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Theorem 2.24. S (a) For any collection fG g 2A of open sets, G = 2A G is open. T (b) For any collection fF g 2A of closed sets, F is closed. 2A Tn (c) For any …nite collection G1 ; : : : ; Gn of open sets, H = i=1 Gi is open. Sn (d) For any …nite collection F1 ; : : : ; Fn of closed sets, i=1 Fi is closed. Proof. (a) Let x 2 G. Then x 2 G for some 2 A: Since G is open, x is an interior point of G : Since G G, x is an interior point of G. We have proved G is open. T S c (b) = F c : For each 2 A, since F is closed, F c is open. 2A F T S 2A c c , we Part (a) implies 2A F 2A F is open. Since this equal to T that conclude that F is closed. 2A (c) Let x 2 H. Then x 2 Gi for each i = 1; : : : ; n: Since each Gi is open, there exists ri > 0 such that Nri (x) Gi . De…ne r = min fr1 ; : : : ; rn g. Since r ri for each i, we have Nr (x) Tn Nri (x), which implies Nr (x) Gi . Since this is true for each i; Nr (x) i=1 Gi = H. We have proved that H is open. Sn Tn c c c (d) ( i=1 F ) = F : Since each Fi is closed, open. By (c) we i i=1 i Tn Sn Fi is c c have that F is open. Since this is equal to ( F ) , we conclude that i i=1 i i=1 Sn F is closed. i i=1 Déjà vu: De…nition A topological space is a set X together with a collection of subsets of X, called open sets, with the following properties: (i) ? and X are open sets. (ii) Any union of open sets is open. (iii) Any …nite intersection of open sets is open. A closed set is de…ned to be the complement of an open set. Given a set E, let E 0 denote the set of limit points of E. The closure is de…ned by E = E [ E 0 . Theorem 2.27. If X is a metric space and E X; then (a) E is closed, (b) E = E if and only if E is closed, (c) E F for every closed set F X such that E F: Proof. (a) We show that (E)c is open. Let x 2 (E)c . Since x 2 = E, we have that x 2 = E and x is not a limit point of E. Thus there exists a neighborhood Nr (x) such that Nr (x) \ E = ?. Claim. Nr (x) \ E 0 = ?. 1 The claim implies that Nr (x) \ E = ?, i.e., Nr (x) implies that (E)c is open. Part (a) follows. (E)c . This, in turn, Proof of the claim: Suppose there is a y 2 Nr (x)\E 0 . Let s = r d (y; x) > 0. Then Ns (y) Nr (x). Since y 2 E 0 , there exists z 2 Ns (y) \ E; which implies z 2 Nr (x) \ E: This contradicts Nr (x) \ E = ?. (b) (i) Suppose E = E. Since E is closed by part (a), we conclude that E is closed. (ii) Suppose E is closed. Then, by de…nition, E 0 E. Hence E = E [ E 0 = E. (c) Let F X be a closed set with E F: Let x 2 E 0 . Since x is a limit point of E and since E F , x is a limit point of F . Since F is closed, x 2 F . Therefore E 0 F . Since E F and E = E [ E 0 , we conclude that E F . 2