A short note on monolithic spaces

In a metrizable space, the density, the network weight and the weight (and several other cardinal functions) always agree (see Theorem 4.1.15 in [2]). This is not the case for topological spaces in general. One handy example is the Sorgenfrey line where the density is $\omega$ (the Sorgenfrey line is separable) and the network weight is continuum (the cardinality of real line). In a monolithic space, the density character and the network weight for any subspace always coincide. Thus metrizable spaces are monolithic. One interesting example of a monolithic space is the $\Sigma$-product of real lines. A compact space is said to be a Corson compact space if it can be embedded in a $\Sigma$-product of real lines. Thus Corson compact spaces are monolithic spaces. As a result, any separable subspace of a Corson compact space is metrizable. On the other hand, any separable non-metrizable compact space cannot be Corson compact. This is an introductory discussion of monolithic spaces and is the first post in a series of posts on Corson compact spaces. A listing of other blog posts on Corson compact spaces is given at the end of this post.

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Density and Network Weight

For any set $A$, the symbol $\lvert A \lvert$ denotes the cardinality of the set $A$. For any space $X$, the density of $X$, denoted by $d(X)$ is the minimum cardinality of a dense subset, i.e., $d(X)$ is the least cardinal number $\kappa$ such that if $Y$ is dense subset of $X$, then $\kappa \le \lvert Y \lvert$. If $X$ is separable, then $d(X)=\omega$.

For any space $X$, a family $\mathcal{N}$ of subsets of $X$ is a network in the space $X$ if for any $x \in X$ and for any open subset $U$ of $X$ with $x \in U$, there exists some $J \in \mathcal{N}$ such that $x \in J \subset U$. In other words, any non-empty open subset of $X$ is the union of elements of the network $\mathcal{N}$. The network weight of $X$, denoted by $nw(X)$, is the minimum cardinality of a network in the space $X$, i.e., $nw(X)$ is the least cardinal number $\kappa$ such that if $\mathcal{N}$ is a network for the space $X$, then $\kappa \le \lvert \mathcal{N} \lvert$.

For any space $X$, the weight of $X$, denoted by $w(X)$, is the minimum cardinality of a base for the space $X$, i.e., $w(X)$ is the least cardinal number $\kappa$ such that if $\mathcal{B}$ is a base for the space $X$, then $\kappa \le \lvert \mathcal{B} \lvert$. If $w(X)=\omega$, then $X$ is a space with a countable base (it is a separable metric space). If $nw(X)=\omega$, $X$ is a space with a countable network. Having a countable network is a strong property, it implies that the space is hereditarily Lindelof (hence hereditarily normal) and hereditarily separable (see this previous post). However, having a countable network is not as strong as having a countable base. The function space $C_p(\mathbb{R})$ has a countable network (see this previous post) and fails to be first countable at every point.

If $\mathcal{N}$ is a network for the space $X$, then picking a point from each set in $\mathcal{N}$ will produce a dense subset of $X$. Then $d(X) \le nw(X)$ for any space $X$. In general $nw(X) \le d(X)$ does not hold, as indicated by the Sorgenfrey line. Monolithic spaces form a class of spaces in which the inequality $nw \le d$ holds for each space in the class and for each subspace of such a space.

Likewise, the inequality $d(X) \le w(X)$ always holds. The inequality $w(X) \le d(X)$ only holds for a restricted class of spaces. On the other hand, it is clear that $nw(X) \le w(X)$ for any space $X$.

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Monolithic Spaces

Let $\tau$ be an infinite cardinal number. A space $X$ is said to be $\tau$-monolithic if for each subspace $Y$ of $X$ with $\lvert Y \lvert \le \tau$, $nw(\overline{Y}) \le \tau$. It is easy to verify that the following two statements are equivalent:

1. $X$ is $\tau$-monolithic for each infinite cardinal number $\tau$.
2. For each subspace $Y$ of $X$, $d(Y)=nw(Y)$.

A space $X$ is monolithic if either statement 1 or statement 2 holds. In a $\omega$-monolithic space, any separable subspace has a countable network.

A space $X$ is said to be strongly $\tau$-monolithic if for each subspace $Y$ of $X$ with $\lvert Y \lvert \le \tau$, $w(\overline{Y}) \le \tau$. It is easy to verify that the following two statements are equivalent:

1. $X$ is strongly $\tau$-monolithic for each infinite cardinal number $\tau$.
2. For each subspace $Y$ of $X$, $d(Y)=w(Y)$.

