# A note on products of sequential fans

Two posts (the previous post and this post) are devoted to discussing the behavior of countable tightness in taking Cartesian products. The previous post shows that countable tightness behaves well in the product operation if the spaces are compact. In this post, we step away from the orderly setting of compact spaces. We examine the behavior of countable tightness in product of sequential fans. In this post, we show that countable tightness can easily be destroyed when taking products of sequential fans. Due to the combinatorial nature of sequential fans, the problem of determining the tightness of products of fans is often times a set-theoretic problem. In many instances, it is hard to determine the tightness of a product of two sequential fans without using extra set theory axioms beyond ZFC. The sequential fans is a class of spaces that have been studied extensively and are involved in the solutions of many problems that were seemingly unrelated. For one example, see [3].

For a basic discussion of countable tightness, see these previous post on the notion of tightness and its relation with free sequences. Also see chapter a-4 on page 15 of [4].

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Sequential fans

Let $S$ be a non-trivial convergent sequence along with its limit point. For convenience, let $\displaystyle S=\left\{0 \right\} \cup \left\{1, 2^{-1}, 3^{-1}, 4^{-1}, \cdots \right\}$, considered as a subspace of the Euclidean real line. Let $\kappa$ be a cardinal number. The set $\kappa$ is usually taken as the set of all the ordinals that precede $\kappa$. The set $\omega$ is the first infinite ordinal, or equivalently the set of all non-negative integers. Let $\omega^\kappa$ be the set of all functions from $\kappa$ into $\omega$.

There are several ways to describe a sequential fan. One way is to describe it as a quotient space. The sequential fan $S(\kappa)$ is the topological sum of $\kappa$ many copies of the convergent sequence $S$ with all non-isolated points identified as one point that is called $\infty$. To make the discussion easier to follow, we also use the following formulation of $S(\kappa)$:

$\displaystyle S(\kappa)=\left\{\infty \right\} \cup (\kappa \times \omega)$

In this formulation, every point is $\kappa \times \omega$ is isolated and an open neighborhood of the point $\infty$ is of the form:

$\displaystyle B_f=\left\{\infty \right\} \cup \left\{(\alpha,n) \in \kappa \times \omega: n \ge f(\alpha) \right\}$ where $f \in \omega^\kappa$.

According to the definition of the open neighborhood $B_f$, the sequence $(\alpha,0), (\alpha,1), (\alpha,2),\cdots$ converges to the point $\infty$ for each $\alpha \in \kappa$. Thus the set $(\left\{\alpha \right\} \times \omega) \cup \left\{\infty \right\}$ is a homeomorphic copy of the convergent sequence $S$. The set $\left\{\alpha \right\} \times \omega$ is sometimes called a spine. Thus the space $S(\kappa)$ is said to be the sequential fan with $\kappa$ many spines.

The point $\infty$ is the only non-isolated point in the fan $S(\kappa)$. The set $\mathcal{B}=\left\{B_f: f \in \omega^\kappa \right\}$ is a local base at the point $\infty$. The base $\mathcal{B}$ is never countable except when $\kappa$ is finite. Thus if $\kappa$ is infinite, the fan $S(\kappa)$ can never be first countable. In particular, for the fan $S(\omega)$, the character at the point $\infty$ is the cardinal number $\mathfrak{d}$. See page 13 in chapter a-3 of [4]. This cardinal number is called the dominating number and is introduced below in the section “The bounding number”. This is one indication that the sequential fan is highly dependent on set theory. It is hard to pinpoint the character of $S(\omega)$ at the point $\infty$. For example, it is consistent with ZFC that $\mathfrak{d}=\omega_1$. It is also consistent that $\mathfrak{d}>\omega_1$.

Even though the sequential fan is not first countable, it has a relatively strong convergent property. If $\infty \in \overline{A}$ and $\infty \notin A$ where $A \subset S(\kappa)$, then infinitely many points of $A$ are present in at least one of the spine $\left\{\alpha \right\} \times \omega$ (if this is not true, then $\infty \notin \overline{A}$). This means that the sequential fan is always a Frechet space. Recall that the space $Y$ is a Frechet space if for each $A \subset Y$ and for each $x \in \overline{A}$, there exists a sequence $\left\{x_n \right\}$ of points of $A$ converging to $x$.

Some of the convergent properties weaker than being a first countable space are Frechet space, sequential space and countably tight space. Let's recall the definitions. A space $X$ is a sequential space if $A \subset X$ is a sequentially closed set in $X$, then $A$ is a closed set in $X$. The set $A$ is sequentially closed in $X$ if this condition is satisfied: if the sequence $\left\{x_n \in A: n \in \omega \right\}$ converges to $x \in X$, then $x \in A$. A space $X$ is countably tight (have countable tightness) if for each $A \subset X$ and for each $x \in \overline{A}$, there exists a countable $B \subset A$ such that $x \in \overline{B}$. See here for more information about these convergent properties. The following shows the relative strength of these properties. None of the implications can be reversed.

First countable space $\Longrightarrow$ Frechet space $\Longrightarrow$ Sequential space $\Longrightarrow$ Countably tight space

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Examples

The relatively strong convergent property of being a Frechet space is not preserved in products or squares of sequential fans. We now look at some examples.

