Fall 2016, Math 409, Section 502
Solutions to some problems of the third assignment
Exercise 1. Find f [A] and f −1 [A], where f : R → R, in the case:
(i) f (x) = 2 − 3x and A = (−1, 2),
2 pts.
Solution. We have x ∈ (−1, 2) ⇔ 2 − 3x ∈ (−4, 5) and hence f [A] = (−4, 5).
Moreover, 2 − 3x ∈ (−1, 2) ⇔ x ∈ (0, 1) and hence f −1 [A] = (0, 1).
Exercise 3. Let A and B be sets and assume that A is uncountable. If
there exists a function g : B → A that is onto, prove that B is uncountable.
Solution. Assume that the conclusion fails, i.e. B is most countable. Then,
by a known proposition, there exists an onto function f : N → B. The
function h : N → A with h = g ◦ f is onto and therefore, by a known
proposition, A must be most countable, which is absurd.
Exercise 4. Let A be a finite set and B ⊆ A. Show that B is finite.
Hint: prove by induction on n the following statement: if F ⊆ {1, . . . , n}
and F 6= ∅, then there exists a natural number k 6 n and a bijection
f : F → {1, . . . , k}.
Solution. We shall prove by induction the statement given in the hint. For
n = 1, if F ⊆ {1} with F 6= ∅, then F = {1}. We deduce that k = 1 and the
identity function satisfy the conclusion. Let n be a natural number for which
the statement holds and let F be a non-empty subset of {1, . . . , n, n + 1}.
Set G = F ∩ {1, . . . , n}. We distinguish three cases.
Case 1: G = ∅. In this case F = {n + 1}. We deduce that k = 1 and the
function g : F → {1} with g(n + 1) = 1 satisfy the conclusion.
Case 2: F = G and hence, F ⊂ {1, . . . , n}. In this case, by the inducive
assumption there already exist k 6 n < n + 1 and a bijection f : F →
{1, . . . , k}.
Case 3: G 6= ∅ and G ( F . In this case, F = G ∪ {n + 1} and by the
inductive assumption there exist k 6 n and a bijection f : G → {1, . . . , k}.
Define g : F → {1, . . . , k + 1} with
(
f (m) if m ∈ G,
g(m) =
k + 1 if m = n + 1.
Then g is the desired bijection to {1, . . . , k + 1}, where k + 1 6 n + 1 because
k 6 n.
To finish the proof, let A be a finite set and B ⊆ A. If B = ∅, then it is
finite. Otherwise, B 6= ∅ and hence A 6= ∅, i.e. there exists n ∈ N and a
bijection φ : A → {1, . . . , n}. Since φ[B] is a non-empty subset of {1, . . . , n},
as we just proved, there exists k 6 n and a bijection f : φ[B] → {1, . . . , k}.
In conclusion, the map h : B → {1, . . . , k}, with h(b) = f (φ(b)) for all b ∈ B,
is a bijection, and hence B is finite.