Andrew van Herick Math 710 Dr. Alex Schuster Sept. 7, 2005 Assignment #1 Section 10.1: 1, 4-8 Section 10.2: 1,2 10.1.1. If a; b X and (a; b) < " for all " > 0; prove that a = b: 2 Proof. Let a; b X and (a; b) < " for all " > 0: By way of contradiction suppose that 2 a = b: Then (a; b) > 0 by the positive de…nite property of metric spaces. By hypothesis 6 this means that (a; b) < (a; b) a clear contradiction. 2 Andrew van Herick 3 10.1.4. (a) Let a X: Prove that if xn = a for every n N; then xn converges. What does 2 2 it converge to? (b) Let X = R with the discrete metric. Prove that xn a as n if and only ! ! 1 if xn = a for large n: Proof. (a) Let xn = a n N and let " > 0: Let n 1: Then 8 2 (xn; a) = (a; a) = 0 < ": By de…nition xn converges to a: Proof. (b) Let X = R with the discrete metric. Suppose that xn a as n ; and choose ! ! 1 N N such that n N implies (xn; a) < 1: In particular this means for n N that 2 (xn; a) = 1: By the de…nition of the discrete metric (xn; a) must be 0; i.e. xn = a for 6 n N: Conversely, suppose that xn = a for large n: Let " > 0: Then M N such that 9 2 m M implies xm = a: By the positive de…nite property of metric spaces this means that for all m M; (xm; a) = 0 < ": and by de…nition establishes that xn a as n : ! ! 1 4 Andrew van Herick 5 10.1.5. (a) Let xn and yn be sequences in X that converge to the same point. Prove f g f g that (xn; yn) 0 as n : ! ! 1 (b) Show that the converse is false. Proof. (a) De…ne : R2 R to be the usual metric for R (i.e. :(x; y) x y ). Since ! 7! j j (x ; y ) is a sequence in the metric space together with ; we’lluse the de…nition n n n N R f g 2 of convergence for metric spaces to show that (xn; yn) 0 as n : ! ! 1 Suppose that xn ; yn both converge to the same point a X: Let " > 0: Choose f g f g 2 N1;N2 N such that 2 " n N1 implies (xn; a) < and 2 " n N2 implies (xn; a) < 2 Suppose that n max (N1;N2) then using the three de…ning properties of metric spaces ( (xn; yn) ; 0) = (xn; yn) 0 = (xn; yn) j j (xn; a) + (a; yn) = (xn; a) + (yn; a) " " < + = ": 2 2 By de…nition (xn; yn) 0 as n : ! ! 1 Proof. (b) Take the converse statement to be: In any metric space X together with ; for all sequences xn ; yn in X; f g f g (xn; yn) 0 as n implies that xn ; yn converge to the same point a X: ! ! 1 f g f g 2 We’llprove that the statement is false in general by establishing a counterexample. Let X = R with the usual metric (i.e. :(x; y) x y ). Then the sequences 7! j j xn ; yn de…ned by xn = yn = n provide a counterexample. Let " > 0 and n 1: Then f g f g ( (xn; yn) ; 0) = (xn; yn) 0 = (xn; yn) j j = (xn; xn) = 0 < " 6 which establishes by de…nition that (xn; yn) 0 as n : It is easy to see that neither ! ! 1 xn nor yn converge to any point a R, thus disproving the converse statement. 2 Andrew van Herick 7 10.1.6. Let xn be Cauchy in X: Prove that xn converges if and only if at least one of f g f g it’ssubsequences converges. Proof. Let xn be Cauchy in X: If xn converges, because xn is by de…nition a subsequence f g f g f g of itself, at least one of it’s subsequences converges. Conversely suppose that at least one subsequence xn converges. Then xn converges to some point a X. Let " > 0: f k g f k g 2 Using the de…nition of convergent sequences choose N1 N such that 2 " k N1 implies (xn ; a) < k 2 Using the de…nition of Cauchy sequences choose N2 N such that 2 " m; n N2 implies (xm; xn) < 2 Let p max (N1;N2) : Since xnk is a subsequence of xn ; it is easy to see that xnp = xq f g f g " for some p q N with q p N1;N2. This means that xn ; a < and xp; xn < 2 p 2 p " : By applying the Triangle Inequality we have 2 " " (xp; a) xp; xn + xn ; a < + = " p p 2 2 establishing that xn is by de…nition convergent. f g 8 Andrew van Herick 9 10.1.7. Prove that the discrete space R is complete. Proof. Let xn be any Cauchy sequence in the discrete space R: Using the de…nition of Cauchy, choose N N such that m; n N implies (xm; xn) < 1: Note that xN R: Let " > 0 2 2 and n N: Then (xN ; xn) < 1: However, since is the discrete metric, this means that (xN ; xn) = 0 < " for all " > 0: By de…nition xn xN as n : Hence every Cauchy ! ! 