A space $X$ is strongly monolithic if either statement 3 or statement 4 holds. In a strongly $\omega$-monolithic space, any separable subspace is metrizable. It is clear that any strongly monolithic space is monolithic. As indicated below, $C_p(\mathbb{R})$ is an example of a monolithic space that is not strongly monolithic. However, the two notions coincide for compact spaces. Note that for any compact space, the weight and network weight coincide. Thus if a compact space is monolithic, it is strongly monolithic.

It is also clear that the property of being monolithic is hereditary. Monolithicity is a notion used in $C_p$-theory and the study of Corson compact spaces (see [1]).

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Examples

Some examples of monolithic spaces are:

• Metrizable spaces.
• Any space with a countable network.
• $\Sigma$-product of separable metric spaces.
• The space $\omega_1$ of countable ordinals.

In fact, with the exception of the spaces with countable networks, the above examples are strongly monolithic. It is well known that the density and weight always agree for metrizable space. $\Sigma$-product of separable metric spaces is strongly monolithic (shown in this subsequent post). In the space $\omega_1$, any countable subset is separable and metrizable and any uncountable subset has both density and weight $=\omega_1$.

If $X$ is a space with a countable network, then for any subspace $Y$, $d(Y)=nw(Y)=\omega$. Thus any space with a countable network is monolithic. However, any space that has a countable network but is not metrizable is not strongly monolithic, e.g., the function space $C_p(\mathbb{R})$. The following proposition about compact monolithic spaces is useful.

Proposition 1
Let $X$ be a compact and monolithic space. Then $X$ is metrizable if and only if $X$ is separable.

Proof of Proposition 1
For the $\Longrightarrow$ direction, note that any compact metrizable space is separable (monolithicity is not needed). For the $\Longleftarrow$ direction, note that any separable monolithic space has a countable network. Any compact space with a countable network is metrizable (see here). $\blacksquare$

Now consider some spaces that are not monolithic. As indicated above, any space in which the density does not agree with the network weight (in the space or in a subspace) is not monolithic. Proposition 1 indicates that any separable non-metrizable compact space is not monolithic. Examples include the Alexandroff double arrow space ( see here) and the product space $I^{\omega_1}$ where $I$ is the closed unit interval $[0,1]$ with the usual Euclidean topology.

Interestingly, “compact” in Proposition 1 can be replaced by pseudocompact because of the following:

Proposition 2
Let $X$ be a separable pseudocompact and monolithic space. Then $X$ is compact.

Proof of Proposition 2
Any separable monolithic space has a countable network. Any space with a countable network is Lindelof (and hence metacompact). Any pseudocompact metacompact space is compact (see here). $\blacksquare$

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Blog posts on Corson compact spaces

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.

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

Pixley-Roy hyperspaces

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

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

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

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

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

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Basic Discussion

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

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

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

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

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

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

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

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

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More Basic Results

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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Examples

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

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

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

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Reference

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

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

Stone-Cech Compactification is Maximal

Let $X$ be a completely regular space. Let $\beta X$ be the Stone-Cech compactification of $X$. In a previous post, we show that among all compactifcations of $X$, the Stone-Cech compactification $\beta X$ is maximal with respect to a partial order $\le$ (see Theorem C2 in Two Characterizations of Stone-Cech Compactification). As a result of the maximality, $\beta X$ is the largest among all compactifications of $X$ both in terms of cardinality and weight. We also establish an upper bound for the cardinality of $\beta X$ and an upper bound for the weight of $\beta X$. As a result, we have upper bounds for cardinalities and weights for all compactifications of $X$. We prove the following points.

Upper Bounds for Stone-Cech Compactification

1. $\lvert \beta X \lvert \le 2^{2^{d(X)}}$.
2. $w(\beta X) \le 2^{d(X)}$.
3. Stone-Cech Compactification is Maximal

4. For every compactification $\alpha X$ of the space $X$, $\lvert \alpha X \lvert \le \lvert \beta X \lvert$.
5. For every compactification $\alpha X$ of the space $X$, $w(\alpha X) \le w(\beta X)$.
6. Upper Bounds for all Compactifications

7. For every compactification $\alpha X$ of the space $X$, $w(\alpha X) \le 2^{d(X)}$.
8. For every compactification $\alpha X$ of the space $X$, $\lvert \alpha X \lvert \le 2^{2^{d(X)}}$.

