The product of the identity map and a quotient map

Posted on April 21, 2023 by Dan Ma

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.

$\text{ }$

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Several ways to define countably tight spaces

Posted on June 1, 2015 by Dan Ma

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 , if condition (b) holds,
then $A$ is closed.
For each , if condition (c) holds,
then $A$ is closed.

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

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

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

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

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

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Generalizations

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

$A \subset X$

$x \in \overline{A}$

$B \subset A$

$\lvert B \lvert \le \tau$

$x \in \overline{B}$

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:

$C \subset X$

$\lvert C \lvert \le \tau$

$C \cap A$

$C \ \ \ \ \ \ \ \ \ \ \ (\beta)$

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

$A \subset X$

$\tau$

$A$

$.\ \ \ \ \ \ \ \ \ \ \ (\chi)$

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.

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Examples

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

$\Longrightarrow$

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

$\Longrightarrow$

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:

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

$i,j \in \mathbb{N}$

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

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

$V_i=V_{i,1} \cup \left\{ x \right\}$ 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:

$\displaystyle \left\{\biggl(\frac{1}{i},0 \biggr) \right\} \cup V_{i,j}$

$j \in \mathbb{N}$

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

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

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

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Remarks

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

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

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Exercises

Exercise 1
This is to verify Example 1. Verify that

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

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

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Reference

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

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

Sequentially compact spaces, II

Posted on September 4, 2010 by Dan Ma

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.

An observation about sequential spaces

Posted on August 22, 2010 by Dan Ma

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.

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.

A note about the Arens’ space

Posted on August 18, 2010 by Dan Ma

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.

k-spaces, II

Posted on August 3, 2010 by Dan Ma

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 $X$ . $\blacksquare$

Sequential spaces, V

Posted on July 21, 2010 by Dan Ma

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

Sequential spaces, III

Posted on July 1, 2010 by Dan Ma

This is a continuation of the discussion on sequential spaces started with the post Sequential spaces, I and Sequential spaces, II and k-spaces, I. The topology in a sequential space is generated by the convergent sequences. The convergence we are interested in is from a topological view point and not necessarily from a metric (i.e. distance) standpoint. In our discussion, a sequence $\left\{x_n\right\}_{n=1}^\infty$ converges to $x$ simply means for each open set $O$ containing $x$ , $O$ contains $x_n$ for all but finitely many $n$ . In any topological space, there are always trivial convergent sequences. These are sequences of points that are eventually constant, i.e. the sequences $\left\{x_n\right\}$ where for some $n$ , $x_n=x_j$ for $j \ge n$ . Any convergent sequence that is not eventually constant is called a non-trivial convergent sequence. We present an example of a space where there are no non-trivial convergent sequences of points. This space is derived from the Euclidean topology on the real line. This space has no isolated point in this space and yet has no non-trivial convergent sequences and has no infinite compact sets. From this example, we make some observations about sequential spaces and k-spaces.

The space we define here is obtained by modifying the Euclidean topology on the real line. Let $\mathbb{R}$ be the real line. Let $\tau_e$ be the Euclidean topology on the real line. Consider the following collection of subsets of the real line:

$\mathcal{B}=\left\{U-C:U \in \tau_e \text{ and } \lvert C \lvert \le \omega\right\}$

It can be verified that $\mathcal{B}$ is a base for a topology $\tau$ on $\mathbb{R}$ . In fact this topology is finer than the Euclidean topology. Denote $\mathbb{R}$ with this finer topology by $X$ . Clearly $X$ is Hausdorff since the Euclidean topology is. Any countable subset of $X$ is closed. Thus $X$ is not separable (no countable set can be dense). This space is a handy example of a hereditarily Lindelof space that is not separable. The following lists some properties of $X$ :

$X$ is herditarily Lindelof.
There are no non-trivial convergent sequences in $X$ .
All compact subsets of $X$ are finite.
$X$ is not a k-space and is thus not sequential.

