# A stroll in Bing’s Example G

In this post we take a leisurely walk in Bing’s Example G, which is a classic example of a normal but not collectionwise normal space. Hopefully anyone who is new to this topological space can come away with an intuitive feel and further learn about it. Indeed this is a famous space that had been extensively studied. This example has been written about in several posts in this topology blog. In this post, we explain how Example G is defined, focusing on intuitive idea as much as possible. Of course, the intuitive idea is solely the perspective of the author. Any reader who is interested in building his/her own intuition on this example can skip this post and go straight to the previous introduction. Other blog posts on various subspaces of Example G are here, here and here. Bing’s Example H is discussed here.

At the end of the post, we will demonstrate that the product of Bing’s Example G with the closed unit interval, $F \times [0,1]$, is a normal space.

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The Product Space Angle

The topology in Example G is tweaked from the product space topology. It is thus a good idea to first examine the relevant product space. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. In other words, $Q$ is the power set of $P$. Consider the product of $\lvert Q \lvert$ many copies of the two element set $\left\{0,1 \right\}$. The usual notation of this product space is $2^Q$. The elements of $2^Q$ are simply the functions from $Q$ into $\left\{0,1 \right\}$. An arbitrary element of $2^Q$ is a function $f$ that maps every subset of $P$ to either 0 or 1.

Though the base set $P$ can be any uncountable set, it is a good idea to visualize clearly what $P$ is. In the remainder of this section, think of $P$ as the real line $\mathbb{R}$. Then $Q$ is simply the collection of all subsets of the real line. The elements of the product space are simply functions that map each set of real numbers to either 0 or 1. Or think of each function as a 2-color labeling of the subsets of the real line, where each subset is either red or green for example. There are $2^c$ many subsets of the real line where $c$ is the cardinality of the continuum.

To further visualize the product space, let’s look at a particular subspace of $2^Q$. For each real number $p$, define the function $f_p$ such that $f_p$ always maps any set of real numbers that contains $p$ to 1 and maps any set of real numbers that does not contain $p$ to 0. For example, the following are several values of the function $f_0$.

$f_0([0,1])=1$

$f_0([1,2])=0$

$f_0(\left\{0 \right\})=1$

$f_0(\mathbb{R}-\left\{0 \right\})=0$

$f_0(\mathbb{R})=1$

$f_0(\varnothing)=0$

$f_0(\mathbb{P})=0$

where $\mathbb{P}$ is the set of all irrational numbers. Consider the subspace $F_P=\left\{f_p: p \in P \right\}$. Members of $F_P$ are easy to describe. Each function in $F_P$ maps a subset of the real line to 0 or 1 depending on whether the subscript belongs to the given subset. Another reason that $F_P$ is important is that Bing’s Example is defined by declaring all points not in $F_P$ isolated points and by allowing all points in $F_P$ retaining the open sets in the product topology.

Any point $f$ in $F_P$ determines $f(q)=0 \text{ or } 1$ based on membership (whether the reference point belongs to the set $q$). Points not in $F_P$ have no easy characterization. It seems that any set can be mapped to 0 or 1. Note that any $f$ in $F_P$ maps equally to 0 or 1. So the constant functions $f(q)=0$ and $f(q)=1$ are not in $F_P$. Furthermore, any $f$ such that $f(q)=1$ for at most countably many $q$ would not be in $F_P$.

Let’s continue focusing on the product space for the time being. When $F_P$ is considered as a subspace of the product space $2^Q$, $F_P$ is a discrete space. For each $p \in P$, there is an open set $W_p$ containing $f_p$ such that $W_p$ contains no other points of $F_P$. So $F_P$ is relatively discrete in the product space $2^Q$. Of course $F_P$ cannot be closed in $2^Q$ since $2^Q$ is a compact space. The open set $W_p$ is defined as follows:

$W_p=\left\{f \in 2^Q: f(\left\{p \right\})=1 \text{ and } f(P-\left\{p \right\})=0 \right\}$

It is clear that $f_p \in W_p$ and that $f_t \notin W_p$ for any real number $t \ne p$.

Two properties of the product space $2^Q$ would be very relevant for the discussion. By the well known Tychonoff theorem, the product space $2^Q$ is compact. Since $P$ is uncountable, $2^Q$ always has the countable chain condition (CCC) since it is the product of separable spaces. A space having CCC means that there can only be at most countably many pairwise disjoint open sets. As a result, the uncountably many open sets $W_p$ cannot be all pairwise disjoint. So there exist at least a pair of $W_p$, say $W_{a}$ and $W_{b}$, with nonempty intersection.

The last observation can be generalized. For each $p \in P$, let $V_p$ be any open set containing $f_p$ (open in the product topology). We observe that there are at least two $a$ and $b$ from $P$ such that $V_a \cap V_b \ne \varnothing$. If there are only countably many distinct sets $V_p$, then there are uncountably many $V_p$ that are identical and the observation is valid. So assume that there are uncountably many distinct $V_p$. By the CCC in the product space, there are at least two $a$ and $b$ with $V_a \cap V_b \ne \varnothing$. This observation shows that the discrete points in $F_P$ cannot be separated by disjoint open sets. This means that Bing’s Example G is not collectionwise Hausdorff and hence not collectionwise normal.

