# 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.

## 2 thoughts on “Sequential spaces, III”

1. Daniel,

Thanks for your site. I have been trying to find the following, and it looks like you might know the answer. Are there locally Hausdorff spaces in which every infinite compact set has nonempty interior?

Aloha,

Wayne Lewis

2. Sorry, Daniel. I’m missing the word compact in my post. It should read:

Are there locally compact Hausdorff spaces in which every infinite compact set has nonempty interior?

Aloha,

Wayne Lewis