Discussion of 1. This follows from the fact that the real line with the Euclidean topology is hereditarily Lindelof and the fact that each open set in $X$ is an Euclidean open set minus a countable set.

Discussion of 2 This follows from the fact that every countable subset of $X$ is closed. If a non-trivial sequence $\left\{x_n\right\}$ were to converge to $x \in X$ , then $\left\{x_n:n=1,2,3,\cdots\right\}$ would be a countable subset of $X$ that is not closed.

Discussion of 3. Let $A \subset X$ be an infinite set. If $A$ is bounded in the Euclidean topology, then there would be a non-trivial convergent sequence of points of $A$ in the Euclidean topology, say, $x_n \mapsto x$ . Let $U_0=X-\left\{x_n:n=1,2,3,\cdots\right\}$ , which is open in $X$ . For $n \ge 1$ , let $U_n$ be Euclidean open such that $x_n \in U_n$ . We also require that all $U_n$ are pairwise disjoint and not contain $x$ . Then $U_0,U_1,U_2,\cdots$ form an open cover of $A$ (in the topology of $X$ ) that has no finite subcover. So any bounded infinite $A$ is not compact in $X$ .

Suppose $A$ is unbounded in the Euclidean topology. Then $A$ contains a closed and discrete subset $\left\{x_1,x_2,x_3,\cdots\right\}$ in the Euclidean topology. We can find Euclidean open sets $U_n$ that are pairwise disjoint such that $x_n \in U_n$ for each $n$ . Let $U_0=X-\left\{x_n:n=1,2,3,\cdots\right\}$ , which is open in $X$ . Then $U_0,U_1,U_2,\cdots$ form an open cover of $A$ (in the topology of $X$ ) that has no finite subcover. So any unbounded infinite $A$ is not compact in $X$ .

Discussion of 4. Note that every point of $X$ is a non-isolated point. Just pick any $x \in X$ . Then $X-\left\{x\right\}$ is not closed in $X$ . However, according to 3, $K \cap (X-\left\{x\right\})$ is finite and is thus closed in $K$ for every compact $K \subset X$ . Thus $X$ is not a k-space.

General Discussion
Suppose $\tau$ is the topology for the space $Y$ . Let $\tau_s$ be the set of all sequentially open sets with respect to $\tau$ (see Sequential spaces, II). Let $\tau_k$ be the set of all compactly generated open sets with respect to $\tau$ (see k-spaces, I). The space $Y$ is a sequential space (a k-space ) if $\tau=\tau_s$ ( $\tau=\tau_k$ ). Both $\tau_s$ and $\tau_k$ are finer than $\tau$ , i.e. $\tau \subset \tau_s$ and $\tau \subset \tau_k$ . When are $\tau_s$ and $\tau_k$ discrete? We discuss sequential spaces and k-spaces separately.

Observations on Sequential Spaces
With respect to the space $(Y,\tau)$ , we discuss the following four properties:

A. $\$ No non-trivial convergent sequences.
B. $\$ $\tau_s$ is a discrete topology.
C. $\$ $\tau$ is a discrete topology.
D. $\$ Sequential, i.e., $\tau=\tau_s$ .

Observation 1
The topology $\tau_s$ is discrete if and only if $Y$ has no non-trivial convergent sequences, i.e. $A \Longleftrightarrow B$ .

If $\tau_s$ is a discrete topology, then every subset of $Y$ is sequentially open and every subset is sequentially closed. Hence there can be no non-trivial convergent sequences. If there are no non-trivial convergent sequences, every subset of the space is sequentially closed (thus every subset is sequentially open).

Observation 2
Given that $Y$ has no non-trivial convergent sequences, $Y$ is not discrete if and only if $Y$ is not sequential. Equivalently, given property A, $C \Longleftrightarrow D$ .

Given that there are no non-trivial convergent sequences in $Y$ , $\tau_s$ is discrete. For $(Y,\tau)$ to be sequential, $\tau=\tau_s$ . Thus for a space $Y$ that has no non-trivial convergent sequences, the only way for $Y$ to be sequential is that it is a discrete space.

