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. 
 A: Your proof of $f\displaystyle\left(\bigcap^\infty_{n=1}{A_n}\right)\subseteq \bigcap^\infty_{n=1}{f(A_n)}$ is okay, and for that side the "one-to-one" condition is not relevant. It only plays a part by proving the other side.
If conversely $y\in\bigcap_{n=1}^{\infty}f(A_n)$ then $y=f(x_n)$ for some $x_n\in A_n$ for every $n$. 
But $f$ is one-to-one so there is a unique $x$ with $f(x)=y$ so that $x=x_n\in A_n$ for every $n$.
Then $x\in\bigcap_{n=1}^{\infty}A_n$ and consequently $y=f(x)\in f(\bigcap_{n=1}^{\infty}A_n)$.

If $f$ is not demanded to be one-to-one then it is not difficult to construct e.g. a situation where $f(A_n)$ is non-empty and is not depending on $n$ while $\bigcap_{n=1}^{\infty}A_n=\varnothing$. 
In that case it is not true that $\bigcap_{n=1}^{\infty}f(A_n)\subseteq f(\bigcap_{n=1}^{\infty}A_n)=f(\varnothing)=\varnothing$.
A: 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$$f(A_1\cap A_2)\varsubsetneq f(A_1)\cap f(A_2).$$
