Fall 2016, Math 409, Section 502
Solutions to some problems of the third assignment
Exercise 5. Let A and B be non-empty subsets of R with the property
a 6 b for all a ∈ A and b ∈ B. Show that sup A and inf B exist and in
particular sup A 6 inf B.
Solution. Fix b in B. Then by assumption, for all a in A we have a 6 b, i.e.
b is an upper bound for A. By the completeness axiom, sup A exists, and
since b is an upper bound for A, we have sup A 6 b.
As b was an arbitrary element of B, sup A is a lower bound for B. By
the completeness property for infima, inf B exists, and since sup A is a lower
bound for B, we have sup A 6 inf B.
Exercise 6. (i) Show that 2n < n!, for all n ∈ N with n > 4.
Solution. For n = 4, we have 24 = 16 < 24 = 4!. Assume now that n > 4
and 2n < n!. Then:
(∗)
2n+1 = 2(2n ) < 2(n!) 6 (n + 1)(n!) = (n + 1)!,
where inequality (∗) follows from the inductive assumption.
Exercise 7. Let a ∈ R with a > 0. The goal is to show that there exists
a non-negative real number
r, with r2 = a. We call r the square root of a
√
and denote it by r = a or r = a1/2 . Define
A = {x ∈ R : 0 6 x and x2 6 a} and B = {x ∈ R : 0 6 x and a 6 x2 }.
√
Without assuming the existence of a prove:
(i) A is non-empty and max{1, a} is an upper bound for A,
(ii) B is non-empty and min{1, a} is a lower bound for B,
(iii) if s = sup A, then s2 6 a (Hint: use the approximation property for
suprema),
(iv) if t = inf B, then a 6 t2 (Hint: use the approximation property for
infima),
(v) s = t and hence s2 = a.
Solution. (i): 0 is in A, i.e. A 6= ∅. To show that max{1, a} is an upper
bound for A we distinguish two cases. If 0 6 a 6 1 we will show that 1
is an upper bound for A. Indeed, if x ∈ A and we assume 1 < x, then
a 6 1 < x < x2 , i.e. a < x2 and x ∈
/ A, which is absurd. In the second case,
a > 1 and we will show that a is an upper bound for a. Indeed, if x ∈ A and
we assume a < x, then x < x2 (because 1 < x) and hence a < x2 , which is
absurd. Note that 0 ∈ A implies 0 6 sup A = s.
(ii): If a 6 1, then 1 ∈ B whereas if a > 1, then a2 > a and hence a ∈ B.
In either case, B 6= ∅. Proving that min{1, a} is a lower bound for B is
similar to the proof of (i). Note that (ii) implies inf B > min{1, a} > 0.
The proofs of (iii) and (iv) are similar, only a that of (iv) is given.
(iv): if the conclusion fails, then 0 < δ = a − t2 . For every x ∈ B we have
(∗)
(1)
0 < δ = a − t2 = (a − x2 ) + (x2 − t2 ) 6 x2 − t2 = (x − t)(x + t),
2
where inequality (∗) follows
from
n
o a − x 6 0, as x ∈ B.
δ
Choose 0 < ε < min 1, 2t+1
and apply the approximation property for
infima to find x ∈ B with t 6 x < t + ε. We conclude
(2)
0 6 x − t < ε and
(3)
0 6 x + t < 2t + ε 6 2t + 1.
Combining (1), (2), and (3):
δ 6 (x − t)(x + t) < ε(2t + 1) 6
δ
(2t + 1) = δ,
2t + 1
i.e. δ < δ which is absurd.
(v) We sill show that t < s and s < t are not possible, and hence s = t. If
we assume t < s, then a 6 t2 < s2 6 a, which is absurd. Assume now that
s < t. Then there is c ∈ R with s < c < t (e.g. c = (s + t)/2). If c2 6 a,
then c ∈ A and sup A = s < c, which is absurd. Hence, c2 > a which means
c ∈ B and c < t = inf B, which is absurd as well. In conclusion, s = t and
a 6 t2 = s2 6 a.