# Collections of Sets, Problem 1(a)

### Solution

Suppose that y is an arbitrary element of $$f(A \cap B)$$. By the definition of the image of a set, there is an element $$x \in A \cap B$$ such that $$f(x)=y$$. Since $$x \in A \cap B$$, we know $$x \in A$$ and $$x \in B$$. Thus, $$y \in f(A)$$ and $$y \in f(B)$$. So $$y \in f(A) \cap f(B)$$. Since y was chosen arbitrarily, we conclude that $$f(A \cap B) \subseteq f(A)\cap f(B)$$.