Andrew van Herick Math 710 Dr. Alex Schuster Sept. 21, 2005 Assignment 3 Section 10.3: 6, 7ab, 8, 9, 10 10.4: 2, 3 10.3.6. Prove (3) : Let E X: Then x =2 E if and only if B r (x) \ E c 6= ; for all all r > 0: Proof. We ll show the forward implication by contradiction. Let x 2 (E ) c ; and suppose that B r (x) \ E c = ; for some r > 0: Note that B r (x) is non-empty as it always contains x as a member. It is not di cult to see that if y 2 B r (x) then y 2 E (otherwise y 2 E c, i.e. B r (x) \ E c 6= ;). This means B r (x) E: As B r (x) is open, by Theorem 10.34(ii) (p. 303, Wade) B r (x) E which implies that x 2 E ; a contradiction. This establishes that if x =2 E ; then B r (x) \ E c 6= ; for all all r > 0: We ll prove the reverse implication by contraposition. Suppose x 2 E : Because E is an open set in X there exists r > 0 such that B r (x) E : In particular B r (x) E, meaning that B r (x) \ E c = ;: This establishes that x 2 E implies 9r > 0 such that B r (x) \ E c = ;: By contraposition if B r (x) \ E c 6= ; for all all r > 0; then x =2 E :
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Andrew van Herick 3 10.3.7. Show that Theorem 10.40 is the best possible in the following sense. (a) There exist set A; B in R such that (A [ B) 6= A [ B : (b) There exist set A; B in R such that (A \ B) 6= A \ B: Proof. (a) Choose A = [ 1; 0] and B = [0; 1] : Then (A [ B) = [ 1; 1] = ( 1; 1) 6= ( 1; 1) n f0g = ( 1; 0) [ (0; 1) = A [ B : Proof. (b) Choose A = ( 1; 0) and B = (0; 1) : Then (A \ B) = ; = ; 6= f0g = [ 1; 0] \ [0; 1] = A \ B:
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Andrew van Herick 5 10.3.8. Let Y be a subspace of X: (a) Show that a set V is open in Y if and only if there is an open set U in X such that V = U \ Y: (b) Show that a set E is closed in Y if and only if there is a closed set A in X such that E = A \ Y: Let Br Y (a) := fy 2 Y : (y; a) < rg : Proof. (a) Suppose that V is open in Y: De ne S := (r; v) 2 R + V : Br Y (v) V Choose U = S (r;v)2s B r (v) : By Theorem 3.31(i) (p. 301, Wade), U is open in X: Now we ll show that V = U \ Y: Let x 2 V. By assumption V Y; so x 2 Y: By the fact that V is open in Y; there exists r 1 > 0 such that Br Y 1 (x) V; meaning (r 1 ; x) 2 S. This means B r1 (x) U; so we have x 2 U: Hence x 2 U \ Y; which means that V U \ Y: To show the reverse containment let y 2 U \ Y: Then y 2 U, so y 2 B r2 (v 2 ) for some (r 2 ; v 2 ) 2 S: As y 2 Y; this means y 2 Br Y 2 (v 2 ) : But Br Y 2 (v 2 ) V; so y 2 V: Hence U \ Y V: This establishes the existence of an open set U in X such that V = U \ Y: Conversely suppose that there is an open set U in X such that V = U \ Y: It is easy to see that V Y: Let a 2 V: Using the fact that a 2 U; an open set in X, choose r > 0 such that B r (a) U: Let y 2 Br Y (a) : By de nition (y; a) < r; so y 2 B r (a) : In particular this means y 2 U and y 2 Y; so y 2 V: Hence, Br Y (a) V: This establishes that for all a 2 V there exists r > 0 such that Br Y (a) V: By de nition V is open in Y: Proof. (b) Suppose that E is closed in Y: This means that E c is open in Y: De ne T := (r; v) 2 R + Y \ E c : Br Y (v) E c Choose A = T (r;v)2t (B r (v)) c : Note that B r (v) c is closed for all (r; v) 2 T: By Theorem 3.31(iii) (p. 301, Wade), A is closed in X: Now we ll show that E = A\Y: Let x 2 E: By de nition x 2 Y: We ll show that x 2 A by contradiction. Suppose x 2 A c : By de Morgan s Laws for sets A c = S (r;v)2a B r (v) : By