Another observation is that any disjoint $A_1, A_2 \subset F_P$ can be separated by disjoint open sets. To see this, define the following two open sets $E_1$ and $E_2$ in the product topology.

$q_1=\left\{p \in P: f_p \in A_1 \right\}$

$q_2=\left\{p \in P: f_p \in A_2 \right\}$

$E_1=\left\{f \in 2^Q: f(q_1)=1 \text{ and } f(q_2)=0 \right\}$

$E_2=\left\{f \in 2^Q: f(q_1)=0 \text{ and } f(q_2)=1 \right\}$

It is clear that $A_1 \subset E_1$ and $A_2 \subset E_2$. Furthermore, $E_1 \cap E_2=\varnothing$. This observation will be the basis for showing that Bing’s Example G is normal.

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The Topology of Bing’s Example G

The topology for Bing’s Example G is obtained by tweaking the product topology on $2^Q$. Let $P$ be any uncountable set. Let $Q$ be the set of all subsets of $P$. The set $F_P$ is defined as above. Bing’s Example G is $F=2^Q$ with points in $F_P$ retaining the open sets in the product topology and with points not in $F_P$ declared isolated. For some reason, in Bing’s original paper, the notation $F$ is used even though the example is identified by G. We will follow Bing’s notation.

The subspace $F_P$ is discrete but not closed in the product topology. However, $F_P$ is both discrete and closed in Bing’s Example G. Based on the discussion in the previous section, one immediate conclusion we can made is that the space $F$ is not collectionwise Hausdorff. This follows from the fact that points in the uncountable closed and discrete set $F_P$ cannot be separated by disjoint open sets. By declaring points not in $F_P$ isolated, the countable chain condition in the original product space $2^Q$ is destroyed. However, there is still a strong trace of CCC around the points in the set $F_P$, which is sufficient to prevent collectionwise Hausdorffness, and consequently collectionwise normality.

To show that $F$ is normal, let $H$ and $K$ be disjoint closed subsets of $F$. To make it easy to follow, let $H=A_1 \cup B_1$ and $K=A_2 \cup B_2$ where

$A_1=H \cap F_P \ \ \ \ B_1=H \cap (F-F_P)$

$A_2=K \cap F_P \ \ \ \ B_2=K \cap (F-F_P)$

In other words, $A$ is the non-isolated part and $B$ is the isolated part of the respective closed set. Based on the observation made in the previous section, obtain the disjoint open sets $E_1$ and $E_2$ where $A_1 \subset E_1$ and $A_2 \subset E_2$. Set the following open sets.

$O_1=(E_1 \cup B_1) - K$

$O_2=(E_2 \cup B_2) - H$

It follows that $O_1$ and $O_2$ are disjoint open sets and that $A_1 \subset O_1$ and $A_2 \subset O_2$. Thus Bing’s Example G is a normal space.

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Bing’s Example G is Countably Paracompact

We discuss one more property of Bing’s Example G. A space $X$ is countably paracompact if every countable countable open cover of $X$ has a locally finite open refinement. In other words, such a space satisfies the property of being a paracompact space but just for countable open covers. A space is countably metacompact if every open cover has a point-finite open refinement (i.e. replacing locally finite in the paracompact definition with point-finite). It is well known that in the class of normal spaces, the two notions are equivalent. Since Bing’s Example G is normal, we only need to show that it is countably metacompact. Note that Bing’s Example G is not metacompact (see here).

Let $\mathcal{U}$ be a countable open cover of $F$. Let $\mathcal{U}^*=\left\{U_1,U_2,U_3,\cdots \right\}$ be the set of all open sets in $\mathcal{U}$ that contain points in $F_P$. For each $i$, let $A_i=U_i \cap F_P$. From the perspective of Bing’s Example G, the sets $A_i$ are discrete closed sets. In any normal space, countably many discrete closed sets can be separated by disjoint open sets (see Lemma 1 here). Let $O_1,O_2,O_3,\cdots$ be disjoint open sets such that $A_i \subset O_i$ for each $i$.

We now build a point-finite open refinement of $\mathcal{U}$. For each $i$, let $V_i=U_i \cap O_i$. Let $V=\cup_{i=1}^\infty V_i$. Consider the following.

$\mathcal{V}=\left\{V_i: i=1,2,3,\cdots \right\} \cup \left\{\left\{ x \right\}: x \in F-V \right\}$

It follows that $\mathcal{V}$ is an open cover of $F$. All points of $F_P$ belong to the open sets $V_i$. Any point that is not in one of the $V_i$ belongs to a singleton open set. It is also clear that $\mathcal{V}$ is a refinement of $\mathcal{U}$. For each $i$, $V_i \subset U_i$ and each singleton set is contained in some member of $\mathcal{U}$. It follows that each point in $F$ belongs to at most finitely many sets in $\mathcal{V}$. In fact, each point belongs to exactly one set in $\mathcal{V}$. Each point in $F_P$ belongs to exactly one $V_i$ since the open sets $O_i$ are disjoint. Any point in $V$ belongs to exactly one singleton open set. What we just show is slightly stronger than countably metacompact. The technical term would be countably 1-bounded metacompact.

