Question

Carefully and rigorously prove the following.

Let X be a metric space. Show X is compact if and only if every sequence contains a convergent subse- quence. Hint for (): Argue by contradiction. If there was a sequence with no convergent subsequence, use that sequence to construct an open cover of X, such that every set in the cover contains only a finite number of elements of the sequence. Then use compactness to get a contradiction. Hint for (): Let U, be an open cover. Consider a minimal subcover of UQ. Minimal means that no U. may be removed from the subcover, if it is to remain a cover of X. Use that every sequence contains a convergent subsequence to argue by contradiction that the minimal subcover must be finite.

Answer 1

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meaning that has the finite intersectiei property for closed sets, so x is sequential compact je every sequence has a convergen subsequence conversely det x is sequentially compact- alt fra CX: KEI} be an open cover of X до эс 70 UA На сли уси (х) и I contained in some Gx. Since x is alquen- tially cumpact, it is totally bounded, to there exist a finite Collection of balls of redius o { B grei) i i=1,2,...ng that covers X. Choose Li EI Sit B8 (2) C G4 "Then & Guit l'aliny I is a finite sub cover of x, shared so X is compact. TCS CamScanner