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

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.

1. $C_p(X)$ is a Frechet-Urysohn space.
2. $C_p(X)$ is a k-space.
3. 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

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

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

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

# 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 – II

This is a continuation of the discussion on network. In the previous post, I showed that the network weight (the minimum cardinality of a network) coincides with the weight for both metrizable spaces and locally compact spaces. In another post, I showed that this is true for compact spaces. I now show that this is also true for the class of Moore spaces. First, some definitions. A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$ and each open set $U \subset X$ with $x \in U$, there is some $n$ such that any open set in $\mathcal{D}_n$ containing the point $x$ is contained in $U$. A developable space is one that has a development. A Moore space is a regular developable space.

For a collection of $\mathcal{G}$ of subsets of a space $X$ and for $x \in X$, define $st(x,\mathcal{G})=\bigcup\lbrace{U \in \mathcal{G}:x \in U}\rbrace$. An equivalent way of defining a development: A sequence $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ of open covers of a space $X$ is a development for $X$ if for each $x \in X$, $\lbrace{st(x,\mathcal{G}_n):n \in \omega}\rbrace$ is a local base at $x$. For a basic introduction to Moore space and the Moore space conjecture, there are numerous places to look in the literature ([1] being one of them).

Theorem. If $X$ is a Moore space, then $nw(X)=w(X)$.

Proof. Since $nw(X) \leq w(X)$ always holds, we only need to show $w(X) \leq nw(X)$. To this end, we exhibit a base $\mathcal{B}$ with $\vert \mathcal{B} \lvert \leq nw(X)$. Let $\lbrace{\mathcal{D}_n}\rbrace_{n<\omega}$ be a development for $X$. Let $\mathcal{N}$ be a network with cardinality $nw(X)$.

For each $N \in \mathcal{N}$, choose open set $O(n,N) \in \mathcal{D}_n$ such that $N \subset O(n,N)$. Let $\mathcal{B}_n=\lbrace{O(n,N):N \in \mathcal{N}}\rbrace$ and $\mathcal{B}=\bigcup_{n<\omega}\mathcal{B}_n$. Note that $\lvert \mathcal{B} \lvert \leq nw(X)$. Because $\mathcal{N}$ is a network, each $\mathcal{B}_n$ is a cover of $X$. To see this, let $x \in X$. Choose some $V \in \mathcal{D}_n$ such that $x \in V$. There is some $N \in \mathcal{N}$ such that $x \in N \subset V$. Then $x \in O(n,N)$. For each $n$, $\mathcal{B}_n \subset \mathcal{D}_n$. The sequence $\lbrace{\mathcal{B}_n}\rbrace$ works like a development. We have just shown that $\mathcal{B}$ is a base for $X$.

Corollary. The example of Butterfly space is not a Moore space.

The example of the Butterfly (or Bow-tie) space is defined in this previous post. This space has a countable network and the weight of this space is continuum. Thus this space cannot be a Moore space.

Reference
[1] Steen, L. A. & Seebach, J. A. [1995] Counterexamples in Topology, Dover Books.

# 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.