $X$ . Let $\mathcal{B}_1=\{ U \times V: U, V \in \mathcal{B} \}$ , which is a base for $X^2$ . Define $\mathcal{U}=\{ \cup F: F \subset \mathcal{B}_1 \text{ and } \lvert F \lvert < \omega \text{ and } \Delta \subset \cup F \}$ . Since $X$ is compact, the diagonal $\Delta$ is compact. As a result, $\mathcal{U} \ne \varnothing$ . Furthermore, $\lvert \mathcal{U} \lvert=\omega_1$ .

We claim that $\mathcal{U}$ is a base for the diagonal $\Delta$ . To show this, let $W$ be an open subset of $X^2$ such that $\Delta \subset W$ . We can assume that $W$ is the union of elements of the base $\mathcal{B}_1$ . Let $W=\cup \mathcal{A}$ for some $\mathcal{A} \subset \mathcal{B}_1$ . Since $\Delta$ is compact, $\Delta \subset \cup \mathcal{A}_0$ for some finite $\mathcal{A}_0 \subset \mathcal{A}$ . Note that $\cup \mathcal{A}_0 \in \mathcal{U}$ and that $\Delta \subset \cup \mathcal{A}_0 \subset W$ .

Enumerate $\mathcal{U}$ as $\mathcal{U}=\{U_\alpha: \alpha<\omega_1 \}$ . Since $X$ is compact and has uncountable weight, $X$ is not metrizable. Hence the diagonal $\Delta$ is not a $G_\delta$ -set. As a result, for each $\alpha<\omega_1$ , $(\bigcap_{\beta<\alpha} U_\beta) \backslash \Delta \ne \varnothing$ . Furthermore, for any countable $\{ y_0, y_1, y_2, \cdots \} \subset X^2 \backslash \Delta$ , $[(\bigcap_{\beta<\alpha} U_\beta) \cap (\bigcap_{n \in \omega} X^2 \backslash \{ y_n \}) ]\backslash \Delta \ne \varnothing$ .

Pick $x_0 \in U_0$ . For any $\alpha<\omega_1$ with $\alpha > 0$ , choose $x_\alpha \in (\bigcap_{\beta<\alpha} U_\beta) \backslash (\Delta \cup \{x_\delta: \delta < \alpha \})$ .

Then the sequence $\{ x_\alpha: \alpha < \omega_1 \}$ converges to the diagonal $\Delta$ . To see this, fix $U_\gamma \in \mathcal{U}$ . From the way the sequence is chosen, $x_\alpha \in U_\gamma$ for all $\alpha > \gamma$ . This concludes the proof that $X$ does not have a small diagonal. $\square$

Fact 2
Let $X$ any compact space with $w(X) \le \omega_1$ . Then if $X$ has a small diagonal, then $w(X)=\omega$ and thus $X$ is metrizable.

Fact 2 is an easy corollary of Fact 1. Fact 2 says that any compact space with “small” weight (no more than $\omega_1$ ) is metrizable if it has a small diagonal. Both Example 1 and Example 2 are compact non-metrizable spaces with small weight. Therefore they do not have small diagonal. The following basic fact is also useful.

Fact 3
Let $X$ any compact space. Then if $X$ has a small diagonal, then $t(X)=\omega$ , i.e. $X$ is countably tight.

Proof of Fact 3
Suppose $X$ is uncountably tight. We show that it does not have a small diagonal. A sequence of points $\{ x_\alpha \in X: \alpha< \tau \}$ is a free sequence of length $\tau$ if for each $\alpha< \tau$ , $\overline{\{ x_\gamma: \gamma < \alpha \}} \cap \overline{\{ x_\gamma: \gamma \ge \alpha \}}=\varnothing$ . For any compact space with tightness at least $\omega_1$ , there exists a free sequence of length $\omega_1$ (see Lemma 2 here). Thus in the space $X$ in question, there exists a free sequence $\{ x_\alpha \in X: \alpha< \omega_1 \}$ .

The main result in [9] says that if a compact space contains a free sequence of length $\omega_1$ , then it contains a free sequence of the same length that is convergent. We now assume that the above free sequence $\{ x_\alpha \in X: \alpha< \omega_1 \}$ is also convergent, i.e. it converges to some point $x \in X$ . Thus every open set containing $x$ contains all but countably many $x_\alpha$ . Consider the sequence $\{ (x_\alpha, x_{\alpha+1}) \in X^2: \alpha< \omega_1 \}$ . Observe that every open set in $X^2$ containing the point $(x,x)$ contains all but countably pairs $(x_\alpha, x_{\alpha+1})$ . This implies that every open set containing the diagonal $\Delta$ contains all but countably many points in the sequence. This shows that the compact $X$ does not have a small diagonal. $\square$

According to Fact 3, the compact ordinal $\omega_1+1=[0, \omega_1]$ does not have a small diagonal since it is uncountably tight at the last point $\omega_1$ .

Example 3

We now consider the double arrow space. Let $D=[0,1] \times \{0, 1 \}$ . The space $D$ consists of two copies of the unit interval, an upper one and a lower one. See the first two diagrams in this previous post. For $0 \le a <1$ , a basic open set containing the point $(a, 1)$ in the upper interval is of the form $\biggl( [a,b) \times \{ 1 \} \biggr) \cup \biggl( (a,b) \times \{0 \} \biggr)$ . For $0<a \le 1$ , a basic open set containing the point $(a, 0)$ in the lower interval is of the form $\biggl( (c,a) \times \{ 1 \} \biggr) \cup \biggl( (c,a] \times \{0 \} \biggr)$ . The rightmost point in the upper interval $(1,1)$ and the leftmost point in the lower interval $(0,0)$ are made isolated points.

The double arrow space $D$ is compact, perfectly normal and not metrizable (discussed here). Thus $D$ does not have a $G_\delta$ -diagonal. We do not have a direct way of showing that it does not have a small diagonal. We rely on a result from [5], which says that every compact metrizably fibered space with a small diagonal is metrizable. In light of this result, we only need to show that the double arrow space $D$ is metrizably fibered. A space $Y$ is metrizably fibered if there is a continuous map from $Y$ onto some metrizable space $M$ such that each point inverse is metrizable.

Starting with the double arrow space $D$ , let $f:D \rightarrow [0,1]$ be defined by $f((a,0))=f((a,1))=a$ for each $0 \le a \le 1$ . This is a two-to-one continuous map from the double arrow space $D$ onto the unit interval. The map $f$ is essentially a quotient map, the result of identifying $\{ (a,0), (a, 1) \}$ as one point $a$ . By the result in [5], the double arrow space $D$ cannot have a small diagonal.

The Open Problem

As mentioned earlier, what makes the property of small diagonal is a related longstanding open problem. The statement “every compact space with a $G_\delta$ -diagonal is metrizable” is true in ZFC. What about the following statement?

(*) Every compact space with a small diagonal is metrizable.

This question whether this statement is true was raised in Husek [7]. As of the writing of this article, the problem is still unsolved. There are partial results, some consistent results and some ZFC results.

Assuming CH, the answer to Hesek’s question is positive. Husek [7] showed that under CH, every compact space $X$ with a small diagonal such that the tightness of $X$ is countable is metrizable. Fact 3 from above states that every compact space with a small diagonal has countble tightness. Combining the two results, it follows that under CH any compact space with a small diagonal is metrizable. Fact 3 followed from a result by Juhasz and Szentmikloss [9]. Dow and Pavlov [3] showed that under PFA, every compact space with a small diagonal is metrizable.

There are partial answers in ZFC. We mention three results. The first one is Fact 2 discussed above, which is that compact spaces with small diagonal are metrizable if there is a weight restriction (weight no more than $\omega_1$ ). If there exists a non-metrizable compact space with a small diagonal, its weight would have to be greater than $\omega_1$ .

Gruenhage [5] showed that every compact metrizably fibered space with a small diagonal is metrizable (this result is discussed in Example 3 above). Dow and Hart [2] showed that every compact space with a small diagonal that is weight $\omega_1$ fibered is metrizable. The notion of weight $\omega_1$ fibered is a generalization of metrizably fibered. A space $X$ is weight $\omega_1$ fibered if there is a continuous surjection $f: X \rightarrow Y$ such that $Y$ and each point inverse $f^{-1}(y)$ have weight at most $\omega_1$ .

The statement “every countably compact space with a $G_\delta$ -diagonal is metrizable” is a true statement in ZFC (see here). How about the statement “every countably compact space with a small diagonal is metrizable”, the statement (*) above with compact replaced by countably compact? Gruenhage [5] showed that this statement is consistent with and independent of ZFC. To prove or disprove this statement extra set theory assumptions beyond ZFC are required. This provides an interesting contrast between compact and countably compact with respect to the open problem. By broadening Husek’s question from compact to countably compact, the statement cannot be settled in ZFC. Dow and Pavlov [3] provided two consistent examples of “countably compact non-metrizable with small diagonal” that are improvements upon examples from Gruenhage.

