Question

Recall that we write f(A) = B to denote the set B that results from applying the function f

on every element a in A.
Prove the following:
(a) Let f : X → Y be a function. For any sets A, B ⊆ X, if f (A ∩ B) = f(A) ∩ f (B) then f
is injective. Hint: use the contrapositive of injectivity
(b) Let f : X Y be a function. If f is injective then f(A) ∩ f(B) = ∩ B). Prove this
in two steps by showing f(A∩B)⊆ g f(A) ∩ f (B) and f (A) ∩ f (B)⊆ g f(A ∩ B).

Answer 1

To prove that a function is injective using the contrapositive statement, we assume that A ≠ B and show that f(A) ≠ f(B). To prove that f(A) ∩ f(B) = ∩ B), we show that f(A∩B) ⊆ g f(A) ∩ f(B) and f(A) ∩ f(B) ⊆ g f(A∩B).

Step-by-step explanation:

(a) To prove that f is injective, we will show that if f(A ∩ B) = f(A) ∩ f(B), then A = B. We will prove this by contrapositive: if A ≠ B, then f is not injective. Let's assume that A ≠ B. This means that there exists an element x in either A or B but not in the other set. Without loss of generality, let's assume x is in A but not in B. This implies that f(x) is in f(A) but not in f(B). Therefore, f(A) ≠ f(B) and f is not injective.

(b) To prove that f(A) ∩ f(B) = ∩ B), we will show that f(A∩B) ⊆ g f(A) ∩ f(B) and f(A) ∩ f(B) ⊆ g f(A∩B). Let's start with f(A∩B) ⊆ g f(A) ∩ f(B). If y is in f(A∩B), this means that there exists an element z in A∩B such that f(z) = y. Since z is in both A and B, it is also in A and B individually. Therefore, y = f(z) is in both f(A) and f(B), so y is in g f(A) ∩ f(B). Hence, f(A∩B) ⊆ g f(A) ∩ f(B). Now let's prove f(A) ∩ f(B) ⊆ g f(A∩B). If y is in f(A) ∩ f(B), this means that y is in both f(A) and f(B). Therefore, there exist elements x1 in A and x2 in B such that f(x1) = f(x2) = y. Since x1 and x2 are in both A and B, they are also in A∩B. Therefore, y = f(x1) = f(x2) is in f(A∩B), so y is in g f(A∩B). Hence, f(A) ∩ f(B) ⊆ g f(A∩B).