Sequential fan and the dominating number

Posted on March 11, 2024 by Dan Ma

The sequential fan $S(\kappa)$ is the quotient space obtained by identifying the limit points of a topological sum of $\kappa$ many convergent sequences. We focus on $S(\omega)$ , the sequential fan derived from countably and infinitely many convergent sequences. Because only countably many convergent sequences are used, $S(\omega)$ is intimately connected to the combinatorics in $\omega^\omega$ , which is the family of all functions from $\omega$ into $\omega$ . In particular, we show that the character at the limit point $\infty$ in $S(\omega)$ equals to the dominating number $\mathfrak{d}$ . The dominating number $\mathfrak{d}$ and the bounding number $\mathfrak{b}$ , introduced below, are quite sensitive to set theoretic assumptions. As a result, pinpointing the precise cardinality of the character of the point $\infty$ in the sequential fan $S(\omega)$ requires set theory beyond ZFC. The fact that the character at $\infty$ in the sequential fan $S(\omega)$ is identical to the dominating number $\mathfrak{d}$ is mentioned in page 13 in chapter a-3 of the Encyclopedia of General Topology [1].

Sequential fans had been discussed previously (see here). See here, here, here, here, and here for previous discussion on the bounding number and the dominating number.

Sequential Fans

As mentioned above, a sequential fan is the quotient space on a disjoint union of convergent sequences with all the limit points of the sequences collapsed to one point called $\infty$ . We first give a working definition. To further provide intuition, we also show that our sequential fan of interest $S(\omega)$ is the quotient space of a subspace of the Euclidean plane (i.e., the countably many convergent sequences can be situated in the plane).

In the discussion that follows, $\omega$ is the set of all non-negative integers. The set $\omega^\omega$ is the family of all functions from $\omega$ into $\omega$ . Let $\kappa$ be an infinite cardinal number. The sequential fan $S(\kappa)$ with $\kappa$ many spines is the set $S(\kappa)=\{ \infty \} \cup (\kappa \times \omega)$ with the topology defined as follows:

Every point in $\kappa \times \omega$ is made an isolated point.
An open neighborhood of the point $\infty$ is of the following form:

$B_f=\{ \infty \} \cup \{ (\alpha,n) \in \kappa \times \omega: n \ge f(\alpha) \}$

$f \in \omega^\omega$

In this formulation of the sequential fan, the set $\{(\alpha, n): n \in \omega \}$ , where $\alpha < \kappa$ , is a sequence converging to $\infty$ . For each such convergent sequence, the open neighborhood $B_f$ contains all but finitely many points.

Our focus is $S(\omega)$ , where $S(\omega)=\{ \infty \} \cup (\omega \times \omega)$ .

A View From the Euclidean Plane

The formulation of the sequential fan $S(\kappa)$ given above is a good working formulation. We now describe how $S(\omega)$ can be derived from the Euclidean plane. Consider the following diagram.

$\displaystyle \begin{aligned} & \\& \text{ } & \text{ } & S_1 & \text{ } & S_2 & \text{ } & S_3 & \text{ } & S_4 & \text{ } & S_5 & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ }& \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \downarrow &\text{ } & \downarrow & \text{ } & \downarrow &\text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & p_1 &\text{ } & p_2 & \text{ } & p_3 &\text{ } & p_4 & \text{ } & p_5 & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \end{aligned}$

In the above diagram, the thick bullets are the points in the space and the tiny dots represent “dot dot dot”, indicating that the points or the sequences are continuing countably infinitely. The arrows in the diagrams point to the direction to which the points are converging. There are countably and infinitely many convergent sequences, named $S_1,S_2,S_3,\cdots$ , with $p_n$ being the limit of the sequence $S_n$ . For convenience, we can let $p_n$ be the point $(n,0)$ in the plane and $S_n$ be a sequence converging downward to $p_n$ . Let $S=S_1 \cup S_2 \cup S_3 \cup \cdots$ and let $P=\{p_1,p_2,p_3,\cdots \}$ . Consider the space $X=S \cup P$ with the topology inherited from the Euclidean plane. Any point in any one of the convergent sequence $S_n$ is an isolated point. An open neighborhood of the limit point $p_n$ consists of $p_n$ and all but finitely many points in $S_n$ .

The diagram and the preceding paragraph set up the scene. We are now ready to collapse points (or define the quotient map). We collapse the set of all limit points $P$ to one point called $\infty$ . The resulting quotient space is $Y=S \cup \{ \infty \}$ . In this quotient space, $S$ is the set of all points in the countably many convergent sequences with each point isolated. An open neighborhood at $\infty$ consists of $\infty$ and all but finitely many points in each convergent sequence. This formulation is clearly equivalent to the sequential fan $S(\omega)$ formulated earlier.

When $\kappa$ is uncountable, the topological sum of $\kappa$ many convergent sequences can no longer viewed in a Euclidean space. However, the topological sum is still a metric space (just not a separable one). We can still collapse the limit points into one point called $\infty$ . The resulting quotient space is identical to $S(\kappa)$ formulated above.

The Combinatorics on the Integers

We begin the combinatorics by defining the order $\le^*$ . Recall that $\omega^\omega$ is the family of all functions from $\omega$ into $\omega$ . For $f,g \in \omega^\omega$ , declare $f \le^* g$ if $f(n) \le g(n)$ for all but finitely many $n \in \omega$ . We write $f \not \le^* g$ if the negation of $f \le^* g$ is true, i.e., $f(n)>g(n)$ for infinitely many $n \in \omega$ . The order $\le^*$ is a reflexive and transitive relation.