Now consider the statement “Lindelof space with a small diagonal must have a $G_\delta$ -diagonal.” Dow and Pavlov [3] provided a consistent negative answer, an example of a Lindelof space with a small diagonal that does not have a $G_\delta$ -diagonal (under negation of CH).

Another broad natural question is: what do compact spaces with small diagonal look like? The main problem is, of course, trying to see if these spaces are metrizable. There have been attempts to explore this general question of what these spaces look like. According to Juhasz and Szentmikloss [9], these spaces have countable tightness (Fact 2 above). However, it is not known if compact spaces with small diagonal have points of countable character. Dow and Hart [2] uncovered a surprising connection that if there is a subset of the real line that is a Luzin set, every compact space with a small diagonal does have points of countable character. Dow and Hart in the same paper also showed that in every compact space with a small diagonal, CCC subspaces have countable $\pi$ -weight.

This is a brief walk through of the open problem based on the statement (*) indicated above. To find out more, consult with the references listed below. Beyond the main open problem, there are many angles to be explored.

Reference

Arhangelskii, A., Bella, A., Few observations on topological spaces with small diagonal, Zb. Rad. Filoz. Fak. Nisu, 6, No. 2, 211-213, 1992.
Dow, A., Hart, P. Elementary chains and compact spaces with a small diagonal, Indagationes Mathematicae, 23, No. 3, 438-447, 2012.
Dow, A., Pavlov, O. More about spaces with a small diagonal, Fund. Math., 191, No. 1, 67-80, 2006.
Dow, A., Pavlov, O. Perfect preimages and small diagonal, Topology Proc., 31, No. 1, 89-95, 2007.
Gruenhage, G., Spaces having a small diagonal, Topology Appl., 22, 183-200, 2002.
Gruenhage, G., Generalized metrizable spaces, Recent Progress in General Topology III (K.P. Hart, J. van Mill, and P. Simon, eds.), Atlantis Press 2014.
Husek, M., Topological spaces without $\kappa$ -accessible diagonal, Comment. Math. Univ. Carolin., 18, No. 4, 777-788, 1977.
Juhasz, I., Cardinals Functions II, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 63-109, 1984.
Juhasz, I., Szentmikloss, Z. Convergent free sequences in compact spaces, Proc. Amer. Math. Soc., 116, No. 4, 1153-1160, 1992.
Zhou, H. X., On The Small Diagonals, Topology Appl., 13, 283-293, 1982.

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Pseudocompact spaces with regular G-delta diagonals

Posted on February 19, 2018 by Dan Ma

This post complements two results discussed in two previous blog posts concerning $G_\delta$ -diagonal. One result is that any compact space with a $G_\delta$ -diagonal is metrizable (see here). The other result is that the compactness in the first result can be relaxed to countably compactness. Thus any countably compact space with a $G_\delta$ -diagonal is metrizable (see here). The countably compactness in the second result cannot be relaxed to pseudocompactness. The Mrowka space is a pseudocompact space with a $G_\delta$ -diagonal that is not submetrizable, hence not metrizable (see here). However, if we strengthen the $G_\delta$ -diagonal to a regular $G_\delta$ -diagonal while keeping pseudocompactness fixed, then we have a theorem. We prove the following theorem.

Theorem 1
If the space $X$ is pseudocompact and has a regular $G_\delta$ -diagonal, then $X$ is metrizable.

All spaces are assumed to be Hausdorff and completely regular. The assumption of completely regular is crucial. The proof of Theorem 1 relies on two lemmas concerning pseudocompact spaces (one proved in a previous post and one proved here). These two lemmas work only for completely regular spaces.

The proof of Theorem 1 uses a metrization theorem. The best metrization to use in this case is Moore metrization theorem (stated below). The result in Theorem 1 is found in [2].

First some basics. Let $X$ be a space. The diagonal of the space $X$ is the set $\Delta=\{ (x,x): x \in X \}$ . When the diagonal $\Delta$ , as a subset of $X \times X$ , is a $G_\delta$ -set, i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$ , the space $X$ is said to have a $G_\delta$ -diagonal.

The space $X$ is said to have a regular $G_\delta$ -diagonal if the diagonal $\Delta$ is a regular $G_\delta$ -set in $X \times X$ , i.e. $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$ . If $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ , then $\Delta=\bigcap_{n=1}^\infty \overline{U_n}=\bigcap_{n=1}^\infty U_n$ . Thus if a space has a regular $G_\delta$ -diagonal, it has a $G_\delta$ -diagonal. We will see that there exists a space with a $G_\delta$ -diagonal that fails to be a regular $G_\delta$ -diagonal.

The space $X$ is a pseudocompact space if for every continuous function $f:X \rightarrow \mathbb{R}$ , the image $f(X)$ is a bounded set in the real line $\mathbb{R}$ . Pseudocompact spaces are discussed in considerable details in this previous post. We will rely on results from this previous post to prove Theorem 1.

The following lemma is used in proving Theorem 1.

Lemma 2
Let $X$ be a pseudocompact space. Suppose that $O_1,O_2,O_2,\cdots$ is a decreasing sequence of non-empty open subsets of $X$ such that $\bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}$ for some point $x \in X$ . Then $\{ O_n \}$ is a local base at the point $x$ .

Proof of Lemma 2
Let $O_1,O_2,O_2,\cdots$ be a decreasing sequence of open subsets of $X$ such that $\bigcap_{n=1}^\infty O_n=\bigcap_{n=1}^\infty \overline{O_n}=\{ x \}$ . Let $U$ be open in $X$ with $x \in U$ . If $O_n \subset U$ for some $n$ , then we are done. Suppose that $O_n \not \subset U$ for each $n$ .

Choose open $V$ with $x \in V \subset \overline{V} \subset U$ . Consider the sequence $\{ O_n \cap (X-\overline{V}) \}$ . This is a decreasing sequence of non-empty open subsets of $X$ . By Theorem 2 in this previous post, $\bigcap \overline{O_n \cap (X-\overline{V})} \ne \varnothing$ . Let $y$ be a point in this non-empty set. Note that $y \in \bigcap_{n=1}^\infty \overline{O_n}$ . This means that $y=x$ . Since $x \in \overline{O_n \cap (X-\overline{V})}$ for each $n$ , any open set containing $x$ would contain a point not in $\overline{V}$ . This is a contradiction since $x \in V$ . Thus it must be the case that $x \in O_n \subset U$ for some $n$ . $\square$

The following metrization theorem is useful in proving Theorem 1.

Theorem 3 (Moore Metrization Theorem)
Let $X$ be a space. Then $X$ is metrizable if and only if the following condition holds.

There exists a decreasing sequence $\mathcal{B}_1,\mathcal{B}_2,\mathcal{B}_3,\cdots$ of open covers of $X$ such that for each $x \in X$ , the sequence $\{ St(St(x,\mathcal{B}_n),\mathcal{B}_n):n=1,2,3,\cdots \}$ is a local base at the point $x$ .

For any family $\mathcal{U}$ of subsets of $X$ , and for any $A \subset X$ , the notation $St(A,\mathcal{U})$ refers to the set $\cup \{U \in \mathcal{U}: U \cap A \ne \varnothing \}$ . In other words, it is the union of all sets in $\mathcal{U}$ that contain points of $A$ . The set $St(A,\mathcal{U})$ is also called the star of the set $A$ with respect to the family $\mathcal{U}$ . If $A=\{ x \}$ , we write $St(x,\mathcal{U})$ instead of $St(\{ x \},\mathcal{U})$ . The set $St(St(x,\mathcal{B}_n),\mathcal{B}_n)$ indicated in Theorem 3 is the star of the set $St(x,\mathcal{B}_n)$ with respect to the open cover $\mathcal{B}_n$ .

Theorem 3 follows from Theorem 1.4 in [1], which states that for any $T_0$ -space $X$ , $X$ is metrizable if and only if there exists a sequence $\mathcal{G}_1, \mathcal{G}_2, \mathcal{G}_3,\cdots$ of open covers of $X$ such that for each open $U \subset X$ and for each $x \in U$ , there exist an open $V \subset X$ and an integer $n$ such that $x \in V$ and $St(V,\mathcal{G}_n) \subset U$ .

Proof of Theorem 1

Suppose $X$ is pseudocompact such that its diagonal $\Delta=\bigcap_{n=1}^\infty \overline{U_n}$ where each $U_n$ is an open subset of $X \times X$ with $\Delta \subset U_n$ . We can assume that $U_1 \supset U_2 \supset \cdots$ . For each $n \ge 1$ , define the following:

$\mathcal{U}_n=\{ U \subset X: U \text{ open in } X \text{ and } U \times U \subset U_n \}$

Note that each $\mathcal{U}_n$ is an open cover of $X$ . Also note that $\{ \mathcal{U}_n \}$ is a decreasing sequence since $\{ U_n \}$ is a decreasing sequence of open sets. We show that $\{ \mathcal{U}_n \}$ is a sequence of open covers of $X$ that satisfies Theorem 3. We establish this by proving the following claims.

Claim 1. For each $x \in X$ , $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$ .