Since among normal spaces, countably paracompactness is equivalent to countably metacompact, we can now say that Bing’s Example G is a topological space that is normal and countably paracompact. By Dowker’s Theorem, we can conclude that the product of Bing’s Example G with the closed unit interval, $F \times [0,1]$, is a normal space.

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Previous Posts

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$\copyright \ 2016 \text{ by Dan Ma}$

# A strategy for finding CCC and non-separable spaces

In this post we present a general strategy for finding CCC spaces that are not separable. As illustration, we give four implementations of this strategy.

In searching for counterexamples in topology, one good place to look is of course the book by Steen and Seebach [2]. There are four examples of spaces that are CCC but not separable found in [2] – counterexamples 20, 21, 24 and 63. Counterexamples 20 and 21 are not Hausdorff. Counterexample 24 is the uncountable Fort space (it is completely normal but not perfectly normal). Counterexample 63 (Countable Complement Extension Topology) is Hausdorff but is not regular. These are valuable examples especially the last two (24 and 63). The examples discussed below expand the offerings in Steen and Seebach.

The discussion of CCC but not separable in this post does not use axioms beyond the usual axioms of set theory (i.e. ZFC). The discussion here does not touch on Suslin lines or other examples that require extra set theory. The existence of Suslin lines is independent of ZFC. A Suslin line would produce an example of a perfectly normal first countable CCC non-separable space. In models of set theory where Suslin lines do not exist, a perfectly normal first countable CCC non-separable space can also be produced using other set-theoretic assumptions. The examples discussed below are not as nice as the set-theoretic examples since they usually are not first countable and perfectly normal.

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The countable chain conditon

A topological space $X$ is said to have the countable chain condition (to have the CCC for short) if $\mathcal{U}$ is a disjoint collection of non-empty open subsets of $X$ (meaning that for any $A,B \in \mathcal{U}$ with $A \ne B$, we have $A \cap B=\varnothing$), then $\mathcal{U}$ is countable. In other words, in a space with the CCC, there cannot be uncountably many pairwise disjoint non-empty open sets. For ease of discussion, if $X$ has the CCC, we also say that $X$ is a CCC space or X is CCC. A space $X$ is separable if there exists a countable subset $A$ of $X$ such that $A$ is dense in $X$ (meaning that if $U$ is a nonempty open subset of $X$, then $U \cap A \ne \varnothing$).

It is clear that any separable space has the CCC. In metric spaces, these two properties are equivalent. Among topological spaces in general, the two properties are not identical. Thus “CCC but not separable” is one way to distinguish between metrizable spaces and non-metrizable spaces. Even in non-metrizable spaces, “CCC but not separable” is also a way to obtain more information about the spaces being investigated.

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The strategy

Here’s the strategy for finding CCC and not separable.

The strategy is to narrow the focus to spaces where “CCC and not separable” is likely to exist. Specifically, look for a space or a class of spaces such that each space in the class has the countable chain condition but is not hereditarily separable. If the non-separable subspace is also a dense subspace of the starting space, it would be “CCC and not separable.”

Any dense subspace of a CCC space always has the CCC. Thus the search focuses on the subspaces in a CCC space that are reliably CCC. The strategy is to find non-separable spaces among these dense subspaces. The search is given an assist if the space or class of spaces in question has a characteristic that delineate the “separable” from the CCC (see Example 3 and Example 4 below).

In the following sections, we illustrate four different ways to apply the strategy.

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Example 1

The first way is a standard example found in the literature. The space to start from is the product space of separable spaces, which is always CCC. By a theorem of Ross and Stone, the product of more than continuum many separable spaces is not separable. Thus one way to get an example of CCC but not separable space is to take the product of more than continuum many separable spaces. For example, if $c$ is the cardinality of continuum, then consider $\left\{0,1 \right\}^{2^c}$, the product of $2^c$ many copies of $\left\{0,1 \right\}$, or consider $X^{2^c}$ where $X$ is your favorite separable space.

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Example 2

The second implementation of the strategy is also from taking the product of separable spaces. This time the number of factors does not have to be more than continuum. Here, we focus on one particular dense subspace of the product space, the $\Sigma$-products. To make this clear, let’s focus on a specific example. Consider $X=\left\{0,1 \right\}^{c}$ where $c$ is the cardinality of continuum. Consider the following subspace.

$\Sigma(\left\{0,1 \right\}^{c})= \left\{x \in X: x(\alpha) \ne 0 \text{ for at most countably many } \alpha < c \right\}$

The subspace $\Sigma(\left\{0,1 \right\}^{c})$ is dense in $X$, thus has CCC. It is straightforward to verify that $\Sigma(\left\{0,1 \right\}^{c})$ is not separable.

