Suppose f is a function with sets A and B.

1. Show that:

$\displaystyle I_{f} \left(A \cap B\right) = I_{f} \left(A\right) \cap I_{f} \left(B\right)$

2. Show by giving a counter example that:

$\displaystyle f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)$

1.

Let c be an element of $\displaystyle I_{f} \left(A \cap B\right)$.

By the definition of $\displaystyle I_{f} \left(A \cap B\right) $, there is a $\displaystyle d\in(A \cap B)$ so that $\displaystyle I_{f}(d)=c$.

Since, $\displaystyle d\in(A \cap B)$, $\displaystyle d \in A & d \in B$. Since $\displaystyle d\inA, I_{f}(d)\in I_{f}(A)$. This follows alongside $\displaystyle d\inB, I_{f}(d)\inI_{f}(B)$.

Since $\displaystyle I_{f}(d)=c \in I_{f}(A) $ and $\displaystyle I_{f}(d)=c \in I_{f}(B), c = I_{f}(A)\capI_{f}(B)$.

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

2.

$\displaystyle f\left(A \cap B\right) \neq f\left(A\right) \cap f \left(B\right)$

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.