# 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.