Let $X$ be a compact Hausdorff topological space.
Is it true that every sequence has a converging subsequence?
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Now that from the answers you know about the terminology sequentially compact space, you can also use pi-base to search for examples. – Martin Sleziak Jan 27 '17 at 07:27
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See also Compactness / sequentially compact and other posts linked there. – Martin Sleziak Jan 27 '17 at 07:40
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No, there are compact Hausdorff spaces without any non-trivial (trivial meaning eventually constant) convergent subsequences,like $\beta \omega$, the Cech-Stone compactification of the natural numbers. compact non-sequentially compact spaces like $[0,1]^I$ for $I$ of size $|\mathbb{R}|$. For so-called sequential spaces, which include the first countable and the metric ones, this implication does hold. The Bolzano-Weierstrass peoperty is now called being sequentially compact, BTW.
Henno Brandsma
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"Every sequence has a convergent subsequence" is the property nowadays called "sequentially compact". It is not equivalent to "compact" for Hausdorff spaces. But it is equivalent for metric spaces.
GEdgar
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