# 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

# Sequentially compact spaces, I

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 counable open cover has a finite subcover, or equivalently, every countably infinite subset has a limit point. The limit points contemplated here are from the topological point of view, i.e. 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. We present examples showing that the notion of sequentially compactness is different from compactness.

Let $\omega_1$ be the first uncountable ordinal. The space of all countable ordinals $W=[0,\omega_1)$ with the ordered topology is sequentially compact and not compact. Let $\left\{w_n\right\}$ be a sequence of points in $W$. Let $A=\left\{w_n:n=1,2,3,\cdots\right\}$. If $A$ is finite, then the sequence $\left\{w_n\right\}$ is eventually constant and thus has a convergent subsequence. So assume $A$ is an infinite set. Then we can choose an increasing sequence of integers $n(1),n(2),n(3),\cdots$ such that $w_{n(1)}. Let $\alpha<\omega_1$ be the least upper bound of all $w_{n(j)}$. Then subsequence $w_{n(j)}$ converges to $\alpha$.

The notion of sequentially compactness is not to be confused with the notion of being a sequential space. The space $[0,\omega_1]=\omega_1+1$, the space of countable ordinals with one additional point $\omega_1$ at the end, is a sequentially compact space for the same reason that $[0,\omega_1)$ is sequentially compact. However, $[0,\omega_1]$ is not sequential. Note that $[0,\omega_1)$ is a sequentially closed set but not closed in $[0,\omega_1]$. On the other hand, being a sequential space does not imply compactness or sequentially compactness, e.g. the real line $\mathbb{R}$.

For discussion of sequential spaces, see Sequential spaces, I, Sequential spaces, II and Sequential spaces, III.

We now present an example of a compact space that is not sequentially compact. Let $I=[0,1]$ be the unit interval. Let $2=\left\{0,1\right\}$, the two-point discrete space. Let $X=2^{I}$ be the product space of uncountably many copies of $2=\left\{0,1\right\}$ indexed by $I$. We show that $X$ is not sequentially compact. To this end, we define a sequence $\left\{f_n\right\}$ that has not convergent subsequence.

For any $y \in \mathbb{R}$, let $[y]$ be the greatest integer less than or equal to $y$. For each $t \in I$ and for each $n=1,2,3,\cdots$, let $t_n$ be:

$t_n=10^n t-[10^n t]$.

For example, if $q=\frac{1}{\sqrt{2}}=0.7071067811 \cdots$, then $q_1=0.071067811 \cdots$, $q_2=0.71067811 \cdots$ and $q_3=0.1067811 \cdots$. For each $n=1,2,3,\cdots$, define $f_n:I \mapsto 2$ by the following:

$\displaystyle f_n(t)=\left\{\begin{matrix}0&\thinspace t_n <0.5 \\{1}&\thinspace t_n \ge 0.5 \end{matrix}\right.$

With the above example, $f_1(q)=0$, $f_2(q)=1$, $f_3(q)=0$, $f_4(q)=0$ and so on. In general, if the $(n+1)^{st}$ decimal place of the number $t$ is less then $5$, then $f_n(t)=0$. Otherwise $f_n(t)=1$.

Let’s observe that if $g_n \in X=2^{I}$ converges to $g \in X$, then $g_n(t) \in X=2^{I}$ converges to $g(t) \in X$ for each $t \in I$ (hence the product topology is called the topology of pointwise convergence). We claim that the sequence $\left\{f_n\right\}$ has no convergence subsequence. To this end, we show that each subsequence of $\left\{f_n\right\}$ does not converge at some $t \in I$.

Let $n(1) be any increasing sequence of positive integers. We define $t \in I$ such that $f_{n(1)}(t),f_{n(2)}(t),f_{n(3)}(t),\cdots$ is an alternating sequence of zeros and ones. Consider $t \in I$ satisfying the following:

For each $j \le n(1)$, the $j^{th}$ decimal place of $t$ is $9$,

For each $n(1), the $j^{th}$ decimal place of $t$ is $1$,

For each $n(2), the $j^{th}$ decimal place of $t$ is $9$ and so on.

For example, if $n(1)=2$, $n(2)=5$ and $n(3)=9$, then let $t=0.9911199991 \cdots$. With this in mind, $f_{n(1)}(t),f_{n(2)}(t),f_{n(3)}(t),\cdots$ is an alternating sequence of zeros and ones.

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, IV

Uncountable product of first countable spaces can never be first countable (see The product of first countable spaces). It turns out that uncountable products of first countable spaces cannot even be sequential. In this post we show that the product of uncountably many Hausdorff spaces, each of which has at least two points, can never be sequential. This follows from the fact that the product space $2^S$ is not sequential where $2=\left\{0,1\right\}$ is the two-point discrete space and $S$ is any uncountable set. The space $2^S$ can be embedded as a closed subspace of a product of uncountably many Hausdorff spaces, each of which has at least two points.

For discussion on this blog about sequential spaces, see Sequential spaces, I, Sequential spaces, II and Sequential spaces, III.

Let $S$ be an uncountable set. Let $X=2^S$ be the product of uncountably many copies of $2=\left\{0,1\right\}$ indexed by the set $S$. Let $Y$ be all points $x \in X$ such that $x(\alpha)=x_\alpha=0$ for all but countably many $\alpha \in S$. In other words, $Y$ is the $\Sigma$-product of $2=\left\{0,1\right\}$. Clearly $Y$ is not closed in $X$. We claim that $Y$ is sequentially closed in $X$.

For each $z \in Y$, let $W_z=\left\{\alpha \in S:z_\alpha \ne 0\right\}$. Note that each $W_z$ is countable. Suppose that $\left\{y_n\right\}$ is a sequence of points of $Y$ such that $y_n \rightarrow y \in X$. We show that $y \in Y$. Let $W$ be the union of all $W_{y_n}$, which is a countable set. For all $\alpha \in S-W$, $y_n(\alpha)=0$ and thus $y(\alpha)=y_\alpha=0$. Thus $y \in Y$. This shows that $Y$ is sequentially closed in $X$.It follows that $2^S$ is not sequential.

As stated at the beginning of the post, any uncountable product of Hausdorff spaces, each of which has at least two points, can never be sequential. Consequently, the property of being a sequential spaces is not preserved in uncountable products.

# 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.