1 sequence in the discreet space R converges to some point in R. By de…nition, the discreet space R is complete. 10 Andrew van Herick 11 10.1.8. (a) Prove that every …nite subset of a metric space X is closed. (b) Prove that Q is not closed in R. Proof. (a) Let A be a …nite subset of X: Then A = a1; : : : ; an for some n N and some c f g 2 a1; : : : ; an X: Let x A and set " = min ( (a1; x) ; : : : ; (an; x)) : Let y B" (x) : 2 2 2 Then (y; x) < " (a1; x) ; : : : ; (an; x) : In particular, this means that (y; x) = 6 c (a1; x) ; : : : ; (an; x) : Because metric spaces are positive de…nite y = a1; : : : ; an; so y A : c c C6 2 This establishes that B" (x) A for all x A : By de…nition A is open, which means 2 that A is closed. Proof. (b) We’ll show that Qc is not open in R. Choose any point a Qc and any " > 0: Then 2 it is easy to see that B" (a) = (a "; a + ") : By the Density of the Rationals q Q such c 9 2 that a " < q < a + "; i.e. q B" (a) but q = Q : This means that B" (a) *R Q for 9 2 2 n any " > 0 or any a Qc; and establishes that Qc is not open. By De…nition 10.8 (p. 292, 2 Wade) Q is closed in R: 12 Andrew van Herick 13 10.2.1. Find all cluster points of the following sets: (a) E = R Q: n (b) E = [a; b) ; a; b R; a < b: 2 n (c) E = ( 1) n : n N : f 2 g (d) E = xn : xn x as n : f ! ! 1g (e) E = 1; 1; 2; 1; 2; 3; 1; 2; 3; 4;::: : f g Proof. (a) We’llshow very point in R is a cluster point of R Q. n Let x R and > 0: Then B" (x0) R Q = (x ; x + ) R Q contains in…nitely 2 \ n \ n many points by the Density of the Irrationals (p. 23, Wade). Proof. (b)[a; b] represents all of the cluster points of [a; b) : Let x0 [a; b] and " > 0: Then x0 + " > a and x0 " < b: In particular this means that 2 B" (x0) [a; b) = (x0 "; x0 + ") [a; b) contains a non-degenerate interval, say (a0; b0) : By \ \ the Density of the Rationals (a0; b0) contains in…nitely many points. This establishes that every point in [a; b] is a cluster point for [a; b) : To show that [a; b) has no cluster points; suppose by way of contradiction that some x1 R [a; b] were a cluster point of [a; b) : Then x1 < a or b > x1: If x1 < a; choose 2 n = a x1 > 0; then B (x1) = (x1 ; x1 + ) = (2x1 a; a) ; so B (x1) [a; b) = a a \ f g …nite set. This means by de…ntion that x1 cannot be a cluster point of [a; b) : An alalygous argument extablishes the same result if x1 > b: Hence no members of R [a; b] are cluster n points of [a; b) : n Proof. (c) E := ( 1) n : n N has no cluster points. f 2 g 1 1 1 Let a R: Choose = : Then B (a) E = a ; a + E has at most one 2 2 \ 2 2 \ member, since E Z. This means that by de…nition that a is not a cluster point of E: n Hence no points in R are cluster points of ( 1) n : n N : f 2 g Proof. (d) The only cluster point of E := xn : xn x as n is x: f ! ! 1g To show that x is a cluster point. Let > 0 and using the de…nition of convergence choose N N such that n N implies (xn; x) < : Then B (x) E = xN ; xN+1; xN+2;::: ; 2 \ f g a set which clearly contains in…nitely many points. y x To show that E has no other cluster points, suppose that y = x.