It is clear that Results 5 and 6 follow from the preceding results. The links for other posts on Stone-Cech compactification can be found toward the end of this post

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Some Cardinal Functions

Let $X$ be a space. The density of $X$ is denoted by $d(X)$ and is defined to be the smallest cardinality of a dense set in $X$. For example, if $X$ is separable, then $d(X)=\omega$. The weight of the space $X$ is denoted by $w(X)$ and is defined to be the smallest cardinality of a base of the space $X$. For example, if $X$ is second countable (i.e. having a countable space), then $w(X)=\omega$. Both $d(X)$ and $w(X)$ are cardinal functions that are commonly used in topological discussion. Most authors require that cardinal functions only take on infinite cardinals. We also adopt this convention here. We use $c$ to denote the cardinality of the continuum (the cardinality of the real line $\mathbb{R}$).

If $\mathcal{K}$ is a cardinal number, then $2^{\mathcal{K}}$ refers to the cardinal number that is the cardinallity of the set of all functions from $\mathcal{K}$ to $2=\left\{0,1 \right\}$. Equivalently, $2^{\mathcal{K}}$ is also the cardinality of the power set of $\mathcal{K}$ (i.e. the set of all subsets of $\mathcal{K}$). If $\mathcal{K}=\omega$ (the first infinite ordinal), then $2^\omega=c$ is the cardinality of the continuum.

If $X$ is separable, then $d(X)=\omega$ (as noted above) and we have $2^{d(X)}=c$ and $2^{2^{d(X)}}=2^c$. Result 5 and Result 6 imply that $2^c$ is an upper bound for the cardinality of all compactifications of any separable space $X$ and $c$ is an upper bound of the weight of all compactifications of any separable space $X$.

In general, Result 5 and Result 6 indicate that the density of $X$ bounds the cardinality of any compactification of $X$ by two exponents and the density of $X$ bounds the weight of any compactification of $X$ by one exponent.

Another cardinal function related to weight is that of the network weight. A collection $\mathcal{N}$ of subsets of the space $X$ is said to be a network for $X$ if for each point $x \in X$ and for each open subset $U$ of $X$ with $x \in U$, there is some set $A \in \mathcal{N}$ with $x \in A \subset U$. Note that sets in a network do not have to be open. However, any base for a topology is a network. The network weight of the space $X$ is denoted by $nw(X)$ and is defined to be the least cardinality of a network for $X$. Since any base is a network, we have $nw(X) \le w(X)$. It is also clear that $nw(X) \le \lvert X \lvert$ for any space $X$. Our interest in network and network weight is to facilitate the discussion of Lemma 2 below. It is a well known fact that in a compact space, the weight and the network weight are the same (see Result 5 in Spaces With Countable Network).
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Some Basic Facts

We need the following two basic results.

Lemma 1
Let $X$ be a space. Let $C(X)$ be the set of all continuous functions $f:X \rightarrow \mathbb{R}$. Then $\lvert C(X) \lvert \le 2^{d(X)}$.

Lemma 2
Let $S$ be a space and let $T$ be a compact space. Suppose that $T$ is the continuous image of $S$. Then $w(T) \le w(S)$.

Proof of Lemma 1
Let $A \subset X$ be a dense set with $\lvert A \lvert=2^{d(X)}$. Let $\mathbb{R}^A$ be the set of all functions from $A$ to $\mathbb{R}$. Consider the map $W:C(X) \rightarrow \mathbb{R}^A$ by $W(f)= f \upharpoonright A$. This is a one-to-one map since $f=g$ whenever $f$ and $g$ agree on a dense set. Thus we have $\lvert C(X) \lvert \le \lvert \mathbb{R}^A \lvert$. Upon doing some cardinal arithmetic, we have $\lvert \mathbb{R}^A \lvert=2^{d(X)}$. Thus Lemma 1 is established. $\blacksquare$

Proof of Lemma 2
Let $g:S \rightarrow T$ be a continuous function from $S$ onto $T$. Let $\mathcal{B}$ be a base for $S$ such that $\lvert \mathcal{B} \lvert=w(S)$. Let $\mathcal{N}$ be the set of all $g(B)$ where $B \in \mathcal{B}$. Note that $\mathcal{N}$ is a network for $T$ (since $g$ is a continuous function). So we have $nw(T) \le \lvert \mathcal{N} \lvert \le \lvert \mathcal{B} \lvert = w(S)$. Since $T$ is compact, $w(T)=nw(T)$ (see Result 5 in Spaces With Countable Network). Thus we have $nw(T)=w(T) \lvert \le w(S)$. $\blacksquare$

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Results 1 and 2

Let $X$ be a completely regular space. Let $I$ be the unit interval $[0,1]$. We show that the Stone-Cech compactification $\beta X$ can be regarded as a subspace of the product space $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$ (the product of $2^{d(X)}$ many copies of $I$). The cardinality of $I^{\mathcal{K}}$ is $2^{2^{d(X)}}$, thus leading to Result 1.

