McGill University
Department of Mathematics and Statistics
MATH 242 Analysis 1, Fall 2022
Midterm 1: Solutions
1. (10 marks) Prove by induction that for all n ∈ N it holds that
1
1 · 2 +
1
2 · 3 +
1
3 · 4 + · · ·+
1
n · (n+ 1) =
n
n+ 1
Solution:
Base Case: 11·2 =
1
2 =
1
1+1 , which is what we had to show.
Inductive Step n→ n+ 1:
Assume that
1
1 · 2 +
1
2 · 3 +
1
3 · 4 + · · ·+
1
n · (n+ 1) =
n
n+ 1
for some n ∈ N. Then
1
1 · 2 +
1
2 · 3 +
1
3 · 4 + · · ·+
1
n · (n+ 1) +
1
(n+ 1) · (n+ 2)
Ind. Hyp.
=
n
n+ 1
+
1
(n+ 1) · (n+ 2)
=
n(n+ 2)
(n+ 1) · (n+ 2) +
1
(n+ 1) · (n+ 2)
=
n2 + 2n+ 1
(n+ 1) · (n+ 2)
=
(n+ 1)2
(n+ 1) · (n+ 2)
=
n+ 1
n+ 2
=
n+ 1
(n+ 1) + 1
We’ve thus successfully completed the inductive step, and we’ve proven the statement for all
n ∈ N.
2. (10 marks) Let D, E be non-empty sets, let A ⊆ D, let B ⊆ E , and let f : D → E be a
function. Prove that
A ∩ f−1(B) ⊆ f−1(f(A) ∩B)
Solution:
1. Solution:
Let x ∈ A ∩ f−1(B). Then
x ∈ A ∩ f−1(B)⇒ x ∈ A ∧ x ∈ f−1(B)⇒ f(x) ∈ f(A) ∧ f(x) ∈ B
⇒ f(x) ∈ f(A) ∩B ⇒ x ∈ f−1(f(A) ∩B)
This proves that A ∩ f−1(B) ⊆ f−1(f(A) ∩B).
2. Solution:
It was shown on assignment 2 that if X,Y ⊆ E, then f−1(X ∩ Y ) = f−1(X) ∩ f−1(Y ). With
X := f(A) and Y := B, we obtain that f−1(f(A) ∩B) = f−1(f(A)) ∩ f−1B.
It was also shown on assignment 2 that for any A ⊆ D it holds that f−1(f(A)) ⊇ A. Combining
both results yields
f−1(f(A) ∩B) = f−1(f(A)) ∩ f−1B ⊇ A ∩ f−1B ⇔ A ∩ f−1(B) ⊆ f−1(f(A) ∩B)
3. (10 marks) Let A and B be non-empty and bounded subsets of R such that sup A < sup B.
Prove the following:
∀a ∈ A∃b ∈ B : a < b
Solution:
1. Solution:
Let a ∈ A be arbitrary. Assume that ∀b ∈ B : b ≤ a. Then a is an upper bound of B. Since
sup B is the least upper bound of B, this implies that sup B ≤ a. However, we also have that
a ≤ sup A, and we conclude that sup B ≤ sup A which is false. Thus there must exist a b ∈ B
such that a < b. This is what we had to show.
2. Solution:
Since sup A < sup B and sup B is the least upper bound of B, it follows that sup A is not an
upper bound of B. Consequently, there exists a b ∈ B such that sup A < b.
Furthermore, since sup A is an upper bound of A, we have that a ≤ sup A for all a ∈ A.
Combining these two results, it follows that a ≤ sup A < b and thus a < b for all a ∈ A. We’ve
thus proven that ∀a ∈ A∃b ∈ B : a < b. In fact, we’ve even proven the stronger statement
∃b ∈ B∀a ∈ A : a < b.
2
4. (a) (2 marks) Prove that
n
n2 + 1
<
1
n
for all n ∈ N.
(b) (8 marks) Let S :=
{
n
n2 + 1
: n ∈ N
}
. Prove that S is bounded and determine sup S and
inf S. Fully justify your answers!
Solution:
(a)
n
n2 + 1
<
n
n2
=
1
n
for all n ∈ N.
(b) S is bounded: For n = 1 we have that
n
n2 + 1
=
1
2
. And for all n ≥ 2 it holds that
n
n2 + 1
(a)
<
1
n
≤ 1
2
. This proves
∀n ∈ N : n
n2 + 1
≤ 1
2
∈ S (1)
We also have
∀n ∈ N : 0 < n
n2 + 1
(2)
It follows immediately from (2) and (1) that S is both bounded from below and bounded
from above and thus bounded. Hence S has both an infimum and a supremum.
sup S: It follows from (1) that S has a maximum and that max S = 12 . Hence sup S =
max S = 12 .
inf S: We will show that inf S = 0 by using the ε-version of the definition of the infimum:
It follows from (2) that 0 is a lower bound of S. Let ε > 0 be arbitrary. We need to show
that 0 + ε = ε is not a lower bound of S. By the Archimedean property there exists an
n ∈ N such that n > 1ε ⇔ 1n < ε. It now follows from (a) that
n
n2 + 1︸ ︷︷ ︸
∈S
<
1
n
< ε
which proves that ε is indeed not a lower bound of S. It now follows from the ε-version of
the definition of the infimum that inf S = 0.