Example 1
Consider the product space $S(\omega) \times S$ where $S$ is the convergent sequence defined above. The first factor is Frechet and the second factor is a compact metric space. We show that $S(\omega) \times S$ is not sequential. To see this, consider the following subset $A$ of $S(\omega) \times S$:

$\displaystyle A_f=\left\{(x,n^{-1}) \in S(\omega) \times S: n \in \omega \text{ and } x=(n,f(n)) \right\} \ \forall \ f \in \omega^\omega$

$\displaystyle A=\bigcup_{f \in \omega^\omega} A_f$

It follows that $(\infty,0) \in \overline{A}$. Furthermore, no sequence of points of $A$ can converge to the point $(\infty,0)$. To see this, let $a_n \in A$ for each $n$. Consider two cases. One is that some spine $\left\{m \right\} \times \omega$ contains the first coordinate of $a_n$ for infinitely many $n \in \omega$. The second is the opposite of the first – each spine $\left\{m \right\} \times \omega$ contains the first coordinate of $a_n$ for at most finitely many $n$. Either case means that there is an open set containing $(\infty,0)$ that misses infinitely many $a_n$. Thus the sequence $a_n$ cannot converge to $(\infty,0)$.

Let $A_1$ be the set of all sequential limits of convergent sequences of points of $A$. With $A \subset A_1$, we know that $(\infty,0) \in \overline{A_1}$ but $(\infty,0) \notin A_1$. Thus $A_1$ is a sequentially closed subset of $S(\omega) \times S$ that is not closed. This shows that $S(\omega) \times S$ is not a sequential space.

The space $S(\omega) \times S$ is an example of a space that is countably tight but not sequential. The example shows that the product of two Frechet spaces does not even have to be sequential even when one of the factors is a compact metric space. The next example shows that the product of two sequential fans does not even have to be countably tight.

Example 2
Consider the product space $S(\omega) \times S(\omega^\omega)$. We show that it is not countably tight. To this end, consider the following subset $A$ of $S(\omega) \times S(\omega^\omega)$.

$\displaystyle S(\omega)=\left\{\infty \right\} \cup (\omega \times \omega)$

$\displaystyle S(\omega^\omega)=\left\{\infty \right\} \cup (\omega^\omega \times \omega)$

$\displaystyle A_f=\left\{(x,y) \in S(\omega) \times S(\omega^\omega): x=(n,f(n)) \text{ and } y=(f,j) \right\} \ \forall \ f \in \omega^\omega$

$\displaystyle A=\bigcup_{f \in \omega^\omega} A_f$

It follows that $(\infty,\infty) \in \overline{A}$. We show that for any countable $C \subset A$, the point $(\infty,\infty) \notin \overline{C}$. Fix a countable $C \subset A$. We can assume that $C=\bigcup_{i=1}^\infty A_{f_i}$. Now define a function $g \in \omega^\omega$ by a diagonal argument as follows.

Define $g(0)$ such that $g(0)>f_0(0)$. For each integer $n>0$, define $g(n)$ such that $g(n)>\text{max} \{ \ f_n(0),f_n(1),\cdots,f_n(n) \ \}$ and $g(n)>g(n-1)$. Let $O=B_g \times S(\omega^\omega)$. The diagonal definition of $g$ ensures that $O$ is an open set containing $(\infty,\infty)$ such that $O \cap C=\varnothing$. This shows that the space $S(\omega) \times S(\omega^\omega)$ is not countably tight.

Example 3
The space $S(\omega_1) \times S(\omega_1)$ is not countably tight. In fact its tightness character is $\omega_1$. This fact follows from Theorem 1.1 in [2].

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The set-theoretic angle

Example 2 shows that $S(\omega) \times S(\omega^\omega)$ is not countably tight even though each factor has the strong property of a Frechet space with the first factor being a countable space. The example shows that Frechetness behaves very badly with respect to the product operation. Is there an example of $\kappa>\omega$ such that $S(\omega) \times S(\kappa)$ is countably tight? In particular, is $S(\omega) \times S(\omega_1)$ countably tight?

First off, if Continuum Hypothesis (CH) holds, then Example 2 shows that $S(\omega) \times S(\omega_1)$ is not countably tight since the cardinality of $\omega^{\omega}$ is $\omega_1$ under CH. So for $S(\omega) \times S(\omega_1)$ to be countably tight, extra set theory assumptions beyond ZFC will have to be used (in fact the extra axioms will have to be compatible with the negation of CH). In fact, it is consistent with ZFC for $S(\omega) \times S(\omega_1)$ to be countably tight. It is also consistent with ZFC for $t(S(\omega) \times S(\omega_1))=\omega_1$. We point out some facts from the literature to support these observations.

Consider $S(\omega) \times S(\kappa)$ where $\kappa>\omega_1$. For any regular cardinal $\kappa>\omega_1$, it is possible that $S(\omega) \times S(\kappa)$ is countably tight. It is also possible for the tightness character of $S(\omega) \times S(\kappa)$ to be $\kappa$ (of course in a different model of set theory). Thus it is hard to pin down the tightness character of the product $S(\omega) \times S(\kappa)$. It all depends on your set theory. In the next section, we point out some facts from the literature to support these observations.

Example 3 points out that the tightness character of $S(\omega_1) \times S(\omega_1)$ is $\omega_1$, i.e. $t(S(\omega_1) \times S(\omega_1))=\omega_1$ (this is a fact on the basis of ZFC only). What is $t(S(\omega_2) \times S(\omega_2))$ or $t(S(\kappa) \times S(\kappa))$ for any $\kappa>\omega_1$? The tightness character of $S(\kappa) \times S(\kappa)$ for $\kappa>\omega_1$ also depends on set theory. We also give a brief explanation by pointing out some basic information from the literature.

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The bounding number

The tightness of the product $S(\omega) \times S(\kappa)$ is related to the cardinal number called the bounding number denoted by $\mathfrak{b}$.