Observation 3
Given $Y$ is not discrete, $Y$ has no non-trivial convergent sequences implies that $Y$ is not sequential, i.e. given $\text{not }C$ , $A \Longrightarrow \text{not }D$ . The converse does not hold.

Observation 3 is a rewording of observation 2. To see that the converse of observation 3 does not hold, consider $Y=[0,\omega_1]=\omega_1+1$ , the successor ordinal to the first uncountable ordinal with the order topology. It is not sequential as the singleton set $\left\{\omega_1\right\}$ is sequentially open and not open.

Observations on k-spaces
The discussion on k-spaces mirrors the one on sequential spaces. With respect to the space $(Y,\tau)$ , we discuss the following four properties:

E. $\$ No infinite compact sets.
F. $\$ $\tau_k$ is a discrete topology.
G. $\$ $\tau$ is a discrete topology.
H. $\$ k-space, i.e., $\tau=\tau_k$ .

Observation 4
The topology $\tau_k$ is discrete if and only if $Y$ has no infinite compact sets, i.e. $E \Longleftrightarrow F$ .

If $\tau_k$ is a discrete topology, then every subset of $Y$ is a compactly generated open set. In particular, for every compact $K \subset Y$ , every subset of $K$ is open in $K$ . This means $K$ is discrete and thus must be finite. Hence there can be no infinite compact sets if $\tau_k$ is discrete. If there are no infinite compact sets, every subset of the space is a compactly generated closed set (thus every subset is a compactly generated open set).

Observation 5
Given that $Y$ has no infinite compact sets, $Y$ is not discrete if and only if $Y$ is not a k-space. Equivalently, given property E, $G \Longleftrightarrow H$ .

Given that there are no infinite compact sets in $Y$ , $\tau_k$ is discrete. For $(Y,\tau)$ to be a k-space, $\tau=\tau_k$ . Thus for a space $Y$ that has no infinite compact sets, the only way for $Y$ to be a k-space is that it is a discrete space.

Observation 6
Given $Y$ is not discrete, $Y$ has no infinite compact sets implies that $Y$ is not a k-space, i.e. given $\text{not }G$ , $E \Longrightarrow \text{not }H$ . The converse does not hold.

Observation 6 is a rewording of observation 5. To see that the converse of observation 6 does not hold, consider the topological sum of a non-k-space and an infinite compact space.

Remark
In the space $X$ defined above by removing countable sets from Euclidean open subsets of the real line, there are no infinite compact sets and no non-trivial convergent sequences. Yet the space is not discrete. Thus it can neither be a sequential space nor a k-space. Another observation we would like to make is that no infinite compact sets implies no non-trivial convergent sequences ( $E \Longrightarrow A$ ). However, the converse is not true. Consider $\beta(\omega)$ , the Stone-Cech compactification of $\omega$ , the set of all nonnegative integers.

Reference

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

k-spaces, I

Posted on June 27, 2010 by Dan Ma

There are many examples in general topology of defining a new topology on a set based on a given topology already defined on the set. For example, given a topology $\tau$ on $X$ , one can define a finer topology $\tau_s$ consisting of all sequentially open subsets of $X$ (based on the original topology $\tau$ ). Sequential spaces are precisely those spaces for which the original topology coincides with $\tau_s$ (see the post Sequential spaces, II). A related concept is the notion of k-spaces. We show that the compactly generated open sets form a finer topology and that k-spaces are precisely those spaces for which the compactly generated topology coincides with the original topology. We also give an external characterization of k-spaces, namely, those spaces that are quotient images of locally compact spaces. All spaces under consideration are Hausdorff.

Let $X$ be a space. We say $A \subset X$ is a compactly generated closed set in $X$ if $A \cap K$ is closed in $K$ for any compact $K \subset X$ . We say $B \subset X$ is a compactly generated open set in $X$ if $X-B$ is a compactly generated closed set in $X$ . The space $X$ is said to be a k-space if

$A \subset X$ is a compactly generated closed set in $X$ if and only if $A$ is a closed set.