A set $F \subset \omega^\omega$ is said to be bounded if $F$ has an upper bound according to the order $\le^*$ , i.e., there exists $g \in \omega^\omega$ such that $f \le^* g$ for all $f \in F$ (Here, $g$ is the upper bound of $F$ ). The set $F$ is said to be unbounded if it is not bounded according to $\le^*$ . That is, $F$ is unbounded if for each $g \in \omega^\omega$ , there exists $f \in F$ such that $f \not \le^* g$ . A set $F \subset \omega^\omega$ is said to be a dominating set if $F$ is cofinal in $\le^*$ , i.e., for each $f \in \omega^\omega$ , there exists $g \in F$ such that $f \le^* g$ . We now define two cardinal numbers as follows:

$\mathfrak{b}=\text{min} \{ \lvert F \lvert: F \subset \omega^\omega \text{ is unbounded} \}$

$\mathfrak{d}=\text{min} \{ \lvert F \lvert: F \subset \omega^\omega \text{ is dominating} \}$

The first number $\mathfrak{b}$ is called the bounding number and the second one $\mathfrak{d}$ is called the dominating number. Both are upper bounded by the continuum $\mathfrak{c}$ , i.e., $\mathfrak{b} \le \mathfrak{c}$ and $\mathfrak{d} \le \mathfrak{c}$ . Using a diagonal argument, we can show that both of these cardinal numbers are not countable. Thus, we have $\omega_1 \le \mathfrak{b},\mathfrak{d} \le \mathfrak{c}$ . How do $\mathfrak{b}$ and $\mathfrak{d}$ relate? We have $\mathfrak{b} \le \mathfrak{d}$ since every dominating set is also an unbounded set.

The Character at Infinity

The sequential fan $S(\omega)$ is not first countable at the point $\infty$ . In other word, there does not exist a countable local base at $\infty$ . To see this, let $\{ B_{f_1},B_{f_2},B_{f_3},\cdots \}$ be a countable collection of open neighborhoods of $\infty$ . Using a diagonal argument, we can find $f \in \omega^\omega$ such that $B_{f_n} \not \subset B_f$ for all $n$ . This shows that no countable collection of open neighborhoods can be a base at $\infty$ . Thus, the character at $\infty$ must be uncountable (the character at a point is the minimum cardinality of a local base at the point). Thus, we have have $\chi(S(\omega),\infty)>\omega$ . Furthermore, we have $\omega_1 \le \chi(S(\omega),\infty) \le \mathfrak{c}$ (character is at least $\omega_1$ but no more than continuum). The range from $\omega_1$ to continuum $\mathfrak{c}$ is a narrow range if continuum hypothesis holds, but can be a large range if continuum hypothesis does not hold. Can we pinpoint the character at $\infty$ more narrowly and more precisely?

Connecting the Dominating Number to the Sequential Fan

We claim the for the sequential fan $S(\omega)$ , the character at the point $\infty$ is the dominating number $\mathfrak{d}$ introduced above. To establish this claim, we set up a different formulation of dominating set. A set $F \subset \omega^\omega$ is said to be a special dominating set if for each $f \in \omega^\omega$ , there exists $g \in F$ such that $f(n) \le g(n)$ for all $n \in \omega$ . We define the cardinal number $\mathfrak{d}_1$ as follows:

$\mathfrak{d}_1=\text{min} \{ \lvert F \lvert: F \subset \omega^\omega \text{ is a special dominating set} \}$

Note that the term “special dominating” is not a standard term. It is simply a definition that facilitates the argument at hand. One key observation is that when $F$ is a special dominating set, the collection $\{B_f: f \in F \}$ becomes a base at the point $\infty$ . Since the cardinal number $\mathfrak{d}_1$ is the minimum cardinality of a base at $\infty$ , we only need to show that $\mathfrak{d}=\mathfrak{d}_1$ . Since every special dominating set is a dominating set, we have $\mathfrak{d} \le$ the cardinality of every special dominating set. Thus, $\mathfrak{d} \le \mathfrak{d}_1$ .

Next we show $\mathfrak{d}_1 \le \mathfrak{d}$ . To this end, we show that $\mathfrak{d}_1 \le$ the cardinality of every dominating set. We claim that for every dominating set $F$ , there exists a special dominating set $F_*$ with $\lvert F_* \lvert=\lvert F \lvert$ . Once this is established, we have $\mathfrak{d}_1 \le$ the cardinality of every dominating set and thus $\mathfrak{d}_1 \le \mathfrak{d}$ .

Let $F$ be a dominating set. For each $n \in \omega$ with $n \ge 1$ , define the following:

$D_n=\{0,1,\cdots,n-1 \}$
$E_n=\{ n,n+1,n+2,\cdots \}$
$A_n=\omega^{D_n}$
$B_n=\omega^{E_n}$

If $h \in A_n$ and $k \in B_n$ , then we take $h \cup k$ to be a function in $\omega^\omega$ . For each $n \ge 1$ and for each $f \in F$ , define the following:

$F_{f,n}=\{h \cup (f \upharpoonright E_n): h \in A_n \}$

with $f \upharpoonright E_n$ representing the function $f$ restricted to the set $E_n$ . Let $F_*=\bigcup \{F_{f,n}: n \ge 1, f \in F \}$ . Note that each $F_{f,n}$ is countable. As a result, $\lvert F_* \lvert=\lvert F \lvert$ . Because $F$ is a dominating set, $F_*$ is a special dominating set. We have just established that $\mathfrak{d}_1 = \mathfrak{d}$ and that the character of the point $\infty$ in the sequential fan $S(\omega)$ is the dominating number $\mathfrak{d}$ .

Remarks

Can we pinpoint the character at $\infty$ ? The answer is a partial yes. We establish that $\chi(S(\omega),\infty)=\mathfrak{d}$ . However, the dominating number and the bounding number as well as other small cardinals are very sensitive to set theory. For example, when continuum hypothesis (CH) holds, The dominating number $\mathfrak{d}$ is continuum. Thus, it is consistent with ZFC that $\chi(S(\omega),\infty)$ is continuum. It is also consistent with ZFC that $\omega_1 \le \mathfrak{b} <\mathfrak{d}<\mathfrak{c}$ . Thus it is consistent that $\chi(S(\omega),\infty)$ is greater than $\omega_1$ and less than continuum. Though the dominating number tells us how big the character at $\infty$ is, we cannot pinpoint precisely where the character is in the range between $\omega_1$ and continuum. For more information about dominating number and other small cardinals, see chapter 3 in the Handbook of Set-Theoretic Topology [2].

