Real Analysis, Sets and Functions Let $f$  be a function, $A, A_1, A_2$ subsets of $X$ and $B, B_1, B_2$ subsets of $Y$.
Prove that if $f$ is one to one then $f(A_1-A_2)=f(A_1)-f(A_2)$. 
This is what I have so far, 
Proof: 
$\subset)$ Assume $y\in f(A_1-A_2)$. Then $\exists x\in A_1-A_2$ s.t $f(x)=y$. 
Since $x\in A_1-A_2$ then $x\in A_1$ and $x\notin A_2$. 
Hence, $y=f(x)\in A_1$ and $ y=f(x)\notin A_2$. 
Thus, $y\in f(A_1)-f(A_2)$ which proves $f(A_1-A_2)\subset f(A_1)-f(A_2).$
The backward way is where I'm stuck. I know this is where I'm supposed to use the 1-1 assumption but not really sure how. 
 A: Your proof is not correct. Let $y\in f(A_1-A_2)$. Then, there is $x\in A_1-A_2$ s.t. $y=f(x)$. In particular, $f(A_1-A_2)\subset f(A_1)$. If there is $u\in A_2$ s.t. $y=f(x)=f(u)$, by injectivity, $x=u$ and thus $x\in A_2$ what is a contradiction. Therefore $y\in f(A_1)-f(A_2)$. 

For a counter example for $f(A_1-A_2)\subset f(A_1)-f(A_2)$, take $A_1=[-1,1]$, $A_2=\{1\}$ and $f(x)=x^2$. Then $f(A_1-A_2)=[0,1]$ and $f(A_1)-f(A_2)=[0,1[$.

For the converse, you don't need injectivity. Let $y\in f(A_1)-f(A_2)$. In particular, there is $x\in A_1-A_2$ s.t. $y=f(x)$. That prove that $y\in f(A_1-A_2)$.
A: Let $b\in f(A_1)-f(A_2)\subseteq f(A_1)$. Then $b=f(a)$ for some $a\in A_1$.
If also $a\in A_2$ then this would lead to $b=f(a)\in f(A_2)$ contradicting that $b\in f(A_1)-f(A_2)$. So we conclude that $a\in A_1-A_2$ and consequently $b=f(a)\in f(A_1-A_2)$.
Here "one-to-one" is not used. But for the converse it is, and your proof of that is not correct. Purely based on $x\notin A_2$ it is wrong to conclude that $y=f(x)\notin f(A_2)$. It might be that some  $x'\in A_2$ exists with $y=f(x')$ and consequently $y\in f(A_2)$. But this cannot happen under the extra condition that $f$ is one-to-one. In that case there is only one $x$ with $y=f(x)$.