Let $C(X,I)$ be the set of all continuous functions $f:X \rightarrow I$. The Stone-Cech compactification $\beta X$ is constructed by embedding $X$ into the product space $\prod \limits_{f \in C(X,I)} I_f$ where each $I_f=I$ (see Embedding Completely Regular Spaces into a Cube or A Beginning Look at Stone-Cech Compactification). Thus $\beta X$ is a subspace of $I^{\mathcal{K}_1}$ where $\mathcal{K}_1=\lvert C(X,I) \lvert$.

Note that $C(X,I) \subset C(X)$. Thus $\beta X$ can be regarded as a subspace of $I^{\mathcal{K}_2}$ where $\mathcal{K}_2=\lvert C(X) \lvert$. By Lemma 1, $\beta X$ can be regarded as a subspace of the product space $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$.

To see Result 2, note that the weight of $I^{\mathcal{K}}$ where $\mathcal{K}= 2^{d(X)}$ is $2^{d(X)}$. Then $\beta X$, as a subspace of the product space, must have weight $\le 2^{d(X)}$. $\blacksquare$

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Results 3 and 4

What drives Result 3 and Result 4 is the following theorem (established in Two Characterizations of Stone-Cech Compactification).

Theorem C2
Let $X$ be a completely regular space. Among all compactifications of the space $X$, the Stone-Cech compactification $\beta X$ of the space $X$ is maximal with respect to the partial order $\le$.

$\text{ }$

To define the partial order, for $\alpha_1 X$ and $\alpha_2 X$, both compactifications of $X$, we say that $\alpha_2 X \le \alpha_1 X$ if there is a continuous function $f:\alpha_1 X \rightarrow \alpha_2 X$ such that $f \circ \alpha_1=\alpha_2$. See the following figure.

Figure 1

In this post, we use $\le$ to denote this partial order as well as the order for cardinal numbers. Thus we need to rely on context to distinguish this partial order from the order for cardinal numbers.

Let $\alpha X$ be a compactification of $X$. Theorem C2 indicates that $\alpha X \le \beta X$ (partial order), which means that there is a continuous $f:\beta X \rightarrow \alpha X$ such that $f \circ \beta=\alpha$ (the same point in $X$ is mapped to itself by $f$). Note that $\alpha X$ is the image of $\beta X$ under the function $f:\beta X \rightarrow \alpha X$. Thus we have $\lvert \alpha X \lvert \le \lvert \beta X \lvert$ (cardinal number order). Thus Result 3 is established.

By Lemma 2, the existence of the continuous function $f:\beta X \rightarrow \alpha X$ implies that $w(\alpha X) \le w(\beta X)$ (cardinal number order). Thus Result 4 is established.

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Blog Posts on Stone-Cech Compactification

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, Heldermann Verlag, Berlin, 1989.
2. Willard, S., General Topology, Addison-Wesley Publishing Company, 1970.

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$\copyright \ \ 2012$

Perfect Image of Separable Metric Spaces

In a previous post on countable network, it was shown that having a countable network is equivalent to being the continuous image of a separable metric space. Since there is an example of a non-metrizable space with countable netowrk, the continuous image of a separable metric space needs not be a separable metric space. However, the perfect image of a separable metrizable space is separable metrizable. First some definitions. A continuous mapping $f:X \rightarrow Y$ is a closed mapping if $f(H)$ is closed in $Y$ for any closed set $H \subset X$. A continuous surjection $f:X \rightarrow Y$ is a perfect mapping if $f$ is closed and $f^{-1}(y)$ is compact for each $y \in Y$.

Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. Because $f$ is a closed mapping, $f(X-B)$ is closed and $f(B)$ is open in $Y$ for each $B \in \mathcal{B}$. We show that $\mathcal{B}_f=\lbrace{f(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$. Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. We have $y \in f(B) \subset U$. Thus the topology on $Y$ can be generated by $\mathcal{B}_f$.

Update (11/24/2009):
The proof in the above paragraph is faulty. Thanks to Dave Milovich for pointing this out. Here’s the corrected proof.

Let me first prove a lemma.

Lemma. Let $f: X \rightarrow Y$ be a closed mapping and let $V \subset X$ be open. Then $f_*(V)=\lbrace{y \in Y:f^{-1}(y) \subset V}\rbrace$ is open in $Y$. Furthermore, $f_*(V) \subset f(V)$.