The fact that the character at $\infty$ in the sequential fan $S(\omega)$ is identical to the dominating number $\mathfrak{d}$ is mentioned in page 13 in chapter a-3 of the Encyclopedia of General Topology [1].

The sequential fan $S(\omega)$ is a space that has a simple definition. After all, the starting point is a subspace of the Euclidean plane with $S(\omega)$ obtained by collapsing the limit points. Though the space is very accessible, the size of the character at the limit point $\infty$ is unknowable if we work only in ZFC. It is a short “distance” from the definition of the sequential fan $S(\omega)$ to the set-theoretic unknowable statement. This makes the sequential fan $S(\omega)$ an interesting example and an excellent entry point of learning more set-theoretic topology.

Reference

Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.
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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Dan Ma Sequential Fan
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$\copyright$ 2024 – Dan Ma

Posted in Convergence Property, Well known examples | Tagged Quotient Map, Quotient space, Sequential fan, Sequential space | Leave a reply

The product of the identity map and a quotient map

Posted on April 21, 2023 by Dan Ma

1

The Cartesian product of the identity map and a quotient map can be a quotient map under one circumstance. We prove the following theorem.

Theorem 1
Let $X$ be a locally compact space. Let $q:Y \rightarrow Z$ be a quotient map. Let the map $f:X \times Y \rightarrow X \times Z$ be defined by $f(x,y)=(x,q(y))$ for each $(x,y) \in X \times Y$ . Then the map $f$ is a quotient map from $X \times Y$ to $X \times Z$ .

This is Theorem 3.3.17 in the Engelking topology text [1]. The theorem is attributed to J. H. C. Whitehead. The mapping $f$ defined in Theorem 1 is the Cartesian product of the identity map from $X$ to $X$ and the quotient map from $Y$ onto $Z$ . The theorem gives one circumstance in which the Cartesian product is also a quotient map. That is, taking the product of the identity map from a locally compact space to itself and a quotient map produces a quotient map. Potentially this gives us information about the product of the locally compact space in question and the space that is the quotient image. We give two natural applications of this theorem. Sequential spaces are precisely spaces that are quotient images of metric spaces (see here). The spaces called k-spaces are precisely the quotient images of locally compact spaces (see here). As corollary of Theorem 1, we show that the product of a locally compact metric space and a sequential space is a sequential space. In another corollary, we show that the product of a locally compact space and a k-space is a k-space. We have the following corollaries.

Corollary 2
Let $X$ be a locally compact metric space. Let $Y$ be a sequential space. Then $X \times Y$ is a sequential space.

Corollary 3
Let $X$ be a compact metric space. Let $Y$ be a sequential space. Then $X \times Y$ is a sequential space.

Corollary 4
Let $X$ be a locally compact space. Let $Y$ be a k-space. Then $X \times Y$ is a k-space.

We give a proof of Theorem 1 and discuss the corollaries. We also give some examples.

Proof of Theorem 1

Let $X$ , $Y$ and $Z$ be the spaces described in the statement of Theorem 1, and let $q$ and $f$ be the mappings described in Theorem 1. To show that the map $f$ is a quotient map, we need to show that for any set $O \subset X \times Z$ , $O$ is an open set in $X \times Z$ if and only if $f^{-1}(O)$ is an open set in $X \times Y$ . Because the mapping $f$ is continuous, if $O$ is open in $X \times Z$ , we know that $f^{-1}(O)$ is an open set in $X \times Y$ . We only need to prove the other direction: if $f^{-1}(O)$ is an open set in $X \times Y$ , then $O$ is an open set in $X \times Z$ . To this end, let $f^{-1}(O)$ be an open set in $X \times Y$ and $(a,b) \in O$ . We proceed to find some open set $U \times V \subset X \times Z$ such that $(a,b) \in U \times V \subset O$ .

We choose $c \in q^{-1}(b)$ and an open set $U \subset X$ with $a \in U$ such that $\overline{U}$ is compact and $\overline{U} \times \{ c \} \subset f^{-1}(O)$ . We make the following observation,

(1) for any $y \in Y$ , $\overline{U} \times \{ y \} \subset f^{-1}(O)$ if and only if $\overline{U} \times q^{-1} q(y) \subset f^{-1}(O)$

Let $V=\{ z \in Z: \overline{U} \times q^{-1}(z) \subset f^{-1}(O) \}$ . By observation (1), we have $\overline{U} \times q^{-1}q(c) \subset f^{-1}(O)$ . Note that $q(c)=b$ . Thus, $b \in V$ . As a result, we have $(a,b) \in U \times V \subset O$ . We now need to show $V$ is an open subset of $Z$ . Since $q$ is a quotient mapping, we know $V$ is open in $Z$ if we can show $q^{-1}(V)$ is open in $Y$ . The set $q^{-1}(V)$ is described as follows:

$\displaystyle \begin{aligned} q^{-1}(V)&=\{ y \in Y: q(y) \in V \} \\&=\{y \in Y: \overline{U} \times q^{-1}q(y) \subset f^{-1}(O) \} \\&=\{y \in Y: \overline{U} \times \{ y \} \subset f^{-1}(O) \} \end{aligned}$

The last equality is due to Observation (1). Let $\pi: \overline{U} \times Y \rightarrow Y$ be the projection map. Since $\overline{U}$ is compact, the projection map $\pi$ is a closed map according to the Kuratowski Theorem (see here for its proof). Since $(\overline{U} \times Y) \backslash f^{-1}(O)$ is closed in $\overline{U} \times Y$ , $C=\pi(\overline{U} \times Y \backslash f^{-1}(O))$ is closed in $Y$ and $Y \backslash C$ is open in $Y$ . It can be verified that $q^{-1}(V)=Y \backslash C$ . Thus, $q^{-1}(V)$ is open in $Y$ . As a result, $V$ is open in $Y$ . Furthermore, we have $(a,b) \in U \times V \subset O$ . This establishes that $O$ is open in $X \times Z$ . With that, the mapping $f$ is shown to be a quotient map. $\square$