To implement this example, find any space $X$ which is a product space of separable spaces, each of which has at least two point (one of the points is labeled 0). The dense subspace is the $\Sigma$-product, which is the subspace consisting of all points that are non-zero at only countably many coordinates. The $\Sigma$-product has the countable chain condition since it is a dense subspace of the CCC space $X$. The $\Sigma$-product is not separable since there are uncountably many factors in the product space $X$ and that each factor has at least two points. This idea had been implemented in this previous post.

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Example 3

The third class of spaces is the class of Pixley-Roy spaces, which are hyperspaces. Given a space $X$, let $\mathcal{F}[X]$ be the set of all non-empty finite subsets of $X$. For $F \in \mathcal{F}[X]$ and for any open subset $U$ of $X$, let $[F,U]=\left\{B \in \mathcal{F}[X]: F \subset B \subset U \right\}$. The sets $[F,U]$ over all $F$ and $U$ form a base for a topology on $\mathcal{F}[X]$. This topology is called the Pixley-Roy topology (or Pixley-Roy hyperspace topology). The set $\mathcal{F}[X]$ with this topology is called a Pixley-Roy space.

The Pixley-Roy hyperspaces are discussed in this previous post. Whenever the ground space $X$ is uncountable, $\mathcal{F}[X]$ is not a separable space. We need to identify the $\mathcal{F}[X]$ that are CCC. According to the previous post on Pixley-Roy hyperspaces, for any space $X$ with a countable network, $\mathcal{F}[X]$ is CCC. Thus for any uncountable space $X$ with a countable network, the Pixley-Roy space $\mathcal{F}[X]$ is a CCC space that is not separable. The following gives a few such examples.

$\mathcal{F}[\mathbb{R}]$

$\mathcal{F}[X]$ where $X$ is any uncountable, separable and metrizable space.

$\mathcal{F}[X]$ where $X$ is uncountable and is the continuous image of a separable metrizable space.

Spaces with countable networks are discussed in this previous post. An example of a space $X$ that is the continuous image of a separable metrizable space is the bow-tie space found this previous post. Another example is any quote space of a separable metrizable space.

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Example 4

For the fourth implementation of the strategy, we go back to the product space of separable spaces in Example 2, with the exception that the focus is on the product of the real line $\mathbb{R}$. Let $X$ be any uncountable completely regular space. The product space $\mathbb{R}^X$ always has the CCC since it is a product of separable space. Now we single out a dense subspace of $\mathbb{R}^X$ for which there is a characterization for separability, namely the subspace $C(X)$, which is the set of all continuous functions from $X$ into $\mathbb{R}$. The subspace $C(X)$ as a topological space is usually denoted by $C_p(X)$. For a basic discussion of $C_p(X)$, see this previous post.

We know precisely when $C_p(X)$ is separable. The following theorem captures the idea.

Theorem 1 – Theorem I.1.3 [1]
The function space $C_p(X)$ is separable if and only if the domain space $X$ has a weaker (or coarser) separable metric topology (in other words, $X$ is submetrizable with a separable metric topology).

Based on the theorem, $C_p(X)$ is separable for any separable metric space $X$. Other examples of separable $C_p(X)$ include spaces $X$ that are created by tweaking the usual Euclidean topology on the real line and at the same time retaining the usual real line topology as a weaker topology, e.g. the Sorgenfrey line and the Michael line. Thus $C_p(X)$ would be separable if $X$ is a space such as the Sorgenfrey line or the Michael line. For our purpose at hand, we need to look for spaces that are not like the Sorgenfrey line or the Michael line. Here’s some examples of spaces $X$ that have no weaker separable metric topology.

• Any compact space $X$ that is not metrizable.
• The space $X=\omega_1$, the first uncountable ordinal with the order topology.
• Any space $X=C_p(Y)$ where $Y$ is not separable.

The function space $C_p(X)$ for any one of the above three spaces has the CCC but is not separable. It is well known that any compact space with a weaker metrizable topology is metrizable. Some examples for compact $X$ are: the first uncountable successor ordinal $\omega_1+1$, the double arrow space, and the product space $\left\{0,1 \right\}^{\omega_1}$.

It can be shown that $C_p(\omega_1)$ is not separable (see this previous post). The last example is due to the following theorem.

Theorem 2 – Theorem I.1.4 [1]
The function space $C_p(Y)$ has a weaker (or coarser) separable metric topology if and only if the domain space $Y$ is separable.

Thus picking a non-separable space $Y$ would guarantee that $C_p(Y)$ has a weaker separable metric topology. As a result, $C_p(C_p(Y))$ is a CCC and not separable space.

Interestingly, Theorem 1 and Theorem 2 show a duality existing between the property of having a weaker separable metric topology and the property of being separable. The two theorems allow us to switch the two properties between the domain space and the function space.

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Remarks

Another interesting point to make is that Theorem 1 and Theorem 2 together allow us to “buy one get one free.” Once we obtain a space $X$ that is CCC and not separable from any one of the avenues discussed here, the function space $C_p(X)$ has no weaker separable metric topology (by Theorem 2) and the function space $C_p(C_p(X))$ is another example of CCC and not separable.

