Math2050A Term1 2016
Tutorial 1, Sept 15
Exercises
1. Let S = [a, b), where a < b. Find inf (S) and sup(S).
2. Let S = { 2nn : n ∈ N} (0 ∈
/ N in our definition). Find inf (S) and
sup(S).
3. (p.45 Q4 in our textbook)
Let A ⊂ R such that A 6= ∅ and bounded above. Let b > 0. Define
bA = {ba : a ∈ A}. Show that sup(bA) = bsup(A). What is sup(bA) if
b < 0 and A is bounded below?
4. (p.45 Q7 in our textbook)
let A ⊂ R, B ⊂ R, be nonempty sets. Show that sup(A + B) =
sup(A) + sup(B) whenever A,B are bounded above and inf (A + B) =
inf (A) + inf (B) whenever A, B are bounded below.
5. (p.45 Q11 in our textbook)
Let X, Y be nonempty sets and let h : X × Y → R have a bounded
range. Define
f (x) = sup{h(x, y) : y ∈ Y }, g(y) = inf {h(x, y) : x ∈ X}
Show that sup{g(y) : y ∈ Y } ≤ inf {f (x) : x ∈ X}
1
Solution (Outline)
1. I only find sup(S) here. We claim that sup(S) = b
Proof:
For s ∈ S, s < b. Therefore, b is an upper bound.
We want to show that if u is an upper bound of S, then u ≥ b:
Since u is an upper bound of S, u ≥ s ∀s ∈ S = [a, b).
This implies also u ≥ s whenever b > s. This says that u ≥ b. Let’s
see:
Suppose u < b, take α = u+b 2
, then b > α but u < α. Contradicts to
”u ≥ s whenever b > s ”. Therefore, u ≥ b.
By definition of supremum, sup(S) = b
You can also argue that ”if v < b, then v cannot be an upper bound of
S” to conclude that b is the least, as what we do in Q2.
2. Since 2n+1 1 n 1 1 n 1
n+1 = 2 ( 2n ) + 2 ( 2n ) is taking mean value of 2n and 2n ,
we have 2n+1 n
n+1 ≤ 2n .
Therefore, 12 is the maximum of S. We have sup(S) = 12
For the infimum, we claim that inf (S) = 0 :
0 is obviously a lower bound of S.
We claim that any postive number cannot be a lower bound of S:
Let > 0 be any postive number.
Since 2n = (1+1)n = 1+n+ n(n−1) 2
+... ≥ n(n−1)
2
∀n ≥ 2, then 2nn ≤ n−1
2
.
By Archimedean property, there is N ∈ N such that 1/N < /2. Take
n = N + 1, we have 2nn ≤ n−1 2
= N2 < . Hence is not a lower bound
of S.
3. bA is a nonempty subset of R bounded above. Hence, sup(bA) exists.
sup(bA) ≥ ba ∀a ∈ A. Then, 1b sup(bA) ≥ a ∀a ∈ A. Since LHS is a
constant and upper bound of A, we have 1b sup(bA) ≥ sup(A), hence
sup(bA) ≥ bsup(A).
One can conclude that for any nonempty subset of R bounded above,
say B, and any postive number, say c, we have sup(cB) ≥ csup(B).
Put c = 1b and B = bA, one can obtain the other inequality sign.
(Check 1b bA = A)
4. I only do the following: Assume sup(A + B) = sup(A) + sup(B) when-
ever A,B is bounded above, we show that inf (A + B) = inf (A) +
inf (B) whenever A,B are bounded below. Define −A = {−a : a ∈ A},
then by assumption, sup((−A) + (−B)) = sup(−A) + sup(−B) and
hence −inf (A + B) = −inf (A) − inf (B). You need to check that −A
is bounded above, sup(−A) = −inf (A) and −(A + B) = (−A) + (−B)
2
5. Since {h(x, y) : y ∈ Y } and {h(x, y) : x ∈ X} are nonempty bounded
subset of R, respectively for each x ∈ X and y ∈ Y , then both f (x)
and g(y) are well-defined.
For each y ∈ Y , g(y) ≤ h(x, y) ∀x ∈ X. For each x ∈ X, h(x, y) ≤
f (x) ∀y ∈ Y . Now, let x0 ∈ X being fixed, then for each y ∈ Y ,
g(y) ≤ h(x0 , y) ≤ f (x0 ) . Since f (x0 ) is a constant and an upper
bound of {g(y) : y ∈ Y }, we have sup{g(y) : y ∈ Y } ≤ f (x0 ). This
holds for all x0 in X and LHS is a constant as well as a lower bound of
{f (x) : x ∈ X}, so sup{g(y) : y ∈ Y } ≤ inf {f (x) : x ∈ X}.
