Inverse Images and Sets (union & intersection)

[NastyAccident]

Homework Statement

Suppose f is a function with sets A and B.
1. Show that:
I_{f} \left(A \cap B\right) = I_{f} \left(A\right) \cap I_{f} \left(B\right)
Inverse Image of F (A intersects B) = Inverse Image of F (A) intersects Inverse Image of B.

2. Show by giving a counter example that:
f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)
F (A intersects B) does not equal F (A) intersects F(B)

Homework Equations

Knowledge of Sets and Inverse Images

The Attempt at a Solution

1.
Let c be an element of I_{f} \left(A \cap B\right).
By the definition of I_{f} \left(A \cap B\right), there is a d\in(A \cap B) so that I_{f}(d)=c.
Since, d\in(A \cap B), d \in A & d \in B. Since d\inA, I_{f}(d)\in I_{f}(A). This follows alongside d\inB, I_{f}(d)\inI_{f}(B).
Since I_{f}(d)=c \in I_{f}(A) and I_{f}(d)=c \in I_{f}(B), c = I_{f}(A)\capI_{f}(B).

Thoughts? Also would I need to show that the I_{f}(A)\capI_{f}(B) \in I_{f} \left(A \cap B\right) to show true equality?

2.
f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)
I'm thinking either the absolute value function or a square function of some sort would show that it is not equal. Though, I'm not sure how to proceed with depicting the counter example.
NA

 

[NastyAccident]

Okay, so my counterexample for No. 2 is:

f(x)=|x| {-3,...,-1} = A and {1,...,200} = B

There is no intersection between A ∩ B. However, there is an intersection with f(A) ∩ f(B) that gives the set {1,3}.

Thus, {null} != {1,3}.

Figured out the wording that I was missing =) Just need help with i now!