Homework 1 Solutions
Math 501
Due September 5, 2014
Exercise 1
Let > 0 be rational. Suppose that there does not exist x ∈ Q such that
x2 < 2 < (x + )2 . That is, suppose that for any x ∈ Q such that x2 < 2 we
have that (x + )2 < 2. Using x = 0, this implies 2 < 2. Using our assumption
again with x = , we get that (2)2 < 2. By induction, we get that (n)2 < 2
for all n ≥ 1. However, this is an obvious contradiction since n = d3−1 e fails
to satisfy this inequality.
Exercise 2
Let B = {x ∈ Q : x2 > 2, x ≥ 0} and let x ∈ B. It is easy to see that if there
exists y ∈ B such that y < x, then since x was arbitrary, B has no smallest
element. Indeed, this is the case. Let
y =x−
x2 − 2
x+2
Note that x2 −2 > 0 and x+2 > 0 since x ∈ B. It follows that y < x. Moreover,
observe that
2(x2 − 2)
y2 − 2 =
.
(x + 2)2
Arguing as above, we see that y 2 − 2 > 0. Finally, it’s clear that y ∈ Q and it’s
not difficult to check that y ≥ 0. Thus, y ∈ B.
Exercise 3
Let S ⊂ R be a bounded nonempty subset and let b = sup S.
(a) Let > 0. Suppose for each s ∈ S, either s < b − or s > b. However, if
s > b for some s ∈ S, this contradicts that b is an upper bound for S. Thus
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it must be that for all s ∈ S, s < b − . However, this implies that b − is an
upper bound for S, which contradicts that b is the least upper bound for S.
Hence it must be the case that there exists s ∈ S such that b − ≤ s ≤ b.
(b) It is certainly not the case that one can always find s ∈ S such that b − <
s < b. Consider S = [0, 1] ∪ {2}. Then b = 2. However, if say = 1/2, there
does not exist s ∈ S such that b − < s < b.
Exercise 4
(a) Let (xn ) be an increasing sequence that is bounded above and let x = sup xn .
We claim that (xn ) converges to x. Indeed, let > 0. By part (a) of Exercise
3, there exists N such that x − ≤ xN ≤ x. Observe that if n ≥ N then
|xn − x| < |xN − x| ≤ since the sequence is increasing.
(b) The corresponding statement for decreasing sequences is as follows:
Any decreasing sequence that is bounded below converges (to its infimum).
Exercise 5
(a) We give the full proof that every sequence (xn ) ⊂ R of real numbers has a
monotone subsequence. Recall that we call n ∈ N a peak of this sequence
if xm < xn for all m > n. Suppose that (xn ) has infinitely many peaks
n1 < n2 < n3 < · · · . Then the subsequence xn1 , xn2 , xn3 , . . . is a decreasing
subsequence. On the other hand, suppose (xn ) has finitely many peaks.
Suppose that the final peak occurs at N . Let n1 = N + 1. Since n1 is not a
peak, there exists n2 > n1 such that xn2 ≥ xn1 . Likewise, since n2 is not a
peak, there exists n3 > n2 such that xn3 ≥ xn2 . Continuing in this fashion,
we are able construct a nondecreasing subsequence xn1 ≤ xn2 ≤ xn3 ≤ · · · .
In either case, we’ve shown that there exists a monotone subsequence.
(b) Let (xn ) ⊂ R be a bounded sequence. By part (a), (xn ) has a bounded
monotone subsequence. By Exercise 4, this subsequence must converge.
Exercise 6
Let > 0 and let
n
X
1
xn =
.
k2
k=1
2
We claim that (xn ) is a Cauchy sequence. Indeed, let N = d2−1 e. For n ≥
m ≥ N observe that
n
X
1 |xn − xm | = k2 k=m+1
n
X
1
<
k(k − 1) k=m+1
n
X
1
1 =
− k − 1 k
k=m+1
1
1
= − m n
2
≤
N
≤ .
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