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Exercises Exercises

1.

Let S be a nonempty subset of R that is bounded below. Let a∈R, and define a+S to be a+S={a+s∣s∈S}.

(a)

Explain why a+inf(S) is a lower bound for a+S. Explain why a+S has an infimum.

(b)

Let b be a lower bound for a+S. Show that a+inf(S)β‰₯b. Then explain why a+inf(S)=inf(a+S).

2.

Let S be a nonempty subset of R.

(a)

Assume that S is bounded above, and let t=sup(S). Show that for every r<t, there is a number s∈S such that r<s≀t.

(b)

Assume that S is bounded below, and let t=inf(S). Show that for every r>t, there is a number s∈S such that t≀s<r.

3.

Let A and B be nonempty subsets of R that are bounded above and below. Let A+B={a+b∣a∈A,b∈B}.

(a)

Follow the steps below to show that sup(A+B)=sup(A)+sup(B).
(i)
Let x=sup(A) and y=sup(B). Show that x+y is an upper bound for A+B.
(ii)
The previous part shows that A+B is bounded above and so has a supremum. Let z=sup(A+B). Explain why z≀x+y.
(iii)
To show that z=x+y we have to prove that z cannot be strictly less than x+y. Suppose to the contrary that z<x+y. Let Ο΅=x+yβˆ’z. Use the result of Exercise 2 to arrive at a contradiction.

4.

Let X=C[a,b], the set of continuous functions from R to R on an interval [a,b]. Define d:X×X→R by
d(f,g)=sup{|f(x)βˆ’g(x)|∣x∈[a,b]}.

(b)

Prove that d is a metric on X. Describe in geometric terms how this metric measures the distance between functions f and g. (This metric is called the supremum metric or the uniform metric on X.)

5.

In this exercise we prove the Archimedean property of the natural numbers. Note that the set of natural numbers, denoted N or Z+, is the set of all positive integers. Let x be a real number.

(a)

Suppose that there is no positive integer N such that N>x. Explain how we can conclude that Z+ is bounded above.

(b)

Assuming that Z+ is bounded above, explain why Z+ must have a least upper bound M.

(c)

Explain why M cannot be a least upper bound for Z+. Explain why this proves the Archimedean property.

6.

In this exercise we prove two statements that are equivalent to the Archimedean property (see Exercise 5). One of the statements is the following theorem:

(b)

A second statement that is equivalent to the Archimedean property is the following. Prove that Theorem 5.6 is equivalent to the Archimedean property.

7.

We can use greatest lower bounds to prove the following theorem. This theorem tells us an important fact β€” that the rational numbers are what is called dense in the set of real numbers. We prove this theorem in this exercise. Let x and y be real numbers and assume x<y. By the Archimedean property of the natural numbers (see Exercises 5 and Exercise 6), there is a positive integer n such that n(yβˆ’x)>1. Let S={k∈Z∣k>nx}.

(b)

Explain why S contains an integer m such that if q∈Z with q<m, then q≀nx. It may be helpful to use the Well-Ordering Principle that states
Every subset of the integers that is bounded below contains its infimum.
(The Well-Ordering Principle is one of many axioms that are equivalent to the Principle of Mathematical Induction. These principles are taken as axioms and are assumed to be true.)

(c)

Explain why m>nx and mβˆ’1≀nx. Use these inequalities, along with n(yβˆ’x)>1, to show that nx<m<ny. Then find a rational number that is strictly between x and y.

8.

Show that every open ball in (R2,dE) contains a point x=(x1,x2) with both x1 and x2 rational.

9.

We are familiar with solving the quadratic equation x2βˆ’2=0 to obtain the solutions Β±2. But do we really know that the number 2 exists? We address that question in this exercise and demonstrate the existence of the number 2 using the greatest lower bound.

(a)

To begin, let S={x∈R+∣x2>2}. Explain why S must have a greatest lower bound m.

(b)

In what follows we demonstrate that m2=2, which makes m=2. We consider the cases m2<2 and m2>2.
(ii)
Suppose m2>2. Show that there is a positive integer n such that
(mβˆ’1n)2 > 2.
Explain why this also cannot happen.

10.

Similar to Exercise 7 we can prove the following theorem.

(a)

The first step is to demonstrate the existence of an irrational number. We will do that by proving that 2 is irrational. Proceed by contradiction and assume that 2 is a rational number. That is, 2=rs for some positive integers r and s such that r and s have no positive common factors other than 1.
(i)
Explain why r2=2s2. Since 2 is prime, it follows that 2 divides r.
(ii)
Show that 2 divides s. Explain how this proves that 2 is an irrational number.

(b)

Let x and y be distinct real numbers. Show that there exists an integer q and a positive integer N such that z=q22N is an irrational number between x and y.
Hint.
Consider the approach in Exercise 7 .

11.

Let (X,d) be a metric space and A a nonempty subset of X. For x,y∈X, prove that d(x,A)≀d(x,y)+d(y,A).

12.

Prove that if (X,d) is a metric space and B and C are nonempty subsets of X, then
d(a,BβˆͺC)=min{d(a,B),d(a,C)}
for every a∈X.

13.

For each of the following, answer true if the statement is always true. If the statement is only sometimes true or never true, answer false and provide a concrete example to illustrate that the statement is false. If a statement is true, explain why. Throughout, let S and T be bounded subsets of R (a subset of R is bounded if it is both bounded above and bounded below).

(b)

If S+T={s+t∣s∈S,t∈T}, then sup(S+T)=max{sup(S),sup(T)}.

(c)

Let S+T={s+t∣s∈S,t∈T}, then inf(S+T)=min{inf(S),inf(T)}.

(d)

If U is a nonempty subset of S, then sup(U)≀sup(S).

(e)

If U is a nonempty subset of S, then inf(S)≀inf(U).

(f)

If A is a subset of R and x∈R with d(x,A)=0, then x∈A.