Recall that $\omega^{\omega}$ is the set of all functions from $\omega$ into $\omega$. For $f,g \in \omega^{\omega}$, define $f \le^* g$ by the condition: $f(n) \le g(n)$ for all but finitely many $n \in \omega$. A set $F \subset \omega^{\omega}$ is said to be a bounded set if $F$ has an upper bound according to $\le^*$, i.e. there exists some $f \in \omega^{\omega}$ such that $g \le^* f$ for all $g \in F$. Then $F \subset \omega^{\omega}$ is an unbounded set if it is not bounded. To spell it out, $F \subset \omega^{\omega}$ is an unbounded set if for each $f \in \omega^{\omega}$, there exists some $g \in F$ such that $g \not \le^* f$.

Furthermore, $F \subset \omega^{\omega}$ is a dominating set if for each $f \in \omega^{\omega}$, there exists some $g \in F$ such that $f \le^* g$. Define the cardinal numbers $\mathfrak{b}$ and $\mathfrak{d}$ as follows:

$\displaystyle \mathfrak{b}=\text{min} \left\{\lvert F \lvert: F \subset \omega^{\omega} \text{ is an unbounded set} \right\}$

$\displaystyle \mathfrak{d}=\text{min} \left\{\lvert F \lvert: F \subset \omega^{\omega} \text{ is a dominating set} \right\}$

The cardinal number $\mathfrak{b}$ is called the bounding number. The cardinal number $\mathfrak{d}$ is called the dominating number. Note that continuum $\mathfrak{c}$, the cardinality of $\omega^{\omega}$, is an upper bound of both $\mathfrak{b}$ and $\mathfrak{d}$, i.e. $\mathfrak{b} \le \mathfrak{c}$ and $\mathfrak{d} \le \mathfrak{c}$. How do $\mathfrak{b}$ and $\mathfrak{d}$ relate? We have $\mathfrak{b} \le \mathfrak{d}$ since any dominating set is also an unbounded set.

A diagonal argument (similar to the one in Example 2) shows that no countable $F \subset \omega^{\omega}$ can be unbounded. Thus we have $\omega < \mathfrak{b} \le \mathfrak{d} \le \mathfrak{c}$. If CH holds, then we have $\omega_1 = \mathfrak{b} = \mathfrak{d} = \mathfrak{c}$. On the other hand, it is also consistent that $\omega < \mathfrak{b} < \mathfrak{d} \le \mathfrak{c}$.

We now relate the bounding number to the tightness of $S(\omega) \times S(\kappa)$. The following theorem is from Theorem 1.3 in [3].

Theorem 1 – Theorem 1.3 in [3]
The following conditions hold:

• For $\omega \le \kappa <\mathfrak{b}$, the space $S(\omega) \times S(\kappa)$ is countably tight.
• The tightness character of $S(\omega) \times S(\mathfrak{b})$ is $\mathfrak{b}$, i.e. $t(S(\omega) \times S(\mathfrak{b}))=\mathfrak{b}$.

Thus $S(\omega) \times S(\kappa)$ is countably tight for any uncountable $\kappa <\mathfrak{b}$. In particular if $\omega_1 <\mathfrak{b}$, then $S(\omega) \times S(\omega_1)$ is countably tight. According to Theorem 5.1 in [6], this is possible.

Theorem 2 – Theorem 5.1 in [6]
Let $\tau$ and $\lambda$ be regular cardinal numbers such that $\omega_1 \le \tau \le \lambda$. It is consistent with ZFC that $\mathfrak{b}=\mathfrak{d}=\tau$ and $\mathfrak{c}=\lambda$.

Theorem 2 indicates that it is consistent with ZFC that the bounding number $\mathfrak{b}$ can be made to equal any regular cardinal number. In the model of set theory in which $\omega_1 <\mathfrak{b}$, $S(\omega) \times S(\omega_1)$ is countably tight. Likewise, in the model of set theory in which $\omega_1 < \kappa <\mathfrak{b}$, $S(\omega) \times S(\kappa)$ is countably tight.

On the other hand, if the bounding number is made to equal an uncountable regular cardinal $\kappa$, then $t(S(\omega) \times S(\kappa))=\kappa$. In particular, $t(S(\omega) \times S(\omega_1))=\omega_1$ if $\mathfrak{b}=\omega_1$.

The above discussion shows that the tightness of $S(\omega) \times S(\kappa)$ is set-theoretic sensitive. Theorem 2 indicates that it is hard to pin down the location of the bounding number $\mathfrak{b}$. Choose your favorite uncountable regular cardinal, there is always a model of set theory in which $\mathfrak{b}$ is your favorite uncountable cardinal. Then Theorem 1 ties the bounding number to the tightness of $S(\omega) \times S(\kappa)$. Thus the exact value of the tightness character of $S(\omega) \times S(\kappa)$ depends on your set theory. If your favorite uncountable regular cardinal is $\omega_1$, then in one model of set theory consistent with ZFC, $t(S(\omega) \times S(\omega_1))=\omega$ (when $\omega_1 <\mathfrak{b}$). In another model of set theory, $t(S(\omega) \times S(\omega_1))=\omega_1$ (when $\omega_1 =\mathfrak{b}$).

One comment about the character of the fan $S(\omega)$ at the point $\infty$. As indicated earlier, the character at $\infty$ is the dominating number $\mathfrak{d}$. Theorem 2 tells us that it is consistent that $\mathfrak{d}$ can be any uncountable regular cardinal. So for the fan $S(\omega)$, it is quite difficult to pinpoint the status of a basic topological property such as character of a space. This is another indication that the sequential fan is highly dependent on additional axioms beyond ZFC.

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The collectionwise Hausdorff property

Now we briefly discuss the tightness of $t(S(\kappa) \times S(\kappa))$ for any $\kappa>\omega_1$. The following is Theorem 1.1 in [2].

Theorem 3 – Theorem 1.1 in [2]
Let $\kappa$ be any infinite regular cardinal. The following conditions are equivalent.