The strategy discussed above unifies all four examples. Undoubtedly there will be other examples that can come from the strategy. To find more examples, find a space or a class of spaces that are reliably CCC and then look for potential non-separable spaces among the dense subspaces of the starting space.

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Exercises

1. Show that in metrizable spaces, CCC and separable are equivalent. The only part that needs to be shown is that if $X$ is metrizable and CCC, then $X$ is separable.
2. Show that any dense subspace of a CCC space is also CCC.
3. Verify that the space $\Sigma(\left\{0,1 \right\}^{c})$ defined in Example 2 is dense in $X$ and is not separable.
4. Verify that the Pixley-Roy space $\mathcal{F}[\mathbb{R}]$ defined in Example 3 is CCC and not separable.
5. Verify that function space $C_p(\omega_1)$ mentioned in Example 4 is not separable. Hint: use the pressing down lemma.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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$\copyright \ 2016 \text{ by Dan Ma}$

# Product Space – Exercise Set 1

This post presents several exercises concerning product spaces. All the concepts involved in the exercises have been discussed in the blog.

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Exercise 1

Exercise 1a
Prove or disprove:
If $X$ and $Y$ are both hereditarily separable, then $X \times Y$ is hereditarily separable.

Exercise 1b
Show that if each $X_\alpha$ is separable, then the product space $\prod_{\alpha < \omega} \ X_\alpha$ is separable.

Exercise 1c
Prove or disprove:
If each $X_\alpha$ is separable, then the product space $\prod_{\alpha < \omega_1} \ X_\alpha$ is not separable.

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Exercise 2

Exercise 2a
Show that if the space $X$ is normal, then every closed subspace of $X$ is a normal space.

Exercise 2b
Prove or disprove:
If the space $X$ is normal, then every dense open subspace of $X$ is a normal space.

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Exercise 3

Consider the product space $\prod_{\alpha \in W} \ X_\alpha$.

Exercise 3a
Suppose that $X_\alpha$ is compact for all but one $\alpha \in W$ such that the non-compact factor is a Lindelof space. Show that the product space $\prod_{\alpha \in W} \ X_\alpha$ is a normal space.

Exercise 3b
Prove or disprove:
Suppose that $X_\alpha$ is compact for all but one $\alpha \in W$ such that the non-compact factor is a normal space. Then the product space $\prod_{\alpha \in W} \ X_\alpha$ is a normal space.

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Exercise 4

Exercise 4a
Let $X$ be a compact space.
Show that if $X^n$ is hereditarily Lindelof for all positive integer $n$, then $X$ is metrizable.

Exercise 4b
Prove or disprove:
If $X^n$ is hereditarily Lindelof for all positive integer $n$, then $X$ is metrizable.

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Exercise 5

Let $Y$ the product of uncountably many copies of the real line $\mathbb{R}$. If a specific example is desired, try $Y=\mathbb{R}^{\omega_1}$ ($\omega_1$ many copies of $\mathbb{R}$) or $Y=\mathbb{R}^{\mathbb{R}}$ (continuum many copies of $\mathbb{R}$). It is also OK to use a larger number of copies of the real line.

Note that the space $Y$ is not normal (see here).

Exercise 5a
Since the product space $Y$ is not normal, it is not Lindelof. As an exercise, find an open cover of $Y$ that proves that $Y$ is not Lindelof, i.e. an open cover $\mathcal{U}$ of $Y$ such that no countable subcollection of $\mathcal{U}$ can cover $Y$.

Exercise 5b
Show that for every open cover $\mathcal{U}$ of the space $Y$, there is a countable $\mathcal{V} \subset \mathcal{U}$ of $Y$ such that $\overline{\mathcal{V}}=Y$, i.e. $\cup \mathcal{V}$ is dense in $Y$. Note that with this property, the space $Y$ is said to be weakly Lindelof.

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Exercise 6

This exercise is about the product $Y=\mathbb{R}^{\mathbb{R}}$ (continuum many copies of $\mathbb{R}$). Show the following.

1. Show that $Y$ is separable by exhibiting a countable dense set.
2. Show that $Y$ is not hereditarily separable by exhibiting a non-separable subspace.
3. Show that the space $Y$ has a closed and discrete subspace of cardinality continuum.
4. Show that $Y$ is not first countable.
5. Show that $Y$ is not a Frechet space.
6. Show that $Y$ is not a countably tight space.

See here for the definition of Frechet space.

See here for the definition of countably tight space.

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Exercise 7

Consider the product space $Y=\mathbb{\omega}^{\omega_1}$. It is not normal (see here).

Exercise 7a
Construct a dense normal subspace of $Y$.

Exercise 7b
Construct a dense Lindelof subspace of $Y$.

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$\copyright \ 2016 \text{ by Dan Ma}$

# Counterexample 106 from Steen and Seebach

As the title suggests, this post discusses counterexample 106 in Steen and Seebach [2]. We extend the discussion by adding two facts not found in [2].