Proof of Lemma. Since $f$ is a closed mapping, $f(X-V)$ is closed. We claim that $f(X-V)=Y-f_*(V)$. It is clear that $f(X-V) \subset Y-f_*(V)$. To show that $Y-f_*(V) \subset f(X-V)$, let $z \in Y-f_*(V)$. Then $f^{-1}(z)$ cannot be a subset of $V$. Choose $x \in f^{-1}(z)-V$. Then we have $z=f(x) \in f(X-V)$. Thus $f(X-V)=Y-f_*(V)$ and $f_*(V)$ is open. It is straitforward to verify that $f_*(V) \subset f(V)$.

Now I prove that the perfect image of a separable metric space is a separable metric space. Let $f:X \rightarrow Y$ be a perfect mapping where $X$ has a countable base $\mathcal{B}$. Assume $\mathcal{B}$ is closed under finite unions. We show that $\mathcal{B}_f=\lbrace{f_*(B):B \in \mathcal{B}}\rbrace$ is a base for $Y$.

Let $y \in Y$ and $U \subset Y$ be open with $y \in U$. For each $x \in f^{-1}(y)$, choose $B_x \in \mathcal{B}$ such that $x \in B_x$ and $f(B_x) \subset U$. Since $f^{-1}(y)$ is compact, we can choose $B_{x(0)},...,B_{x(n)}$ that cover $f^{-1}(y)$. Let $B=B_{x(0)} \cup ... \cup B_{x(n)}$, which is in $\mathcal{B}$. Since $f^{-1}(y) \subset B$, we have $y \in f_*(B)$. We also have $f_*(B) \subset f(B) \subset U$. Thus the topology on $Y$ can be generated by the countable base $\mathcal{B}_f$.

Metrization Theorems for Compact Spaces

In this blog I have already presented two metrization theorems for compact spaces: (1) any compact space with a countable network is metrizable (see the post), (2) any compact space with a $G_\delta-$diagonal is metrizable (see the post). I now present another classic theorem: any countably compact space with a point-countable base is metrizable. This theorem is a classic result of Miscenko ([1]). All spaces are at least Hausdorff and regular. We have the following three metrization theorems for compact spaces. In subsequent posts, I will discuss generalizations of these theorems and discuss related concepts.

Thoerem 1. Any compact space with a countable network is metrizable.
The proof is in this post.

Thoerem 2. Any compact space with a $G_\delta-\text{diagonal}$ is metrizable.
The proof is in this post.

Thoerem 3. Any countably compact space with a point-countable base is metrizable.

A base $\mathcal{B}$ for a space $X$ is a point-countabe base if every point in $X$ belongs to at most countably elements of $\mathcal{B}$.

Proof of Theorem 3. Let $\mathcal{B}$ be a point-countable base for the countably compact space $X$. We show that $X$ is separable. Once we have a countable dense subset, the base $\mathcal{B}$ has to be a countable base. So we inductively define a sequence of countable sets $\lbrace{D_0,D_1,...}\rbrace$ such that $D=\bigcup_{n<\omega}D_n$ is dense in $X$.

Let $D_0=\lbrace{x_0}\rbrace$ be a one-point set to start with. For $n>0$, let $E_n=\bigcup_{i. Let $\mathcal{B}_n=\lbrace{B \in \mathcal{B}:B \cap E_n \neq \phi}\rbrace$. For each finite $T \subset \mathcal{B}_n$ such that $X - \bigcap T \neq \phi$, choose a point $x(T) \in X - \bigcup T$. Let $D_n$ be the union of $E_n$ and the set of all points $x(T)$. Let $D=\bigcup_{n<\omega}D_n$.

We claim that $\overline{D}=X$. Suppose we have $x \in X-\overline{D}$. Let $\mathcal{A}=\lbrace{B \in \mathcal{B}:B \cap D \neq \phi \phantom{X} \text{and} \thinspace x \notin B}\rbrace$. We know that $\mathcal{A}$ is countable since every element of $\mathcal{A}$ contains points of the countable set $D$. We also know that $\mathcal{A}$ is an open cover of $\overline{D}$. By the countably compactness of $\overline{D}$, we can find a finite $T \subset \mathcal{A}$ such that $\overline{D} \subset \bigcup T$. The finite set $T$ must have appeared during the induction process of selecting points for $D_n$ for some $n$ (i.e. $T \subset \mathcal{B}_n$). So a point $x(T)$ has been chosen such that $x(T) \notin \bigcup T$ (thus we have $x(T) \in D_n \subset \overline{D}$). On the other hand, since $\overline{D} \subset \bigcup T$, we observe that $x(T) \notin \overline{D}$, producing a contradiction. Thus the countable set $D$ is dense in $X$, making the point-countable base $\mathcal{B}$ a countable base.