To prove the claim, let $x \ne y$ . There is an integer $n$ such that $(x,y) \notin \overline{U_n}$ . Choose open sets $U$ and $V$ such that $(x,y) \in U \times V$ and $(U \times V) \cap \overline{U_n}=\varnothing$ . Note that $(x,y) \notin U_k$ and $(U \times V) \cap U_n=\varnothing$ .

We want to show that $V \cap St(x,\mathcal{U}_n)=\varnothing$ , which implies that $y \notin \overline{St(x,\mathcal{U}_n)}$ . Suppose $V \cap St(x,\mathcal{U}_n) \ne \varnothing$ . This means that $V \cap W \ne \varnothing$ for some $W \in \mathcal{U}_n$ with $x \in W$ . Then $(U \times V) \cap (W \times W) \ne \varnothing$ . Note that $W \times W \subset U_n$ . This implies that $(U \times V) \cap U_n \ne \varnothing$ , a contradiction. Thus $V \cap St(x,\mathcal{U}_n)=\varnothing$ . Since $y \in V$ , $y \notin \overline{St(x,\mathcal{U}_n)}$ . We have established that for each $x \in X$ , $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$ .

Claim 2. For each $x \in X$ , $\{ St(x,\mathcal{U}_n) \}$ is a local base at the point $x$ .

Note that $\{ St(x,\mathcal{U}_n) \}$ is a decreasing sequence of open sets such that $\bigcap_{n=1}^\infty \overline{St(x,\mathcal{U}_n)}=\{ x \}$ . By Lemma 2, $\{ St(x,\mathcal{U}_n) \}$ is a local base at the point $x$ .

Claim 3. For each $x \in X$ , $\bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}=\{ x \}$ .

Let $x \ne y$ . There is an integer $n$ such that $(x,y) \notin \overline{U_n}$ . Choose open sets $U$ and $V$ such that $(x,y) \in U \times V$ and $(U \times V) \cap \overline{U_n}=\varnothing$ . It follows that $(U \times V) \cap \overline{U_t}=\varnothing$ for all $t \ge n$ . Furthermore, $(U \times V) \cap U_t=\varnothing$ for all $t \ge n$ . By Claim 2, choose integers $i$ and $j$ such that $St(x,\mathcal{U}_i) \subset U$ and $St(y,\mathcal{U}_j) \subset V$ . Choose an integer $k \ge \text{max}(n,i,j)$ . It follows that $(St(x,\mathcal{U}_i) \times St(y,\mathcal{U}_j)) \cap U_k=\varnothing$ . Since $\mathcal{U}_k \subset \mathcal{U}_i$ and $\mathcal{U}_k \subset \mathcal{U}_j$ , it follows that $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k=\varnothing$ .

We claim that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$ . Suppose not. Choose $w \in St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)$ . It follows that $w \in B$ for some $B \in \mathcal{U}_k$ such that $B \cap St(x,\mathcal{U}_k) \ne \varnothing$ and $B \cap St(y,\mathcal{U}_k) \ne \varnothing$ . Furthermore $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap (B \times B)=\varnothing$ . Note that $B \times B \subset U_k$ . This means that $(St(x,\mathcal{U}_k) \times St(y,\mathcal{U}_k)) \cap U_k \ne \varnothing$ , contradicting the fact observed in the preceding paragraph. It must be the case that $St(y,\mathcal{U}_k) \cap St(St(x,\mathcal{U}_k), \mathcal{U}_k)=\varnothing$ .

Because there is an open set containing $y$ , namely $St(y,\mathcal{U}_k)$ , that contains no points of $St(St(x,\mathcal{U}_k), \mathcal{U}_k)$ , $y \notin \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n)}$ . Thus Claim 3 is established.

Claim 4. For each $x \in X$ , $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n)) \}$ is a local base at the point $x$ .

Note that $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \}$ is a decreasing sequence of open sets such that $\bigcap_{n=1}^\infty \overline{St(St(x,\mathcal{U}_n),\mathcal{U}_n))}=\{ x \}$ . By Lemma 2, $\{ St(St(x,\mathcal{U}_n),\mathcal{U}_n) \}$ is a local base at the point $x$ .

In conclusion, the sequence $\mathcal{U}_1,\mathcal{U}_2,\mathcal{U}_3,\cdots$ of open covers satisfies the properties in Theorem 3. Thus any pseudocompact space with a regular $G_\delta$ -diagonal is metrizable. $\square$

Example

Any submetrizable space has a $G_\delta$ -diagonal. The converse is not true. A classic example of a non-submetrizable space with a $G_\delta$ -diagonal is the Mrowka space (discussed here). The Mrowka space is also called the psi-space since it is sometimes denoted by $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal family of almost disjoint subsets of $\omega$ . Actually $\Psi(\mathcal{A})$ would be a family of spaces since $\mathcal{A}$ is any maximal almost disjoint family. For any maximal $\mathcal{A}$ , $\Psi(\mathcal{A})$ is a pseudocompact non-submetrizable space that has a $G_\delta$ -diagonal. This example shows that the requirement of a regular $G_\delta$ -diagonal in Theorem 1 cannot be weakened to a $G_\delta$ -diagonal. See here for a more detailed discussion of this example.

Reference

Gruenhage, G., Generalized Metric Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 423-501, 1984.
McArthur W. G., $G_\delta$ -Diagonals and Metrization Theorems, Pacific Journal of Mathematics, Vol. 44, No. 2, 613-317, 1973.

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Dan Ma math

Daniel Ma mathematics

$\copyright$ 2018 – Dan Ma

Looking for spaces in which every compact subspace is metrizable

Posted on June 11, 2017 by Dan Ma

Once it is known that a topological space is not metrizable, it is natural to ask, from a metrizability standpoint, which subspaces are metrizable, e.g. whether every compact subspace is metrizable. This post discusses several classes of spaces in which every compact subspace is metrizable. Though the goal here is not to find a complete characterization of such spaces, this post discusses several classes of spaces and various examples that have this property. The effort brings together many interesting basic and well known facts. Thus the notion “every compact subspace is metrizable” is an excellent learning opportunity.

Several Classes of Spaces

The notion “every compact subspace is metrizable” is a very broad class of spaces. It includes well known spaces such as Sorgenfrey line, Michael line and the first uncountable ordinal $\omega_1$ (with the order topology) as well as Moore spaces. Certain function spaces are in the class “every compact subspace is metrizable”. The following diagram is a good organizing framework.

$\displaystyle \begin{aligned} &1. \ \text{Metrizable} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&2. \ \text{Submetrizable} \Longleftarrow 5. \ \exists \ \text{countable network} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&3. \ \exists \ G_\delta \text{ diagonal} \\&\ \ \ \ \ \ \ \ \ \Downarrow \\&4. \ \text{Every compact subspace is metrizable} \end{aligned}$

Let $(X, \tau)$ be a space. It is submetrizable if there is a topology $\tau_1$ on the set $X$ such that $\tau_1 \subset \tau$ and $(X, \tau_1)$ is a metrizable space. The topology $\tau_1$ is said to be weaker (coarser) than $\tau$ . Thus a space $X$ is submetrizable if it has a weaker metrizable topology.

Let $\mathcal{N}$ be a set of subsets of the space $X$ . $\mathcal{N}$ is said to be a network for $X$ if for every open subset $O$ of $X$ and for each $x \in O$ , there exists $N \in \mathcal{N}$ such that $x \in N \subset O$ . Having a network that is countable in size is a strong property (see here for a discussion on spaces with a countable network).

The diagonal of the space $X$ is the subset $\Delta=\left\{(x,x): x \in X \right\}$ of the square $X \times X$ . The space $X$ has a $G_\delta$ -diagonal if $\Delta$ is a $G_\delta$ -subset of $X \times X$ , i.e. $\Delta$ is the intersection of countably many open subsets of $X \times X$ .

The implication $1 \Longrightarrow 2$ is clear. For $5 \Longrightarrow 2$ , see Lemma 1 in this previous post on countable network. The implication $2 \Longrightarrow 3$ is left as an exercise. To see $3 \Longrightarrow 4$ , let $K$ be a compact subset of $X$ . The property of having a $G_\delta$ -diagonal is hereditary. Thus $K$ has a $G_\delta$ -diagonal. According to a well known result, any compact space with a $G_\delta$ -diagonal is metrizable (see here).

None of the implications in the diagram is reversible. The first uncountable ordinal $\omega_1$ is an example for $4 \not \Longrightarrow 3$ . This follows from the well known result that any countably compact space with a $G_\delta$ -diagonal is metrizable (see here). The Mrowka space is an example for $3 \not \Longrightarrow 2$ (see here). The Sorgenfrey line is an example for both $2 \not \Longrightarrow 5$ and $2 \not \Longrightarrow 1$ .

To see where the examples mentioned earlier are placed, note that Sorgenfrey line and Michael line are submetrizable, both are submetrizable by the usual Euclidean topology on the real line. Each compact subspace of the space $\omega_1$ is countable and is thus contained in some initial segment $[0,\alpha]$ which is metrizable. Any Moore space has a $G_\delta$ -diagonal. Thus compact subspaces of a Moore space are metrizable.