The direction $\Leftarrow$ of the above statement always holds. So a space is a k-space if it satisfies the direction $\Rightarrow$ in the above statement. We also want to mention that in the above definition, “closed” can be replaced by “open”.

Suppose $\tau$ is the topology of the space $X$ . Then define $\tau_k$ as the set of all compactly generated open sets in $X$ . It can be easily verified that $\tau_k$ is a topology defined on the set $X$ and that $\tau_k$ is a finer topology than the original topology $\tau$ , i.e. $\tau \subset \tau_k$ . It follows that $X$ is a k-space if and only if $\tau=\tau_k$ .

Much of the discussion here mirrors the one in Sequential spaces, II. In a sequential space, the topology coincides with $\tau_s$ , the open sets generated by convergent sequences (a particular type of compact sets). In a k-space, the topology conincides with $\tau_k$ , the open sets generated by the compact sets. A sequential space is the quotient space of a topological sum of disjoint convergent sequences. Any k-space is the quotient space of a topological sum of disjoint compact sets.

We have the following theorem.

Theorem
For any space $X$ , the following conditions are equivalent:

$X$ is a k-space.
$X$ is a quotient space of a locally compact space.

Proof. $1 \Rightarrow 2$ Let $X$ be a k-space. Let $\mathcal{K}$ be the set of all compact subsets of $X$ . Let $Y=\oplus_{K \in \mathcal{K}}K$ be the topological sum of all $K \in \mathcal{K}$ where each $K \in \mathcal{K}$ has the relative topology inherited from the space $X$ . Then $Y$ is a locally compact space. There is a natural mapping we can define on $Y$ onto $X$ . The space $Y$ is a disjoint union of all compact subsets of $X$ . We can map each compact set $K \in \mathcal{K}$ onto the corresponding compact subset $K$ of $X$ by the identity map. Let $f:Y \mapsto X$ be this natural mapping. We claim that the quotient topology generated by this mapping coincides with the original topology on $X$ .

Let $\tau$ be the given topology on $X$ and let $\tau_f$ be the quotient topology generated on the set $X$ . More precisely, $\tau_f$ is the set of all $O \subset X$ such that $f^{-1}(O)$ is open in $Y$ . Clearly, $\tau \subset \tau_f$ since $f$ is a continuous map. We need to show that $\tau_f \subset \tau$ . Let $O \in \tau_f$ . Since $X$ is a k-space, if we can show that $O$ is a compactly generated open set, then $O \in \tau$ .

Let $K \subset X$ be compact. We need to show that $O \cap K$ is open in $K$ . Since $O \in \tau_f$ , $f^{-1}(O)$ is open in $Y$ . Since $K$ is open in $Y$ , $f^{-1}(O) \cap K$ is open in $Y$ . It is also the case that $f^{-1}(O) \cap K$ is open in $K$ ( $K$ as a subset of $Y$ ). We can consider $f^{-1}(O) \cap K$ as a subset of $K$ with $K$ being a subset of $X$ . As a subset of $K \subset X$ , we have $f^{-1}(O) \cap K=O \cap K$ . Thus $O \cap K$ is open in $K$ and $O \in \tau$ .

$2 \Rightarrow 1$ Let $Y$ be locally compact and let $f:Y \mapsto X$ be a quotient map. We show that $X$ is a k-space. To this end, we show that if $A \subset X$ is a compactly generated closed set in $X$ , $A$ is closed in $X$ . Or equivalently, if $A$ is not closed in $X$ , then $A$ is not a compactly generated closed set in $X$ . Under the quotient map $f$ , $A$ is closed in $X$ if and only if $f^{-1}(A)$ is closed in $Y$ .