The counterexample 106 is the space $X=\omega_1 \times I^I$, which is the product of $\omega_1$ with the interval topology and the product space $I^I=\prod_{t \in I} I$ where $I$ is of course the unit interval $[0,1]$. The notation of $\omega_1$, the first uncountable ordinal, in Steen and Seebach is $[0,\Omega)$.

Another way to notate the example $X$ is the product space $\prod_{t \in I} X_t$ where $X_0$ is $\omega_1$ and $X_t$ is the unit interval $I$ for all $t>0$. Thus in this product space, all factors except for one factor is the unit interval and the lone non-compact factor is the first uncountable ordinal. The factor of $\omega_1$ makes this product space an interesting example.

The following lists out the basic topological properties of the space that $X=\omega_1 \times I^I$ are covered in [2].

• The space $X$ is Hausdorff and completely regular.
• The space $X$ is countably compact.
• The space $X$ is neither compact nor sequentially compact.
• The space $X$ is neither separable, Lindelof nor $\sigma$-compact.
• The space $X$ is not first countable.
• The space $X$ is locally compact.

All the above bullet points are discussed in Steen and Seebach. In this post we add the following two facts.

• The space $X$ is not normal.
• The space $X$ is a dense subspace that is normal.

It follows from these bullet points that the space $X$ is an example of a completely regular space that is not normal. Not being a normal space, $X$ is then not metrizable. Of course there are other ways to show that $X$ is not metrizable. One is that neither of the two factors $\omega_1$ or $I^I$ is metrizable. Another is that $X$ is not first countable.

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The space $X$ is not normal

Now we are ready to discuss the non-normality of the example. It is a natural question to ask whether the example $X=\omega_1 \times I^I$ is normal. The fact that it was not discussed in [2] could be that the tool for answering the normality question was not yet available at the time [2] was originally published, though we do not know for sure. It turns out that the tool became available in the paper [1] published a few years after the publication of [2]. The key to showing the normality (or the lack of) in the example $X=\omega_1 \times I^I$ is to show whether the second factor $I^I$ is a countably tight space.

The main result in [1] is discussed in this previous post. Theorem 1 in the previous post states that for any compact space $Y$, the product $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight. Thus the normality of the space $X$ (or the lack of) hinges on whether the compact factor $I^I=\prod_{t \in I} I$ is countably tight.

A space $Y$ is countably tight (or has countable tightness) if for each $S \subset Y$ and for each $x \in \overline{S}$, there exists some countable $B \subset S$ such that $x \in \overline{B}$. The definitions of tightness in general and countable tightness in particular are discussed here.

To show that the product space $I^I=\prod_{t \in I} I$ is not countably tight, we let $S$ be the subspace of $I^I$ consisting of points, each of which is non-zero on at most countably many coordinates. Specifically $S$ is defined as follows:

$S=\Sigma_{t \in I} I=\left\{y \in I^I: y(t) \ne 0 \text{ for at most countably many } t \in I \right\}$

The set $S$ just defined is also called the $\Sigma$-product of copies of unit interval $I$. Let $g \in I^I$ be defined by $g(t)=1$ for all $t \in I$. It follows that $g \in \overline{S}$. It can also be verified that $g \notin \overline{B}$ for any countable $B \subset S$. This shows that the product space $I^I=\prod_{t \in I} I$ is not countably tight.

By Theorem 1 found in this link, the space $X=\omega_1 \times I^I$ is not normal.

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The space $X$ has a dense subspace that is normal

Now that we know $X=\omega_1 \times I^I$ is not normal, a natural question is whether it has a dense subspace that is normal. Consider the subspace $\omega_1 \times S$ where $S$ is the $\Sigma$-product $S=\Sigma_{t \in I} I$ defined in the preceding section. The subspace $S$ is dense in the product space $I^I$. Thus $\omega_1 \times S$ is dense in $X=\omega_1 \times I^I$. The space $S$ is normal since the $\Sigma$-product of separable metric spaces is normal. Furthermore, $\omega_1$ can be embedded as a closed subspace of $S=\Sigma_{t \in I} I$. Then $\omega_1 \times S$ is homeomorphic to a closed subspace of $S \times S$. Since $S \times S \cong S$, the space $\omega_1 \times S$ is normal.

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Reference

1. Nogura, T., Tightness of compact Hausdorff space and normality of product spaces, J. Math. Soc. Japan, 28, 360-362, 1976
2. Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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

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

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

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Pseudocompactness

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

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

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

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

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The exercise

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

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

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

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

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

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

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Reference

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

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

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

# Products of compact spaces with countable tightness

In the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.

The following theorems are the main results in this post.

Theorem 1
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is compact, then $X \times Y$ is countably tight.

Theorem 2
The product of finitely many compact countably tight spaces is countably tight.

Theorem 3
Suppose that $X_1, X_2, X_3, \cdots$ are countably many compact spaces such that each $X_i$ has at least two points. If each $X_i$ is a countably tight space, then the product space $\prod_{i=1}^\infty X_i$ is countably tight.

