7/28/2019 Answers to q Quiz

1/4

Math 521 Spring 2010

Some Answers to the Homework

May 12, 2010

(1.a) Prove that if f : X Y, g : Y Z and W Z, then

(g f)1(W) = f1

g1(W)

.

Answer: This is simply a matter of unravelling the definitions. For sets A and Bthe notation A B means that every element of A is an element of B, i.e.

A B means x [x A = x B] .

Also A = B means A B and B A, i.e.

A = B if and only ifx [x A x B] .

A proof that A B looks like this:

Choose x A. Bla Bla Bla Therefore x B.

A proof that A = B looks like this:

Choose x A. Bla Bla Bla Therefore x B. Conversely choosex B/ Bla Bla Bla Therefore x B

In the case at hand every step in the proof that A B is an if and only if sowe can prove both halves at the same time:

x (g f)1

(W) (g f)(x) W x h1

(W) means h(x) W g(f(x)) W as (g f)(x) = g(f(x))

f(x) g1(W) x f1(V) means f(x) V

x f1(g1(W)) y g1(W) means g(y) W .

1

7/28/2019 Answers to q Quiz

2/4

In these steps I used the following substitutions: h = g f, V = g1(W), y = f(x).

This argument does not require that f be continuous.2

Some students are still not using set theoretic notation correctly. It is impor-tant to distinguish between sets (e.g. X, Y, Z, W) and elements (e.g. x, y).Also f1 (as a map from Y X) is only defined if f is bijective (one-one onto).However if we define P(X) to be the set of all subsets of X so that

U P(X) U X

then every map f : X Y determines a map f1 : P(Y) P(X) defined by

f1(V) := {x X : f(x) V}

for V Y.

(1.b) Using (1.a) and the characterization of continuity in terms of preimagesof open sets, prove that if f and g are continuous so is g f.

Answer: Choose W Z relatively open in Z. Then g1(W) is relatively openbecause g is continuous. Hence f1(g1(W)) is relatively open in X becausef is continuous. Therefore (g f)1(W) is relatively open as (g f)1(W) =f1(g1(W)). Hence g f is continuous. 2

Here are the definitions used in this argument. The word set means subsetof some Rn. For p X Rn

BX(p, ) := B(p, ) X := {q X : |p q| < }.

A subset U X to be relatively open in X iff for every p U there is a > 0

such that BX(p, ) U. Proposition 14.7 page 34 of the notes says that a map

f : X Y is continuous if and only if for every relatively open set V Y the

preimage f1(V) is a relatively open subset of X.

(2.a) Assume that f : X Y and S Y. Prove that

f1(Y \ S) = X \ f1(S).

Answer: As in (1.a) x f1(Y\S) f(x) Y\S f(x) Y and f(x) /

S x f1(Y \ S) 2

(2.b) True or false? A map is continuous if and only if the preimage of anyclosed subset of its target is a closed subset of its source.

Answer: There are two correct answers to this question depending on how youinterpret the word closed. (Part of the purpose of this question was to encourage

2

7/28/2019 Answers to q Quiz

3/4

you to think about the distinction.) By analogy with the previous question you

might argue as follows: Say that S Y is relatively closed in Y iff Y \ S isrelatively open. As Y \ (Y \ V) = V for V Y, the relatively open sets in Yare precisely the sets Y \ S where S is relatively closed. If we interpret open tomean relatively open and closed to mean relatively closed we have

f : X Y is continuous f1(V) is open for every open V Y f1(Y \ S) is open for every closed S Y f1(Y) \ f1(S) is open for every closed S Y X\ f1(S) is open for every closed S Y f1(S) is closed in X for every closed S Y

If on the other hand the word closed is taken to mean closed in Rn ratherthan relatively closed in X then the assertion is false. Take X to be the open

interval (1, 1), Y = R, f(x) = 0 for all x X, S = {0}. Then S is closed butf1(S) = (1, 1) is not closed.

Remark: If Y is closed subset ofRm then a subset S Y is relatively closedin Y if and only if it is closed in Rm, i.e. if and only ifRm \ S is open (prove this!)so if both X Rn and Y Rm are closed, the two interpretations agree. 2

Some students said that the constant map f : R R with (say) f(x) = 0 for

all x is a counterexample as {0} is closed but f1({0}) = R is open. The reason

is bad: it is true that R is open but it is also true that R is closed.

(3) Let A = {(x, y) R2 : x > 0, y = sin(1/x)}, B = {(x, y) R2 : x = 0},and C = A B. Which of the sets A,B,C are closed?

Answer: According to Theorem 14.11 on page 35 of the notes a set S R2 isclosed if and only if it is closed under limits, i.e. iff whenever limnpn = p andpn S for all n we have p S. Suppose that

(x, y) = p = limn

pn =

limn

xn, limn

yn

.

Then

A. If 1/xn = n and yn = sin(1/xn), then pn A and yn = 0, but p = (0, 0) =limnpn / A . Hence A is not closed.

B. If pn = (xn, yn) B then yn = 0 so y = 0 so p B. Hence B is closed.

C. Finally assume pn C := A B. If x > 0 then xn > 0 for sufficiently largen so yn = sin(1/xn) so y = sin(1/x) by continuity so p A. If x 0 thenx = 0 (as xn 0) so p B. In either case p C so C is closed. 2

3

7/28/2019 Answers to q Quiz

4/4

(4) A function f defined on an open interval is called analytic iff for every

a in its domain there is a power series such that

f(x) =k=0

ck(x a)k

for all x in some open interval containing a. Give two proofs that the functionf(x) = x1 defined for x > 0 is analytic.

First: Write f(x) = Pn(a, x) + Rn(a, x) where

Pn(a, x) :=n

k=0

f(k)(a)(x a)k

k!

and show that limnRn(a, x) = 0 for x sufficiently near a.

Answer: Choose a > 0. By Theorem 19.1 page 49 of the notes we have that

Rn(a, x) =

x

a

f(n+1)(t)

n!(x t)n dt.

By induction f(k)(t) = (1)kk! t(k+1) so

Rn(a, x) =

x

a

(1)n+1(n + 1)t(n+2)(x t)n dt

Choose m > 0 with m < min(a, x). Then for t between a and x we have both theinequalities t(n+1) m(n+1) and |x t| |x a|. Hence

|Rn(a, x)| (n + 1)m(n+1)

x

a

(x a)n dt

= (n + 1)x am

n+1

.

Hence limnRn(a, x) = 0 if the power on the right is