Function Spaces

We now look at some function spaces that are in the class “every compact subspace is metrizable.” For any Tychonoff space (completely regular space) $X$ , $C_p(X)$ is the space of all continuous functions from $X$ into $\mathbb{R}$ with the pointwise convergence topology (see here for basic information on pointwise convergence topology).

Theorem 1
Suppose that $X$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.

Proof
The proof here actually shows more than is stated in the theorem. We show that $C_p(X)$ is submetrizable by a separable metric topology. Let $Y$ be a countable dense subspace of $X$ . Then $C_p(Y)$ is metrizable and separable since it is a subspace of the separable metric space $\mathbb{R}^{\omega}$ . Thus $C_p(Y)$ has a countable base. Let $\mathcal{E}$ be a countable base for $C_p(Y)$ .

Let $\pi:C_p(X) \longrightarrow C_p(Y)$ be the restriction map, i.e. for each $f \in C_p(X)$ , $\pi(f)=f \upharpoonright Y$ . Since $\pi$ is a projection map, it is continuous and one-to-one and it maps $C_p(X)$ into $C_p(Y)$ . Thus $\pi$ is a continuous bijection from $C_p(X)$ into $C_p(Y)$ . Let $\mathcal{B}=\left\{\pi^{-1}(E): E \in \mathcal{E} \right\}$ .

We claim that $\mathcal{B}$ is a base for a topology on $C_p(X)$ . Once this is established, the proof of the theorem is completed. Note that $\mathcal{B}$ is countable and elements of $\mathcal{B}$ are open subsets of $C_p(X)$ . Thus the topology generated by $\mathcal{B}$ is coarser than the original topology of $C_p(X)$ .

For $\mathcal{B}$ to be a base, two conditions must be satisfied – $\mathcal{B}$ is a cover of $C_p(X)$ and for $B_1,B_2 \in \mathcal{B}$ , and for $f \in B_1 \cap B_2$ , there exists $B_3 \in \mathcal{B}$ such that $f \in B_3 \subset B_1 \cap B_2$ . Since $\mathcal{E}$ is a base for $C_p(Y)$ and since elements of $\mathcal{B}$ are preimages of elements of $\mathcal{E}$ under the map $\pi$ , it is straightforward to verify these two points. $\square$

Theorem 1 is actually a special case of a duality result in $C_p$ function space theory. More about this point later. First, consider a corollary of Theorem 1.

Corollary 2
Let $X=\prod_{\alpha<c} X_\alpha$ where $c$ is the cardinality continuum and each $X_\alpha$ is a separable space. Then every compact subspace of $C_p(X)$ is metrizable.

The key fact for Corollary 2 is that the product of continuum many separable spaces is separable (this fact is discussed here). Theorem 1 is actually a special case of a deep result.

Theorem 3
Suppose that $X=\prod_{\alpha<\kappa} X_\alpha$ is a product of separable spaces where $\kappa$ is any infinite cardinal. Then every compact subspace of $C_p(X)$ is metrizable.

Theorem 3 is a much more general result. The product of any arbitrary number of separable spaces is not separable if the number of factors is greater than continuum. So the proof for Theorem 1 will not work in the general case. This result is Problem 307 in [2].

A Duality Result

Theorem 1 is stated in a way that gives the right information for the purpose at hand. A more correct statement of Theorem 1 is: $X$ is separable if and only if $C_p(X)$ is submetrizable by a separable metric topology. Of course, the result in the literature is based on density and weak weight.

The cardinal function of density is the least cardinality of a dense subspace. For any space $Y$ , the weight of $Y$ , denoted by $w(Y)$ , is the least cardinaility of a base of $Y$ . The weak weight of a space $X$ is the least $w(Y)$ over all space $Y$ for which there is a continuous bijection from $X$ onto $Y$ . Thus if the weak weight of $X$ is $\omega$ , then there is a continuous bijection from $X$ onto some separable metric space, hence $X$ has a weaker separable metric topology.

There is a duality result between density and weak weight for $X$ and $C_p(X)$ . The duality result:

The density of $X$ coincides with the weak weight of $C_p(X)$ and the weak weight of $X$ coincides with the density of $C_p(X)$ . These are elementary results in $C_p$ -theory. See Theorem I.1.4 and Theorem I.1.5 in [1].

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Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Tkachuk V. V., A $C_p$ -Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

An exercise gleaned from the proof of a theorem on pseudocompact space

Posted on August 2, 2015 by Dan Ma

Filling in the gap is something that is done often when following a proof in a research paper or other published work. In fact this is necessary since it is not feasible for authors to prove or justify every statement or assertion in a proof (or define every term). The gap could be a basic result or could be an older result from another source. If the gap is a basic result or a basic fact that is considered folklore, it may be OK to put it on hold in the interest of pursuing the main point. Then come back later to fill the gap. In any case, filling in gaps is a great learning opportunity. In this post, we focus on one such example of filling in the gap. The example is from the book called Topological Function Spaces by A. V. Arkhangelskii [1].

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Pseudocompactness

The exercise we wish to highlight deals with continuous one-to-one functions defined on pseudocompact spaces. We first give a brief backgrounder on pseudocompact spaces with links to earlier posts.

All spaces considered are Hausdorff spaces. A space $X$ is a pseudocompact space if every continuous real-valued function defined on $X$ is bounded, i.e., if $f:X \rightarrow \mathbb{R}$ is a continuous function, then $f(X)$ is a bounded set in the real line. Compact spaces are pseudocompact. In fact, it is clear from definitions that

$\text{compact} \Longrightarrow \text{countably compact} \Longrightarrow \text{pseudocompact}$

None of the implications can be reversed. An example of a pseudocompact space that is not countably compact is the space $\Psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$ (see here for the details). Some basic results on pseudocompactness focus on the conditions to add in order to turn a pseudocompact space into countably compact or even compact. For example, for normal spaces, pseudocompact implies countably compact. This tells us that when looking for pseudocompact space that is not countably compact, do not look among normal spaces. Another interesting result is that pseudocompact + metacompact implies compact. Likewise, when looking for pseudocompact space that is not compact, look among non-metacompact spaces. On the other hand, this previous post discusses when a pseudocompact space is metrizable. Another two previous posts also discuss pseudocompactness (see here and here).

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The exercise

Consider Theorem II.6.2 part (c) in pp. 76-77 in [1]. We do not state the theorem because it is not the focus here. Instead, we focus on an assertion in the proof of Theorem II.6.2.

The exercise that we wish to highlight is stated in Theorem 2 below. Theorem 1 is a standard result about continuous one-to-one functions defined on compact spaces and is stated here to contrast with Theorem 2.

Theorem 1
Let $Y$ be a compact space. Let $g: Y \rightarrow Z$ be a one-to-one continuous function from $Y$ onto a space $Z$ . Then $g$ is a homeomorphism.

Theorem 2
Let $Y$ be a pseudocompact space. Let $g: Y \rightarrow Z$ be a one-to-one continuous function from $Y$ onto $Z$ where $Z$ is a separable and metrizable space. Then $g$ is a homeomorphism.

Theorem 1 says that any continuous one-to-one map from a compact space onto another compact space is a homeomorphism. To show a given map between two compact spaces is a homeomorphism, we only need to show that it is continuous in one direction. Theorem 2, the statement used in the proof of Theorem II.6.2 in [1], says that the standard result for compact spaces can be generalized to pseudocompactness if the range space is nice.

The proof of Theorem II.6.2 part (c) in [1] quoted [2] as a source for the assertion in our Theorem 2. Here, we leave both Theorem 1 and Theorem 2 as exercise. One way to prove Theorem 2 is to show that whenever there exists a map $g$ as described in Theorem 2, the domain $Y$ must be compact. Then Theorem 1 will finish the job.

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Reference

Arkhangelskii A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
Arkhangelskii A. V., Ponomarev V. I., Fundamental of general topology: problems and exercises, Reidel, 1984. (Translated from the Russian).

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$\copyright \ 2015 \text{ by Dan Ma}$

Looking for non-normal subspaces of the square of a compact X

Posted on April 14, 2015 by Dan Ma

A theorem of Katetov states that if $X$ is compact with a hereditarily normal cube $X^3$ , then $X$ is metrizable (discussed in this previous post). This means that for any non-metrizable compact space $X$ , Katetov’s theorem guarantees that some subspace of the cube $X^3$ is not normal. Where can a non-normal subspace of $X^3$ be found? Is it in $X$ , in $X^2$ or in $X^3$ ? In other words, what is the “dimension” in which the hereditary normality fails for a given compact non-metrizable $X$ (1, 2 or 3)? Katetov’s theorem guarantees that the dimension must be at most 3. Out of curiosity, we gather a few compact non-metrizable spaces. They are discussed below. In this post, we motivate an independence result using these examples.

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Katetov’s theorems

First we state the results of Katetov for reference. These results are proved in this previous post.

Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

The factor $X$ is perfectly normal.
Every countable and infinite subset of the factor $Y$ is closed.

Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.

Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.