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Finite products

Before proving Theorem 1 and Theorem 2, we prove the following results.

Theorem 4
Let $f:Y_1 \rightarrow Y_2$ be a continuous and closed map from the space $Y_1$ onto the space $Y_2$. Suppose that the space $Y_2$ is countably tight and that each fiber of the map $f$ is countably tight. Then the space $Y_1$ is countably tight.

Proof of Theorem 4
Let $x \in Y_1$ and $x \in \overline{A}$ where $A \subset Y_1$. We proceed to find a countable $W \subset Y_1$ such that $x \in \overline{W}$. Choose $y \in Y_2$ such that $y=f(x)$.

Let $M$ be the fiber of the map $f$ at the point $y$, i.e. $M=f^{-1}(y)$. By assumption, $M$ is countably tight. Call a point $w \in M$ countably reached by $A$ if there is some countable $C \subset A$ such that $w \in \overline{C}$. Let $G$ be the set of all points in $M$ that are countably reached by $A$. We claim that $x \in \overline{G}$.

Let $U \subset Y_1$ be open such that $x \in U$. Because the space $Y_1$ is regular, choose open $V \subset U$ such that $x \in V$ and $\overline{V} \subset U$. Then $V \cap A \ne \varnothing$. Furthermore, $x \in \overline{V \cap A}$. Let $C=f(V \cap A)$. By the continuity of $f$, we have $y \in \overline{C}$. Since $Y_2$ is countably tight, there exists some countable $D \subset C$ such that $y \in \overline{D}$. Choose a countable $E \subset V \cap A$ such that $f(E)=D$. It follows that $y \in \overline{f(E)}$.

We show that that $\overline{E} \cap M \ne \varnothing$. Since $E \subset \overline{E}$, we have $f(E) \subset f(\overline{E})$. Note that $f(\overline{E})$ is a closed set since $f$ is a closed map. Thus $\overline{f(E)} \subset f(\overline{E})$. As a result, $y \in f(\overline{E})$. Then $y=f(t)$ for some $t \in \overline{E}$. We have $t \in \overline{E} \cap M$.

By the definition of the set $G$, we have $\overline{E} \cap M \subset G$. Furthermore, $\overline{E} \cap M \subset \overline{V} \subset U$. Note that the arbitrary open neighborhood $U$ of $x$ contains points of $G$. This establishes the claim that $x \in \overline{G}$.

Since $M$ is a fiber of $f$, $M$ is countably tight by assumption. Choose some countable $T \subset G$ such that $x \in \overline{T}$. For each $t \in T$, choose a countable $W_t \subset A$ with $t \in \overline{W_t}$. Let $W=\bigcup_{t \in T} W_t$. Note that $W \subset A$ and $W$ is countable with $x \in \overline{W}$. This establishes the space $Y_1$ is countably tight at $x \in Y_1$. $\blacksquare$

Lemma 5
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a compact space, then $f$ is a closed map.

Proof of Lemma 5
Let $A$ be a closed subset of $X \times Y$. Suppose that $f(A)$ is not closed. Let $y \in \overline{f(A)}-f(A)$. It follows that no point of $X \times \left\{y \right\}$ belongs to $A$. For each $x \in X$, choose open subset $O_x$ of $X \times Y$ such that $(x,y) \in O_x$ and $O_x \cap A=\varnothing$. The set of all $O_x$ is an open cover of the compact space $X \times \left\{y \right\}$. Then there exist finitely many $O_x$ that cover $X \times \left\{y \right\}$, say $O_{x_i}$ for $i=1,2,\cdots,n$.

Let $W=\bigcup_{i=1}^n O_{x_i}$. We have $X \times \left\{y \right\} \subset W$. Since $X$ is compact, we can then use the Tube Lemma which implies that there exists open $G \subset Y$ such that $X \times \left\{y \right\} \subset X \times G \subset W$. It follows that $G \cap f(A) \ne \varnothing$. Choose $t \in G \cap f(A)$. Then for some $x \in X$, $(x,t) \in A$. Since $t \in G$, $(x,t) \in W$, implying that $W \cap A \ne \varnothing$, a contradiction. Thus $f(A)$ must be a closed set in $Y$. This completes the proof of the lemma. $\blacksquare$

Proof of Theorem 1
Let $X$ be the factor that is compact. let $f: X \times Y \rightarrow Y$ be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space $Y$ is countably tight by assumption. Each fiber of the projection map $f$ is of the form $X \times \left\{y \right\}$ where $y \in Y$, which is countably tight. Then use Theorem 4 to establish that $X \times Y$ is countably tight. $\blacksquare$

Proof of Theorem 2
This is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight. $\blacksquare$

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Countable products

Our proof to establish that the product space $\prod_{i=1}^\infty X_i$ is countably tight is an indirect one and makes use of two non-trivial results. We first show that $\omega_1 \times \prod_{i=1}^\infty X_i$ is a closed subspace of a $\Sigma$-product that is normal. It follows from another result that the second factor $\prod_{i=1}^\infty X_i$ is countably tight. We now present all the necessary definitions and theorems.