Reference

1. Miscenko, A., Spaces with a point-countable base, Dokl. Acad. Nauk SSSR, 144 (1962), 985-988. (English translation: Soviet Math. Dokl. 3 (1962), 1199-1202).

Network Weight of Topological Spaces – I

In the previous post, I discussed the notion of network of a topological space. It was noted that for any space $X$, the network weight (the least cardinality of a network for $X$) is always $\leq$ the weight (the least cardinality of a base for $X$). When $X$ is compact, the network weight and weight would coincide. Are there other classes of spaces for which network weight = weight? I would like to discuss two other classes of spaces where network weight and weight coincide, namely metrizable spaces and locally compact spaces. The following two theorems are proved. For a basic discussion on network, see the previous post.

Theorem 1. If $X$ is metrizable, then $nw(X)=w(X)$.

Proof. For the case of $w(X)=\omega$, we have $nw(X)=\omega$. Now consider the case that $w(X)$ is an uncountable cardinal. Based on the Bing-Nagata-Smirnov metrization theorem, any metrizable space has a $\sigma-$discrete base. Let $\mathcal{B}=\bigcup_{n<\omega} \mathcal{B}_n$ be a $\sigma-$discrete base for the metrizable space $X$. Now let $\mathcal{K}=\lvert \mathcal{B} \lvert$. Because each $\mathcal{B}_n$ is a discrete collection of open sets, any network woulld have cardinality at least as big as $\lvert \mathcal{B}_n \lvert$ for each $n$. If $\mathcal{K}=\lvert \mathcal{B}_n \lvert$ for some $n$, then $\mathcal{K} \leq nw(X)$. If $\mathcal{K}$ is the least upper bound of $\lvert \mathcal{B}_n \lvert$, then $\mathcal{K} \leq nw(X)$. Both cases imply $w(X) \leq nw(X)$. Since $nw(X) \leq w(X)$ always hold,  we have $w(X)=nw(X)$.

Theorem 2. If $X$ is a locally compact space, then $nw(X)=w(X)$.

Proof. Let $X$ be locally compact where $nw(X)=\mathcal{K}$. The idea is that we can obtain a base for $X$ of cardinality $\leq \mathcal{K}$ (i.e. $w(X) \leq nw(X)$). Let $\mathcal{N}$ be a network whose cardinality is $\mathcal{K}$. Here’s a sketch of the proof. Each point in $X$ has an open neighborhood whose closure is compact. For the compact closure of such open neighborhood, the weight would coincide with the network weight. Thus we can find a base of size $\leq \mathcal{K}$ within such open neighborhood. Because $\lvert \mathcal{N} \lvert=\mathcal{K}$, we only need to consider $\mathcal{K}$ many such open neighborhoods with compact closure. Thus we can obtain a base for $X$ of cardinality $\leq \mathcal{K}$. To make this sketch more precise, consider the following three claims.

Claim 1
The collection of all $N \in \mathcal{N}$, where $\overline{N}$ is compact, is a cover of the space $X$.

Claim 2
For every compact set $A \subset X$, there is an open set $U$ such that $A \subset U$ and $\overline{U}$ is compact.

Claim 3
If $U \subset X$ is open with $\overline{U}$ compact, then we can obtain a base $\mathcal{B}_U$ for the open subspace $U$ with $\vert \mathcal{B}_U \vert \leq \mathcal{K}$.

For each $N \in \mathcal{N}$ in Claim 1, we can select an open $U$ (as in Claim 2)such that $\overline{N} \subset U$ and $\overline{U}$ is compact. Let $\mathcal{B}$ be the union of all the $\mathcal{B}_U$ in Claim 3 over all such $U$. There are $\leq \mathcal{K}$ many $N$ in Claim 1. Thus $\vert \mathcal{B} \lvert \leq \mathcal{K}$. Note that $\mathcal{B}$ would form a base for the whole space $X$.

Both Claim 1 and Claim 2 are direct consequence of locally compactness. To see Claim 3, let $U$ be open such that $\overline{U}$ is compact. We have $nw(\overline{U}) \leq nw(X)=\mathcal{K}$ (the network weight of a subspace cannot exceed the original network weight). By the result in the previous post, we have $w(\overline{U})=nw(\overline{U})$. We now have $w(U) \leq w(\overline{U})$ (the weight of a subspace cannot exceed the weight of the space containing it). So the weight of any open subspace with compact closure cannot exceed $\mathcal{K}$.