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Examples of compact non-metrizable spaces

The set-theoretic result presented here is usually motivated by looking at Theorem 3. The question is: Can $X^3$ in Theorem 3 be replaced by $X^2$ ? We take a different angle of looking at some standard compact non-metric spaces and arrive at the same result. The following is a small listing of compact non-metrizable spaces. Each example in this list is defined in ZFC alone, i.e. no additional axioms are used beyond the generally accepted axioms of set theory.

One-point compactification of the Tychonoff plank.
One-point compactification of $\psi(\mathcal{A})$ where $\mathcal{A}$ is a maximal almost disjoint family of subsets of $\omega$ .
The first compact uncountable ordinal, i.e. $\omega_1+1$ .
The one-point compactification of an uncountable discrete space.
Alexandroff double circle.
Double arrow space.
Unit square with the lexicographic order.

Since each example in the list is compact and non-metrizable, the cube of each space must not be hereditarily normal according to Theorem 3 above. Where does the hereditary normality fail? For #1 and #2, $X$ is a compactification of a non-normal space and thus not hereditarily normal. So the dimension for the failure of hereditary normality is 1 for #1 and #2.

For #3 through #7, $X$ is hereditarily normal. For #3 through #5, each $X$ has a closed subset that is not a $G_\delta$ set (hence not perfectly normal). In #3 and #4, the non- $G_\delta$ -set is a single point. In #5, the the non- $G_\delta$ -set is the inner circle. Thus the compact space $X$ in #3 through #5 is not perfectly normal. By Theorem 2, the dimension for the failure of hereditary normality is 2 for #3 through #5.

For #6 and #7, each $X^2$ contains a copy of the Sorgenfrey plane. Thus the dimension for the failure of hereditary normality is also 2 for #6 and #7.

In the small sample of compact non-metrizable spaces just highlighted, the failure of hereditary normality occurs in “dimension” 1 or 2. Naturally, one can ask:

Question

$X$

$X^2$

$X^3$

Such a space $X$ would be an example to show that the condition “ $X^3$ is hereditarily normal” in Theorem 3 is necessary. In other words, the hypothesis in Theorem 3 cannot be weakened if the example just described were to exist.

The above list of compact non-metrizable spaces is a small one. They are fairly standard examples for compact non-metrizable spaces. Could there be some esoteric example out there that fits the description? It turns out that there are such examples. In [1], Gruenhage and Nyikos constructed a compact non-metrizable $X$ such that $X^2$ is hereditarily normal. The construction was done using MA + not CH (Martin’s Axiom coupled with the negation of the continuum hypothesis). In that same paper, they also constructed another another example using CH. With the examples from [1], one immediate question was whether the additional set-theoretic axioms of MA + not CH (or CH) was necessary. Could a compact non-metrizable $X$ such that $X^2$ is hereditarily normal be still constructed without using any axioms beyond ZFC, the generally accepted axioms of set theory? For a relatively short period of time, this was an open question.

In 2001, Larson and Todorcevic [3] showed that it is consistent with ZFC that every compact $X$ with hereditarily normal $X^2$ is metrizable. In other words, there is a model of set theory that is consistent with ZFC in which Theorem 3 can be improved to assuming $X^2$ is hereditarily normal. Thus it is impossible to settle the above question without assuming additional axioms beyond those of ZFC. This means that if a compact non-metrizable $X$ is constructed without using any axiom beyond ZFC (such as those in the small list above), the hereditary normality must fail at dimension 1 or 2. Numerous other examples can be added to the above small list. Looking at these ZFC examples can help us appreciate the results in [1] and [3]. These ZFC examples are excellent training ground for general topology.

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Reference

Gruenhage G., Nyikos P. J., Normality in $X^2$ for Compact $X$ , Trans. Amer. Math. Soc., Vol 340, No 2 (1993), 563-586
Katetov M., Complete normality of Cartesian products, Fund. Math., 35 (1948), 271-274
Larson P., Todorcevic S., KATETOV’S PROBLEM, Trans. Amer. Math. Soc., Vol 354, No 5 (2001), 1783-1791

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$\copyright \ 2015 \text{ by Dan Ma}$

When a product space is hereditarily normal

Posted on April 14, 2015 by Dan Ma

When the spaces $X$ and $Y$ are normal spaces, the product space $X \times Y$ is not necessarily normal. Even if one of the factors is metrizable, there is still no guarantee that the product is normal. So it is possible that the normality of each of the factors $X$ and $Y$ can have no influence on the normality of the product $X \times Y$ . The dynamics in the other direction are totally different. When the product $X \times Y$ is hereditarily normal, the two factors $X$ and $Y$ are greatly impacted. In this post, we discuss a theorem of Katetov, which shows that the hereditary normality of the product can impose very strict conditions on the factors, which lead to many interesting results. This theorem also leads to an interesting set-theoretic result, and thus can possibly be a good entry point to the part of topology that deals with consistency and independence results – statements that cannot be proved true or false based on the generally accepted axioms of set theory (ZFC). In this post, we discuss Katetov’s theorem and its consequences. In the next post, we discuss examples that further motivate the set-theoretic angle.

A subset $W$ of a space $X$ is said to be a $G_\delta$ -set in $X$ if $W$ is the intersection of countably many open subsets of $X$ . A space $X$ is perfectly normal if it is normal and that every closed subset of $X$ is a $G_\delta$ -set. Some authors use other statements to characterize perfect normality (here is one such characterization). Perfect normality implies hereditarily normal (see Theorem 6 in this previous post). The implication cannot be reversed. Katetov’s theorem implies that the hereditary normality of the product $X \times Y$ will in many cases make one or both of the factors perfectly normal. Thus the hereditary normality in the product $X \times Y$ is a very strong property.

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Katetov’s theorems

Theorem 1
If $X \times Y$ is hereditarily normal (i.e. every one of its subspaces is normal), then one of the following condition holds:

The factor $X$ is perfectly normal.
Every countable and infinite subset of the factor $Y$ is closed.

Proof of Theorem 1
The strategy we use is to define a subspace of $X \times Y$ that is not normal after assuming that none of the two conditions is true. So assume that $X$ has a closed subspace $W$ that is not a $G_\delta$ -set and assume that $T=\left\{t_n: n=1,2,3,\cdots \right\}$ is an infinite subset of $Y$ that is not closed. Let $p \in Y$ be a limit point of $T$ such that $p \notin T$ . The candidate for a non-normal subspace of $X \times Y$ is:

$M=X \times Y-W \times \left\{p \right\}$

Note that $M$ is an open subspace of $X \times Y$ since it is the result of subtracting a closed set from $X \times Y$ . The following are the two closed sets that demonstrate that $M$ is not normal.

$H=W \times (Y-\left\{p \right\})$

$K=(X-W) \times \left\{p \right\}$

It is clear that $H$ and $K$ are closed subsets of $M$ . Let $U$ and $V$ be open subsets of $M$ such that $H \subset U$ and $K \subset V$ . We show that $U \cap V \ne \varnothing$ . To this end, define $U_j=\left\{x \in X: (x,t_j) \in U \right\}$ for each $j$ . It follows that for each $j$ , $W \subset U_j$ . Furthermore each $U_j$ is an open subspace of $X$ . Thus $W \subset \bigcap_j U_j$ . Since $W$ is not a $G_\delta$ -set in $X$ , there must exist $t \in \bigcap_j U_j$ such that $t \notin W$ . Then $(t, p) \in K$ and $(t, p) \in V$ .

Since $V$ is open in the product $X \times Y$ , choose open sets $A \subset X$ and $B \subset Y$ such that $(t,p) \in A \times B$ and $A \times B \subset V$ . With $p \in B$ , there exists some $j$ such that $t_j \in B$ . First, $(t,t_j) \in V$ . Since $t \in U_j$ , $(t,t_j) \in U$ . Thus $U \cap V \ne \varnothing$ . This completes the proof that the subspace $M$ is not normal and that $X \times Y$ is not hereditarily normal. $\blacksquare$

Let’s see what happens in Theorem 1 when both factors are compact. If both $X$ and $Y$ are compact and if $X \times Y$ is hereditarily normal, then both $X$ and $Y$ must be perfect normal. Note that in any infinite compact space, not every countably infinite subset is closed. Thus if compact spaces satisfy the conclusion of Theorem 1, they must be perfectly normal. Hence we have the following theorem.

Theorem 2
If $X$ and $Y$ are compact and $X \times Y$ is hereditarily normal, then both $X$ and $Y$ are perfectly normal.

Moe interestingly, Theorem 1 leads to a metrization theorem for compact spaces.

Theorem 3
Let $X$ be a compact space. If $X^3=X \times X \times X$ is hereditarily normal, then $X$ is metrizable.

Proof of Theorem 3
Suppose that $X^3$ is hereditarily normal. By Theorem 2, the compact spaces $X^2$ and $X$ are perfectly normal. In particular, the following subset of $X^2$ is a $G_\delta$ -set in $X^2$ .

