Let $f: A \rightarrow B$ be a map of sets, and let {$X_i$}$_{i\in I}$ be an indexed collection of subsets of A. Prove that $f(\cap_{i\in I}X_i)\subset\cap_{i\in I}f(X_i)$
From a second year calculus course. Have only done Basic Set Theory and Injective/Surjective functions. I'm thinking the proof has something to do with the injectivity of $f(\cap_{i\in I}X_i)$? 
Any help would be appreciated, thanks.

Well it should be clear that if $y\in f (\cap_i X_i) $,  that is to say, $y =f (x)$ for some $x \in \cap_i X_i $, then $x \in X_i $ for all $X_i $ so $f (x) \in f (X_i) $ for all $f (X_i) $.  So $y = f (x) \in \cap_i (X_i) $ and therefore $f (\cap_i X_i) \subset \cap_i f (X_i) $.
That is clear.  What isn't clear is why it isn't an equality. 
Consider $f(x)= \sin x $.  Let $X_i = [2i\pi, 2i\pi + 2\pi] $.  Then $f (X_i) = [-1,1] $ and $\cap_i f (X_i) =[-1,1] $.
But $\cap_i X_i = \emptyset $ and $f (\emptyset) = \emptyset $.
So $f (\cap_i X_i) = \emptyset \subsetneq [-1,1] = \cap_i f (X_i) $.