Corollary. For both metrizable spaces and locally compact spaces $X$, $w(X) \leq \lvert X \lvert$.

Spaces With Countable Network

The concept of network is a useful tool in working with generalized metric spaces. A network is like a base for a topology, but the members of a network do not have to be open. After a brief discussion on network, the focus here is on the spaces with networks that are countably infinite in size. The following facts are presented:

1. Any space with a countable network is separable and Lindelof.
2. The property of having a countable network is hereditary. Thus any space with a countable network is hereditarily separable and hereditarily Lindelof.
3. The property of having a countable network is preserved by taking countable product.
4. The Sorgenfrey Line is an example of a hereditarily separable and hereditarily Lindelof space that has no countable network.
5. For any compact space $X$, $nw(X)=w(X)$. In particular, any compact space with a countable network is metrizable.
6. As a corollary to 5, $w(X) \leq \vert X \vert$ for any compact $X$.
7. A space $X$ has a countable network if and only if it is the continuous impage of a separable metric space (hence such a space is sometimes called cosmic).
8. Any continuous image of a cosmic space is cosmic.
9. Any continuous image of a compact metric space is a compact metric space.
10. As a corollary to 2, any space with countable network is perfectly normal.
11. An example is given to show that the continuous image of a separable metric space needs not be metric (i.e. an example of a cosmic space that is not metrizable).

All spaces in this discussion are at least $T_3$ (Hausdorff and regular). Let $X$ be a space. A collection $\mathcal{N}$ of subsets of $X$ is said to be a network for $X$ if for each $x \in X$ and for each open $U \subset X$ with $x \in U$, then we have $x \in N \subset U$ for some $N \in \mathcal{N}$. The network weight of a space $X$, denoted by $nw(X)$, is defined as the minimum cardinality of all the possible $\vert \mathcal{N} \vert$ where $\mathcal{N}$ is a network for $X$. The weight of a space $X$, denoted by $w(X)$, is defined as the minimum cardinality of all possible $\vert \mathcal{B} \vert$ where $\mathcal{B}$ is a base for $X$. Obviously any base is also a network. Thus $nw(X) \leq w(X)$. For any compact space $X$, $nw(X)=w(X)$. On the other hand, the set of singleton sets is a network. Thus $nw(X) \leq \vert X \vert$.

Our discussion is based on an important observation. Let $\mathcal{T}$ be the topology for the space $X$. Let $\mathcal{K}=nw(X)$. We can find a base $\mathcal{B}_0$ that generates a weaker (coarser) topology such that $\lvert \mathcal{B}_0 \lvert=\mathcal{K}$. We can also find a base $\mathcal{B}_1$ that generates a finer topology such that $\lvert \mathcal{B}_1 \lvert=\mathcal{K}$. These are restated as lemmas.

Lemma 1. We can define base $\mathcal{B}_0$ that generates a weaker (coarser) topology $\mathcal{S}_0$ on $X$ such that $\lvert \mathcal{B}_0 \lvert=\mathcal{K}$. Thus $w(X,\mathcal{S}_0) \leq nw(X)$.

Proof. Let $\mathcal{N}$ be a network for $(X,\mathcal{T})$ such that $\vert \mathcal{N} \vert=nw(X,\mathcal{T})$. Consider all pairs $N_0,N_1 \in \mathcal{N}$ such that there exist disjoint $O_0,O_1 \in \mathcal{T}$ with $N_0 \subset O_0$ and $N_1 \subset O_1$. Such pairs exist because we are working in a Hausdorff space. Let $\mathcal{B}_0$ be the collection of all such open sets $O_0,O_1$ and their finite interections. This is a base for a topology and let $\mathcal{S}_0$ be the topology generated by $\mathcal{B}_0$. Clearly, $\mathcal{S}_0 \subset \mathcal{T}$ and this is a Hausdorff topology. Note that $w(X,\mathcal{S}_0) \leq \vert \mathcal{B}_0 \vert =\vert \mathcal{N} \vert$.

Lemma 2. We can define base $\mathcal{B}_1$ that generates a finer topology $\mathcal{S}_1$ on $X$ such that $\lvert \mathcal{B}_1 \lvert=\mathcal{K}$. Thus $w(X,\mathcal{S}_1) \leq nw(X)$.

