I want to prove that if $X$ is not countably compact, then there exists a countable subset $\{x_n:n\in\mathbb{N}\}$ and has no accumulation points.

If $X$ is not CC, then there exists an open cover of $X$, $\{A_i:i\in\mathbb{N}\}$ without a finite subcover of $X$. If we choose $x_i\in A_i$, can we say that $\{x_i:i\in\mathbb{N}\}$ has no accumulation points?


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    $\begingroup$ It's not so easy. You need to choose your $x_i$ appropriately, arbitrary choice of $x_i$ may produce a set with accumulation points. $\endgroup$ – Daniel Fischer Feb 13 '14 at 15:32

You should start by choosing a point $x_1\in A_1$. Then choose a point $x_2\in A_{m_2}-A_1$ where $m_2$ is the minimal number so that $A_{m_2}\not\subseteq A_1$ (This exists because there is no finite subcover). By induction we can choose a sequence $x_1,x_2,...$ such that $x_n\in A_{m_n}-\bigcup_{i<m_n} A_{i}$. Assume that this sequence has a cluster point $c$. Can you get a contradiction by using that fact that $(A_i)_i$ is an open cover and every neighborhood of $c$ contains infinitely many $x_j$ ?


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