Corollaries

Proof of Corollary 2
Let $X$ be a locally compact metric space and $Y$ be a sequential space. According to the theorem shown here, $Y$ is the quotient space of a metric space. There exists a metric space $M$ such that $Y$ is the quotient image of $M$ . Let $q:M \rightarrow Y$ be a quotient map from $M$ onto $Y$ . Consider the mapping $f:X \times M \rightarrow X \times Y$ defined by $f(x,y)=(x,q(y))$ for all $(x,y) \in X \times M$ . By Theorem 1, $f$ is a quotient map. Since $X \times Y$ is the quotient image of the metric space $X \times M$ , we establish that $X \times Y$ is a sequential space. $\square$

Corollary 3 follows from Corollary 2 since any compact space is a locally compact space.

Proof of Corollary 4
Let $X$ be a locally compact space and let $Y$ be a k-space. According to the theorem shown here, there is a locally compact space $W$ such that $Y$ is the quotient image of $W$ . Let $q:W \rightarrow Y$ be a quotient map from $W$ onto $Y$ . Define $f:X \times W \rightarrow X \times Y$ by letting $f(x,y)=(x,q(y))$ for all $(x,y) \in X \times W$ . According to Theorem 1, $f$ is a quotient map. Since $X \times Y$ is the quotient image of the locally compact space $X \times W$ , we establish that $X \times Y$ is a k-space. $\square$

Sequential Fans

We illustrate the above corollaries using sequential fans. Sequential fans are sequential spaces. Products of sequential fans may no longer be sequential, in fact, may no longer be countably tight. In some cases, the tightness of a product of two sequential fans is dependent of your favorite set theory axiom (see here). However, the product of a sequential fan and a compact metric space is sequential.

Let $S$ be a convergent sequence including its limit. For convenience, denote $S=\{ q_0,q_1,q_2,\cdots \} \cup \{ q \}$ such that each $q_n$ is isolated and an open neighborhood of $q$ consists of the point $q$ and all but finitely many $q_n$ . To make things more concrete, we can also let $S=\{ 1,\frac{1}{2},\frac{1}{3},\cdots \} \cup \{ 0 \}$ with the usual Euclidean topology. Let $\kappa$ be an infinite cardinal number. Let $M(\kappa)$ be the topological sum of $\kappa$ many copies of $S$ . The space $S(\kappa)$ is defined as $M(\kappa)$ with all the sequential limit points identified as one point called $\infty$ . The space $S(\kappa)$ is called the sequential fan with $\kappa$ many spines. In $S(\kappa)$ , there are $\kappa$ many copies of $S \cup \{ \infty \}$ , which is called a spine.

Note that $M(\kappa)$ is a metric space. Because $S(\kappa)$ is the quotient image of $M(\kappa)$ , the sequential fan $S(\kappa)$ is a sequential space. In fact, $S(\kappa)$ is a Frechet space since it is a sequential space that does not contain a copy of the Arens’ space (see here). For the discussion of the Arens’ space, see here.

According to Corollary 3, the product of the sequential fan $S(\kappa)$ and a compact metric space is a sequential space. In particular, the product $S(\kappa) \times S$ is always a sequential space. According to Corollary 4, $S(\kappa) \times S$ is a k-space. The fact that the product is both a sequential space and a k-space is not surprising. Whenever the spaces $X$ and $Y$ are sequential spaces, the product $X \times Y$ is a sequential space if and only if it is a k-space (see Theorem 2.2 [2]).

Reference

Engelking R., General Topology, Revised and Completed edition, Elsevier Science Publishers B. V., Heldermann Verlag, Berlin, 1989.

Tanaka Y., On Quasi-k-Spaces, Proc. Japan Acad., 46, 1074-1079, Berlin, 1970.

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Posted in Mappings | Tagged k-space, Quotient Map, Quotient space, Sequential fan, Sequential space | 1 Reply

Defining Arens’ space using diagrams

Posted on April 16, 2023 by Dan Ma

2

One way to define the Arens’ space is a 2-step approach, which is the quotient space approach. The first step is to identify an Euclidean plane consisting of convergent sequences (usually conveniently situated in the two-dimensional plane). The second step is to collapse certain points to make it a quotient space. Another way is to define the space directly, usually using an appropriate subset of the plane (of course, the resulting space is not an Euclidean space). We demonstrate both approaches using diagrams. In the first approach, we use two diagrams, the first one showing what the Euclidean space should look like, the second showing the resulting Arens’ space after certain points are identified. In the second approach, only one diagram is used (the standalone approach). The two-step approach is actually more informative since the quotient space of a separable metric space is a sequential space.

The following diagrams define the spaces without identifying specific points or locations in the Euclidean plane. The diagrams only indicate how the points relate to one another. For a definition of Arens’ space using the quotient space approach using specific points in the plane, see here. For a definition without connection to quotient space, see here. The red diagram and the blue diagram are for the quotient space approach (two-step). The pink diagram is the standalone approach.

The Arens’ space as discussed here is related to the Arens-Fort space, example 26 in Counterexamples in Topology [2].

The Red Diagram – The Euclidean Space

$\displaystyle \begin{aligned} & \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & S_5 & \text{ } & S_4 & \text{ } & S_3 & \text{ } & S_2 & \text{ } & S_1 \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } &\bullet & \text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet & \text{ } &\bullet & \text{ } & \bullet & \text{ } &\bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot &\text{ } &\cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot &\text{ } & \cdot & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & q_5 &\text{ } & q_4 & \text{ } & q_3 &\text{ } & q_2 & \text{ } & q_1 \\& \text{ } & \text{ } & \text{ } &\text{ } & \text{ } & \text{ } & \text{ } &\text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\& \bullet & \leftarrow & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet \\& p & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & p_5 &\text{ } & p_4 & \text{ } & p_3 &\text{ } & p_2 & \text{ } & p_1 \end{aligned}$

In all 3 diagrams, the thick bullets represent points in the space and the tiny dots represent “dot dot dot”, indicating that the points or the sequences are continuing countably infinitely. The arrows in the diagrams point to the direction to which the points are converging.