Suppose $A$ is not closed in $X$ . Then $f^{-1}(A)$ is not closed in $Y$ . Then there is $y \in \overline{f^{-1}(A)}-f^{-1}(A)$ . Let $U \subset Y$ be open in $Y$ such that $y \in U$ and $\overline{U}$ is compact. Then $f(\overline{U})$ is compact. It follows that $A \cap f(\overline{U})$ is not closed in $f(\overline{U})$ . Note that $f(y) \in f(\overline{U})$ and $f(y) \notin A \cap f(\overline{U})$ . However, $f(y) \in \overline{A \cap f(\overline{U})}$ . To see this, let $V \subset X$ be open with $f(y) \in V$ . Then $f^{-1}(V)$ is open in $Y$ . Since $y \in f^{-1}(V)$ and $y \in U$ , $U \cap f^{-1}(V) \cap f^{-1}(A) \ne \varnothing$ with $t$ being one element of the intersection. This means that $f(t) \in V \cap A \cap f(\overline{U})$ . Thus $A$ is not a compactly generated closed set in $X$ . $\blacksquare$

Reference

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

Posted: June 27, 2010
Revised: April 18, 2023

Sequential spaces, I

Posted on June 21, 2010 by Dan Ma

Any topological space where there is a countable base at every point is said to satisfy the first axiom of countability or to be first countable. In this post we discuss several properties weaker than the first axiom of countability. All spaces under consideration are Hausdorff. Any first countable space $X$ satisfies each of the following conditions:

If $A \subset X$ and $x \in \overline{A}$ then there is a sequence $\left\{x_n\right\}$ of points in $A$ such that the sequence converges to $x$ .
The set $A \subset X$ is closed in $X$ if $A$ is sequentially closed in $X$ , which means that: if $\left\{x_n\right\}$ is a sequence of points in $A$ such that $\left\{x_n\right\}$ converges to $x \in X$ , then $x \in A$ .
The set $A \subset X$ is closed in $X$ if this condition holds: if $K \subset X$ is compact, then $A \cap K$ is closed in $K$ .

Spaces satisfying:

condition 1 are called Frechet spaces,
condition 2 are called sequential spaces,
condition 3 are called k-spaces.

All three of these conditions hold in first countable spaces. We have the following implications:

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

This post is an introductory discussion of these notions. Each of the above implications is not reversible (see the section below on examples). After we discuss sequential spaces, we take a look at the behavior of these four classes of spaces in terms of whether the property can be passed onto subspaces (the property being hereditary) and in terms of images under quotient maps. For previous posts in this blog on first countable spaces and quotient spaces, see the links at the end of the post. Excellent texts on general topology are [1] and [2].

For a subsequent discussion on sequential space, see Sequential spaces, II.

The Forward Implications

First countable $\Rightarrow \$ Frechet
Suppose $x \in \overline{A}$ and $A \subset X$ . Let $U_1,U_2,\cdots$ be a local base at $x$ . Then choose $x_n \in A \cap U_n$ and we have $x_n \mapsto x$ .

Frechet $\Rightarrow \$ Sequential
Let $A \subset X$ be sequentially closed in $X$ . Suppose $A$ is not closed in $X$ . Then there is $x \in \overline{A}$ such that $x \notin A$ . By Frechet, there is a sequence $x_n \in A$ such that $x_n \mapsto x$ . Since $A$ is sequentially closed, $x \in A$ , a contradiction. So any sequentially closed set in a Frechet space must be a closed set.

Sequential $\Rightarrow \$ k-space
Suppose $A \subset X$ is not closed in $X$ . Since $X$ is sequential, there is a sequence $x_n \in A$ such that $x_n \mapsto x$ and $x \notin A$ . The set $K=\left\{x_n:n=1,2,3,\cdots\right\} \cup \left\{x\right\}$ is a compact set. Note that $A \cap K$ is not closed in $K$ . This shows that $X$ is a k-space.

