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.
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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.
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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:
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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. Let " < j j : 6 2 Using the de…nition of convergence choose M such that m M implies (xm; x) < ": In 14
particular, B" (x) E = xM ; xM+1; xM+2;::: . It is not di¢ cult to see that \ f g y x y x y x y x B" (x) B" (y) = x j j; x + j j y j j; y + j j = \ 2 2 \ 2 2 ;
Hence, B" (y) E = E (B" (x) E) = x1; : : : ; xM 1 is a …nite set, which means by \ n \ f g de…nition that y is not a cluster point. This establishes that the only cluster point of
xn : xn x as n is x: f ! ! 1g Proof. (e) 1; 1; 2; 1; 2; 3; 1; 2; 3; 4;::: has no cluster points. f g + + Notice that 1; 1; 2; 1; 2; 3; 1; 2; 3; 4;::: = Z : Note that for any x R;B1=2 (x) Z = 1 1 f g 2 \ x ; x + Z+ has a cardinality of at most 1. By de…nition no x R is a cluster 2 2 \ 2 point. Andrew van Herick 15
10.1.2. Find all cluster points of the following sets:
(a) A point a in a metric space X is said to be isolated if and only if there is an
r > 0 so small that Br (a) = a : Show that a point a X is not a cluster point f g 2 if and only if a is isolated. (b) Prove that the discrete space has no cluster points.
Proof. (a)
Let a X: 2
( ) If a is not a cluster point, then by de…nition > 0 such that B (a) = n for some ) 9 j j n N. (Note that B (a) = ; since (a; a) = 0 < implies a B (a)). If B (a) = 1 2 6 ; 2 j j then B (a) = a and a is isolated. If n > 1; then B (a) = a; a1; : : : ; an for some distinct f g f g a1; : : : ; an X where a1; : : : ; an = a. Choose r = min ( (a1; a) ; : : : ; (an; a)) : 2 6
Note that r > 0; since a1; : : : ; an = a: Now we’ll show that Br (a) = a : It is 6 f g obvious that a Br (a). To show the reverse containment let x B (a) and observe f g 2 (x; a) < r < which means that x B (a) : Note that x = a1; : : : ; an; otherwise x = ai 2 6 for some i 1; : : : ; n 2
(x; a) = (ai; a) min ( (a1; a) ; : : : ; (an; a)) = r < (x; a) ;
a contradiction. This means that x = a a ; i.e. Br (a) a : By de…nition a is 2 f g f g isolated.
( ) If a is isolated, then by de…nition it is not a cluster point. (
Proof. (b)
Let a R (with the discrete metric ). We’ll show that a is isolated. Choose r = 1: 2 Observe that a Br (x) since (a; a) = 0 < 1; so a Br (a) : If x Br (a) then 2 f g 2 (x; a) < 1; i.e. (x; a) = 1: By the de…nition of the discreet metric, (x; a) must be 0: 6 This means that x = a a ; in particular that Br (a) a : By de…nition a is isolated. 2 f g f g By the above proof a is not a cluster point. Thus the discrete space has no cluster points.
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