The points in the red diagram form a subspace of the Euclidean plane. There are convergent sequences $S_n$ going downward and going across from right to left. The point $q_n$ is the limit of the sequence $S_n$ . The sequence of points $q_n$ converges to a point, which is ignored and not shown in the diagram. The points $p_n$ are situated below the points $q_n$ and converge to the point $p$ . In this Euclidean space, the points in the sequences $S_n$ are isolated points. An open set of the point $q_n$ consists of $q_n$ and all but finitely many points in the sequence $S_n$ . Each point $p_n$ is isolated. An open set of the point $p$ consists of $p$ and all but finitely many $p_n$ .

The Blue Diagram – The Arens’ Space as a Quotient Space

$\displaystyle \begin{aligned} & \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & S_5 & \text{ } & S_4 & \text{ } & S_3 & \text{ } & S_2 & \text{ } & S_1 \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } &\bullet & \text{ } & \bullet & \text{ } & \bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet & \text{ } &\bullet & \text{ } & \bullet & \text{ } &\bullet & \text{ } & \bullet \\& \text{ } & \text{ } & \cdot &\text{ } &\cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot &\text{ } & \cdot & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow & \text{ } & \downarrow \\& \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet \\& p & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & p_5 &\text{ } & p_4 & \text{ } & p_3 &\text{ } & p_2 & \text{ } & p_1 \end{aligned}$

The blue diagram is established from the red diagram. The blue diagram is obtained by identifying the points $q_n$ and $p_n$ in the red diagram as one point called $p_n$ . The resulting quotient space is the Arens’ space. Let $S$ be the set of all points in the sequences $S_n$ . Let $P$ be the set of all points $p_n$ . The Arens’ space is $X=S \cup P \cup \{ p \}$ with the quotient topology. With the quotient topology, an open set containing the point $p$ is obtained by removing finitely many vertical columns (a column is $S_n$ plus the point $p_n$ ) from $X$ and by removing finitely many points in the remaining columns. An open neighborhood of the point $p_n$ consists of $p_n$ and all but finitely many points in the sequence $S_n$ . Points in the sequences $S_n$ continue to be isolated points.

The Arens’ space is a sequential space since it is the quotient image of a separable metric space.

The Pink Diagram – The Arens’ Space as a Standalone Space

$\displaystyle \begin{aligned} & \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ }& \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot &\text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & S_1 & \text{ } & S_2 & \text{ } & S_3 & \text{ } & S_4 & \text{ } & S_5 & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& \text{ } & \text{ } & \text{ } &\text{ } & \text{ } & \text{ } & \text{ } &\text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\& \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet &\text{ } & \bullet & \text{ } & \bullet & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \\& p & \text{ } & p_1 &\text{ } & p_2 & \text{ } & p_3 &\text{ } & p_4 & \text{ } & p_5 & \text{ } & \cdot & \text{ } & \cdot & \text{ } & \cdot & \text{ } \end{aligned}$

In the pink diagram, the bottom rows are the limit points. The point $p_n$ is the sequential limit of the sequence $S_n$ . The sequence $S_n$ is displayed vertically. The convergence of the sequence $S_n$ is not exhibited in the diagram and follows from how the open sets are defined.

Let $S$ be the set of all points in the sequences $S_n$ . Let $P$ be the set of all points $p_n$ . The Arens’ space is $X=S \cup P \cup \{ p \}$ with the topology defined as follows. Each point in $S$ is an isolated point. An open neighborhood of $p_n \in P$ consists of the point $p_n$ and all but finitely many points in $S_n$ . An open neighbood containing the point $p$ is obtained by removing finitely many vertical columns (a column is $S_n$ plus the point $p_n$ ) from $X$ and by removing finitely many points in the remaining columns.

Remarks

The blue diagram and the pink diagram are both representations of the Arens’ space. The space consists of countably many convergent sequences and their limits points plus one additional limit point called $p$ . Both diagrams present the essential ideas without being tied to specific points or sequences in the Euclidean plane. Perhaps the diagrams will make it easier to think about the Arens’ space and remember the definition.

As mentioned earlier, the Arens’ space is a sequential space since it is the quotient space of a metric space (see the theorem here). Recall from above that the Arens’ space is $X=S \cup P \cup \{ p \}$ . Clearly, $p \in \overline{S}$ . Note that no sequence of points in $S$ can converge to the point $p$ . Thus, the Arens’ space is an example of a sequential space that is not a Frechet space. In fact, in order to know whether a sequential space $W$ is Frechet, all we need to do is to determine if $W$ contains a copy of the Arens’ space (see the theorem here). Thus, any space that is sequential but not Frechet contains a copy of the Arens’ space. In this case, Frechetness is characterized the absence of an Arens’ subspace. The Arens’ space is a canonical quotient space that appears in the characterization of other properties. See [1] for an example.

The property of being a sequential space is not hereditary. Consider the subspace of the Arens’ space $Y=S \cup \{ p \}$ . As observed in the preceding paragraph, no sequence of points in $S$ can converge to $p$ . Thus, the set $S$ is a sequentially closed set but not closed in $Y$ . This means that $Y$ is not a sequential space. Thus, the Arens’ space is a sequential that is not hereditarily sequential. In fact, a space is a Frechet space if and only if it is a hereditarily sequential space (see Theorem 1 here).

The subspace $Y=S \cup \{ p \}$ discussed in the preceding paragraph is the Arens-Fort space, which is the example 26 in Steen and Seebach [2].

Reference

Lin, S., A note on the Arens’ space and sequential fan, Topology Appl, 81, 185-196, 1997.

Steen, L. A., Seebach, J. A., Counterexamples in Topology, Dover Publications, Inc., New York, 1995.

