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MATH 104, SUMMER 2006, HOMEWORK 2 SOLUTION

BENJAMIN JOHNSON

Due July 5

Assignment:

Section 4: 4.1(d)(e)(h)(t)(u)(v); 4.7, 4.12

B1: Prove that an ordered field F is Archimedean if and only if satisfies (a F)(n N)(n > a)Section 5: 5.1, 5.2, 5.6

Section 6: 6.1, 6.3, 6.4

Section 7: 7.3(f)(g)(j)(t), 7.4

Section 4

4.1 For each set bounded above, list 3 upper bounds for the set. Otherwise, write NBA.

(d) {, e}: , 4, 5(e) { 1

n: n N}: 1, 2, 3

(h)

n=1[2n, 2n + 1]: NBA

(t) {x R : x3 < 8}: 2, 3, 4(u) {x2 : x R}: NBA(v) {cos( n

3) : n N}: 1, 2, 3

4.7 Let S and T be nonempty bounded subsets ofR.

(a) Prove that ifS T then infT infS sup S sup T.Proof. Note that S, T bounded and non-empty implies infT, infS, sup S, sup T all exist. Let

x S. We prove the three inequalities separately.Since S T, we have x T, and since infT is a lower bound for T, we have infT x. Sincex was arbitrary in S, the above argument shows infT is a lower bound for S. Since infS is

the greatest lower bound for S, we have infT infS.Since infS is a lower bound for S, we have infS x. Since sup S is an upper bound for Swe have x sup S. So we have infS x sup S. By transitivity infS sup S.Since S T, we have x T, and since sup T is an upper bound for T, we have x sup T.Since x was arbitrary in S, the above argument shows sup T is an upper bound for S. Since

sup S is the least upper bound for S, we have sup S sup T.

(b) Prove that sup(S T) = max{sup S, sup T}.Proof. Let x S T. Either x S or x T (or both). If x S, then x sup S max{sup S, sup T}. Similarly if x T, then x sup T max{sup S, sup T}. We see that inany case we have x max{sup S, sup T}. This shows max{sup S, sup T} is an upper bound forS T.Let B be any upper bound for S T. Then B is an upper bound for S, and since sup S is theleast upper bound for S, we have sup S B. Similarly, B is an upper bound for T, and since

Date: July 5, 2006. 1

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2 BENJAMIN JOHNSON

sup T is the least upper bound for T, we have sup T B. Thus we have max{sup S, sup T} B.This shows max{sup S, sup T} B for all upper bounds B for S T.Weve shown max{sup S, sup T} satisfies the definition for supremum of the set S T. Sincesuprema are unique, sup(S T) = max{sup S, sup T}.

4.12 Let I be the set of real numbers that are not rational. Prove that if a < b, then there exists x Isuch that a < x < b.

Proof. First we show that ifc is rational, then c+

2 is irrational. Proof by contradiction. Suppose

c is rational. If c +

2 is also rational, then since rational numbers are closed under subtraction

(c +

2) c =

2 is rational. But weve shown

2 is irrational, a contradiction.

Continuing with the main assertion, let a, b R with a < b. By Denseness ofQ in R, there isa rational number c with a

2 < c < b

2. Then we have a < c +

2 < b. And c +

2 is

irrational by what we proved above.

B1 Prove that an ordered field F is Archimedean if and only if satisfies (

a

F)(

n

N)(n > a).

Proof. To say that F is Archimedean (according to the books definition) means that a, b F :a, b > 0)(n N)(na > b). Call this (1), and call the other property listed above (2).

(1) (2)Assume (1). Let a F. If a 0, then take n = 1, and we have n > a. If a > 0, then applying (1)substituting a for b and 1 for a yields (n N)(n 1 > a). This proves (a F)(n N)(n > a),i.e. (2).

(2) (1)Assume (2). Let a, b F Assume a, b > 0. Then b

a F. Using (2), substituting b

afor a, we have

(n N)(n > ba

). Note that since ba> 0 such an n must actually be in N. Also, for a, b > 0, n > b

a

is equivalent to n a > b. Thus we have proven (a, b F : a, b > 0)(n N)(na > b), i.e. (1).