$\Delta=\left\{(x,x): x \in X \right\}$

The set $\Delta$ is said to be the diagonal of the space $X$ . It is a well known result that any compact space whose diagonal is a $G_\delta$ -set in the square is metrizable (discussed here). $\blacksquare$

The results discussed here make it clear that hereditary normality in product spaces is a very strong property. One obvious question is whether Theorem 3 can be improved by assuming only the hereditary normality of $X^2$ . This was indeed posted by Katetov himself. This leads to the discussion in the next post.

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Reference

Engelking R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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$\copyright \ 2015 \text{ by Dan Ma}$

Compact metrizable scattered spaces

Posted on April 1, 2015 by Dan Ma

A scattered space is one in which there are isolated points found in every subspace. Specifically, a space $X$ is a scattered space if every non-empty subspace $Y$ of $X$ has a point $y \in Y$ such that $y$ is an isolated point in $Y$ , i.e. the singleton set $\left\{y \right\}$ is open in the subspace $Y$ . A handy example is a space consisting of ordinals. Note that in a space of ordinals, every non-empty subset has an isolated point (e.g. its least element). In this post, we discuss scattered spaces that are compact metrizable spaces.

Here’s what led the author to think of such spaces. Consider Theorem III.1.2 found on page 91 of Arhangelskii’s book on topological function space [1], which is Theorem 1 stated below:

Thereom 1
For any compact space $X$ , the following conditions are equivalent:

The function space $C_p(X)$ is a Frechet-Urysohn space.
The function space $C_p(X)$ is a k space.
$X$ is a scattered space.

Let’s put aside the Frechet-Urysohn property and the k space property for the moment. For any Hausdorff space $X$ , let $C(X)$ be the set of all continuous real-valued functions defined on the space $X$ . Since $C(X)$ is a subspace of the product space $\mathbb{R}^X$ , a natural topology that can be given to $C(X)$ is the subspace topology inherited from the product space $\mathbb{R}^X$ . Then $C_p(X)$ is simply the set $C(X)$ with the product subspace topology (also called the pointwise convergence topology).

Let’s say the compact space $X$ is countable and infinite. Then the function space $C_p(X)$ is metrizable since it is a subspace of $\mathbb{R}^X$ , a product of countably many lines. Thus the function space $C_p(X)$ has the Frechet-Urysohn property (being metrizable implies Frechet-Urysohn). This means that the compact space $X$ is scattered. The observation just made is a proof that any infinite compact space that is countable in cardinality must be scattered. In particular, every infinite compact and countable space must have an isolated point. There must be a more direct proof of this same fact without taking the route of a function space. The indirect argument does not reveal the essential nature of compact metric spaces. The essential fact is that any uncountable compact metrizable space contains a Cantor set, which is as unscattered as any space can be. Thus the only scattered compact metrizable spaces are the countable ones.

The main part of the proof is the construction of a Cantor set in a compact metrizable space (Theorem 3). The main result is Theorem 4. In many settings, the construction of a Cantor set is done in the real number line (e.g. the middle third Cantor set). The construction here is in a more general setting. But the idea is still the same binary division process – the splitting of a small open set with compact closure into two open sets with disjoint compact closure. We also use that fact that any compact metric space is hereditarily Lindelof (Theorem 2).

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Compact metrizable spaces

We first define some notions before looking at compact metrizable spaces in more details. Let $X$ be a space. Let $A \subset X$ . Let $p \in X$ . We say that $p$ is a limit point of $A$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$ . So the notion of limit point here is from a topology perspective and not from a metric perspective. In a topological space, a limit point does not necessarily mean that it is the limit of a convergent sequence (however, it does in a metric space). The proof of the following theorem is straightforward.

Theorem 2
Let $X$ be a hereditarily Lindelof space (i.e. every subspace of $X$ is Lindelof). Then for any uncountable subset $A$ of $X$ , all but countably many points of $A$ are limit points of $A$ .

We now discuss the main result.

Theorem 3
Let $X$ be a compact metrizable space such that every point of $X$ is a limit point of $X$ . Then there exists an uncountable closed subset $C$ of $X$ such that every point of $C$ is a limit point of $C$ .

Proof of Theorem 3
Note that any compact metrizable space is a complete metric space. Consider a complete metric $\rho$ on the space $X$ . One fact that we will use is that if there is a sequence of closed sets $X \supset H_1 \supset H_2 \supset H_3 \supset \cdots$ such that the diameters of the sets $H$ (based on the complete metric $\rho$ ) decrease to zero, then the sets $H_n$ collapse to one point.

The uncountable closed set $C$ we wish to define is a Cantor set, which is constructed from a binary division process. To start, pick two points $p_0,p_1 \in X$ such that $p_0 \ne p_1$ . By assumption, both points are limit points of the space $X$ . Choose open sets $U_0,U_1 \subset X$ such that

$p_0 \in U_0$ ,
$p_1 \in U_1$ ,
$K_0=\overline{U_0}$ and $K_1=\overline{U_1}$ ,
$K_0 \cap K_1 = \varnothing$ ,
the diameters for $K_0$ and $K_1$ with respect to $\rho$ are less than 0.5.

Note that each of these open sets contains infinitely many points of $X$ . Then we can pick two points in each of $U_0$ and $U_1$ in the same manner. Before continuing, we set some notation. If $\sigma$ is an ordered string of 0’s and 1’s of length $n$ (e.g. 01101 is a string of length 5), then we can always extend it by tagging on a 0 and a 1. Thus $\sigma$ is extended as $\sigma 0$ and $\sigma 1$ (e.g. 01101 is extended by 011010 and 011011).

Suppose that the construction at the $n$ th stage where $n \ge 1$ is completed. This means that the points $p_\sigma$ and the open sets $U_\sigma$ have been chosen such that $p_\sigma \in U_\sigma$ for each length $n$ string of 0’s and 1’s $\sigma$ . Now we continue the picking for the $(n+1)$ st stage. For each $\sigma$ , an $n$ -length string of 0’s and 1’s, choose two points $p_{\sigma 0}$ and $p_{\sigma 1}$ and choose two open sets $U_{\sigma 0}$ and $U_{\sigma 1}$ such that

$p_{\sigma 0} \in U_{\sigma 0}$ ,
$p_{\sigma 1} \in U_{\sigma 1}$ ,
$K_{\sigma 0}=\overline{U_{\sigma 0}} \subset U_{\sigma}$ and $K_{\sigma 1}=\overline{U_{\sigma 1}} \subset U_{\sigma}$ ,
$K_{\sigma 0} \cap K_{\sigma 1} = \varnothing$ ,
the diameters for $K_{\sigma 0}$ and $K_{\sigma 1}$ with respect to $\rho$ are less than $0.5^{n+1}$ .

For each positive integer $m$ , let $C_m$ be the union of all $K_\sigma$ over all $\sigma$ that are $m$ -length strings of 0’s and 1’s. Each $C_m$ is a union of finitely many compact sets and is thus compact. Furthermore, $C_1 \supset C_2 \supset C_3 \supset \cdots$ . Thus $C=\bigcap \limits_{m=1}^\infty C_m$ is non-empty. To complete the proof, we need to show that

$C$ is uncountable (in fact of cardinality continuum),
every point of $C$ is a limit point of $C$ .

To show the first point, we define a one-to-one function $f: \left\{0,1 \right\}^N \rightarrow C$ where $N=\left\{1,2,3,\cdots \right\}$ . Note that each element of $\left\{0,1 \right\}^N$ is a countably infinite string of 0’s and 1’s. For each $\tau \in \left\{0,1 \right\}^N$ , let $\tau \upharpoonright n$ denote the string of the first $n$ digits of $\tau$ . For each $\tau \in \left\{0,1 \right\}^N$ , let $f(\tau)$ be the unique point in the following intersection:

$\displaystyle \bigcap \limits_{n=1}^\infty K_{\tau \upharpoonright n} = \left\{f(\tau) \right\}$

This mapping is uniquely defined. Simply conceptually trace through the induction steps. For example, if $\tau$ are 01011010…., then consider $K_0 \supset K_{01} \supset K_{010} \supset \cdots$ . At each next step, always pick the $K_{\tau \upharpoonright n}$ that matches the next digit of $\tau$ . Since the sets $K_{\tau \upharpoonright n}$ are chosen to have diameters decreasing to zero, the intersection must have a unique element. This is because we are working in a complete metric space.

It is clear that the map $f$ is one-to-one. If $\tau$ and $\gamma$ are two different strings of 0’s and 1’s, then they must differ at some coordinate, then from the way the induction is done, the strings would lead to two different points. It is also clear to see that the map $f$ is reversible. Pick any point $x \in C$ . Then the point $x$ must belong to a nested sequence of sets $K$ ‘s. This maps to a unique infinite string of 0’s and 1’s. Thus the set $C$ has the same cardinality as the set $\left\{0,1 \right\}^N$ , which has cardinality continuum.

