Solutions to Assignment 2, Math 220
1 Let
S = {1 − (−1)n /n : n ∈ N} .
Find inf S and sup S and prove your answers.
Solution We claim that inf S = 1/2 and sup S = 2. Note that, if n is
odd, 1 − (−1)n /n = 1 + 1/n, while, if n is even, 1 − (−1)n /n = 1 − 1/n.
It follows, if n is odd, that 1 − (−1)n /n > 1 > 1/2. If n ≥ 2 is even,
1 − (−1)n /n = 1 − 1/n ≥ 1 − 1/2 = 1/2.
Arguing similarly, 1 − (−1)n /n ≤ 2 and so 1/2 and 2 are, respectively,
lower and upper bounds for S. Since 1/2 ∈ S, there cannot be a lower
bound m > 1/2 and so 1/2 is the greatest lower bound for S, i.e.
inf S = 1/2. Since 2 ∈ S, there cannot be a upper bound M < 2 and
so 2 is the least upper bound for S, i.e. sup S = 2.
2 Let S ⊆ R and suppose that s∗ = sup S belongs to S. If u 6∈ S,
show that
sup (S ∪ {u}) = sup{s∗ , u}.
Solution Since s∗ ∈ S and u 6∈ S, it follows that s∗ 6= u and so either
s∗ > u or s∗ < u. In the first case, sup{s∗ , u} = s∗ , while u is not an
upper bound for S and so sup (S ∪ {u}) = sup S = s∗ . In the second
case, sup{s∗ , u} = u and s∗ < u implies that u is an upper bound for S.
It follows that sup (S ∪ {u}) = u. In either case, we have the desired
conclusion.
3 Show that a nonempty finite set S ⊆ R contains its supremum.
Hint: Use Mathematical Induction and the previous question.
Solution We use induction upon n, the number of elements of S. If S
has a single element, say S = {x}, then sup S = x ∈ S. Suppose that
the assertion is true for sets with exactly k elements for k ∈ N. If S
has k + 1 elements, let x ∈ S. Then S1 = S \ {x} has k elements and
hence s∗ = sup S1 ∈ S1 . By the previous result,
sup S = sup (S1 ∪ {x}) = sup{s∗ , x} = max{s∗ , x}.
Since s∗ ∈ S1 ⊆ S, we have that s∗ ∈ S and also x ∈ S. It follows that
max{s∗ , x} ∈ S and so sup S ∈ S, as claimed.
1
2
4 Question 12.4, page 113 of Lay (4th edition).
Solution
a:
S = {1, 3}, inf S = min S = 1.
b:
S = {π, 3}, inf S = min S = 3.
c:
S = [0, 4], inf S = min S = 0.
d:
S = (0, 4), inf S = 0, min S does not exist.
e:
S = {1/n : n ∈ N} , inf S = 0, min S does not exist.
f:
S = {1 − 1/n : n ∈ N} , inf S = min S = 0.
g:
S=
n
: n ∈ N , inf S = min S = 1/2.
n+1
h:
S = {(−1)n (1 + 1/n) : n ∈ N} , inf S = min S = −2.
i:
S = {n + (−1)n /n : n ∈ N} , inf S = min S = 0.
j:
S = (−∞, 4), inf S, min S do not exist.
k:
l:
S=
∞ \
1
1
S=
1 − ,1 +
= {1}, inf S = min S = 1.
n
n
n=1
∞ [
1
n=1
1
,2 −
= (0, 2), inf S = 0, min S does not exist.
n
n
m:
S = {r ∈ Q : r < 5} , inf S, min S do not exist.
n:
√
S = r ∈ Q : r 2 ≤ 5 , inf S = − 5, min S does not exist.
3
5 Suppose that a and b are rational numbers with a < b. Give an
explicit example of an irrational number α with a < α < b.
Solution If a, b ∈ Q, a < b, let us define x(t) = a + (b − a) · t. Notice
that x(0) = a, x(1) = b and, if 0 < t < 1, then a < x(t) <√b. Also,
if t 6∈ Q, then x(t) 6∈ Q, so if we take, for example, t = 1/ 2, then
x(t) 6∈ Q and a < x(t) < b; i.e. α = a + √12 (b − a) is an example of an
irrational number with a < α < b.
6 Question 13.4, page 121 of Lay (4th edition).
Solution
a:
S = {1/n : n ∈ N} , bd S = S ∪ {0}.
b:
c:
d:
S = [0, 3] ∪ (3, 5) = [0, 5), bd S = {0, 5}.
n
√ o
√
S = r ∈ Q : 0 < r < 2 , bd S = [0, 2].
n
√ o
√
S = r ∈ Q : r ≥ 2 , bd S = [ 2, ∞).
e:
S = [0, 2] ∩ [2, 4] = {2}, bd S = S = {2}.