Section 5

5.1 Write the following sets in interval notation:

(a) {x R : x < 0} = (, 0)(b) {x R : x3 8} = (, 2](c) {x2 : x R} = [0,)(d) {x R : x2 < 8} = (

8,

8)

5.2 Give the infimum and supremum of each set listed in Exercise 5.1.

(a) inf= , sup = 0(b) inf= , sup = 2(c) inf= 0, sup = (d) inf=

8, sup =

8

5.6 Let S and T be nonempty subsets ofR such that S T. Prove that infT infS sup S sup T.Proof. Regardless of whether the values are allowed, inf(S) is the unique extended numbersatisfying

(1) inf(S) x for every x S, and(2) for every lower bound b ofS (where b could also be ), we have b inf(A).

Similarly, sup(S) is the unique extended real number satisfying

(1) x sup(S) for every x S, and(2) for every upper bound B ofS, (where B could also be

+), we have inf(S) B.

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MATH 104, SUMMER 2006, HOMEWORK 2 SOLUTION 3

With this in mind, the proofs for each of the three inequalities from Exercise 4.7(a) (where we

did not allow ) go through without modification. (Copied below).Let x SSince S T, we have x T, and since infT is a lower bound for T, we have infT x. Since

x was arbitrary in S, the above argument shows infT is a lower bound for S. Since infS is the

greatest lower bound for S, we have infT infS.Since infS is a lower bound for S, we have infS x. Since sup S is an upper bound for S we

have x sup S. So we have infS x sup S. By transitivity infS sup S.Since S T, we have x T, and since sup T is an upper bound for T, we have x sup T. Since

x was arbitrary in S, the above argument shows sup T is an upper bound for S. Since sup S is the

least upper bound for S, we have sup S sup T.

Section 6

6.1 Consider s, t Q. Show that(a) s t if and only if s t

Proof. Assume s t. Let x s. Then x < s. So x < t. So x t. This shows s t.Assume s t. Proceed by contradiction. Assume t< s. Then there is some rational numberc with t< c < s. So c s but c t This contradicts s t

(b) s = t if and only if s = t

Proof. Assume s = t. Then {r Q : r< s} = {r Q : r< t}. So s = t.Assume s = t. Proceed by contradiction. If s t, then one of s < t or t< s holds. Withoutloss of generality, assume that s < t holds. Then there is a rational number c with s < c < t.

We have c t but c s. This contradicts s = t. Conclude s = t. (c) (s + t) = s + t

Proof. Let x (s + t). Then x Q and x < s + t. By algebra, x s < t. By density, there is arational number c with x s < c < t. By algebra x c < s. Because x and c are rational, so isxc. We have c t, xc s, and x = c+ (xc). So x s+ t. This proves (s+ t) s+ t

Let x s + t. Then there are rational numbers a s and b t with x = a + b. Since a, bare rational so is x. Since a < s and b < t, we have x < s + t. So x (s + t) This provess+ t

(s + t

). Hence (s + t

)= s

+ t

.

6.3 (a) Show that + 0 = for all Dedekind cuts .

Proof. Let be any Dedekind cut.

Let x + 0. Then there exist r and s 0 with x = r+ s. Since s < 0, we have x < r.since is closed under

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4 BENJAMIN JOHNSON

(b) How would you define ? = {r Q : r }

6.4 Let and be Dedekind cuts and define the product: = {r1r2 : r1 and r2 }.(a) Calculate some products of Dedekind cuts using the Dedekind cuts 0, 1, and (1).

0

0 =

{x

Q : x > 0

} 0 1 = Q 0 (1) = {x Q : x > 0} 1 1 = Q 1 (1) = Q (1) (1) = {x Q : x > 1}

(b) Discuss why the definition of product is totally unsatisfactory for defining multiplication in

R.

Aside from the less consequential shortcomings of not handling 0 and 1 correctly with respect

to several of the ordered field axioms, the product function defined above does not even

output a Dedekind cut when given Dedekind cuts as inputs. Always giving a Dedekind cut

as an output would be the first prerequisite for any binary function on the set of all Dedekindcuts. A product function of this type would normally be expected to be a binary function.

Section 7

7.3 For each sequence below, determine whether it converges, and if it converges, give its limit.

(f) sn = 21n : The limit is 1.

(g) yn = n!: Does not converge.

(j) 7n3+8n

2n331 : The limit is72

.

(t) 6n+49n2+7

: The limit is 0.

7.4 Give examples of

(a) a sequencexn

of irrational numbers having a limit that is a rational number.

xn =

2n

defines a sequence of irrational numbers with rational limit 0.

(b) a sequence rn of rational numbers having a limit that is an irrational number.rn = (1 +

1n

)n defines a sequence of rational numbers with irrational limit e.