Math 317 HW #2 Solutions
1. Exercise 1.3.3.
(a) Let A be bounded below, and define B = {b ∈ R : b is a lower bound for A}. Show that
sup B = inf A.
Proof. First, note that B is bounded above by every element of A, so, by the Axiom of
Completeness, B has a supremum. Moreover, inf A ∈ B since inf A is a lower bound for
A and B is the set of all lower bounds for A.
Therefore, since sup B is an upper bound for B, it must be the case that sup B ≥ inf A.
If it were the case that sup B > inf A, then
:= sup B − inf A > 0,
so, by Lemma 1.3.7, there exists b ∈ B such that
b > sup B − = sup B − (sup B − inf A) = inf A.
But b ∈ B means that b is a lower bound for A, so this means that inf A is not the
greatest lower bound for A. This is plainly impossible, so it must not be the case that
sup B > inf A.
The only remaining possibility is that sup B = inf A, as desired.
Note: By definition of the infimum, inf A ≥ b for all b ∈ B, since B is precisely the set
of lower bounds of A. If you had Exercise 1.3.7 in hand, this would complete the proof
(and, really, the bulk of the above proof is essentially the proof of Exercise 1.3.7).
[Added 09.12.10] Here’s a better proof, borrowed (with minor modifications) from
Zach:
Proof. If b ∈ B, then b is a lower bound for A, meaning that b ≤ a for all a ∈ A.
Therefore, every element of A is an upper bound for B, so B is bounded above and thus,
by the Axiom of Completeness, has a least upper bound sup B. We want to see that
sup B satisfies the two conditions for being the greatest lower bound of A.
i. If sup B is not a lower bound for A, then there exists a ∈ A such that a < sup B.
Then := sup B − a > 0 and so, by Lemma 1.3.7, there exists b ∈ B such that
b > sup B − = sup B − (sup B − a) = a.
However, this means that b is not a lower bound for A, so b ∈
/ B. From this
contradiction, we conclude that sup B is a lower bound for A.
ii. Since sup B is an upper bound for B, sup B ≥ b for all b ∈ B. In turn, since B
is the set of lower bounds of A, this implies that if b is a lower bound for A, then
sup B ≥ b.
Therefore, since sup B satisfies the definition of the greatest lower bound, we conclude
that sup B = inf A.
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(b) Use (a) to explain why there is no need to assert that greatest lower bounds exist as
part of the Axiom of Completeness.
Answer. Suppose A is bounded below, and define B as in (a). By the Axiom of Completeness, sup B exists and, by part (a), sup B = inf A, meaning that inf A exists. Therefore,
if A is bounded below, A has a greatest lower bound in R.
(c) Propose another way to use the Axiom of Completeness to prove that sets bounded
below have greatest lower bounds.
Answer. Another approach would be, for any A ⊆ R which is bounded below, to define
−A := {−x ∈ R : x ∈ A}. Then the set −A is bounded above and so, by the Axiom of
Completeness, has a least upper bound s. Then one can prove that inf A = −s, so A
has a greatest lower bound.
2. Exercise 1.3.6. Compute, without proofs, the suprema and infima of the following sets:
(a) A = {n ∈ N : n2 < 10}.
Answer. A = {0, 1, 2, 3}, so sup A = 3 and inf A = 0 (or, using the book’s convention,
inf A = 1).
(b) A = {n/(m + n) : m, n ∈ N}.
n
n
≤ 1+n
<
Answer. Using the book’s convention that N = {1, 2, 3, . . .}, it’s clear that m+n
n
1. Of course 1+n gets arbitrarily close to 1, so sup A = 1. Likewise, since the numerator
n
1
is always positive, m+n
> 0, but, when n = 1, the fraction m+1
gets arbitrarily close to
0, so inf A = 0.
(c) A = {n/(2n + 1) : n ∈ N}.
n
Answer. Note that 2n+1
< 12 ; the left hand side gets arbitrarily close to 12 , so sup A = 21 .
Also (again using the convention N = {1, 2, 3, . . .}), we know that
clearly inf A = 31 .
n
2n+1
≥ 13 ; since
1
3
∈ A,
(d) A = {n/m : m, n ∈ N with m + n ≤ 10}.
