Exam 2 : Thursday, October 17, 2013 1

Problem 1. Let x ≥ −1. Prove that for all n ∈ ℕ we have (1 + x)^{n} ≥ 1 + xn. Answer:

Problem 2. Let X and Y be sets and f : X → Y be a function. Consider the following

argument: Suppose that f : X → Y is not injective. Then there exist elements x, z ∈ X with f(x) = f(z) and x ≠ z. Let A = {x}. Note that z ∈ X \ A so

f(z) ∈ f(X \ A).

Since x ∈ A, f(x) ∈ f(A), so

f(x) ∉ Y \ f(A).

Since f(x) = f(z), we’ve found an element in f(X \ A) that is not in Y \ f(A). Therefore, f(X \ A) ⊈ Y \ f(A).

Which of the following propositions does the argument above prove?

(a) If f is injective, then for all A ⊆ X we have f(X \ A) ⊆ Y \ f(A).

(b) If f is injective, then there exists a set A ⊆ X with f(X \ A) ⊆ Y \ f(A).

(c) If for all subsets A ⊆ X we have f(X \ A) ⊆ Y \ f(A), then f is injective.

(d) If f is injective, then for all A ⊆ X we have Y \ f(A) ⊆ f(X \ A).

(e) If for all sets A ⊆ X we have Y \ f(A) ⊆ f(X \ A), then f is injective.

(f) If f is not injective, then for all A ⊆ X we have f(X \ A) ⊈ Y \ f(A).

Exam 2 : Thursday, October 17, 2013 3

Problem 3. Give an example of a function f : X → Y and a set A ⊆ X for which

f^{−1}(f(A)) ≠ A.

Answer:

Exam 2 : Thursday, October 17, 2013 4

Problem 4. True or False. Right answer +1, wrong answer −1, no answer 0.

(a) The field axioms of ℝ imply that 1 + 1 ≠ 0.

(b) For all functions f : X → Y and for all sets C ⊆ Y, we have f^{−1}(Y \C) = X \f^{−1}(C).

(c) For all functions f : X → Y and for all sets C ⊆ Y, we have f(f^{−1}(C)) = C.

(d) For all injective functions f : X → Y and for all sets A ⊆ X we have f^{−1}(f(A)) = A.

EXAM

Exam 2 Math 157 Thursday, October 17, 2013

• You have one hour to complete this exam.

• Problems 1 and 3 are worth 4 points, Problem 2 is worth 2 points for a total of 10 possible points. For a bonus point, define ”lapidate.”

Success!