# Several ways to define countably tight spaces

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

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Several definitions

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

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

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

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

then $A$ is closed.

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

then $A$ is closed.

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

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

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

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

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

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Generalizations

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

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

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

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

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

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

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

then $A$ is closed.

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

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

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

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

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

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

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

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

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Examples

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

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

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

Sequential $\Longrightarrow$ countably tight.

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

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

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

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

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

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

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

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

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

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

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

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

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Remarks

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

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

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Exercises

Exercise 1
This is to verify Example 1. Verify that

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

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

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Reference

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

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

# Sequentially compact spaces, II

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

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

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

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

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

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

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

Reference

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

# An observation about sequential spaces

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

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Henkel, D. Solution to Monthly Problem 5698, American Mathematical Monthly 77, p. 896, 1970
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

# A note about the Arens’ space

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

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

# k-spaces, II

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:

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

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, III

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 $X$:

1. $X$ is herditarily Lindelof.
2. There are no non-trivial convergent sequences in $X$.
3. All compact subsets of $X$ are finite.
4. $X$ 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

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