Proof. As before, let $\mathcal{N}$ be a network for $(X,\mathcal{T})$ such that $\vert \mathcal{N} \vert=nw(X,\mathcal{T})$. Since we are working in a regular space, we can assume that the sets in $\mathcal{N}$ are closed. If not, take closures of the elements of $\mathcal{N}$ and we still have a network. Consider $\mathcal{B}_1$ to be the set of all finite intersections of elements in $\mathcal{N}$. This is a base for a topology on $X$. Let $\mathcal{S}_1$ be the topology generated by this base. Clearly, $\mathcal{T} \subset \mathcal{S}_1$. It is also clear that $w(X,\mathcal{S}_1) \leq nw(X)$. The only thing left to show is that the finer topology is regular. Note that the network $\mathcal{N}$ consists of closed sets in the topology $\mathcal{T}$. Thus the sets in the base $\mathcal{B}_1$ also consists of closed sets with respect to $\mathcal{T}$ and the sets in $\mathcal{B}_1$ are thus closed in the finer topology. Since $\mathcal{B}_1$ is a base consisting of cloased and open sets, the topology $\mathcal{S}_1$ regular.

Discussion of 1, 2, and 3
Points 1, 2 and 3 are basic facts about countable network and they are easily verified based on definitions. They are called out for the sake of having a record.

Discussion of 4
The Sorgenfrey Line does not have a countable network for the same reason that the Sorgenfrey Plane is not Lindelof. If the Sorgenfrey Line has a countable netowrk, then the Sorgenfrey plane would have a countable network and hence Lindelof.

Discussion of 5
In general, $nw(X) \leq w(X)$. In a compact Hausdorff space, any weaker Hausdorff topology must conincide with the original topology. So the weaker topology produced in Lemma 1 must coincide with the original topology. In the countable case, any compact space with a countable network has a weaker topology with a countable base. This weaker topology must coincide with the original topology.

Discussion of 6
Note that $nw(X) \leq \lvert X \lvert$ always holds. For compact spaces, we have $w(X)=nw(X) \leq \lvert X \lvert$.

Discussion of 7
Let $X$ be a space with a countable network. By Lemma 2, $X$ has a finer topology that has a countable base. Let $Y$ denote $X$ with this finer second countable topology. Then the identity map from $Y$ onto $X$ is continuous.

For the other direction, let $f:Y \rightarrow X$ be a continuous function mapping a separable metric space $Y$ onto $X$. Let $\mathcal{B}$ be a countable base for $Y$. Then $\lbrace{f(B):B \in \mathcal{B}}\rbrace$ is a network for $X$.

Discussion of 8
This is easily verified. Let $X$ is the continuous image of a cosmic space $Y$. Then $Y$ is the continuous image of some separable metric space $Z$. It follows that $X$ is the continuous image of $Z$.

Discussion of 9
Let $X$ be compact metrizable and let $Y$ be a continuous image of $X$. Then $Y$ is compact. By point 7, $Y$ has a countable network. By point 5, $Y$ is metrizable.

Discussion of 10
A space is perfectly normal if it is normal and that every closed subset is a $G_\delta-$set. Let $X$ be a space with a countable network. The normality of $X$ comes from the fact that it is regular and Lindelof. Note that $X$ is also hereditarily Lindelof. In a hereditarily Lindelof and regular space, every open subspace is an $F_\sigma-$set (thus every closed set is a $G_\delta-$set.

Discussion of 11 (Example of cosmic but not separable metrizable space)
This is the “Butterfly” space or “Bow-tie” space due to L. F. McAuley. I found this example in [Michael]. Let $Y=T \cup S$ where
$T=\lbrace{(x,y) \in \mathbb{R}^2:y>0}\rbrace$ and
$S=\lbrace{(x,y) \in \mathbb{R}^2:y=0}\rbrace$.

Points in $T$ have the usual plane open neighborhoods. A basic open set at $p \in S$ is of the form $B_c(p)$ where $B_c(p)$ consists of $p$ and all points $q \in Y$ having distance $ from $p$ and lying underneath either one of the two straight lines in $Y$ which emanate from $p$ and have slopes $+c$ and $-c$, respectively.

It is clear that $Y$ is a Hausdorff and regular space. The relative “Bow-tie” topologies on $T$ and $S$ coincide with the usual topology on $T$ and $S$, respectively. Thus the union of the usual countable bases on $T$ and $S$ would be a countable network for $Y$. On the other hand, $Y$ is separable but cannot have a countable base (hence not metrizable).

Reference
[Michael]
Michael, E., $\aleph_0-$spaces, J. Math. Mech. 15, 983-1002.