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Posted in Well known examples | Tagged Arens' Space, Quotient space, Sequential space | 2 Replies

A note on products of sequential fans

Posted on July 1, 2015 by Dan Ma

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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].

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

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 a sequential space. In general, the product of any sequential space and a compact metric space is a sequential space (see Corollary 3 here). The next example shows that the product of two sequential fans does not 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].

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.

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.

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].

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.

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.

Reference

Bella A., van Mill J., Tight points and countable fan-tightness, Topology Appl., 76, (1997), 1-27.

Eda K., Gruenhage G., Koszmider P., Tamano K., Todorčeviće S., Sequential fans in topology, Topology Appl., 67, (1995), 189-220.

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.

Hart, K. P., Nagata J. I., Vaughan, J. E., editors, Encyclopedia of General Topology, First Edition, Elsevier Science Publishers B. V, Amsterdam, 2003.

LaBerge T., Landver A., Tightness in products of fans and psuedo-fans, Topology Appl., 65, (1995), 237-255.

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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Posted: July 1, 2015
Revised: April 21, 2023

Posted in Basic Topology, Convergence Property | Tagged Countably tight space, Frechet space, General topology, Point-set topology, Quotient space, Sequential fan, Sequential space, Tightness, Topology | 2 Replies

Several ways to define countably tight spaces

Posted on June 1, 2015 by Dan Ma

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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.

For each $A \subset X$ , the set equality (a) holds. $\text{ }$

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

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.

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.

____________________________________________________________________

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.

$t(X) \le \tau$ .

The space $X$ is $\tau$ -generated.

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.

____________________________________________________________________

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.

____________________________________________________________________

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)

____________________________________________________________________

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.

____________________________________________________________________

Reference

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

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

Posted in Basic Topology, Convergence Property | Tagged Convergence Property, Countably tight space, First countable space, Frechet space, General topology, k-space, Point-set topology, Sequential space, Topology | 3 Replies

Sequentially compact spaces, II

Posted on September 4, 2010 by Dan Ma

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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
An observation about sequential spaces

Reference

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

Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970

Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Posted in Basic Topology, Compact space | Tagged Compact space, Countably compact space, First countable space, Frechet space, General topology, k-space, Lindelof space, Point-set topology, Sequential space, Topology | Leave a reply

An observation about sequential spaces

Posted on August 22, 2010 by Dan Ma

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This post is about an observation about sequential spaces. In a sequential space, non-trivial convergent sequences abound. Thus in the extreme case of there being no trivial convergent convergent sequences, the space in question must not be sequential. Specifically we observe that if $X$ is a Hausdorff sequential space and if $p \in X$ is a non-isolated point (i.e. the singleton set $\left\{p\right\}$ is not open), there is a convergent sequence $p_n$ of points of $X-\left\{p\right\}$ such that $p_n \mapsto p$ . Thus it is necessary condition that in a sequential space, there exist non-trivial convergent sequences at every non-isolated point. We present examples showing that this condition is not a sufficient condition for a space being sequential. As the following examples show, the property that there are non-trivial convergent sequences at every non-isolated is a rather weak property.

The first example is from the problem section of Mathematical Monthly in 1970 (see [2]). Let $\mathbb{R}$ be the real line and let $\mathbb{P}$ be the set of all irrational numbers. Let $\mathbb{Q}=\mathbb{R}-\mathbb{P}$ . Let $X=\mathbb{R}$ and define a new topology on $X$ by calling a subset $U \subset X$ open if and only if $U=W-H$ where $W$ is a usual open subset of the real line and $H$ is a subset of $\mathbb{P}$ that is at most countable. This new topology on the real line is finer than the Euclidean topology. Thus $X$ is a Hausdorff space. Every point of $X$ is a non-isolated point and is the the sequential limit of a sequence of rational numbers, satisfying the condition that every non-isolated point is the sequential limit of a non-trivial convergent sequence.

In the topology for $X$ , every countably infinite subset of the set $\mathbb{P}$ is closed in $X$ . Thus no sequence of points of $\mathbb{P}$ can converge to a point not in $\mathbb{P}$ . Therefore $\mathbb{P}$ is sequentially closed and non-closed in $X$ , making $X$ not a sequential space.

Not only that every countably infinite subset of $\mathbb{P}$ is closed in $X$ , every countably infinite subset of $\mathbb{P}$ is relatively discrete. Then it follows that for every compact $K \subset X$ , $K \cap \mathbb{P}$ is finite (and is thus closed in $K$ ). Thus $X$ is also not a k-space.

Another example is that of a product space. Any uncountable product where each factor has at least two points is not sequential. This follows from the fact that $2^{\omega_1}$ is not sequential (see Sequential spaces, IV). Furthermore, in any product space with infinitely many factors each of which has at least two points, every point is the sequential limit of a non-trivial convergent sequence. Thus any product space with uncountably many factors, each of which has at least two points, is another example of a non-sequential space where there are non-trivial convergent sequences at every point.

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

Reference

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

Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970

Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Posted in Basic Topology, The real line | Tagged First countable space, Frechet space, General topology, k-space, Point-set topology, Sequential space, Topology | Leave a reply

A note about the Arens’ space

Posted on August 18, 2010 by Dan Ma

8

The Arens’ space is a canonical example of a sequential space that is not a Frechet space. It also has a subspace that is not sequential (thus the notion of being a sequential is not hereditary). We show that any space that is sequential but not Frechet contains a copy of the Arens’ space. For previous discussion on sequential spaces and Frechet spaces, see the links at the end of this post. Also see [1] and [2].

Let $\omega$ be the set of all nonnegative integers. Let $\mathbb{N}$ be the set of all positive integers. In one formulation, the Arens’ space is the set $X=\left\{\infty\right\} \cup \mathbb{N} \cup (\mathbb{N} \times \mathbb{N})$ with the open neighborhoods defined by:

The points in $\mathbb{N} \times \mathbb{N}$ are isolated;

The neighborhoods at each $n \in \mathbb{N}$ are of the form $B_{n,m}=\left\{n\right\} \cup \left\{(n,j) \in \mathbb{N} \times \mathbb{N}:j \ge m\right\}$ for some $m \in \mathbb{N}$ ;

The neighborhoods at $\infty$ are obtained by removing from $X$ finitely many $B_{n,1}$ and by removing finitely many isolated points in each of the remaining $B_{n,1}$ .

