# A note on products of sequential fans

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

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

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Sequential fans

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

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

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

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

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

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

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

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

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

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

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Examples

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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The set-theoretic angle

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

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

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

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

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The bounding number

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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The collectionwise Hausdorff property

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

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

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

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

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Remarks

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

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Exercise

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

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

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Reference

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

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

# 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, I

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:

1. $X$ is a k-space.
2. $X$ 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. Call this mapping $f$ where $f:Y \mapsto X$. 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 $X$. Clearly, $\tau \subset \tau_f$. 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 $K$. We can consider $f^{-1}(O) \cap K$ as a subset of $K \subset Y$ and as a subset of $K \subset X$. 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, $f(y) \in \overline{A \cap f(\overline{U})}$. Thus $A$ is not a compactly generated closed set in $X$. $\blacksquare$

Reference

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

# Sequential spaces, I

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:

1. 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 $x$.
2. 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 $x \in A$.
3. 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.

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.

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.

Reference

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

# An example of a quotient space, II

This post presents another example of a quotient space of a first countable space. The resulting quotient space is not first countable. After we first define the space without referring to the concepts of quotient space, we show that this example is a quotient space. This example will be further discussed when we discuss sequential spaces.

For a previous discussion on quotient space in this blog, see An example of a quotient space, I. For more information on quotient spaces in general, see [2].

Let $\mathbb{N}$ be the set of all positive integers. For each $i \in \mathbb{N}$ and $j \in \mathbb{N}$, let $V_{i,j}=\left\{(\frac{1}{i},\frac{1}{k}):k \ge j\right\}$. Let $V=\bigcup \limits_{i=1}^{\infty} V_{i,1}$. Let $H=\left\{(\frac{1}{i},0):i \in \mathbb{N}\right\}$. For each $i \in \mathbb{N}$, let $V_i=V_{i,1} \cup \left\{(\frac{1}{i},0)\right\}$. Define the space $Y$ as:

$Y=\left\{(0,0)\right\} \cup H \cup V$.

The set $H$ is the horizontal part of the space and the set $V$ is the vertical part of the space. The origin $(0,0)$ is an additional point. In the topology for the space $Y$, each point in $V$ is isolated. Each point $(\frac{1}{i},0) \in H$ has an open neighborhood of the form

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

We wish to discuss $Y$ and its subspace $Z=\left\{(0,0)\right\} \cup V$. If we consider $Z$ as a topological space in its own right (open sets in $Z$ are of the form $U \cap Z$ where $U$ is open in $Y$), then in $Z$ there are no infinite compact sets (all compact subsets of $Z$ are finite). In particular, no sequence of points $\left\{x_n \in V:n \in \mathbb{N}\right\}$ can converge to the point $(0,0)$. However, with respect to $Z$, $(0,0)$ is a limit point of $V$. This implies that the space $Y$ is not first countable.

A space $W$ is said to be a Frechet space if $w \in \overline{A}$ where $A \subset W$, then there is some sequence $\left\{w_n \in A:n \in \mathbb{N}\right\}$ that converges to the point $w$. Any first countable space is a Frechet space. If there is a local base $\left\{U_n: n \in \mathbb{N}\right\}$ at $w$ and if $w \in \overline{A}$, then whenever we pick $w_n \in A \cap U_n$, $w_n$ would converge to $w$.

The space $Y$ defined above is not a Frechet space. Note that $(0,0) \in \overline{V}$ and no sequence of points in $V$ can converge to $(0,0)$.

Though the space $Y$ defined above is not a first countable space, $Y$ is a quotient space of a first countable space $X$ (in fact, a subspace of the Euclidean plane). Let $E=H \cup V$ where $H$ and $V$ are defined as in the definition of the space $Y$ above. Let $H_{-1}$ be the set $H_{-1}=\left\{(\frac{1}{i},-1):i \in \mathbb{N}\right\}$. Let $X=E \cup H_{-1} \cup \left\{(0,-1)\right\}$ with the Euclidean topology inherited from the Euclidean plane.

Define a quotient space from $X$ by collapsing each pair of points $\left\{(\frac{1}{i},0),(\frac{1}{i},-1)\right\}$ into one point $\left\{(\frac{1}{i},0)\right\}$. This is done for each $i \in \mathbb{N}$. For convenience, the point $\left\{(0,-1)\right\}$ is shifted to $\left\{(0,0)\right\}$. The resulting quotient space is the same as the space $Y$ defined above.

The quotient space we just describe can also be described by the quotient map $f:X \mapsto Y$:

$f((0,-1))=(0,0)$,

$f((\frac{1}{i},-1))=(\frac{1}{i},0)$ for each $(\frac{1}{i},-1) \in H_{-1}$,

$f((\frac{1}{i},0))=(\frac{1}{i},0)$ for each $(\frac{1}{i},0) \in H$,

$f(x)=x$ for each $x \in V$.

