Problems in Advanced Calculus, Math 409 Fall 2016
Homework 5 - Solutions
Prof.: Thomas Schlumprecht
Problem 1. Let X and Y be non empty sets and let f : X → Y .
a) For any family (Bi )i∈I of subsets of Y it follows that
[ [
\ \
f −1
Bi =
f −1 (Bi ) and f −1
Bi =
f −1 (Bi ).
i∈I
i∈I
i∈I
i∈I
b) For any family (Ai )i∈I of subsets of X it follows that
[ [
\ \
f
Ai =
f (Ai ) and f
Ai ⊂
f (Ai ).
i∈I
i∈I
i∈I
i∈I
c) Show that the inclusion in (b) could be strict.
Proof. (a) Let x ∈ X. Then
[ [
Bi ⇐⇒ f (x) ∈
Bi
x ∈ f −1
i∈I
i∈I
⇐⇒ ∃ ∈ I
f (x) ∈ Bi
⇐⇒ ∃ ∈ I
x ∈ f −1 (Bi ) ⇐⇒ x ∈
[
f −1 (Bi )
i∈I
and
x ∈ f −1
\
Bi
⇐⇒ f (x) ∈
i∈I
\
Bi
i∈I
⇐⇒ ∀i ∈ I
f (x) ∈ Bi
⇐⇒ ∀i ∈ I
x ∈ f −1 (Bi ) ⇐⇒ x ∈
\
f −1 (Bi )
i∈I
(b)
y∈f
[
Ai
⇐⇒ ∃x ∈
i∈I
[
Ai
y = f (x)
i∈I
⇐⇒ ∃i ∈ I ∃x ∈ Ai
⇐⇒ ∃i ∈ I
y = f (x)
y ∈ f (Ai ) ⇐⇒ y ∈
[
f (Ai )
i∈I
and
y∈f
\
i∈I
Ai
⇐⇒ ∃x ∈
\
Ai
y = f (x)
i∈I
=⇒ ∀i ∈ I ∃x ∈ Ai
⇐⇒ ∀i ∈ I
y = f (x) (this implication is not reversible)
\
y ∈ f (Ai ) ⇐⇒ y ∈
f (Ai ).
i∈I
1
2
(c) Let X = {1, 2}, Y = {1}, and f : X → Y , 1 7→ 1, and 2 7→ 1 let A1 = {1}
and A2 = {2}. Then
f (A1 ∩ A2 ) = f (∅) = ∅, but f (A1 ) ∩ f (A2 ) = {1}.
Problem 2. Let X and Y be non empty sets and let f : X → Y .
a) For A, B ⊂ X, show that f (A) \ f (B) ⊂ f (A \ B),
b) For A, B ⊂ X, show that f (A) \ f (B) = f (A ∩ B) if f is injective.
c) Give an example that the conclusion of (b) may fail if f is not injective.
d) For C, D ⊂ X, show that f −1 (A) \ f −1 (B) = f −1 (A ∩ B),
Proof. (a) for y ∈ Y
y ∈ f (A) \ f (B) ⇒ ∃x ∈ A,
f (x) = y and x 6∈ B
⇒ ∃x ∈ A \ By = f (x) ⇒ y ∈ f (A \ B).
(b) Now assume that f is injective, and assume y ∈ f (A \ B). Thus, theres
is an x ∈ A, with x 6∈ B and y = f (x). Since f is injective there is no z ∈ B
so that y = f (z) (otherwise f (x) = y = f (z) and x 6= z). This implies that
y 6∈ f (B), and thus y ∈ f (A) \ f (B)
(c) Take X, Y and f : X → Y as in 1(c) and A = {1} and B = {2}.
(d) for x ∈ X
x ∈ f −1 (C \ D) ⇐⇒ f (x) ∈ C \ D
⇐⇒ f (x) ∈ C and f (x) 6∈ D
⇐⇒ x ∈ f −1 (C) and x 6∈ f −1 (D) ⇐⇒ x ∈ f −1 (C) \ f −1 (D).
Problem 3. Assume that X is a countable infinite set. Prove that there
are subsets A and B of X so that A ∩ B = ∅, A ∪ B = X , and A and B are
countable infinite.
Proof. By assumption there exists a bijection f : N → X. Put
A = {f (2n) : n ∈ N} and B = {f (2n − 1) : n ∈ N}
and define
g : N → A,
n 7→ f (2n), and h : N → B,
n 7→ f (2n − 1).
From the definition of A and B it follows that g and h are surjective. From
the injectivity of f it follows that g and h are injective. Indeed,
g(m) = g(n) ⇒ f (2m) = f (2n) ⇒ 2m = 2n ⇒ m = n and
h(m) = h(n) ⇒ f (2m − 1) = f (2n − 1) ⇒ 2m − 1 = 2n − 1 ⇒ m = n.
Since f is surjective and N is the union of the odd and even natural numbers
it follows that A ∪ B = X.
From the injectivity of f it follows that A ∩ B = ∅. Indeed. assume that
x ∈ A ∩ B, then x = f (2m) = f (2n − 1), for some m, n ∈ N. But this would
3
yield (by injectivity of f ) that 2m = 2n − 1 which is not possible since a
number cannot be even and odd at the same time.
Problem 4. Let X be an infinite set.
a) Prove that X contains an infinite countable subset.
b) Let x0 6∈ X, and put Y = X ∪ {x0 }. Show that X and Y have the
same cardinality.
Hint: first take X = N, and then use (a) to prove the general case.
Proof. (a) Recursively we choose for each k ∈ N, and element xk ∈ X so
that xk 6= xl , for all l = 1, 2, 3 . . . k − 1.
Induction start: Since X 6= ∅, we pick x1 ∈ X.
Induction step: k → k + 1 Assume x1 , x2 , . . . xk have been chosen. Since
X is infinite X \ {x1 , x2 , . . . xk } must be also infinite, and thus not empty.
Therefore we can pick xk+1 ∈ X \ {x1 , x2 , . . . xk }.
Thus we have chosen a sequence (xk )∞
k=1 ⊂ X, for which xk 6= xl , whenever k 6= l. It follows therefore that the map
f : N → {xn : n ∈ N},
n 7→ xn
is a bijection.
(b) By part (a) We can write X as a disjoint union X = A ∪ B. Where A
is countable infinite. Let f : N → A be a bijection, the define F : Y → X,
as follows


if y ∈ B
y
F (y) = f (1)
if y = x0


−1
f (1 + f (y)) if y ∈ A
(meaning x0 is mapped to f (1), f (1) to f (2), f (2) to f (3) etc.)
Problem 5. Fo a set X let P (X) be the Powerset of X, i.e., the set of all
subsets of X,
P (X) = {A : A ⊆ X}.
Show that there is no surjection f : S → P (S) (and thus S does not have
the same cardinality as P(S)
Hint: Assume that f : S → P (S) is a surjection. Consider the set
S = {s ∈ X : s 6∈ f (s)}.
Proof. Assume that f : S → P (S) is any map. We will show that it cannot
be surjective, by showing that the set S = {s ∈ X : s 6∈ f (s)} is not in the
range of f .
Suppose it is in the range, and there is an x ∈ X so that S = f (x).
There two cases, both will lead to a contradiction.
Case 1. x ∈ S. But this means x 6∈ f (x) = S, which is a contradiction.
Case 2. x 6∈ S. This means that x ∈ f (x) = S, which is also a contradiction.