• There exists a first countable $< \kappa$-collectionwise Hausdorff space which fails to be a $\kappa$-collectionwise Hausdorff space.
• $t(S(\kappa) \times S(\kappa))=\kappa$.

The existence of the space in the first condition, on the surface, does not seem to relate to the tightness character of the square of a sequential fan. Yet the two conditions were proved to be equivalent [2]. The existence of the space in the first condition is highly set-theory sensitive. Thus so is the tightness of the square of a sequential fan. It is consistent that a space in the first condition exists for $\kappa=\omega_2$. Thus in that model of set theory $t(S(\omega_2) \times S(\omega_2))=\omega_2$. It is also consistent that there does not exist a space in the first condition for $\kappa=\omega_2$. Thus in that model, $t(S(\omega_2) \times S(\omega_2))<\omega_2$. For more information, see [3].

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Remarks

Sequential fans and their products are highly set-theoretic in nature and are objects that had been studied extensively. This is only meant to be a short introduction. Any interested readers can refer to the small list of articles listed in the reference section and other articles in the literature.

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Exercise

Use Theorem 3 to show that $t(S(\omega_1) \times S(\omega_1))=\omega_1$ by finding a space $X$ that is a first countable $< \omega_1$-collectionwise Hausdorff space which fails to be a $\omega_1$-collectionwise Hausdorff space.

For any cardinal $\kappa$, a space $X$ is $\kappa$-collectionwise Hausdorff (respectively $< \kappa$-collectionwise Hausdorff) if for any closed and discrete set $A \subset X$ with $\lvert A \lvert \le \kappa$ (repectively $\lvert A \lvert < \kappa$), the points in $A$ can be separated by a pairwise disjoint family of open sets.

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Reference

1. Bella A., van Mill J., Tight points and countable fan-tightness, Topology Appl., 76, (1997), 1-27.
2. Eda K., Gruenhage G., Koszmider P., Tamano K., Todorčeviće S., Sequential fans in topology, Topology Appl., 67, (1995), 189-220.
3. Eda K., Kada M., Yuasa Y., Tamano K., The tightness about sequential fans and combinatorial properties, J. Math. Soc. Japan, 49 (1), (1997), 181-187.
4. Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
5. LaBerge T., Landver A., Tightness in products of fans and psuedo-fans, Topology Appl., 65, (1995), 237-255.
6. Van Douwen, E. K., The Integers and Topology, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, (1984) 111-167.

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

# Several ways to define countably tight spaces

This post is an introduction to countable tight and countably generated spaces. A space being a countably tight space is a convergence property. The article [1] lists out 8 convergence properties. The common ones on that list include Frechet space, sequential space, k-space and countably tight space, all of which are weaker than the property of being a first countable space. In this post we discuss several ways to define countably tight spaces and to discuss its generalizations.

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Several definitions

A space $X$ is countably tight (or has countable tightness) if for each $A \subset X$ and for each $x \in \overline{A}$, there is a countable $B \subset A$ such that $x \in \overline{B}$. According to this Wikipedia entry, a space being a countably generated space is the property that its topology is generated by countable sets and is equivalent to the property of being countably tight. The equivalence of the two definitions is not immediately clear. In this post, we examine these definitions more closely. Theorem 1 below has three statements that are equivalent. Any one of the three statements can be the definition of countably tight or countably generated.

Theorem 1
Let $X$ be a space. The following statements are equivalent.

1. For each $A \subset X$, the set equality (a) holds.$\text{ }$
• $\displaystyle \overline{A}=\cup \left\{\overline{B}: B \subset A \text{ and } \lvert B \lvert \le \omega \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (a)$

2. For each $A \subset X$, if condition (b) holds,
For all countable $C \subset X$, $C \cap A$ is closed in $C \ \ \ \ \ \ \ \ (b)$

then $A$ is closed.

3. For each $A \subset X$, if condition (c) holds,
For all countable $B \subset A$, $\overline{B} \subset A \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (c)$

then $A$ is closed.

Statement 1 is the definition of a countably tight space. The set inclusion $\supset$ in (a) is always true. We only need to be concerned with $\subset$, which is the definition of countable tightness given earlier.

Statement 2 is the definition of a countably generated space according to this Wikipedia entry. This definition is in the same vein as that of k-space (or compactly generated space). Note that a space $X$ is a k-space if Statement 2 holds when “countable” is replaced with “compact”.

Statement 3 is in the same vein as that of a sequential space. Recall that a space $X$ is a sequential space if $A \subset X$ is a sequentially closed set then $A$ is closed. The set $A$ is a sequentially closed set if the sequence $x_n \in A$ converges to $x \in X$, then $x \in A$ (in other words, for any sequence of points of $A$ that converges, the limit must be in $A$). If the sequential limit in the definition of sequential space is relaxed to be just topological limit (i.e. accumulation point), then the resulting definition is Statement 3. Thus Statement 3 says that for any countable subset $B$ of $A$, any limit point (i.e. accumulation point) of $B$ must be in $A$. Thus any sequential space is countably tight. In a sequential space, the closed sets are generated by taking sequential limit. In a space defined by Statement 3, the closed sets are generated by taking closures of countable sets.

All three statements are based on the countable cardinality and have obvious generalizations by going up in cardinality. For any set $A \subset X$ that satisfies condition (c) in Statement 3 is said to be an $\omega$-closed set. Thus for any cardinal number $\tau$, the set $A \subset X$ is a $\tau$-closed set if for any $B \subset A$ with $\lvert B \lvert \le \tau$, $\overline{B} \subset A$. Condition (c) in Statement 3 can then be generalized to say that if $A \subset X$ is a $\tau$-closed set, then $A$ is closed.

