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Homesequentially compact
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sequentially compact
A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.
Every sequentially compact space is countably compact. Conversely, every first countable countably compact space is sequentially compact. The ordinal space $W(2\omega_{1})$ is sequentially compact but not first countable, since $\omega_{1}$ has not countable local basis.
Next, compactness and sequential compactness are not compatible. In other words, neither one implies the other. Here’s an example of a compact space that is not sequentially compact. Let $X=I^{I}$, where $I$ is the closed unit interval (with the usual topology), and $X$ is equipped with the product topology. Then $X$ is compact (since $I$ is, together with Tychonoff theorem). However, $X$ is not sequentially compact. To see this, let $f_{n}:I\to I$ be the function such that for any $r\in I$, $f(r)$ is the $n$th digit of $r$ in its binary expansion. But the sequence $f_{1},\ldots,f_{n},\ldots$ has no convergent subsequences: if $f_{{n_{1}}},\ldots,f_{{n_{k}}},\ldots$ is a subsequence, let $r\in I$ such that its binary expansion has its $k$th digit $0$ iff $k$ is odd, and $1$ otherwise. Then $f_{{n_{1}}}(r),\ldots,f_{{n_{k}}}(r),\ldots$ is the sequence $0,1,0,1,\ldots$, and is clearly not convergent. The ordinal space $\Omega_{0}:=W(\omega_{1})$ is an example of a sequentially compact space that is not compact, since the cover $\{W(\alpha)\mid\alpha\in\Omega_{0}\}$ has no finite subcover.
When $X$ is a metric space, the following are equivalent:

$X$ is sequentially compact.

$X$ is limit point compact.

$X$ is compact.

$X$ is totally bounded and complete.
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Comments
don't you need separability?
Are you sure that
"Sequential compactness is equivalent to compactness when $X$ is a metric space."?
I can prove this claim only when $X$ is metric and separable.
Em.
Re: don't you need separability?
Munkres proves this as part of Theorem 28.2 in Topology, 2nd ed. (pp. 179180).
In particular, Munkres proves that if $X$ is sequentially compact metric, then $X$ is totally bounded, that is, given a positive $\epsilon$, there is a finite covering of $X$ by $\epsilon$balls. Hence $X$ is Lindel\"{o}f. Because $X$ is also metric, this implies that $X$ is separable. So the assumption that $X$ is separable is not necessarily to show that sequential compactness implies compactness.
The proof that compactness implies sequential compactness does not use separability.
Re:
I try to prove that "Sequentially compactness is equivalent to compactness when X is metric space", if you interested in this problem, can you give me some help?