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 is a limit point of if every open subset of containing contains a point of distinct from . On the other hand, a space is sequentially compact if every sequence of points of has a subsequence that converges. We present examples showing that the notion of sequentially compactness is different from compactness.

Let be the first uncountable ordinal. The space of all countable ordinals with the ordered topology is sequentially compact and not compact. Let be a sequence of points in . Let . If is finite, then the sequence is eventually constant and thus has a convergent subsequence. So assume is an infinite set. Then we can choose an increasing sequence of integers such that . Let be the least upper bound of all . Then subsequence converges to .

The notion of sequentially compactness is not to be confused with the notion of being a sequential space. The space , the space of countable ordinals with one additional point at the end, is a sequentially compact space for the same reason that is sequentially compact. However, is not sequential. Note that is a sequentially closed set but not closed in . On the other hand, being a sequential space does not imply compactness or sequentially compactness, e.g. the real line .

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 be the unit interval. Let , the two-point discrete space. Let be the product space of uncountably many copies of indexed by . We show that is not sequentially compact. To this end, we define a sequence that has not convergent subsequence.

For any , let be the greatest integer less than or equal to . For each and for each , let be:

.

For example, if , then , and . For each , define by the following:

With the above example, , , , and so on. In general, if the decimal place of the number is less then , then . Otherwise .

Let’s observe that if converges to , then converges to for each (hence the product topology is called the topology of pointwise convergence). We claim that the sequence has no convergence subsequence. To this end, we show that each subsequence of does not converge at some .

Let be any increasing sequence of positive integers. We define such that is an alternating sequence of zeros and ones. Consider satisfying the following:

For each , the decimal place of is ,

For each , the decimal place of is ,

For each , the decimal place of is and so on.

For example, if , and , then let . With this in mind, is an alternating sequence of zeros and ones.

*Reference*

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

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.