To see the second point, pick $x \in C$ . Suppose $x=f(\tau)$ where $\tau \in \left\{0,1 \right\}^N$ . Consider the open sets $U_{\tau \upharpoonright n}$ for all positive integers $n$ . Note that $x \in U_{\tau \upharpoonright n}$ for each $n$ . Based on the induction process described earlier, observe these two facts. This sequence of open sets has diameters decreasing to zero. Each open set $U_{\tau \upharpoonright n}$ contains infinitely many other points of $C$ (this is because of all the open sets $U_{\tau \upharpoonright k}$ that are subsets of $U_{\tau \upharpoonright n}$ where $k \ge n$ ). Because the diameters are decreasing to zero, the sequence of $U_{\tau \upharpoonright n}$ is a local base at the point $x$ . Thus, the point $x$ is a limit point of $C$ . This completes the proof. $\blacksquare$

Theorem 4
Let $X$ be a compact metrizable space. It follows that $X$ is scattered if and only if $X$ is countable.

Proof of Theorem 4
$\Longleftarrow$
In this direction, we show that if $X$ is countable, then $X$ is scattered (the fact that can be shown using the function space argument pointed out earlier). Here, we show the contrapositive: if $X$ is not scattered, then $X$ is uncountable. Suppose $X$ is not scattered. Then every point of $X$ is a limit point of $X$ . By Theorem 3, $X$ would contain a Cantor set $C$ of cardinality continuum.

$\Longrightarrow$
In this direction, we show that if $X$ is scattered, then $X$ is countable. We also show the contrapositive: if $X$ is uncountable, then $X$ is not scattered. Suppose $X$ is uncountable. By Theorem 2, all but countably many points of $X$ are limit points of $X$ . After discarding these countably many isolated points, we still have a compact space. So we can just assume that every point of $X$ is a limit point of $X$ . Then by Theorem 3, $X$ contains an uncountable closed set $C$ such that every point of $C$ is a limit point of $C$ . This means that $X$ is not scattered. $\blacksquare$

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Remarks

A corollary to the above discussion is that the cardinality for any compact metrizable space is either countable (including finite) or continuum (the cardinality of the real line). There is nothing in between or higher than continuum. To see this, the cardinality of any Lindelof first countable space is at most continuum according to a theorem in this previous post (any compact metric space is one such). So continuum is an upper bound on the cardinality of compact metric spaces. Theorem 3 above implies that any uncountable compact metrizable space has to contain a Cantor set, hence has cardinality continuum. So the cardinality of a compact metrizable space can be one of two possibilities – countable or continuum. Even under the assumption of the negation of the continuum hypothesis, there will be no uncountable compact metric space of cardinality less than continuum. On the other hand, there is only one possibility for the cardinality of a scattered compact metrizable, which is countable.

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Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2015 \text{ by Dan Ma}$

Cp(omega 1 + 1) is monolithic and Frechet-Urysohn

Posted on August 4, 2014 by Dan Ma

This is another post that discusses what $C_p(X)$ is like when $X$ is a compact space. In this post, we discuss the example $C_p(\omega_1+1)$ where $\omega_1+1$ is the first compact uncountable ordinal. Note that $\omega_1+1$ is the successor to $\omega_1$ , which is the first (or least) uncountable ordinal. The function space $C_p(\omega_1+1)$ is monolithic and is a Frechet-Urysohn space. Interestingly, the first property is possessed by $C_p(X)$ for all compact spaces $X$ . The second property is possessed by all compact scattered spaces. After we discuss $C_p(\omega_1+1)$ , we discuss briefly the general results for $C_p(X)$ .

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Initial discussion

The function space $C_p(\omega_1+1)$ is a dense subspace of the product space $\mathbb{R}^{\omega_1}$ . In fact, $C_p(\omega_1+1)$ is homeomorphic to a subspace of the following subspace of $\mathbb{R}^{\omega_1}$ :

$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$

The subspace $\Sigma(\omega_1)$ is the $\Sigma$ -product of $\omega_1$ many copies of the real line $\mathbb{R}$ . The $\Sigma$ -product of separable metric spaces is monolithic (see here). The $\Sigma$ -product of first countable spaces is Frechet-Urysohn (see here). Thus $\Sigma(\omega_1)$ has both of these properties. Since the properties of monolithicity and being Frechet-Urysohn are carried over to subspaces, the function space $C_p(\omega_1+1)$ has both of these properties. The key to the discussion is then to show that $C_p(\omega_1+1)$ is homeopmophic to a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ .

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Connection to $\Sigma$ -product

We show that the function space $C_p(\omega_1+1)$ is homeomorphic to a subspace of the $\Sigma$ -product of $\omega_1$ many copies of the real lines. Let $Y_0$ be the following subspace of $C_p(\omega_1+1)$ :

$Y_0=\left\{f \in C_p(\omega_1+1): f(\omega_1)=0 \right\}$

Every function in $Y_0$ has non-zero values at only countably points of $\omega_1+1$ . Thus $Y_0$ can be regarded as a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ .

By Theorem 1 in this previous post, $C_p(\omega_1+1) \cong Y_0 \times \mathbb{R}$ , i.e, the function space $C_p(\omega_1+1)$ is homeomorphic to the product space $Y_0 \times \mathbb{R}$ . On the other hand, the product $Y_0 \times \mathbb{R}$ can also be regarded as a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ . Basically adding one additional factor of the real line to $Y_0$ still results in a subspace of the $\Sigma$ -product. Thus we have:

$C_p(\omega_1+1) \cong Y_0 \times \mathbb{R} \subset \Sigma(\omega_1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Thus $C_p(\omega_1+1)$ possesses all the hereditary properties of $\Sigma(\omega_1)$ . Another observation we can make is that $\Sigma(\omega_1)$ is not hereditarily normal. The function space $C_p(\omega_1+1)$ is not normal (see here). The $\Sigma$ -product $\Sigma(\omega_1)$ is normal (see here). Thus $\Sigma(\omega_1)$ is not hereditarily normal.

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A closer look at $C_p(\omega_1+1)$

In fact $C_p(\omega_1+1)$ has a stronger property that being monolithic. It is strongly monolithic. We use homeomorphic relation in (1) above to get some insight. Let $h$ be a homeomorphism from $C_p(\omega_1+1)$ onto $Y_0 \times \mathbb{R}$ . For each $\alpha<\omega_1$ , let $H_\alpha$ be defined as follows:

$H_\alpha=\left\{f \in C_p(\omega_1+1): f(\gamma)=0 \ \forall \ \alpha<\gamma<\omega_1 \right\}$

Clearly $H_\alpha \subset Y_0$ . Furthermore $H_\alpha$ can be considered as a subspace of $\mathbb{R}^\omega$ and is thus metrizable. Let $A$ be a countable subset of $C_p(\omega_1+1)$ . Then $h(A) \subset H_\alpha \times \mathbb{R}$ for some $\alpha<\omega_1$ . The set $H_\alpha \times \mathbb{R}$ is metrizable. The set $H_\alpha \times \mathbb{R}$ is also a closed subset of $Y_0 \times \mathbb{R}$ . Then $\overline{A}$ is contained in $H_\alpha \times \mathbb{R}$ and is therefore metrizable. We have shown that the closure of every countable subspace of $C_p(\omega_1+1)$ is metrizable. In other words, every separable subspace of $C_p(\omega_1+1)$ is metrizable. This property follows from the fact that $C_p(\omega_1+1)$ is strongly monolithic.

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Monolithicity and Frechet-Urysohn property

As indicated at the beginning, the $\Sigma$ -product $\Sigma(\omega_1)$ is monolithic (in fact strongly monolithic; see here) and is a Frechet-Urysohn space (see here). Thus the function space $C_p(\omega_1+1)$ is both strongly monolithic and Frechet-Urysohn.

Let $\tau$ be an infinite cardinal. A space $X$ is $\tau$ -monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$ , we have $nw(\overline{A}) \le \tau$ . A space $X$ is monolithic if it is $\tau$ -monolithic for all infinite cardinal $\tau$ . It is straightforward to show that $X$ is monolithic if and only of for every subspace $Y$ of $X$ , the density of $Y$ equals to the network weight of $Y$ , i.e., $d(Y)=nw(Y)$ . A longer discussion of the definition of monolithicity is found here.

A space $X$ is strongly $\tau$ -monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$ , we have $w(\overline{A}) \le \tau$ . A space $X$ is strongly monolithic if it is strongly $\tau$ -monolithic for all infinite cardinal $\tau$ . It is straightforward to show that $X$ is strongly monolithic if and only if for every subspace $Y$ of $X$ , the density of $Y$ equals to the weight of $Y$ , i.e., $d(Y)=w(Y)$ .

In any monolithic space, the density and the network weight coincide for any subspace, and in particular, any subspace that is separable has a countable network. As a result, any separable monolithic space has a countable network. Thus any separable space with no countable network is not monolithic, e.g., the Sorgenfrey line. On the other hand, any space that has a countable network is monolithic.

In any strongly monolithic space, the density and the weight coincide for any subspace, and in particular any separable subspace is metrizable. Thus being separable is an indicator of metrizability among the subspaces of a strongly monolithic space. As a result, any separable strongly monolithic space is metrizable. Any separable space that is not metrizable is not strongly monolithic. Thus any non-metrizable space that has a countable network is an example of a monolithic space that is not strongly monolithic, e.g., the function space $C_p([0,1])$ . It is clear that all metrizable spaces are strongly monolithic.