Then the following topology coincides with the topology on $Y$ that we define earlier in the post:

$\tau_f=\left\{U \subset Y: f^{-1}(U) \text{ is open in } X\right\}$

Reference

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

# An example of a quotient space, I

It is well known that in a compact Hausdorff space, if every point in the space is a $G_\delta$ point, then the space is first countable (see The cardinality of compact first countable spaces, II). Outside of compact spaces, this is not true. We present an example of a space where every point is a $G_\delta$ point but has no countable local base at any one point. This is an opportunity to introduce the notion of quotient space. Our example can be viewed as the image of a first countable space under a quotient map. We first define the example and then present a brief discussion of quotient spaces.

Let $\mathbb{N}$ be the set of all positive integers. Let $W=\left\{\frac{1}{n}: n \in \mathbb{N}\right\}$. Let $S_1=\mathbb{N} \times W$ and $S_2=\mathbb{N} \times \left\{0\right\}$. Consider $X=S_1 \cup S_2$ as a subspace of the Eucliean plane. In the Euclidean topology, each point is $S_1$ is isolated and an open neighborhood at each point $(i,0)$ in $S_2$ is of the form:

$\left\{(i,0)\right\} \cup \left\{(i,\frac{1}{k}):k \ge n\right\}$ for some $n \in \mathbb{N}$

We use $X$ to define a space $Y$ by considering $S_2$ as one point (call this point $p$). Thus $Y=S_1 \cup \left\{p\right\}$. Points in $S_1$ remain isolated. An open neighborhood of $p$ is of the form $\left\{p\right\} \cup U^-$ where $U^-=U-S_2$ and $U$ is a Euclidean open subset of $X$ such that $S_2 \subset U$.

To facilitate the discussion, we describe the open sets at the point $p$ in more details. Let $\mathbb{N}^{\mathbb{N}}$ be the set of all functions $f:\mathbb{N} \mapsto \mathbb{N}$. For each $i \in \mathbb{N}$ and $j \in \mathbb{N}$, let $V_{i,j}=\left\{(i,\frac{1}{k}): k \ge j\right\}$. An open neighborhood of $p$ is of the following form:

$B_f=\left\{p\right\} \cup \bigcup \limits_{i=1}^{\infty} V_{i,f(i)}$

To describe in words, each open neighborhood at $p$ is obtained by removing finitely many points in each vertical segment $V_{i,1}$.

The resulting space $Y$ is not first countable. We show that any countable set of open neighborhoods at $p$ cannot be a local base at $p$. To this end, we consider $B_{f_1},B_{f_2},B_{f_3},\cdots$ where each $f_i \in \mathbb{N}^{\mathbb{N}}$. We wish to find an open set $O$ such that $p \in O$ and each $B_{f_i}$ is not a subset of $O$.

For each $i \in \mathbb{N}$, pick some $g(i) \in \mathbb{N}$ such that $g(i)>f_i(i)$. Consider $O=B_g$. Note that for each $i \in \mathbb{N}$, $(i,\frac{1}{f_i(i)}) \in B_{f_i}$ and $(i,\frac{1}{f_i(i)}) \notin O=B_g$. This shows that no countable number of open neighborhoods can form a base at the point $p$.

However, every point in the space $Y$ is a $G_\delta$ point. For the point $p$, we have $\left\{p\right\}=\bigcap \limits_{i=1}^\infty B_{h_i}$ where $h_i \in \mathbb{N}^{\mathbb{N}}$ is a constant function mapping to $i$.

Quotient Spaces
We now present one definition of quotient spaces. Let $(X,\tau)$ be a topological space. Let $q:X \mapsto Y$ be a surjection, i.e. $q(X)=Y$. Consider the following collection of subsets of $Y$:

$\tau_q=\left\{O \subset Y: q^{-1}(O) \text{ is open in }X\right\}$

The set $\tau_q$ is a topology for the set $Y$. With this topology $\tau_q$ on $Y$, the function $q$ is continuous. In fact, $\tau_q$ is the finest (largest or strongest) topology on $Y$ that makes the function $q$ continuous, i.e. if $\tau_1$ is another topology on $Y$ making $q$ a continuous function, then $\tau_1 \subset \tau_q$.

The topology $\tau_q$ is called the quotient topology induced on $Y$ by the mapping $q$. When $Y$ is given such a quotient topology, it is called a quotient space of $X$ (most of the times just called quotient space). The induced map $q$ is called a quotient map. If $Y$ has a quotient topology defined by a quotient map, then $Y$ is said to be the quotient image of $X$. We plan to discuss quotient space and quotient topology in greater details in this blog. For further information, see [2].

In the example discussed in this post, the quotient map is to map each point in $S_1$ to itself and to map each point in $S_2$ to a single point $p$. In essence, we collapse the whole x-axis $S_2$ into one single point. The open sets for the point $p$ are simply the Euclidean open sets containing $S_2$. This example shows that the quotient image of a first countable space needs not be first countable. Though the quotient image of a first countable space may not be first countable, it has the property that sequences suffice to define the topology (these are called sequential spaces, see [1]).

Reference

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