Finding a correspondence between $\{0,1\}^A$ and $\mathcal P(A)$ I got this question in homework:

Let $\{0,1\}^A$ the set of all functions from A (not necessarily a finite set)
      to $\{0,1\}$. Find a correspondence (function) between $\{0,1\}^A$ and
      $\mathcal P(A)$ (The power set of $A$).
      Prove that this correspondence is one-to-one and onto.

I don't know where to start, so I need a hint. What does it mean to find a correspondence?
I'm not really supposed to define a function, right? 
I guess once I have the correspondence defined somehow, the proof will be easier.
Any ideas? Thanks!
 A: This is essentially the same as Martin and yuone's answers:
Fix a set $A$. For a function $f$ from $A$ to $\{0,1\}$, let $  A_f$ be the set of elements of $A$ that are mapped to 1 by $f$. That is, $a\in  A_f$ if and only if $f(a)=1$.
Consider the map $\Phi(f) =A_f$.
Now  if $f\ne g$, there is an $a\in A$ with $f(a)=0$ and $g(a)=1$ (or  $f(a)=1$ and $g(a)=0$). 
Then $A_f\ne A_g$. So $\Phi$ is one-to-one.  
Now let $B\in{\cal P}(A)$. Define $f(x)=\cases{1,&x\in B\cr 0,&x\notin B }$
Then $\Phi(f)=B$.  This shows that $\Phi$ is onto ${\cal P}(A)$
A: For any function $f\in\{0,1\}^A$ try associating it with $f^{-1}(1)$ in $\mathcal{P}(A)$, that is, the subset of $A$ whose elements map to $1$ under $f$.
A: What about $f:{\mathcal P (A)}\to {\{0,1\}^A}$ and  $g:{\{0,1\}^A}\to{\mathcal P(A)}$ given by
$$
\begin{align}
&f(X)=\chi_X &\text{ for $X\subseteq A$,}\\
&g(h)=\{x\in A; h(x)=1\} &\text{ for $h:A\to{\{0,1\}}$,}
\end{align}
$$
where $\chi_X(x)=1$ for $x\in X$ and $\chi_X(x)=0$ for $x\notin X$,
i.e. $\chi_X$ is the characteristic function of the set $X$.
Can you show that these two functions are inverse to each other?
NOTE: This is basically the same thing as yunone's answer, I've just added the inverse function too. 
A: I'll try to say this without all the technicalities that accompany some of the earlier answers.
Let $B$ be a member of $\mathcal{P}(A).$
That means $B\subseteq A$.
You want to define a function $f$ corresponding to the set $B$.  If $x\in A$, then what is $f(x)$?  It is: $f(x)=1$ if $x\in B$ and $f(x) = 0$ if $x\not\in B$.
After that, you need to show that this correspondence between $B$ and $f$ is really a one-to-one correspondence between the set of all subsets of $A$ and the set of all functions from $A$ into $\{0,1\}$.  If has to be "one-to-one in both directions"; i.e. you need to check both, and you need to check that the word "all" is correct in both cases.
