HW4 Solutions
February 21, 2011
Derek Olson
(1.12) Let S be a set of real numbers, and let x be a real number. Show that
x = sup S if and only if
(a) x is an upper bound of S, and
(b) For every > 0, there exists s ∈ S such that x − < s.
First assume x = sup S. By the definition of the supremum of a set, x is an upper
bound of S so (a) is satisfied. To prove that (b) is satisfied, we will use a proof by
contradiction. That is, assume there exists some > 0 such that for every s ∈ S,
x − ≥ s. But this implies x − is an upper bound for S. Moreover, x − < x since
> 0, and this contradicts the definition of x as the least upper bound of S.
Conversely, assume x ∈ R satisfies both (a) and (b). Showing x = sup S is equivalent to showing x is an upper bound for S, and if x0 is any other upper bound, then
x ≤ x0 . By property (a), we can immediately conclude x is an upper bound for S.
Now suppose there exists an upper bound x0 of S with x0 < x. This implies x − x0 > 0
so taking = x − x0 and applying property (b) shows the existence and an s ∈ S such
that s > x − () = x − (x − x0 ) = x0 . This contradicts the fact that x0 was an upper
bound for S. Hence x ≤ x0 and x = sup S.
(2.8) Show that there does not exist a rational number x such that x2 = 3.
First we prove the following lemma:
Lemma 1 Let m be an integer. If 3 divides m2 , then 3 divides m. (Note this is
actually an if and only if statement, but we only need one direction for this problem).
Every integer m can be written as one of either m = 3k1 , m = 3k2 + 1, or m = 3k3 + 2
(i.e. we divide m by three and add the remainder). Observe that
(3k1 )2 = 9k12 = 3(3k12 )
(3k2 + 1)2 = 9k22 + 6k2 + 1 = 3(3k22 + 2k2 ) + 1
(3k3 + 2)2 = 9k32 + 12k3 + 4 = 3(3k32 + 4k3 + 1) + 1.
Of these possibilities for m2 , only in the case of m = 3k1 is m2 divisible by 3 (simply
note the remainders of each upon division by 3). Thus m2 being divisible by three
implies m is divisible by 3.
Now we prove the result. Assume to the contrary that (m/n)2 = 3 for some
integers m, n which we may assume share no common factors. Thus m2 /n2 = 3 only
if m2 = 3n2 implying 3 divides m2 . By the preceding lemma, m is divisible by 3.
That is m = 3k1 for some integer k1 . Substituting this gives
3n2 = m2 = (3k1 )2 = 9k12 only if n2 = 3k12 .
This shows n is also divisible by 3 by our lemma. But this is impossible since m and
√
n were taken to share no common factors. Hence no such integers m, n exist and 3
1
(if it exists) must be irrational.
(2.11) Suppose x > 0, y > 0, and L < xy. Show that there is some n such that
1
1
L< x−
(y − ).
n
n
We will prove this using a chain of if and only if statements. Observe that
1
1
L< x−
(y − )
↔
n
n
1
x y
L < xy − − + 2
↔
n n n 1
1
L < xy −
x+y−
↔
n
n
1
1
L − xy < −
x+y−
↔
n
n
1
1
x+y−
∗∗.
xy − L >
n
n
Since this is a string of if and only if statements, if we can show the final inequality
holds for some integer n, then we will have the result. Notice that xy − L > 0 since
xy > 0 so there exists some integer n0 such that xy − L > n10 > 0. Our goal is now
to find an integer n such that n10 > n1 x + y − n1 if and only if nn0 > x + y − n1 .
Next notice that x + y > x + y − n1 so it will suffice to prove the existence of an
integer n0 such that nn0 > x + y > x + y − n1 . If no such integer existed, then
n
≤ x + y only if n ≤ n0 (x + y) for all natural n. But this is false since the natn0
ural numbers are not bounded
above. Consequently, there is an integer n such that
xy − L > n10 > n1 x + y − n1 , which shows ** holds and completes the proof.
(2.12) Let A and B be subsets of [0, ∞), and let
AB = {xy : x ∈ A, y ∈ B} .
Let a = sup A and b = sup B. Show that AB is bounded and sup AB = ab.
First we show AB is bounded by ab. By the definition of the supremum of a
set, x ≤ a for all x ∈ A and y ≤ b for all y ∈ B. Since elements of A and B are
nonnegative by hypothesis, then for any x ∈ A, y ∈ B, we have
xy ≤ ay ≤ ab.
This shows AB is bounded above by ab. To be complete, we should also note that
AB is bounded below by 0.
Now we claim that sup AB = ab. We have already shown that ab is an upper
bound of AB. It remains to show that if w is any other upper bound, then ab ≤ w.
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Suppose then that w is an upper bound, and assume to get a contradiction that
w < ab. By the previous exercise, there exists an integer n such that
1
1
w < a−
b−
.
n
n
Next, take a0 ∈ A such that a ≥ a0 > a − n1 . Such an a0 must exist; for otherwise
a − n1 would be an upper bound of A strictly smaller than sup A = a. The exact same
reasoning proves the existence of b0 ∈ B such that b ≥ b0 > b − n1 . This gives
1
1
w <w < a−
b−
< a0 b0 ∈ AB.
n
n
Hence w cannot be an upper bound AB, which is a contradiction. This means ab ≤ w,
and shows ab = sup AB.
(2.13) Omitted. Use the variant of Theorem 2.4 that I proved in class, and mimic
the proof of Theorem 2.5 exactly substituting the variant of 2.4 for theorem 2.4 and
replacing rational numbers by dyadic rationals. As in class, explain why the other
three cases hold. See me or Professor Ciocan-Fontanine if you have questions.
(2.21) Let A be a nonempty subset of R which has a lower bound. Let B be the set
of all lower bounds of A.
(a) Show that B has a least upper bound.
Note that B is nonempty since A is assumed to be bounded above. Furthermore,
B is bounded above by any element of A by the definition of the set B. Since A is
nonempty, the set B is therefore bounded above. By the Completeness axiom, sup B
exists.
(b) Formulate a plausible statement about sup B and lower bounds of A.
Let B be the set of lower bounds of A. We claim that sup B = inf A. We have
already shown that the supremum of B exists under the hypotheses of part (a). Now
we must show that sup B satisfies the requirements for being the infimum of A.
Suppose that sup B is not a lower bound of A. Then there is an element a0 ∈ A
such that a0 < sup B. Take b0 ∈ B such that a0 < b0 ≤ sup B. If no such b0 existed,
then a0 would be an upper bound of B strictly less than sup B which is impossible.
But this is likewise impossible since b0 is supposed to be a lower bound of A. The
moral of the argument is that sup B must be a lower bound of A.
Not take x0 to be a lower bound of A and suppose to get a contradiction that
x0 > sup B. But this means sup B cannot be an upper bound for the set of lower
bounds of A since, in particular, x0 is a lower bound of A strictly greater than sup B.
This gives yet another impossibility, from which we conclude that sup B = inf A.
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