Another formulation is that of a quotient space. For each $n \in \omega$ , let $K_n=\left\{x_{n,j}:j \in \mathbb{N}\right\} \cup \left\{y_n\right\}$ be a convergent sequence such that $y_n$ is the limit. Let $G$ be a topological sum of the convergent sequences $K_n$ . We then identify $\left\{x_{0,j},y_j\right\}$ for each $j \in \mathbb{N}$ . The Arens’ space is the resulting quotient space and let $Y$ denote this space (in the literature $S_2$ is used). Note that the Arens’ space has been previously defined in this blog (see An example of a quotient space, II). Note that the quotient space $Y$ is topologically identical to $X$ . In the remainder of this note, we work with $X$ in discussing the Arens’ space.

The Arens’ space is sequential since it is a quotient space of a first countable space. The subspace $\left\{\infty\right\} \cup (\mathbb{N} \times \mathbb{N})$ is not sequential, proving that the Arens’ space is not a Frechet space.

We now show that any sequential space that is not Frechet contains a copy of the Arens’ space. We have the following theorem.

Theorem
Let $W$ be a sequential space. Then $W$ is Frechet if and only $W$ does not contain a copy of the Arens’ space.

Proof
$\Longrightarrow$ This direction is clear since the Frechet property is hereditary.

$\Longleftarrow$ For any $T \subset W$ , let $T^s$ be the set of limits of sequences of points of $T$ . Suppose $W$ is not Frechet. Then for some $A \subset W$ , there exists $x \in \overline{A}$ such that $x \notin A^s$ . Since $A^s$ is non-closed in $W$ and since $W$ is sequential, there is a sequence $w_n$ of points of $A^s$ converging to $z_0 \notin A^s$ . We can assume that $w_n \notin A$ for all but finitely many $n$ (otherwise $z_0 \in A^s$ ). Thus without loss of generality, assume $w_n \notin A$ for all $n$ .

For each $n \in \mathbb{N}$ , there is a sequence $z_{n,j}$ of points of $A$ converging to $w_n$ . It is OK to assume that all $w_n$ are distinct and all $z_{n,j}$ are distinct across the two indexes. Let $W_0=\left\{z_0\right\} \cup W_1 \cup W_2$ where $W_1=\left\{w_n: n \in \mathbb{N}\right\}$ and $W_2=\left\{z_{n,j}:n,j \in \mathbb{N}\right\}$ . Then $W_0$ is a homeomorphic copy of the Arens’ space. $\blacksquare$

Remark
The above theorem is not valid outside of sequential spaces. Let $Z$ be a countable space with only one non-isolated point where $Z$ is not sequential (for example, the subspace $Z=\left\{\infty\right\} \cup (\mathbb{N} \times \mathbb{N})$ of the Arens’ space). Clearly $Z$ contains no copy of the Arens’ space. Yet $Z$ is not Frechet (it is not even sequential).

Previous posts on sequential spaces and Frechet spaces:
Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I
k-spaces, II

Reference

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

Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

Posted in Basic Topology, Well known examples | Tagged First countable space, Frechet space, General topology, k-space, Point-set topology, Quotient space, Sequential space, Topology | 8 Replies

k-spaces, II

Posted on August 3, 2010 by Dan Ma

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A space $X$ is a k-space if for each $A \subset X$ , $A$ is closed in $X$ if and only if $K \cap A$ is closed in $K$ for all compact $K \subset X$ . A space $X$ is a sequential space if for each $A \subset X$ , $A$ is closed in $X$ if and only if $A$ is a sequentially closed set in $X$ . A set $A \subset X$ is sequentially closed in the space $X$ if whenever we have $x_n \in A$ and the sequence $x_n$ converges to $x \in X$ , we have $x \in A$ . A set $A \subset X$ is sequentially open in $X$ if $X-A$ is sequentially closed in $X$ . In both of these definitions, we can replace “closed” with “open” and the “only if” part of the definition always hold. Thus in working with these definitions, we only need to be concerned with the “if” part. Every sequential space is a k-space. The converse does not hold. In this short note, we show that the converse holds if every point in the space is a $G_\delta$ -set. This is a basic fact about k-spaces. For other basic facts on k-spaces and sequential spaces, see the following:

Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
Sequential spaces, V
k-spaces, I

In a given space $X$ , $W \subset X$ is a $G_\delta$ -set in $X$ if $W=\bigcap \limits_{i=1}^\infty U_i$ where each $U_i$ is open in $X$ , i.e. $W$ is the intersection of countably many open sets. A point $x \in X$ is a $G_\delta$ -set in $X$ if the singleton $\left\{x\right\}$ is the intersection of countably many open subsets of $X$ . It is a well known fact in general topology that in a compact Hausdorff space $X$ , if $x \in X$ is a $G_\delta$ -set in $X$ , then there is a countable local base at $x$ . It follows that if every point of a compact Hausdorff space $X$ is a $G_\delta$ -set in $X$ , then $X$ is first countable (see The cardinality of compact first countable spaces, II).

Theorem
Let $X$ be a space in which every point is a $G_\delta$ -set in $X$ . Then if $X$ is a k-space then $X$ is a sequential space.

Proof. Suppose $A \subset X$ is not closed in $X$ . We show that $A$ is not sequentially closed in $X$ , i.e. there is a sequence $x_n \in A$ such that $x_n \mapsto x \in X$ and $x \notin A$ .

Since $X$ is a k-space and $A$ is not closed, there is a compact $K \subset X$ such that $K \cap A$ is not closed in $K$ . Every point of $K$ is a $G_\delta$ -set in $X$ and thus a $G_\delta$ -set in $K$ . It follows that $K$ is first countable.

