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Dragonfall
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How do I prove that every metric space that is sequentially compact and separable is compact? I can't seem to use either hypotheses.
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Dragonfall said:X is sequentially compact and separable, and hence if an open cover A does not have a finite subcover, it has a countable subcover since we can map a dense subset into a countable subcover of A.
Take an element that is in X\Ai for i=1 to infty and form a sequence, that sequence has a convergent subsequence that cannot converge to anything in the subcover, and hence outside X.
Sequential compactness is a property of a topological space, which means that every sequence in the space has a convergent subsequence.
Sequential compactness only requires that every sequence has a convergent subsequence, while compactness requires that every open cover has a finite subcover. Therefore, sequential compactness is a weaker condition than compactness.
Examples of sequentially compact spaces include closed and bounded intervals of real numbers, compact subsets of Euclidean spaces, and the Cantor set.
Yes, every compact space is also sequentially compact. This is a consequence of the Bolzano-Weierstrass theorem, which states that every bounded sequence in Euclidean space has a convergent subsequence.
Sequential compactness is an important concept in mathematics, particularly in the study of topological spaces and analysis. It is used to prove many important theorems, such as the Heine-Borel theorem, and is also a key property in the definition of a complete metric space.