Sequential Spaces

Sequential spaces are ones in which the topology can be completely described by convergent sequences. Let $X$ be a space. Let $A \subset X$ . The set $A$ is said to be sequentially closed in $X$ if whenever we have a convergent sequence of points in $A$ , the sequential limit must be in $A$ . In other words, $A$ contains all the limits of the convergent sequences of points in $A$ . For $U \subset X$ , $U$ is sequentially open in $X$ if this condition holds: if $\left\{x_n \in X: n=1,2,3,\cdots\right\}$ is a sequence of points converging to some $x \in U$ , then $x_n \in U$ for all but finitely many $n$ . It can be verified that:

$U$ is sequentially open in $X$ if and only if $X-U$ is sequentially closed in $X$ .

Theorem 1
A space $X$ is Frechet if and only if every subspace of $X$ is sequential.

Proof
$\Rightarrow$ Suppose $W \subset X$ is not a sequential space. Then there is $A \subset W$ such that $A$ is sequentially closed in $W$ but $A$ is not closed in $W$ . We show that $X$ is not Frechet. There is a point $w \in W$ such that $w$ is a limit point of $A$ (in the subspace $W$ ) and $w \notin A$ . Since $A$ is sequentially closed, no sequence of points in $A$ can converge to $w$ (otherwise $w \in A$ ).

Clearly, the point $w$ is also a limit point of $A$ with respect to the toplology of $X$ , i.e. $w \in \overline{A}$ with respect to $X$ . Since no sequence of points in $A$ can converge to $w$ , $X$ is not Frechet.

$\Leftarrow$ Suppose $X$ is not Frechet. Then there is $x \in \overline{A}$ such that $A \subset X$ and no sequence of points in $A$ can converge to $x$ . Consider the subspace $Y=A \cup \left\{x\right\}$ . The set $Y-\left\{x\right\}$ is sequentially closed in $Y$ but is not closed in $Y$ . $\blacksquare$

Theorem 2
Every quotient space of a sequential space is always a sequential space.

Proof. Let $X$ be a sequential space. Let $f:X \mapsto Y$ be a quotient map. We show that $Y$ is a sequential space. Suppose that $B \subset Y$ is sequentially closed. We need to show that $B$ is closed in $Y$ . Because $f$ is a quotient map, $B$ is closed in $Y$ if and only if $f^{-1}(B)$ is closed in $X$ . So we need to show $f^{-1}(B)$ is closed in $X$ .

Suppose $x_n \in f^{-1}(B)$ for each $n=1,2,3,\cdots$ and the sequence $x_n$ converges to $x \in X$ . Then $f(x_n) \in B$ . Since the map $f$ is continuous, $f(x_n) \mapsto f(x)$ . Since $B$ is sequentially closed, $f(x) \in B$ . This means $x \in B$ . Thus $f^{-1}(B)$ is sequentially closed in $X$ . Since $X$ is sequential, $f^{-1}(B)$ is closed in $X$ . $\blacksquare$

Corollary 3
Every quotient space of a first countable space is sequential. Every quotient space of a Frechet space is sequential.

Some Examples

Example 1. First countable $\nLeftarrow \$ Frechet
The example is defined in the post An example of a quotient space, I. This is a non-first countable example. There is only one non-isolated point $p$ in the space. It is easy to verify it is a Frechet space.

Example 2. Frechet $\nLeftarrow \$ Sequential
We consider the space $Y$ defined in the post An example of a quotient space, II. Note that the space $Y$ is the quotient image of a first countable space. Thus $Y$ is sequential by Corollary 3. Consider the subspace $Z=\left\{(0,0)\right\} \cup V$ . Within $Z$ , no sequence of points in $V$ can converge to the point $(0,0)$ . However, $(0,0)$ is a limit point of $V$ . Thus $Z$ is not sequential. By Theorem 1, $Y$ is not Frechet.

Example 3. Sequential $\nLeftarrow \$ k-space
Any compact space is a k-space. Let $\omega_1$ be the first uncountable ordinal. Then $\omega_1+1=[0,\omega_1]$ with the ordered topology is compact. Note that $[0,\omega_1)$ is sequentially closed but not closed. Thus $[0,\omega_1]$ is not sequential.

