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A first countable, countably compact space is sequentially compact

José Carlos Santos's answer

Let $(x_n)_{n\in\mathbb N}$ be a sequence of elements of $X$. There are two possibilities:

  1. There is some $x\in X$ such that $x_n=x$ for infinitely many $n$'s. Then it is obvious that $x$ is a limit of a subsequence of the sequence $(x_n)_{n\in\mathbb N}$.
  2. For every $x\in X$, the equality $x_n=x$ holds for finitely many $n$'s only. In that case, the set $\{x_n\,|\,n\in\mathbb N\}$ is infinite. Since $X$ is limit point compact, there is some $x\in X$ which is a limit of a sequence of elements of $\{x_n\,|\,n\in\mathbb N\}$. Therefore, $x$ is a limit of some subsequence of the sequence $\{x_n\,|\,n\in\mathbb N\}$.

I cannot really see if we have one more assumption such as Compactness not Countably Compactness, how this assumption ease the proof?

Jale'de jaled
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1 Answers1

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The proof for a compact space is exactly the same, we need only that for a countably infinite subset $A$ there is a strong limit point i. e. a $p \in X$ such that for every neighbourhood $O$ of $p$ we have that $A\cap O$ is infinite. It is well-known that countable compactness is equivalent to this fact.

In either case we then proceed to exploit the first countability at $p$ to construct the required convergent subsequence to $p$.

Compactness does not help the proof at all. But as compactness implies countable compactness the theorem is stated with the weakest assumption only.

Henno Brandsma
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