Consider a product space $Y=\prod_{\alpha<\kappa} Y_\alpha$ where $\kappa$ is an infinite cardinal number. Fix a point $p \in Y$. The $\Sigma$-product of the spaces $Y_\alpha$ with $p$ as the base point is the following subspace of the product space $Y=\prod_{\alpha<\kappa} Y_\alpha$:

$\displaystyle \Sigma_{\alpha<\kappa} Y_\alpha=\left\{y \in \prod_{\alpha<\kappa} Y_\alpha: y_\alpha \ne p_\alpha \text{ for at most countably many } \alpha < \kappa \right\}$

The definition of the space $\Sigma_{\alpha<\kappa} Y_\alpha$ depends on the base point $p$. The discussion here is on properties of $\Sigma_{\alpha<\kappa} Y_\alpha$ that hold regardless of the choice of base point. If the factor spaces are indexed by a set $A$, the notation is $\Sigma_{\alpha \in A} Y_\alpha$.

If all factors $Y_\alpha$ are identical, say $Y_\alpha=Z$ for all $\alpha$, then we use the notation $\Sigma_{\alpha<\kappa} Z$ to denote the $\Sigma$-product. Once useful fact is that if there are $\omega_1$ many factors and each factor has at least 2 points, then the space $\omega_1$ can be embedded as a closed subspace of the $\Sigma$-product.

Theorem 6
For each $\alpha<\omega_1$, let $Y_\alpha$ be a space with at least two points. Then $\Sigma_{\alpha<\omega_1} Y_\alpha$ contains $\omega_1$ as a closed subspace. See Exercise 3 in this previous post.

Now we discuss normality of $\Sigma$-products. This previous post shows that if each factor is a separable metric space, then the $\Sigma$-product is normal. It is also well known that if each factor is a metric space, the $\Sigma$-product is normal. The following theorem handles the case where each factor is a compact space.

Theorem 7
For each $\alpha<\kappa$, let $Y_\alpha$ be a compact space. Then the $\Sigma$-product $\Sigma_{\alpha<\kappa} Y_\alpha$ is normal if and only if each factor $Y_\alpha$ is countably tight.

Theorem 7 is Theorem 7.5 in page 821 of [1]. Theorem 7.5 in [1] is stated in a more general setting where each factor of the $\Sigma$-product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.

Theorem 8
Let $Y$ be a compact space. Then the product space $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight.

We now prove Theorem 3.

Proof of Theorem 3
Let $\omega_1=\cup \left\{A_n: n \in \omega \right\}$, where for each $n$, $\lvert A_n \lvert=\omega_1$ and that $A_n \cap A_m=\varnothing$ if $n \ne m$. For each $n=1,2,3,\cdots$, let $S_n=\Sigma_{\alpha \in A_n} X_n$. By Theorem 7, each $S_n$ is normal. Let $S_0=\Sigma_{\alpha \in A_0} X_1$, which is also normal. By Theorem 6, the space $\omega_1$ of countable ordinals is a closed subspace of $S_0$. Let $T=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots$. We have the following derivation.

\displaystyle \begin{aligned} T&=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots \\&\subset S_0 \times S_1 \times S_2 \times S_3 \times \cdots \\&\cong W=\Sigma_{\alpha<\omega_1} W_\alpha \end{aligned}

Recall that $\omega_1=\cup \left\{A_n: n \in \omega \right\}$. The space $W=\Sigma_{\alpha<\omega_1} W_\alpha$ is defined such that for each $n \ge 1$ and for each $\alpha \in A_n$, $W_\alpha=X_n$. Furthermore, for $n=0$, for each $\alpha \in A_0$, let $W_\alpha=X_1$. Thus $W$ is a $\Sigma$-product of compact countably tight spaces and is thus normal by Theorem 7. The space $T=\omega_1 \times \prod_{n=1}^\infty X_n$ is a closed subspace of the normal space $W$. By Theorem 8, the product space $\prod_{n=1}^\infty X_n$ must be countably tight. $\blacksquare$

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Remarks

Theorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let $X=X_1 \times X_2 \times \cdots \times X_n$.

\displaystyle \begin{aligned} X&=X_1 \times X_2 \times \cdots \times X_n \\&\cong X_1 \times X_2 \times \cdots \times X_n \times \left\{t_{n+1} \right\} \times \left\{t_{n+2} \right\} \times \cdots \\&\subset X_1 \times X_2 \times \cdots \times X_n \times X_{n+1} \times X_{n+2} \times \cdots \end{aligned}

In the above derivation, $t_m$ is a point of $X_m$ for all $m >n$. When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.

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Exercise

Exercise 1
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a countably compact space and $Y$ is a Frechet space, then $f$ is a closed map.

Exercise 2
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is a countably compact space and the other space is a Frechet space, then $X \times Y$ is countably tight.

Exercise 2 is a variation of Theorem 1. One factor is weakened to “countably compact”. However, the other factor is strengthened to “Frechet”.

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Reference

1. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.

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