Let $x \in \overline{K \cap A}$ such that $x \notin A$ (the closure is taken in $K$ ). Since $K$ is first countable, there is a sequence $x_n \in K \cap A$ such that $x_n \mapsto x$ . This means $A$ is not sequentially closed in $X$ . $\blacksquare$

Posted in Basic Topology, Compact space | Tagged Compact space, First countable space, General topology, k-space, Point-set topology, Sequential space, Topology | Leave a reply

Sequential spaces, V

Posted on July 21, 2010 by Dan Ma

2

In the previous post Sequential spaces, IV, we show that the uncountable product of sequential spaces is not sequential (e.g. the product $2^{\omega_1}$ is not sequential). What is more remarkable is that the product of two sequential spaces needs not be sequential. We present an example of a first countable space and a Frechet space whose product is not a k-space (thus not sequential). For the previous discussion on this blog on sequential spaces and k-spaces, see the links at the end of this post.

Let $\mathbb{R}$ be the real line and let $\mathbb{N}$ be the set of all positive integers. Let $X$ be the space $\mathbb{R}-\left\{1,\frac{1}{2},\frac{1}{3},\cdots\right\}$ with the topology inherited from the usual topology on the real line. Let $Y=\mathbb{R}$ with the positive integers identified as one point (call this point $p$ ). We claim that $X \times Y$ is not a k-space and thus not a sequential space. To this end, we define a non-closed $A \subset X \times Y$ such that $K \cap A$ is closed in $K$ for all compact $K \subset X \times Y$ .

Let $A=\bigcup \limits_{i=1}^\infty A_i$ where for each $i \in \mathbb{N}$ , the set $A_i$ is defined by the following:

$\displaystyle A_i =\left\{\biggl(\frac{1}{i}+\frac{a_i}{j},i+\frac{0.5}{j} \biggr) \in X \times Y:j \in \mathbb{N}\right\}$

where $\displaystyle a_i=\biggl(\frac{1}{i}-\frac{1}{i+1} \biggr) 10^{-i}$ .

Clearly $A$ is not closed as $(0,p) \in \overline{A}-A$ . In fact in the product space $X \times Y$ , the point $(0,p)$ is the only limit point of the set $A$ . Another observation is that for each $n \in \mathbb{N}$ , $(0,p)$ is not a limit point of $\bigcup \limits_{i=1}^n A_i$ . Furthermore, if $z_i \in A_i$ for each $i \in S$ where $S$ is an infinite subset of $\mathbb{N}$ , then $(0,p)$ is not a limit point of $\left\{z_i:i \in S\right\}$ . It follows that no infinite subset of $A$ is compact. Consequently, $K \cap A$ is finite for each compact $K \subset X \times Y$ . Thus $X \times Y$ is not a k-space. To see that $X \times Y$ is not sequential directly, observe that $A$ is sequentially closed.

Previous posts on sequential spaces and k-spaces

Sequential spaces, I
Sequential spaces, II
Sequential spaces, III
Sequential spaces, IV
k-spaces, I

Posted in Basic Topology, Convergence Property, Product space | Tagged First countable space, General topology, k-space, Point-set topology, Product space, Sequential space, Topology | 2 Replies

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Introductory topology

Lindelof space

Mappings

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Michael Line

Moore space

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Normality in Product

Omega 1

Paracompact space

Product of unit interval

Product space

Selection Principles

Separable metrizable

Set-Theoretic Topology

The real line

The Sorgenfrey Line

Uncategorized

Well known examples

Recent Posts

Sequential fan and the dominating number

The product of the identity map and a quotient map

Defining Arens’ space using diagrams

Perfect preimages of Lindelof spaces

Perfect images of separable metric spaces

The Bow-Tie Space

Kuratowski Theorem

A Pixley-Roy Meta-Lindelof Space

Examples of Meta-Lindelof Spaces

Divergence of the sum of the reciprocals of the primes

Top Posts
The First Uncountable Ordinal
A Short Note About The Sorgenfrey Line
Looking for a closed and discrete subspace of a product space
Adding up to a non-meager set
Weakly Lindelof spaces
The ideal of bounded sets
The product of first countable spaces
A short note on monolithic spaces
A Note On The Sorgenfrey Line
Stone-Cech Compactification of the Integers - Basic Facts

Almost disjoint family

Baire Category Theorem

Baire Space

Bing's G

Bounding Number

Bow-tie space

Cantor set

CCC Space

Closed and discrete sets

Collectionwise Hausdorff

Collectionwise normal

Compact space

Completely Regular Space

Convergence Property

Corson compact space

Countable chain condition

Countable extent

Countable network

Countably compact space

Countably paracompact space

Countably tight space

Density of a Topological Space

Double arrow space

Dowker space

Euclidean topology

Eventual domination order

Extent

First countable space

Frechet-Urysohn Space

Frechet space

Function space

G-delta diagonal

G-delta sets

General topology

Hereditarily lindelof

Hereditarily normal space

Hereditarily separable

Ideal

Introductory topology

k-space

Limit Point

Lindelof space

Meager sets

Menger Spaces

Metacompact space

Metrizable spaces

Michael Line

Monolithic space

Moore space

Network

Network weight

Non-normal product

Normal P-Space

Normal space

Omega 1

Paracompact space

Perfectly normal space

Perfect mapping

Point-set topology

Pointwise convergence topology

Product of unit interval

Product space

Pseudocompact Space

Quotient space

Separable metrizable space

Separable space

Sequential fan

Sequential space

Sigma products

Stone-Cech Compactification

The real line

The Sorgenfrey Line

Tightness

Topology

Weight of a topological space

Dan Ma's Topology Blog

Expository Articles In General Topology

Tag Archives: Sequential space

Sequential fan and the dominating number

The product of the identity map and a quotient map

Defining Arens’ space using diagrams

A note on products of sequential fans

Several ways to define countably tight spaces

Sequentially compact spaces, II

An observation about sequential spaces

A note about the Arens’ space

k-spaces, II

Sequential spaces, V