6 similar logic to part (a) ; we can conclude that x 2 Br Y 3 (v 3 ) for some (r 3 ; v 3 ) 2 T: However Br Y 3 (v 3 ) E c ; meaning that x 2 E c : This contradicts our initial assumption that x 2 E: Hence x must belong to A: This means x 2 A \ Y and establishes that E A \ Y: Now we ll show the reverse containment, again by contradiction. Assume y 2 A \ Y and y 2 E c : Because E c is open in Y and y 2 Y; E c there exists r 4 > 0 such that Br Y 4 (y) E c, in particular (r 4 ; y) 2 T: As B r4 (y) S (r;v)2a B r (v) ; we have y 2 A c : This contradicts the assumption that y 2 A \ Y; meaning that y 2 E and establishing that A \ Y E: This proves the existence of a closed set A in X such that E = A \ Y: For the proof of the converse suppose there is a closed set A in X such that E = A \ Y: To show that E is closed in Y; we need to prove that E c \Y is open in Y: Clearly E c \Y Y: Let a 2 E c \ Y and note that E c \ Y = (A c [ Y c ) \ Y = (A c \ Y ) [ (Y \ Y c ) = A c \ Y Hence a 2 A c \Y: Using the fact that A c is open in X choose r 5 > 0 such that B r5 (a) A c : Note that Br Y 5 (a) B r5 (a) A c and that Br Y 5 (a) Y. Hence Br Y 5 (a) A c \ Y = E c \ Y: This establishes that for all a 2 E c \ Y there exists r > 0 such that Br Y (a) E c \ Y; or equivalently that E c is open in Y and E is closed in Y:
Andrew van Herick 7 10.3.9. Let f : R! R: Prove that f is continuous on R if and only if f 1 (I) is open in R for every open interval I: Proof. Let f : R! R: ()) Suppose that f is continuous in R: Let I be an open interval in R: If f 1 (I) = ; then f 1 (I) is open in R: So lets assume that f 1 (I) 6= ;: If I is a nite open interval, then I = (u; v), u < v for some u; v 2 R: Let a 2 f 1 (I) and de ne " = min (f (a) u; v f (a)) ; then " > 0: Using the de nition of continuity choose > 0 such that jx aj < and x 2 R implies jf (x) f (a)j < " jf (y) Now we ll show that B (a) f 1 (I) : Let y 2 B (a) : Then jy f (a)j < ": This means aj < and y 2 R; so u < " + f (a) < f (y) < " + f (a) < v i.e. f (y) 2 I and y 2 f 1 (I) : This establishes that B (a) f 1 (I) : Thus for all a 2 f 1 (I) there exists > 0 such that B (a) f 1 (I) ; meaning that I is open in R. Similar reasoning establishes the same result if one of u; v is nite and the other in nite, i.e. I = ( 1; v) or I = (u; 1) : If neither u nor v are nite, then I = ( 1; 1) = R, meaning that f 1 (I) = R is open in R: This covers all cases and establishes that if f is continuous on R; then f 1 (I) is open in R for every open interval I: (() Conversely, suppose that f 1 (I) is open in R for every open interval I: Let a 2 R and " > 0: Then I " := (f (a) "; f (a) + ") is an open interval. By hypothesis f 1 (I " ) is open in R. As a 2 f 1 (I " ) ;(recall that f (a) 2 I e ) choose > 0 such that B (a) f 1 (I " ) : Assume that jx aj < and x 2 R: Then x 2 B (a) f 1 (I " ) ; so f (x) 2 I " : Hence f (a) " < f (x) < f (a) + " " < f (x) f (a) < "
