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)}<w_{n(2)}<w_{n(3)}<\cdots. 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)<n(2)<n(3)<\cdots 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)<j \le n(2), the j^{th} decimal place of t is 1,

For each n(2)<j \le n(3), 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.


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

One thought on “Sequentially compact spaces, I

  1. Thank you for the article. There is a minor error in the second paragraph. The sequence need not be eventually constant even if the set A is finite (consider a sequence of alternating 0’s and 1’s). What can be said is that some element of A is hit infinitely-often, which gives us a constant subsequence.

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