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DESCRIPTION
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Math 2510: Real Analysis I Homework Solutions 12Fall 2008
12 The Completeness Axiom
12.3 For each subset of R, give its supremum and its maximum, if they exist. Otherwise, write none.
(a) {1, 3}
sup{1, 3} = max{1, 3} = 3
(c) [0, 4]
sup[0, 4] = max[0, 4] = 4
(e){
1n | n N
}Notice that
{1n | n N
}={1, 12 ,
13 ,
14 , . . .
}.
sup{
1n | n N
}= max
{1n | n N
}= 1
(g){
nn+1 | n N
}Notice that
{n
n+1 | n N}={
12 ,
23 ,
34 ,
45 , . . .
}.
sup{
nn+1 | n N
}= 1, maximum is none.
(i){
n + (1)n
n | n N}
Notice that{
n + (1)n
n | n N}={0, 2 + 12 , 3
13 , 4 +
14 , 5
15 , . . .
}.
The set is not bounded above, so the supremum is none and the maximum is none.
12.4 Repeat Exercise 12.3 for the infimum and the minimum of each set.
(a) {1, 3}
inf{1, 3} = min{1, 3} = 1
(c) [0, 4]
inf[0, 4] = min[0, 4] = 0
(e){
1n | n N
}Notice that
{1n | n N
}={1, 12 ,
13 ,
14 , . . .
}.
inf{
1n | n N
}= 0, minimum is none
(g){
nn+1 | n N
}Notice that
{n
n+1 | n N}={
12 ,
23 ,
34 ,
45 , . . .
}.
inf{
nn+1 | n N
}= min
{n
n+1 | n N}= 12
(i){
n + (1)n
n | n N}
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2
Notice that{
n + (1)n
n | n N}={0, 2 + 12 , 3
13 , 4 +
14 , 5
15 , . . .
}.
inf{
n + (1)n | n N}= min
{n + (1)n | n N
}= 0
12.6a Let S be a non-empty bounded subset of R. Prove that supS is unique.
Suppose that x and y are both suprema for S. Then
(1) x is an upper bound for S.(2) If z is any upper bound for S, then x z.(3) y is an upper bound for S.(4) If z is any upper bound for S, then y z.By (1), x is an upper bound for S and so by (4), y x. By (3), y is an upper bound for Sand so by (2), x y. Therefore x y and y x, so x = y by Trichotomy.
12.7a Let S be a non-empty bounded subset of R and let k be in R. Define kS = { ks | s S }. Prove thefollowing: If k 0, then sup(kS) = k supS and inf(kS) = k inf S.
To prove that sup(kS) = k supS there are two cases.If k = 0, then kS = 0 S = {0} and so
sup(kS) = sup{0} = 0 = 0 supS = k supS.
Now suppose that k > 0. The number sup(kS) is the unique real number with the properties
(a) sup(kS) is an upper bound for kS and(b) if y is any upper bound for kS, then sup(kS) yTherefore, because sup(kS) is the only number for which properties (a) and (b) are true, toshow sup(kS) = k supS it is sufficient to show that (a) k supS is an upper bound for kSand (b) if y is any upper bound for kS, then k supS y.To show that k supS is an upper bound for kS, suppose that x is in kS. Then there isan s in S so that x = ks. Then s supS since supS is an upper bound for S and sox = ks k supS. Since x was arbitrary, k supS is an upper bound for kS.To show (b), suppose that y is an upper bound for kS. Then ks y for every s in S, sos y/k for every s in S. Therefore, y/k is an upper bound for S and so supS y/k.Multiplying by k we see that k supS y as desired.To prove that inf(kS) = k inf S there are two cases.If k = 0, then kS = 0 S = {0} and so
inf(kS) = inf{0} = 0 = 0 inf S = k inf S.
Now suppose that k > 0. To show inf(kS) = k inf S it is sufficient to show that (c) k inf S isa lower bound for kS and (d) if y is any lower bound for kS, then y k inf S.To show that k inf S is a lower bound for kS, suppose that x is in kS. Then there is an s inS so that x = ks. Then s inf S since inf S is a lower bound for S and so x = ks k inf S.Since x was arbitrary, k inf S is a lower bound for kS.To show (d), suppose that y is a lower bound for kS. Then ks y for every s in S, so s y/kfor every s in S. Therefore, y/k is a lower bound for S and so inf S y/k. Multiplying byk we see that k inf S y as desired.
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12.10a Prove:If x and y are real numbers with x < y, then there are infinitely many rational numbers in theinterval [x, y].
In class.