8 This establishes that for all " > 0 there exists > 0 such that jx aj < and x 2 R implies jf (x) f (a)j < " meaning that f is continuous at a: By generalization f is continuous at every a 2 R; so f is continuous on R:
Andrew van Herick 9 10.3.10. Let V be a subset of a metric space X: (a) Prove that V is open in X if and only if there is a collection of open balls fb : 2 Ag such that V = [ 2A B : (b) What happens to this result if "open" is replaced by "closed"? Proof. (a) Suppose that V is open in X: Let A := (r; x) 2 R + X : B r (x) V Choose a collection of open balls fb r (x) : (r; x) 2 Ag : Now we ll show that V = S (r;x)2a B r (x) : Suppose that v 2 V: Because V is open there exists > 0 such that B (v) V: By de nition (; v) 2 A; so B (v) [ B r (x) : (r;x)2a As v 2 B (v) ; v 2 S (r;x)2a B r (x) ; establishing that V S 2A B : To show the reverse containment, let u 2 S (r;x)2a B r (x) : Then u 2 B r0 (x 0 ) for some (r 0 ; x 0 ) 2 A: By the requirements for membership of (r 0 ; x 0 ) in A we have B r0 (x 0 ) V: This means that u 2 V and establishes that S (r;x)2a B r (x) V: As every open ball is open in X, the converse follows by Theorem 3.31 (p. 301, Wade). Proof. The result if false if "open" is replaced by "closed". Let X = R and V = N: Suppose by way of contradiction the result were true for this set assignment, then because N is closed in R, N = [ B 2A for some index set A 6= ; (otherwise N = S 2; B = ;). This means that there exists B 2 fb : 2 Ag : But B = B r (x) for some r > 0; and some x 2 R: Using the density of the irrationals we can choose z 2 RnQ; such that z 2 B r (x) : Then z 2 S 2A B = N; meaning that z is a rational number, a clear contradiction.
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Andrew van Herick 11 10.4.2. Let A; B be compact subsets of X: Prove that A [ B and A \ B are compact. Proof. Let V = fv g 2 be an open covering of A[B: Because A A[B V; V is also an open covering of A: As A is compact there exists a nite subset 0 of such that S 2 0 V A: Similarly there exists 1 such that S 2 1 V B: As 0; 1 are both nite set, it follows that 0 [ 1 is nite, that 0 [ 1 : To show that A [ B S 2 0 [ 1 V ; let x 2 A [ B: If x 2 A then x 2 V for some 2 0 : As 2 0 [ 1 ; V S 2 0 [ 1 V ; so x 2 S 2 0 [ 1 : Similar logic establishes the same result if x 2 B: Note that V is are open in X for all 2 ; i.e. for all 2 0 [ 1 This means that fv g 2 0 [ 1 is a nite open subcover of V, establishing every open covering of A[ B has a nite subcover. Equivalently A [ B is compact. Now we ll show that A \ B is compact. By Theorem 10.46 (p. 308, Wade) both A and B are closed. By Theorem 10.31(iii) (p. 301, Wade) A \ B is closed. As A \ B A; a compact set, by Remark 10.45 A \ B is compact.
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Andrew van Herick 13 10.4.3. Suppose that E R is compact and non-empty. Prove that sup E; inf E 2 E: Proof. By Theorem 10.46 (p. 308, Wade) E is closed and bounded. By Theorem 2.26 (p. 47, Wade) R satis es the Bolzano-Weierstrass Property. It is easy to see that f : E! R de ned by f (x) = x is continuous on E: By Exercise 10.2.9(c) there exist points x M ; x m 2 E such that f (x M ) = sup f (x) and f (x m ) = inf f (x) : x2e x2e But f (x M ) = x M and f (x m ) = x m ; establishing that sup E; inf E 2 E: