Math 415 Assignment 11
Section 2.2: 1,3,5; Section 2.3: 3,4,5,6
Rough Answers
√
√
p. 26 #1. (a) The supremum is 2, the infimum is − 2.
√
√
(b) The supremum is 2, the infimum is − 2.
√
(c) The supremum is +∞, the infimum is 2.
(d) The supremum is +∞, the infimum is −∞.
p. 26 #3. Let A and B be nonempty subsets of R with A ⊆ B. Show that
inf B ≤ inf A ≤ sup A ≤ sup B.
Proof. . Let β = inf B, then β ≤ b for all b ∈ B. Because A ⊆ B, it follows that β ≤ a
for all a ∈ A. Therefore, β is a lower bound of A. Consequently, β ≤ inf A, because inf A is
the greatest lower bound of A. Now, fix any a ∈ A, then inf A ≤ a and a ≤ sup A, therefore
inf A ≤ sup A. Now let α = sup B. Then b ≤ α for all b ∈ B. Therefore, a ≤ α for all a ∈ A
because A ⊆ B. Consequently, α is an upper bound of A. Because sup A is the least upper
bound of A, it follows that sup A ≤ α.
p. 26 # 5. Let S be a nonempty set that is bounded above. Prove that inf(−S) = − sup S
where −S = (−1)S = {−s : s ∈ S}.
Proof. Let α = sup S. Then s ≤ α for all s ∈ S. Therefore, −α ≤ −s for all s ∈ S.
Consequently, −α is a lower bound of −S. Now let b be any lower bound of −S. Then, b ≤ −s
for all s ∈ S, and so −b ≥ s for all s ∈ S. Consequently, −b ≥ α since α was the least upper
bound for S. Thus −(−b) ≤ −α, that is b ≤ −α. This shows that any lower bound of b of
S satisfies b ≤ −α. Consequently, −α = − sup S is the greatest lower bound, of −S. That is
inf(−S) = − sup S.
p. 30 #3. (a) sup{1 − 1/n : n ∈ N} = 1, inf{1 − 1/n : n ∈ N} = 0
(b) sup Q = +∞, inf Q = −∞
(c) sup{n − 1/n : n ∈ N} = +∞, inf{n − 1/n : n ∈ N} = 0
√
√
(d) sup{x ∈ Q : x2 < 2} = 2, inf{x ∈ Q : x2 < 2} = − 2
p. 30 #4. The irrational numbers are dense in R.
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Proof. Given any a, b ∈ R such that a < b, it suffices to show there is an irrational number t
with a < t < b. So let a, b ∈ R satisfy a < b. By Theorem 2.2, there is a rational number r
b−r
such that a < r < b. Now
> 0, so by the Archimedian principle, we can choose n ∈ N
2
√
b−r
2
2
1
. Then t is irrational,
for which <
. Conseqently, a < r + < b. Now let t = r +
n
n
2 √
n
for if it weren’t, then 2 = n(t − r) would be rational. Also, a < r < t < r + n2 < b, and so t
is an irrational number within the interval (a, b).
p. 30 #5. Let A and B be nonempty subsets of R that are bounded below. Prove that
inf(A + B) = inf A + inf B.
Proof. Because −A and −B are bounded above, p. 26 #5 ensures that inf A = − sup(−A)
and inf B = − sup(−B). Then sup(−A − B) = sup(−A) + sup(−B) by Proposition 2.6. Also,
inf(A + B) = − sup(−A − B) by p. 26 #5. Putting this all together, we get
inf(A + B) = − sup(−A − B)
= − sup(−A) + − sup(−B)
= inf A + inf B
as desired.
Alternatively, the previous exercise could have been completed by appropriately modifying the
proof of Proposition 2.6
p. 31 #6. Let f and g be bounded functions from a nonempty set X into R. Show that
sup{f (x) + g(x) : x ∈ X} ≤ sup{f (x) : x ∈ X} + sup{g(x) : x ∈ X}.
Show by examples that equality and strict inequality can occur.
Proof. By the least upper bound principle, we let α = sup{f (x) : x ∈ X} and β = sup{g(x) :
x ∈ X}. Then f (x) ≤ α and g(x) ≤ β for all x ∈ X. Consequently,
(f + g)(x) = f (x) + g(x) ≤ α + β.
Therefore, α + β is an upper bound for {f (x) + g(x) : x ∈ X}. Consequently,
sup{f (x) + g(x) : x ∈ X} ≤ sup{f (x) : x ∈ X} + sup{g(x) : x ∈ X}.
For an example where equality occurs, consider X = [0, 2], f (x) = x and g(x) = x2 . Then
sup{f (x) + g(x) : x ∈ X} = 6 and sup{f (x) : x ∈ X} + sup{g(x) : x ∈ X} = 2 + 4 = 6. For an
example where there is a strict inequality, consider X = [0, 2], f (x) = x and g(x) = 2−x. Then
sup{f (x) + g(x) : x ∈ X} = 2 while sup{f (x) : x ∈ X} + sup{g(x) : x ∈ X} = 2 + 2 = 4.
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