Example 4. A space that is not a k-space
This example is also defined in the post An example of a quotient space, II. Consider the subspace $Z=\left\{(0,0)\right\} \cup V$ . Every compact subset of $Z$ is finite. So $V \cap K$ is closed in $K$ for every compact $K \subset Z$ . But $V$ is not closed.

Comments About Subspaces

Which of the four properties discussed here are preserved in subspaces? Or which of them are hereditary? It is fairly straightforward to verify that first countability is hereditary and so is the property of being Frechet. By Theorem 1, for any sequential space that is not Frechet has a subspace that is not sequential. Thus the property of being a sequential space is not hereditary. However, closed subspaces and open subspaces of a sequential space are sequential.

The property of being a k-space is also not hereditary. The space $Y$ defined in An example of a quotient space, II is a sequential space (thus a k-space). Yet the subspace $Z=\left\{(0,0)\right\} \cup V$ is not a k-space.

Comments About Quotient Mappings

Continuous image of a first countable space needs not be first countable. The other three properties (Frechet, sequential and k-space) are also not necessarily preserved by continuous mappings. A quick example is to consider any space $X$ that does not have any one of the four properties. Then consider $D=X$ with the discrete topology. Then the indentity map from $D$ onto $X$ is continuous.

Example 1 shows that the property of being first countable is not preserved by quotient map. Example 2 shows that the Frechet property is not preserved by quotient map. Theorem 2 shows that the property of being sequential space is preserved by quotient map. We have the following theorem about k-spaces under quotient map.

k-spaces

The spaces that are k-spaces are called compactly generated spaces. In a k-space, the closed sets and open sets are generated by compact sets. For example, for a k-space $X$ , $A \subset X$ is closed in $X$ if and only if $A \cap K$ is closed in $K$ for every compact $K \subset X$ . Let’s take another look at sequential spaces. The following definition is equivalent to the definition of sequential space given above:

$A \subset X$ is closed in $X$ if and only if $A \cap K$ is closed in $K$ for every compact $K \subset X$ of the form $\left\{x\right\} \cup \left\{x_1,x_2,x_3,\cdots\right\}$ where the $x_n$ are a convergent sequence and $x$ is the sequential limit.

Thus the sequential spaces are compactly generated by a special type of compact sets, namely the convergent sequences.

Theorem 4
Quotient images of k-spaces are always k-spaces.

Proof. Let $X$ be a k-space. Let $f:X \mapsto Y$ be a quotient map. We wish to show that $Y$ is a k-space. Suppose $B \subset Y$ is not closed in $Y$ . Since $f$ is a quotient mapping, $f^{-1}(B)$ is not closed in $X$ . Since $X$ is a k-space, there is a compact $K \subset X$ such that $f^{-1}(B) \cap K$ is not closed in $K$ . Let $x \in K$ such that $x \in \overline{f^{-1}(B) \cap K}-(f^{-1}(B) \cap K)$ . We have just produced a compact set $f(K)$ in $Y$ such that $B \cap f(K)$ is not closed in $f(K)$ . Note that $f(x) \in f(K)$ and $f(x)$ is a limit point of $B \cap f(K)$ . This implies that if $B \cap C$ is closed in $C$ for every compact $C \subset Y$ , then $B$ must be closed in $Y$ (i.e. $Y$ is a k-space). $\blacksquare$

The discussion on sequential space continues with the post Sequential spaces, II.

Links to Previous Posts

An example of a quotient space, II
An example of a quotient space, I
The cardinality of compact first countable spaces, III
The cardinality of compact first countable spaces, II
The cardinality of compact first countable spaces, I

Reference

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

Dan Ma's Topology Blog

Expository Articles In General Topology

Tag Archives: k-space

The product of the identity map and a quotient map

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

Sequential spaces, III

k-spaces, I

Sequential spaces, I