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I am reading this Bolzano-Weierstrass property saying that a metric space X is sequentially compact if every sequence in X has a convergent subsequence. I also read a theorem saying that sequentially compact is equivalent to compact.

So I am thinking of this sequence$a_{2k-1}=1, a_{2k}=n$. this sequence obviously has a convergent subsequence, but it is not bounded above. How to prove that an unbounded space is compact? given that compact preserves boundedness in real metric.

Halk
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  • " I also read a theorem saying that sequentially compact is equivalent to compact." This is true for metric spaces but false (in both directions) for general topological spaces. See https://math.stackexchange.com/questions/152447/compactness-sequentially-compact – Anne Bauval Apr 14 '23 at 01:10
  • What you call "Bolzano-Weierstrass" "property" is the definition of sequential compactness, for general topological spaces (not necessarily metrizable). – Anne Bauval Apr 14 '23 at 01:17
  • Your last sentence "compact preserves boundedness in real metric" makes no sense. – Anne Bauval Apr 14 '23 at 01:20

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An unbounded metric space is not compact, as given any $x \in X$, the open balls $B(x, n)$ for $n \in \mathbb{N}$ is an open cover that does not have a finite subcover.

The sequence above does indeed have a convergent subsequence, but to apply the property that sequentially compact is equivalent to compact for metric spaces, you need all sequences to have a convergent subsequence, not just that particular one.

David Lui
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