The function space $C_p(\omega_1+1)$ is not separable. Since it is strongly monolithic, every separable subspace of $C_p(\omega_1+1)$ is metrizable. We can see this by knowing that $C_p(\omega_1+1)$ is a subspace of the $\Sigma$ -product $\Sigma(\omega_1)$ , or by using the homeomorphism $h$ as in the previous section.

For any compact space $X$ , $C_p(X)$ is countably tight (see this previous post). In the case of the compact uncountable ordinal $\omega_1+1$ , $C_p(\omega_1+1)$ has the stronger property of being Frechet-Urysohn. A space $Y$ is said to be a Frechet-Urysohn space (also called a Frechet space) if for each $y \in Y$ and for each $M \subset Y$ , if $y \in \overline{M}$ , then there exists a sequence $\left\{y_n \in M: n=1,2,3,\cdots \right\}$ such that the sequence converges to $y$ . As we shall see below, $C_p(X)$ is rarely Frechet-Urysohn.

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General discussion

For any compact space $X$ , $C_p(X)$ is monolithic but does not have to be strongly monolithic. The monolithicity of $C_p(X)$ follows from the following theorem, which is Theorem II.6.8 in [1].

Theorem 1
Then the function space $C_p(X)$ is monolithic if and only if $X$ is a stable space.

See chapter 3 section 6 of [1] for a discussion of stable spaces. We give the definition here. A space $X$ is stable if for any continuous image $Y$ of $X$ , the weak weight of $Y$ , denoted by $ww(Y)$ , coincides with the network weight of $Y$ , denoted by $nw(Y)$ . In [1], $ww(Y)$ is notated by $iw(Y)$ . The cardinal function $ww(Y)$ is the minimum cardinality of all $w(T)$ , the weight of $T$ , for which there exists a continuous bijection from $Y$ onto $T$ .

All compact spaces are stable. Let $X$ be compact. For any continuous image $Y$ of $X$ , $Y$ is also compact and $ww(Y)=w(Y)$ , since any continuous bijection from $Y$ onto any space $T$ is a homeomorphism. Note that $ww(Y) \le nw(Y) \le w(Y)$ always holds. Thus $ww(Y)=w(Y)$ implies that $ww(Y)=nw(Y)$ . Thus we have:

Corollary 2
Let $X$ be a compact space. Then the function space $C_p(X)$ is monolithic.

However, the strong monolithicity of $C_p(\omega_1+1)$ does not hold in general for $C_p(X)$ for compact $X$ . As indicated above, $C_p([0,1])$ is monolithic but not strongly monolithic. The following theorem is Theorem II.7.9 in [1] and characterizes the strong monolithicity of $C_p(X)$ .

Theorem 3
Let $X$ be a space. Then $C_p(X)$ is strongly monolithic if and only if $X$ is simple.

A space $X$ is $\tau$ -simple if whenever $Y$ is a continuous image of $X$ , if the weight of $Y$ $\le \tau$ , then the cardinality of $Y$ $\le \tau$ . A space $X$ is simple if it is $\tau$ -simple for all infinite cardinal numbers $\tau$ . Interestingly, any separable metric space that is uncountable is not $\omega$ -simple. Thus $[0,1]$ is not $\omega$ -simple and $C_p([0,1])$ is not strongly monolithic, according to Theorem 3.

For compact spaces $X$ , $C_p(X)$ is rarely a Frechet-Urysohn space as evidenced by the following theorem, which is Theorem III.1.2 in [1].

Theorem 4
Let $X$ be a compact space. Then the following conditions are equivalent.

$C_p(X)$ is a Frechet-Urysohn space.
$C_p(X)$ is a k-space.
The compact space $X$ is a scattered space.

A space $X$ is a scattered space if for every non-empty subspace $Y$ of $X$ , there exists an isolated point of $Y$ (relative to the topology of $Y$ ). Any space of ordinals is scattered since every non-empty subset has a least element. Thus $\omega_1+1$ is a scattered space. On the other hand, the unit interval $[0,1]$ with the Euclidean topology is not scattered. According to this theorem, $C_p([0,1])$ cannot be a Frechet-Urysohn space.

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Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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

Cp(X) is countably tight when X is compact

Posted on August 3, 2014 by Dan Ma

Let $X$ be a completely regular space (also called Tychonoff space). If $X$ is a compact space, what can we say about the function space $C_p(X)$ , the space of all continuous real-valued functions with the pointwise convergence topology? When $X$ is an uncountable space, $C_p(X)$ is not first countable at every point. This follows from the fact that $C_p(X)$ is a dense subspace of the product space $\mathbb{R}^X$ and that no dense subspace of $\mathbb{R}^X$ can be first countable when $X$ is uncountable. However, when $X$ is compact, $C_p(X)$ does have a convergence property, namely $C_p(X)$ is countably tight.

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Tightness

Let $X$ be a completely regular space. The tightness of $X$ , denoted by $t(X)$ , is the least infinite cardinal $\kappa$ such that for any $A \subset X$ and for any $x \in X$ with $x \in \overline{A}$ , there exists $B \subset A$ for which $\lvert B \lvert \le \kappa$ and $x \in \overline{B}$ . When $t(X)=\omega$ , we say that $Y$ has countable tightness or is countably tight. When $t(X)>\omega$ , we say that $X$ has uncountable tightness or is uncountably tight. Clearly any first countable space is countably tight. There are other convergence properties in between first countability and countable tightness, e.g., the Frechet-Urysohn property. The notion of countable tightness and tightness in general is discussed in further details here.

The fact that $C_p(X)$ is countably tight for any compact $X$ follows from the following theorem.

Theorem 1
Let $X$ be a completely regular space. Then the function space $C_p(X)$ is countably tight if and only if $X^n$ is Lindelof for each $n=1,2,3,\cdots$ .

Theorem 1 is the countable case of Theorem I.4.1 on page 33 of [1]. We prove one direction of Theorem 1, the direction that will give us the desired result for $C_p(X)$ where $X$ is compact.

Proof of Theorem 1
The direction $\Longleftarrow$
Suppose that $X^n$ is Lindelof for each positive integer. Let $f \in C_p(X)$ and $f \in \overline{H}$ where $H \subset C_p(X)$ . For each positive integer $n$ , we define an open cover $\mathcal{U}_n$ of $X^n$ .

Let $n$ be a positive integer. Let $t=(x_1,\cdots,x_n) \in X^n$ . Since $f \in \overline{H}$ , there is an $h_t \in H$ such that $\lvert h_t(x_j)-f(x_j) \lvert <\frac{1}{n}$ for all $j=1,\cdots,n$ . Because both $h_t$ and $f$ are continuous, for each $j=1,\cdots,n$ , there is an open set $W(x_j) \subset X$ with $x_j \in W(x_j)$ such that $\lvert h_t(y)-f(y) \lvert < \frac{1}{n}$ for all $y \in W(x_j)$ . Let the open set $U_t$ be defined by $U_t=W(x_1) \times W(x_2) \times \cdots \times W(x_n)$ . Let $\mathcal{U}_n=\left\{U_t: t=(x_1,\cdots,x_n) \in X^n \right\}$ .

For each $n$ , choose $\mathcal{V}_n \subset \mathcal{U}_n$ be countable such that $\mathcal{V}_n$ is a cover of $X^n$ . Let $K_n=\left\{h_t: t \in X^n \text{ such that } U_t \in \mathcal{V}_n \right\}$ . Let $K=\bigcup_{n=1}^\infty K_n$ . Note that $K$ is countable and $K \subset H$ .

We now show that $f \in \overline{K}$ . Choose an arbitrary positive integer $n$ . Choose arbitrary points $y_1,y_2,\cdots,y_n \in X$ . Consider the open set $U$ defined by

$U=\left\{g \in C_p(X): \forall \ j=1,\cdots,n, \lvert g(y_j)-f(y_j) \lvert <\frac{1}{n} \right\}$

We wish to show that $U \cap K \ne \varnothing$ . Choose $U_t \in \mathcal{V}_n$ such that $(y_1,\cdots,y_n) \in U_t$ where $t=(x_1,\cdots,x_n) \in X^n$ . Consider the function $h_t$ that goes with $t$ . It is clear from the way $h_t$ is chosen that $\lvert h_t(y_j)-f(x_j) \lvert<\frac{1}{n}$ for all $j=1,\cdots,n$ . Thus $h_t \in K_n \cap U$ , leading to the conclusion that $f \in \overline{K}$ . The proof that $C_p(X)$ is countably tight is completed.

The direction $\Longrightarrow$
See Theorem I.4.1 of [1].

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Remarks

As shown above, countably tightness is one convergence property of $C_p(X)$ that is guaranteed when $X$ is compact. In general, it is difficult for $C_p(X)$ to have stronger convergence properties such as the Frechet-Urysohn property. It is well known $C_p(\omega_1+1)$ is Frechet-Urysohn. According to Theorem II.1.2 in [1], for any compact space $X$ , $C_p(X)$ is a Frechet-Urysohn space if and only if the compact space $X$ is a scattered space.

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Reference

Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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