What are the elements of $2^A$ if $A$ is a set I have always seen $2^A$ as an alternative notation for the power set of $A$, so I always assumed $2^A$ must contain the same elements as does $P(A).$
Having just seen the proof of bijection between $2^A$ and $P(A)$, I am not so sure anymore. That proof (using characteristic function) only(?) makes sense if $2^A$ contains only two elements. But then I can't seem to make sense of the elements in $2^A$.
My thinking is $2 = \{\emptyset, \{\emptyset\}\}$ if $0 = \emptyset$ and $1 = \{\emptyset \}$. So, then $2^A = \{\emptyset, \{\emptyset\}\}^A$, but what does that even mean? Please, elaborate on this. Thanks.
 A: For any two sets $A$ and $B$, $B^A$ denotes the collection of functions from $A$ to $B$.
When mapping $A$ to the set $2=\{0,1\}$ you can interpret each function $f\in 2^A$ as a subset $F$ of $A$, where $f(a)=1$ indicates that $a\in F$ and $f(a)=0$ indicates $a\not\in F$. 
A: This is actually an abuse of notation. The "correct" (or at least most common) notation for the power set of $A$ is $P(A)$ (or $\mathscr{P}(A)$, $\mathbb{P}(A)$,etc...).
Given sets $A$ and $B$, it is common to denote by $B^A$ the set of all functions from $A$  to $B$. The characteristic functions of subsets of $A$ are precisely those functions defined on $A$ with values in $\left\{0,1\right\}$. In other words
$$\left\{0,1\right\}^A=\left\{\text{characteristic functions of subsets of }A\right\}$$
Now we also know that there exists a natural bijection between $\left\{0,1\right\}^A$ and $P(A)$, namely the map $P(A)\to\left\{0,1\right\}^A$, which maps a subset $C\subseteq A$ to its characteristic function $\chi_C$, is a bijection (and its inverse is given by $f\mapsto f^{-1}$).
It is commmon practice to identify two mathematical objects which are in natural correspondence, so one might say that $\left\{0,1\right\}^A$ and $P(A)$ are "equal".
Finally, the usual construction of the natural numbers, inside a model of ZF, for example, is given recursively by $0=\varnothing$ and $n+1=\left\{n\right\}\cup n$. Intuitively, "$n$ is a canonical set with $n$ elements", and moreover this construction indeed yields $2=\left\{0,1\right\}$.
Putting all this togheter, we have a natural identification between $2^A$ and $P(A)$. Although they are not really equal as sets, laziness (or whatever other reason you prefer) allows us to write $2^A=P(A)$.
A: Set notation is that $B^A$ is the set of all functions $f:A\to B$. So, $2^A$ is the set of all functions $f:A\to 2$, $2$ being the two element set. This can be interpreted as $2=\{\emptyset,\{\emptyset\}\}$, but it's usually just written as $\{0,1\}$.
