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.