The proof of Theorem 1 is handled in the next section where we look at the generalizations of all three statements and prove their equivalence.

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Generalizations

The definition in Statement 1 in Theorem 1 above can be generalized as a cardinal function called tightness. Let $X$ be a space. By $t(X)$ we mean the least infinite cardinal number $\tau$ such that the following holds:

For all $A \subset X$, and for each $x \in \overline{A}$, there exists $B \subset A$ with $\lvert B \lvert \le \tau$ such that $x \in \overline{B}$.

When $t(X)=\omega$, the space $X$ is countably tight (or has countable tightness). In keeping with the set equality (a) above, the tightness $t(X)$ can also be defined as the least infinite cardinal $\tau$ such that for any $A \subset X$, the following set equality holds:

$\displaystyle \overline{A}=\cup \left\{\overline{B}: B \subset A \text{ and } \lvert B \lvert \le \tau \right\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\alpha)$

Let $\tau$ be an infinite cardinal number. To generalize Statement 2, we say that a space $X$ is $\tau$-generated if the following holds:

For each $A \subset X$, if the following condition holds:

For all $C \subset X$ with $\lvert C \lvert \le \tau$, the set $C \cap A$ is closed in $C \ \ \ \ \ \ \ \ \ \ \ (\beta)$

then $A$ is closed.

To generalize Statement 3, we say that a set $A \subset X$ is $\tau$-closed if for any $B \subset A$ with $\lvert B \lvert \le \tau$, $\overline{B} \subset A$. A generalization of Statement 3 is that

For any $A \subset X$, if $A \subset X$ is a $\tau$-closed set, then $A$ is closed $.\ \ \ \ \ \ \ \ \ \ \ (\chi)$

Theorem 2
Let $X$ be a space. Let $\tau$ be an infinite cardinal. The following statements are equivalent.

1. $t(X) \le \tau$.
2. The space $X$ is $\tau$-generated.
3. For each $A \subset X$, if $A \subset X$ is a $\tau$-closed set, then $A$ is closed.

Proof of Theorem 2
$1 \rightarrow 2$
Suppose that (2) does not hold. Let $A \subset X$ be such that the set $A$ satisfies condition $(\beta)$ and $A$ is not closed. Let $x \in \overline{A}-A$. By (1), the point $x$ belongs to the right hand side of the set equality $(\alpha)$. Choose $B \subset A$ with $\lvert B \lvert \le \tau$ such that $x \in \overline{B}$. Let $C=B \cup \left\{x \right\}$. By condition $(\beta)$, $C \cap A=B$ is closed in $C$. This would mean that $x \in B$ and hence $x \in A$, a contradiction. Thus if (1) holds, (2) must holds.

$2 \rightarrow 3$
Suppose (3) does not hold. Let $A \subset X$ be a $\tau$-closed set that is not a closed set in $X$. Since (2) holds and $A$ is not closed, condition $(\beta)$ must not hold. Choose $C \subset X$ with $\lvert C \lvert \le \tau$ such that $B=C \cap A$ is not closed in $C$. Choose $x \in C$ that is in the closure of $C \cap A$ but is not in $C \cap A$. Since $A$ is $\tau$-closed, $\overline{B}=\overline{C \cap A} \subset A$, which implies that $x \in A$, a contradiction. Thus if (2) holds, (3) must hold.

$3 \rightarrow 1$
Suppose (1) does not hold. Let $A \subset X$ be such that the set equality $(\alpha)$ does not hold. Let $x \in \overline{A}$ be such that $x$ does not belong to the right hand side of $(\alpha)$. Let $A_0=\overline{A}-\left\{x \right\}$. Note that the set $A_0$ is $\tau$-closed. By (3), $A_0$ is closed. Furthermore $x \in \overline{A_0}$, leading to $x \in A_0=\overline{A}-\left\{x \right\}$, a contradiction. So if (3) holds, (1) must hold. $\blacksquare$

Theorem 1 obviously follows from Theorem 2 by letting $\tau=\omega$. There is another way to characterize the notion of tightness using the concept of free sequence. See the next post.

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Examples

Several elementary convergence properties have been discussed in a series of blog posts (the first post and links to the other are found in the first one). We have the following implications and none is reversible.

First countable $\Longrightarrow$ Frechet $\Longrightarrow$ Sequential $\Longrightarrow$ k-space

Where does countable tightness place in the above implications? We discuss above that

Sequential $\Longrightarrow$ countably tight.

How do countably tight space and k-space compare? It turns out that none implies the other. We present some supporting examples.

Example 1
The Arens’ space is a canonical example of a sequential space that is not a Frechet space. A subspace of the Arens’ space is countably tight and not sequential. The same subspace is also not a k-space. There are several ways to represent the Arens’ space, we present the version found here.

Let $\mathbb{N}$ be the set of all positive integers. Define the following:

$\displaystyle V_{i,j}=\left\{\biggl(\frac{1}{i},\frac{1}{k} \biggr): k \ge j \right\}$ for all $i,j \in \mathbb{N}$

$V=\bigcup_{i \in \mathbb{N}} V_{i,j}$

$\displaystyle H=\left\{\biggl(\frac{1}{i},0 \biggr): i \in \mathbb{N} \right\}$

$V_i=V_{i,1} \cup \left\{ x \right\}$ for all $i \in \mathbb{N}$

Let $Y=\left\{(0,0) \right\} \cup H \cup V$. Each point in $V$ is an isolated point. Open neighborhoods at $(\frac{1}{i},0) \in H$ are of the form:

$\displaystyle \left\{\biggl(\frac{1}{i},0 \biggr) \right\} \cup V_{i,j}$ for some $j \in \mathbb{N}$

The open neighborhoods at $(0,0)$ are obtained by removing finitely many $V_i$ from $Y$ and by removing finitely many isolated points in the $V_i$ that remain. The open neighborhoods just defined form a base for a topology on the set $Y$, i.e. by taking unions of these open neighborhoods, we obtain all the open sets for this space. The space $Y$ can also be viewed as a quotient space (discussed here).

