# Inverse Images and Sets (union & intersection)

1. Sep 7, 2010

### NastyAccident

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

Knowledge of Sets and Inverse Images

3. 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.

Sincerely,

NA

Last edited: Sep 7, 2010
2. Sep 7, 2010

### 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!