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Every sequentially compact space is countably compact.

The most that I can get out of this $(\Rightarrow)$ is that ${x_{n_j}} \subseteq \bigcup_{i=1}^{n_0}O_{x_i}$, where $O_{x_i}$ is open neighborhoods of the finite sequence points and $O_{x_{n_0}}$ is the open set containing the infinitely many points beyond $x_{n_0}.$ I'm a bit stuck beyond this. I tried relating this to: $X$ is countably compact iff the intersection of every sequence of nonempty nested closed proper subsets of $X$ is nonempty, but I couldn't figure out how to show a contradiction.

Anyone have any ideas?

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    $\begingroup$ You’ll find a proof in the statement of this question; ignore the title to the question. $\endgroup$ Jun 7 '16 at 22:13
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The finite case is trivial. Consider {N(x_1),N(x_2),N(x_3),...} an arbitrary countable open cover. {x_1,x_2,x_3,...} partitions into finitely many subse uences all of which are convergent. Without loss of generality {x_1,x_2,x_3,...} converges to x_infinity. But x_infinity is in N(x_k) for some k. It follows that we have a finite subcover.

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  • $\begingroup$ Is $A$ the topological space? $\endgroup$
    – Oliver G
    Jun 8 '16 at 2:04
  • $\begingroup$ ^Irrelevant. The variable is only used once. Deleted. $\endgroup$ Jun 8 '16 at 2:42
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Use the fact that $X$ is countably compact iff every infinite set $A$ has a point $p$ such that for every open neighbourhood $O$ of $p$, $O \cap A$ is infinite.

For a proof, see your own question here. Such a $p$ is called an ($\omega$-)accumulation point of $A$.

So if $A$ is infinite, pick $a_n, n \in \mathbb{N}$, all different, in $A$. This defines a sequence $(a_n)_n$ in $X$, so it has a convergent subsequence $(a_{n_k})_k$ by sequential compactness of $X$, and say that $p$ is a limit of this convergent subsequence.

Now if $O$ is any open neighbourhood of $p$, it will contain all $a_{n_k}$ for $k \ge K_0$ for some $K_0 \in \mathbb{N}$, by convergence of the subsequence. In particular, $O \cap A$ is infinite. The above characterisation shows we are done: $X$ is countably compact, as we can find an accumulation point $p$ for every infinite $A \subseteq X$.

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Let $X=\bigcup\limits_n V_n$ and suppose there is no finite subcover. Then for all $n$ we have $V_1\cup V_2\cup\dots\cup V_n\subsetneq X$. So pick $x_n\in X\setminus V_1\cup V_2\cup\dots\cup\ V_n$. Then $(x_n)$ is a sequence satisfying $x_n\notin V_m$ for all $m\leq n$. Let $(x_{n_k})$ be a convergent subsequence converging to some $x$. There is some $V_N$ containing $x$. So all but finitely many terms of $(x_{n_k})$ lie in $V_N$ which is a contradiction.

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