If X is countably compact, then every sequence in X has a cluster point?

Question

It is the first part of problem 4.40 in Folland's Real Analysis. I found some answers online, but all of them regard this as a collory of "Countably Infinite Set in Countably Compact Space has $\omega$-Accumulation Point". For example, the proofwiki: https://proofwiki.org/wiki/Countably_Infinite_Set_in_Countably_Compact_Space_has_Omega-Accumulation_Point

I think this is not what Folland want us to do. Anyone has a direct approach to this? Thanks!

Answer 1

Folland does want you to do something like that, but that result by itself isn’t quite enough: it covers only one of two cases. Here’s a sketch of the argument; I’ll leave you to fill in the details.

Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in the countably compact space $X$.

• Show that if there is an $x\in X$ such that $\{n\in\Bbb N:x_n=x\}$ is infinite, then $x$ is a cluster point of $\sigma$.
• Otherwise, show that $\sigma$ has an infinite subsequence $\langle x_{n_k}:k\in\Bbb Z^+\rangle$ of distinct terms, so that $A=\{x_{n_k}:k\in\Bbb Z^+\}$ is an infinite set. Then use the countable compactness of $X$ to show that the set $A$ has an accumulation point $p\in X$ and verify that $p$ is a cluster point of the sequence $\sigma$.

Answer 2

If you remember the proof of the equivalent condition of compact space in the language of nets.（Th4.29）You 'll find that you can just copy that proof. You only need to show that countably compact is equivalent to every countable family of closed sets which has the finite intersection property has nonempty intersection.