The space $Y$ is a sequential space that is not Frechet. The subspace $Z=\left\{(0,0) \right\} \cup V$ is not sequential. Since $Y$ is a countable space, the space $Z$ is by default a countably tight space. The space $Z$ is also not an k-space. These facts are left as exercises below.

Example 2
Consider the product space $X=\left\{0,1 \right\}^{\omega_1}$. The space $X$ is compact since it is a product of compact spaces. Any compact space is a k-space. Thus $X$ is a k-space (or compactly generated space). On the other hand, $X$ is not countably tight. Thus the notion of k-space and the notion of countably tight space do not relate.

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Remarks

There is another way to characterize the notion of tightness using the concept of free sequence. See the next post.

The notion of tightness had been discussed in previous posts. One post shows that the function space $C_p(X)$ is countably tight when $X$ is compact (see here). Another post characterizes normality of $X \times \omega_1$ when $X$ is compact (see here)

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Exercises

Exercise 1
This is to verify Example 1. Verify that

• The space $Y$ is a sequential space that is not Frechet.
• $Z=\left\{(0,0) \right\} \cup V$ is not sequential.
• The space $Z$ is not an k-space.

Exercise 2
Verify that any compact space is a k-space. Show that the space $X$ in Example 2 is not countably tight.

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Reference

1. Gerlits J., Nagy Z., Products of convergence properties, Commentationes Mathematicae Universitatis Carolinae, Vol 23, No 4 (1982), 747–756

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

# Comparing two function spaces

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. Furthermore consider these ordinals as topological spaces endowed with the order topology. It is a well known fact that any continuous real-valued function $f$ defined on either $\omega_1$ or $\omega_1+1$ is eventually constant, i.e., there exists some $\alpha<\omega_1$ such that the function $f$ is constant on the ordinals beyond $\alpha$. Now consider the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$. Thus individually, elements of these two function spaces appear identical. Any $f \in C_p(\omega_1)$ matches a function $f^* \in C_p(\omega_1+1)$ where $f^*$ is the result of adding the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$. This fact may give the impression that the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$ are identical topologically. The goal in this post is to demonstrate that this is not the case. We compare the two function spaces with respect to some convergence properties (countably tightness and Frechet-Urysohn property) as well as normality.

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Tightness

One topological property that is different between $C_p(\omega_1)$ and $C_p(\omega_1+1)$ is that of tightness. The function space $C_p(\omega_1+1)$ is countably tight, while $C_p(\omega_1)$ is not countably tight.

Let $X$ be a space. The tightness of $X$, denoted by $t(X)$, is the least infinite cardinal $\kappa$ such that for any $A \subset X$ and for any $x \in X$ with $x \in \overline{A}$, there exists $B \subset A$ for which $\lvert B \lvert \le \kappa$ and $x \in \overline{B}$. When $t(X)=\omega$, we say that $X$ has countable tightness or is countably tight. When $t(X)>\omega$, we say that $X$ has uncountable tightness or is uncountably tight.

First, we show that the tightness of $C_p(\omega_1)$ is greater than $\omega$. For each $\alpha<\omega_1$, define $f_\alpha: \omega_1 \rightarrow \left\{0,1 \right\}$ such that $f_\alpha(\beta)=0$ for all $\beta \le \alpha$ and $f_\alpha(\beta)=1$ for all $\beta>\alpha$. Let $g \in C_p(\omega_1)$ be the function that is identically zero. Then $g \in \overline{F}$ where $F$ is defined by $F=\left\{f_\alpha: \alpha<\omega_1 \right\}$. It is clear that for any countable $B \subset F$, $g \notin \overline{B}$. Thus $C_p(\omega_1)$ cannot be countably tight.

The space $\omega_1+1$ is a compact space. The fact that $C_p(\omega_1+1)$ is countably tight follows from the following theorem.

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

Theorem 1 is a special case of Theorem I.4.1 on page 33 of [1] (the countable case). One direction of Theorem 1 is proved in this previous post, the direction that will give us the desired result for $C_p(\omega_1+1)$.

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The Frechet-Urysohn property

In fact, $C_p(\omega_1+1)$ has a property that is stronger than countable tightness. The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space (see this previous post). Of course, $C_p(\omega_1)$ not being countably tight means that it is not a Frechet-Urysohn space.

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Normality

The function space $C_p(\omega_1+1)$ is not normal. If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ would have countable extent. However, there exists an uncountable closed and discrete subset of $C_p(\omega_1+1)$ (see this previous post). On the other hand, $C_p(\omega_1)$ is Lindelof. The fact that $C_p(\omega_1)$ is Lindelof is highly non-trivial and follows from [2]. The author in [2] showed that if $X$ is a space consisting of ordinals such that $X$ is first countable and countably compact, then $C_p(X)$ is Lindelof.

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Embedding one function space into the other

The two function space $C_p(\omega_1+1)$ and $C_p(\omega_1)$ are very different topologically. However, one of them can be embedded into the other one. The space $\omega_1+1$ is the continuous image of $\omega_1$. Let $g: \omega_1 \longrightarrow \omega_1+1$ be a continuous surjection. Define a map $\psi: C_p(\omega_1+1) \longrightarrow C_p(\omega_1)$ by letting $\psi(f)=f \circ g$. It is shown in this previous post that $\psi$ is a homeomorphism. Thus $C_p(\omega_1+1)$ is homeomorphic to the image $\psi(C_p(\omega_1+1))$ in $C_p(\omega_1)$. The map $g$ is also defined in this previous post.

