An observation about sequential spaces

Posted on August 22, 2010 by Dan Ma

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

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

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

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

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

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

Reference

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

A note about the Arens’ space

Posted on August 18, 2010 by Dan Ma

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

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

The points in $\mathbb{N} \times \mathbb{N}$ are isolated;
The neighborhoods at each $n \in \mathbb{N}$ are of the form $B_{n,m}=\left\{n\right\} \cup \left\{(n,j) \in \mathbb{N} \times \mathbb{N}:j \ge m\right\}$ for some $m \in \mathbb{N}$ ;
The neighborhoods at $\infty$ are obtained by removing from $X$ finitely many $B_{n,1}$ and by removing finitely many isolated points in each of the remaining $B_{n,1}$ .

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

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

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

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

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

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

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

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

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

Reference

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

k-spaces, II

Posted on August 3, 2010 by Dan Ma

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

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

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

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

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

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

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

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Expository Articles In General Topology

Monthly Archives: August 2010

An observation about sequential spaces

A note about the Arens’ space

k-spaces, II