76 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

as ∑∞k=1 2−kak where ak is either 0 or 1. Let F be ∪nP ({1,2, · · · ,n}) . Explain why

F is countable. Then let S ≡P (N) \F . Explain why S is uncountable. Let Cbe all points of the form ∑

mk=1 2−kak where ak is 0 or 1. Explain why C is countable.

Let J = [0,1]\C. Now let θ : S → J be given by θ (S) = ∑k∈S 2−k. Explain why θ

is one to one onto J. If [0,1] is countable, show there are onto mappings as indicatedN→ [0,1]→ J→S showing that S is countable after all.

15. Using the above problem as needed, let B be a countable set of real numbers. SayB = {bn}∞

n=1. Let

fn (t)≡{

1 if t ∈ {b1, · · · ,bn}0 otherwise

Let g(t) ≡ ∑∞k=1 2−k fk (t) . Explain why g is continuous on R\B and discontinuous

on B. Note that B could be the rational numbers.

16. Consider R\{0} . Show this is not connected.

17. Show S ≡{(

x,sin( 1

x

))if x > 0

}∪ {(0,y) : |y| ≤ 1} is connected but not arcwise

connected.

18. Let A be an m×n matrix. Then A∗, called the adjoint matrix, is obtained from A bytaking the transpose and then the conjugate. For example, i 1

1+ i 23 1− i

∗ = ( −i 1− i 31 2 1+ i

)

Formally, (A∗)i j = A ji. Show (Ax,y) = (x,A∗y) and (x,By) = (B∗x,y). The innerproduct is described in the chapter. Recall (x,y)≡ ∑ j x jy j.

19. Let X be a subspace of Fm having dimension d and let y ∈ Fm. Show that x ∈X is closest to y in the Euclidean norm |·| out of all vectors in X if and only if(y−x,u) = 0 for all u ∈ X . Next show there exists such a closest point and it equals∑

dj=1(y,u j

)u j for

{u j}d

j=1 an orthonormal basis for X .

20. Let A : Fn → Fm be an m× n matrix. (Note how it is being considered as a lineartransformation.) Show Im(A) ≡ {Ax : x ∈ Fn} is a subspace of Fm. If y ∈ Fm isgiven, show that there exists x such that y−Ax is as small as possible (Ax is the pointof Im(A) closest to y) and it is a solution to the least squares equation A∗Ax = A∗y.Hint: You might want to use Problem 18.

21. Show that the usual norm in Fn given by |x|= (x,x)1/2 satisfies the following iden-tities, the first of them being the parallelogram identity and the second being thepolarization identity.

|x+y|2 + |x−y|2 = 2 |x|2 +2 |y|2

Re(x,y) =14

(|x+y|2−|x−y|2

)Show that these identities hold in any inner product space, not just Fn. By definition,an inner product space is just a vector space which has an inner product.

7615.16.17.18.19.20.21.CHAPTER 2. BASIC TOPOLOGY AND ALGEBRAas Pe} 2a, where a, is either 0 or 1. Let ¥ be Un F ({1,2,--- ,n}). Explain whyF is countable. Then let. Y = A (N)\.¥. Explain why -7 is uncountable. Let Cbe all points of the form )°7"_, 2-*a, where a, is 0 or 1. Explain why C is countable.Let J = [0,1] \\C. Now let 6: .Y — J be given by 0 (S) = Yyes2~*. Explain why 0is one to one onto J. If [0, 1] is countable, show there are onto mappings as indicatedN = [0,1] + J > 7 showing that .Y is countable after all.Using the above problem as needed, let B be a countable set of real numbers. SayB= {b,}*_,. LetLift € {by,---, bn}0 otherwisemi ={Let g(t) = Vy, 2 -*f (t). Explain why g is continuous on R \ B and discontinuouson B. Note that B could be the rational numbers.Consider R \ {0}. Show this is not connected.Show S = {(x,sin(4)) ifx>0}U {(0,y) : |y| < 1} is connected but not arcwiseconnected.Let A be an m x n matrix. Then A*, called the adjoint matrix, is obtained from A bytaking the transpose and then the conjugate. For example,*i 1-—i l-i 31+i 2 =( /)3 li 1 2 1+iFormally, (A*);; = Aji. Show (Ax,y) = (x,A*y) and (x,By) = (B*x,y). The innerproduct is described in the chapter. Recall (x,y) = )jxjyj-Let X be a subspace of F” having dimension d and let y € F”. Show that x €X is closest to y in the Euclidean norm |-| out of all vectors in X if and only if(y —x,u) = 0 for all u € X. Next show there exists such a closest point and it equalsY4_, (y,u;) uy for {uj}‘_, an orthonormal basis for X.Let A: F” > F” be an m x n matrix. (Note how it is being considered as a lineartransformation.) Show Im(A) = {Ax:x € F”} is a subspace of F”. If y € F” isgiven, show that there exists x such that y — Ax is as small as possible (Ax is the pointof Im (A) closest to y) and it is a solution to the least squares equation A*Ax = A*y.Hint: You might want to use Problem 18.Show that the usual norm in F” given by |x| = (x,x)!/ * satisfies the following iden-tities, the first of them being the parallelogram identity and the second being thepolarization identity.2|x|* +2 Iy/?1i (Ix+yl?—x—yl’)Ix+y/+|x—yl?Re(x,y)Show that these identities hold in any inner product space, not just F”. By definition,an inner product space is just a vector space which has an inner product.