The homeomposhism $\psi$ tells us that the function space $C_p(\omega_1)$, though Lindelof, is not hereditarily normal.

On the other hand, the function space $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. Note that $C_p(\omega_1+1)$ is countably tight, which is a hereditary property.

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Remark

There is a mapping that is alluded to at the beginning of the post. Each $f \in C_p(\omega_1)$ is associated with $f^* \in C_p(\omega_1+1)$ which is obtained by appending the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$. It may be tempting to think of the mapping $f \rightarrow f^*$ as a candidate for a homeomorphism between the two function spaces. The discussion in this post shows that this particular map is not a homeomorphism. In fact, no other one-to-one map from one of these function spaces onto the other function space can be a homeomorphism.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Buzyakova, R. Z., In search of Lindelof $C_p$‘s, Comment. Math. Univ. Carolinae, 45 (1), 145-151, 2004.

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

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

# Cp(X) is countably tight when X is compact

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

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Tightness

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

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

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

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

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

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

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

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

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

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

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

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Remarks

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

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Reference

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

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$\copyright \ 2014 - 2015 \text{ by Dan Ma}$

# Sigma-products of first countable spaces

A product space is never first countable if there are uncountably many factors. For example, $\prod_{\alpha < \omega_1}\mathbb{R}=\mathbb{R}^{\omega_1}$ is not first countable. In fact any dense subspace of $\mathbb{R}^{\omega_1}$ is not first countable. In particular, the subspace of $\mathbb{R}^{\omega_1}$ consisting of points which have at most countably many non-zero coordinates is not first countable. This subspace is called the $\Sigma$-product of $\omega_1$ many copies of the real line $\mathbb{R}$ and is denoted by $\Sigma_{\alpha<\omega_1} \mathbb{R}$. However, this $\Sigma$-product is a Frechet space (or a Frechet-Urysohn space). In this post, we show that the $\Sigma$-product of first countable spaces is a Frechet space.

Consider the product space $X=\prod_{\alpha \in A} X_\alpha$. Fix a point $a \in X$. Consider the following subspace of $X$:

$\Sigma_{\alpha \in A} X_\alpha(a)=\left\{x \in X: x_\alpha \ne a_\alpha \text{ for at most countably many } \alpha \in A \right\}$

The above subspace of $X$ is called the $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ about the base point $a$. When the base point is understood, we simply say the $\Sigma$-product of the spaces $\left\{X_\alpha: \alpha \in A \right\}$ and use the notation $\Sigma_{\alpha \in A} X_\alpha$ to denote the space.

For each $y \in \Sigma_{\alpha \in A} X_\alpha$, define $S(y)$ to be the set of all $\alpha \in A$ such that $y_\alpha \ne a_\alpha$, i.e., the support of the point $y$. Another notion of support is that of standard basic open sets in the product topology. A standard basic open set is a set $O=\prod_{\alpha \in A} O_\alpha$ where each $O_\alpha$ is an open subset of $X_\alpha$. The support of $O$, denoted by $supp(O)$ is the finite set of all $\alpha \in A$ such that $O_\alpha \ne X_\alpha$.

A space $Y$ is said to be first countable if there exists a countable local base at each point in $Y$. A space $Y$ is said to be 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 $\left\{y_n: n=1,2,3,\cdots \right\}$ of points of $M$ such that the sequence converges to $y$. Frechet spaces also go by the name of Frechet-Urysohn spaces. Clearly, any first countable space is Frechet. The converse is not true (see Example 1 in this post). We prove the following theorem.

Theorem 1

Suppose each factor $X_\alpha$ is a first countable space. Then the $\Sigma$-product $\Sigma_{\alpha \in A} X_\alpha$ is a Frechet space.

Proof of Theorem 1
Let $\Sigma=\Sigma_{\alpha \in A} X_\alpha$. Let $M \subset \Sigma$ and let $x \in \overline{M}$. We proceed to define a sequence of points $t_n \in M$ such that the sequence $t_n$ converges to $x$. For each $\alpha \in A$, choose a countable local base $\left\{B_{\alpha,j}: j=1,2,3,\cdots \right\}$ at the point $x_\alpha \in X_\alpha$. Assume that $B_{\alpha,1} \supset B_{\alpha,2} \supset B_{\alpha,3} \supset \cdots$. Then enumerate the countable set $S(x)$ by $S(x)=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\cdots \right\}$. Let $C_1=\left\{\beta_{1,1} \right\}$. The following set $O_1$ is an open subset of $\Sigma$.

$O_1=\biggl(\prod_{\alpha \in C_1} B_{\alpha,1} \times \prod_{\alpha \in A-C_1} X_\alpha \biggr) \cap \Sigma$

Note that $O_1$ is an open set containing $x$. Choose $t_2 \in O_1 \cap M$. Enumerate the support $S(t_2)$ by $S(t_2)=\left\{\beta_{2,1},\beta_{2,2},\beta_{2,3},\cdots \right\}$. Form the finite set $C_2$ by picking the first two points of $S(x)$ and the first two points of $S(t_2)$, i.e., $C_2=\left\{\beta_{1,1},\beta_{1,2},\beta_{2,1},\beta_{2,2} \right\}$. Then form the following open subset of $\Sigma$.

