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I really need help with this question: Prove that a metric space which contains a sequence with no convergent subsequence also contains an cover by open sets with no finite subcover.

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    Do you know what a contrapositive is? Are you familiar with the various characterizations of compactness?2012-06-10
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    Just so you know, a topological space is *sequentially compact* if every sequence has a convergent subsequence, and *compact* if every open cover admits a finite subcover. In the case of a metric space, these are equivalent.2012-06-10

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