Answer. Again assume N = {1, 2, 3, . . .}. The set A is actually a finite set, so we could
list all the elements. If we did so, the biggest element (and hence the sup) will be 9, and
the smallest element (and hence the inf) will be 1/9.
3. Exercise 1.3.8. If sup A < sup B, then show that there exists an element b ∈ B that is an
upper bound for A.
Proof. Suppose sup A < sup B. Define := sup B − sup A, which is a positive number. Then,
by Lemma 1.3.7, there exists b ∈ B such that
b > sup B − = sup B − (sup B − sup A) = sup A.
Hence, for any a ∈ A,
a ≤ sup A < b,
so b is an upper bound for A, as desired.
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4. Exercise 1.3.9. Without worrying about formal proofs for the moment, decide if the following
statements about suprema and infima are true or false. For any that are false, supply an
example where the claim in question does not appear to hold.
(a) A finite, nonempty set always contains its supremum.
Answer. True.
(b) If a < L for every element a in the set A, then sup A < L.
Answer. False. Let A = (0, 1). Then a < 1 for all a ∈ A, but sup A = 1.
(c) If A and B are sets with the property that a < b for every a ∈ A and every b ∈ B, then
it follows that sup A < inf B.
Answer. False. Let A = (−1, 0) and B = (0, 1). Then a < b for all a ∈ A and b ∈ B, but
sup A = 0 = inf B.
(d) If sup A = s and sup B = t, then sup(A + B) = s + t. The set A + B is defined as
A + B = {a + b : a ∈ A and b ∈ B}.
Answer. True.
(e) If sup A ≤ sup B, then there exists an element b ∈ B that is an upper bound for A.
Answer. False. Let A = B = (0, 1). Then sup A = sup B, but no element of B is an
upper bound for A.
5. Exercise 1.4.2. Recall that I stands for the set of irrational numbers.
(a) Show that if a, b ∈ Q, then ab and a + b are elements of Q as well.
Proof. Suppose a, b ∈ Q. Then there exist p1 , p2 , q1 , q2 ∈ N with q1 6= 0, q2 6= 0 and
a = pq11 , b = pq22 . Hence,
p1 p2
p1 p2
ab =
=
,
q1 q2
q1 q2
which is an element of Q since both numerator and denominator are integers.
Likewise,
p1 p2
p 1 q 2 + p2 q 1
a+b=
+
=
,
q1
q2
q1 q2
which is also an element of Q since both numerator and denominator are integers.
Hence, we’ve shown that ab ∈ Q and a + b ∈ Q, as desired.
(b) Show that if a ∈ Q and t ∈ I, then a + t ∈ I and at ∈ I as long as a 6= 0.
Proof. By contradiction. Suppose there exist a ∈ Q and t ∈ I such that a + t ∈ Q. Then
−a ∈ Q and part (a) implies that
t = (a + t) + (−a) ∈ Q.
This contradicts the fact that t ∈ I; from this contradiction we conclude that a + t ∈ Q
for all a ∈ Q and t ∈ I.
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Likewise, we prove at ∈ I whenever a 6= 0 by contradiction. To derive the contradiction,
suppose a ∈ Q and t ∈ I such that a 6= 0 and at ∈ Q. Then, by definition of Q and since
a 6= 0, there exist p, q ∈ N with p 6= 0, q 6= 0 such that a = pq . Hence,
q
1
= ∈ Q.
a
p
By part (a), then,
1
t=
(at) ∈ Q,
a
which contradicts the fact that t ∈ I. From this contradiction, we conclude that at ∈ I
for all a ∈ Q and t ∈ I with a 6= 0.
(c) Part (a) can be summarized by saying that Q is closed under addition and multiplication.
Is I closed under addition and multiplication? Given two irrational numbers s and t,
what can we say about s + t and st?
Answer. The set I is not closed under either√addition or√multiplication. To see that it
is not closed under addition, note that both 2 and 1 − 2 are irrational (the latter by
part (b) above), but
√
√
2 + (1 − 2) = 1 ∈ Q.
In other words, the sum of irrational numbers can be rational.
√
Likewise, to see that I is not closed under multiplication, observe that 2 ∈ I but
√
2·
√
2=
√
2
2 = 2 ∈ Q.
In other words the product of irrational numbers can be rational.
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