$O_2=\biggl(\prod_{\alpha \in C_2} B_{\alpha,2} \times \prod_{\alpha \in A-C_2} X_\alpha \biggr) \cap \Sigma$

Choose $t_3 \in O_2 \cap M$. Enumerate the support $S(t_3)$ by $S(t_3)=\left\{\beta_{3,1},\beta_{3,2},\beta_{3,3},\cdots \right\}$. Then let $C_3=\left\{\beta_{1,1},\beta_{1,2},\beta_{1,3},\ \beta_{2,1},\beta_{2,2},\beta_{2,3},\ \beta_{3,1},\beta_{3,2},\beta_{3,3} \right\}$, i.e., picking the first three points of $S(x)$, the first three points of $S(t_2)$ and the first three points of $S(t_3)$. Now, form the following open subset of $\Sigma$.

$O_3=\biggl(\prod_{\alpha \in C_3} B_{\alpha,3} \times \prod_{\alpha \in A-C_3} X_\alpha \biggr) \cap \Sigma$

Choose $t_4 \in O_2 \cap M$. Let this inductive process continue and we would obtain a sequence $t_2,t_3,t_4,\cdots$ of points of $M$. We claim that the sequence converges to $x$. Before we prove the claim, let’s make a few observations about the inductive process of defining $t_2,t_3,t_4,\cdots$. Let $C=\bigcup_{j=1}^\infty C_j$.

• Each $C_j$ is the support of the open set $O_j$.
• The sequence of open sets $O_j$ is decreasing, i.e., $O_1 \supset O_2 \supset O_3 \supset \cdots$. Thus for each integer $j$, we have $t_k \in O_j$ for all $k \ge j$.
• The support of the point $x$ is contained in $C$, i.e., $S(x) \subset C$.
• The support of the each $t_j$ is contained in $C$, i.e., $S(t_j) \subset C$.
• In fact, $C=S(x) \cup S(t_2) \cup S(t_3) \cup \cdots$.
• The previous three bullet points are clear since the inductive process is designed to use up all the points of these supports in defining the open sets $O_j$.
• Consequently, for each $j$, $x_\alpha=(t_j)_\alpha=a_\alpha$ for each $\alpha \in A-C$. In other words, $x$ and each $t_j$ agree (and agree with the base point $a$) on the coordinates outside of the countable set $C$.

Let $U=\prod_{\alpha \in A} U_\alpha$ be a standard open set in the product space $X=\prod_{\alpha \in A} X_\alpha$ such that $x \in U$. Let $U^*=U \cap \Sigma$. We show that for some $n$, $t_j \in U^*$ for all $j \ge n$.

Let $F=supp(U)$ be the support of $U$. Let $F_1=F \cap C$ and $F_2=F \cap (A-C)$. Consider the following open set:

$U^{**}=\biggl(\prod_{\alpha \in C} U_\alpha \times \prod_{\alpha \in A-C} X_\alpha \biggr) \cap \Sigma$

Note that $supp(U^{**})=F_1$. For each $\alpha \in F_1$, choose $B_{\alpha,k(\alpha)} \subset U_\alpha$. Let $m$ be the maximum of all $k(\alpha)$ where $\alpha \in F_1$. Then $B_{\alpha,m} \subset U_\alpha$ for each $\alpha \in F_1$. Choose a positive integer $p$ such that:

$F_1 \subset W=\left\{\beta_{i,j}: i \le p \text{ and } j \le p \right\}$

Let $n=\text{max}(m,p)$. It follows that there exists some $n$ such that $O_n \subset U^{**}$. Then $t_j \in U^{**}$ for all $j \ge n$. It is also the case that $t_j \in U^{*}$ for all $j \ge n$. This is because $x=t_j$ on the coordinates not in $C$. $\blacksquare$

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

# Sequentially compact spaces, II

All spaces under consideration are Hausdorff. Countably compactness and sequentially compactness are notions related to compactness. A countably compact space is one in which every countable open cover has a finite subcover, or equivalently, every countably infinite subset has a limit point. For a space $X$, the point $p \in X$ is a limit point of $A \subset X$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$. On the other hand, a space $X$ is sequentially compact if every sequence $\left\{x_n:n=1,2,3,\cdots\right\}$ of points of $X$ has a subsequence that converges. Any sequentially compact space is countably compact. The converse is not true. The product space $2^I$ where $I=[0,1]$ is not sequentially compact (see Sequentially compact spaces, I) . However, for sequential spaces (first countable spaces in particular), the notion of sequentially compactness and countably compactness are equivalent. For previous discussion in this blog about sequential spaces, see the links below.

Lemma
Any countably compact space that is countable in size is metrizable and thus first countable.

Proof. Let $X$ be countably compact such that $\lvert X \lvert=\aleph_0$. Then $X$ is compact (any Lindelof countably compact space is compact). In any countable space, the set of all singleton sets is a countable network. Any compact Hausdorff space with a countable network is metrizable and thus first countable. See Spaces With Countable Network. $\blacksquare$

Theorem
Let $X$ be a sequential space. Then $X$ is countably compact if and only if $X$ is sequentially compact.

Proof. The direction $\Leftarrow$ always holds without the space being sequential.

$\Rightarrow$ Suppose $X$ is countably compact. Suppose that $X$ is not sequentially compact. Then there is a sequence $\left\{x_n\right\}$ of points of $X$ with no convergent subsequence. Let $A$ be the set of all terms in this sequence, i.e. $A=\left\{x_n:n=1,2,3,\cdots\right\}$. Note that $A$ is sequentially closed. Since $X$ is sequential, $A$ is closed in $X$. As a closed subset of a countably compact space, $A$ is countably compact. By the lemma, $A$ is first countable. Since $A$ is an infinite compact space, $A$ has a non-isolated point $x$. This means some sequence of points of $A$ converges to $x$, contradicting the assumption that $\left\{x_n\right\}$ has no convergent subsequence. Therefore $X$ must be sequentially compact. $\blacksquare$

Previous posts on sequential spaces and k-spaces:
Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I
k-spaces, II
A note about the Arens’ space