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.
Prove that a metric space which contains a sequence with no convergent subsequence also contains an cover by open sets with no finite subcover.
3
$\begingroup$
real-analysis
general-topology
metric-spaces
compactness
-
3Do you know what a contrapositive is? Are you familiar with the various characterizations of compactness? – 2012-06-10
-
0Just 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