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來點難的!Let (M,d) be a complete metric space and let K be a compact set in M(a)Prove that every closed subset of K is compact(b)If C is closed in M,Prove that C∩K is compact

2006-08-04 19:54:27 · 1 個解答 · 發問者 ? 7 in 科學 數學

1 個解答

Proof.(a)Let F be a closed subset of K, and let Γ = {Vα} be an open cover of F (meaning each Vα is open in M and ∪α Vα contains F).Since M∖F is open and (∪α Vα)∪(M∖F) contains K, the collectionΓ2 = {M∖F}∪Γis an open cover for K. K is compact, so by definition there exists a finite subcollection Γ3 of Γ2 which covers K. Since M∖F and F are disjoint, Γ4 = Γ3 ∖ {M∖F}still covers F, and Γ4 is a finite subcollection of Γ. We have found a finite subcover for the arbitrary open cover Γ of F. Hence, F is compact.∴ Every closed subset of K is compact.(b)C∩K is a closed subset of K, so by (a), C∩K is compact. ∎

2006-08-05 02:08:06 · answer #1 · answered by ? 6 · 0 0

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