Functions: Induced Set functions Let $f:X\longrightarrow Y$ be a function, $A,A_1,A_2$ be subsets of $X$ and $B,B_1,B_2$ subsets of $Y$.
Prove that if $f$ is one-to-one then $f\displaystyle\left(\bigcap^\infty_{n=1}{A_n}\right)= \bigcap^\infty_{n=1}{f(A_n)}$ 
This is what I have so far, I'm pretty sure I'm right up until this point...
Proof: Suppose $y\in f\displaystyle\left(\bigcap^\infty_{n=1}{A_n}\right)$. Then $y=f(x)$ for some $x\in\displaystyle\bigcap^\infty_{n=1}{A_n}$. Thus $x\in A_n  \forall n\in\mathbb{N}$. Since $y=f(x), y\in f(A_n),\forall n\in\mathbb{N}$. Therefore $y\in\displaystyle\bigcap^\infty_{n=1}{f(A_n)}$. 
This proves $f\displaystyle\left(\bigcap^\infty_{n=1}{A_n}\right)\subset \bigcap^\infty_{n=1}{f(A_n)}$. 
Firstly, please let me know if that's right. Secondly you will notice that I never used the one to one assumption. I'm just not sure where exactly it fits in. I'm thinking that by using the one to one assumption then I can write the entire proof with iff's and thus making them equal in the end, which is what I ultimately want. Is that correct, and if so, I still don't know where to place the one to one. 

You you've done so far is correct. It's only while proving the reverse inclusion that you'll need to use the one-to-one hypothesis. In fact, take the null function from $\mathbb Z$ into itself, take $A_1=\{0\}$ and take $A_2=\{1\}$. Then $A_1\cap A_2=\emptyset$ and $f(A_1)\cap f(A_2)=\{0\}$. So, in this case$$f(A_1\cap A_2)\varsubsetneq f(A_1)\cap f(A_2).$$
