What I am struggling with is understanding the part where we prove that the function is onto.
Letting f be a element of $2^A$ means that in this particular example f could be either of the $f_1 f_2 f_3 f_4$, now defining the set S={a,b},
but then they say: Hence $f_s=f$, but they just stated that s is a set with elements (a,b), and $f_{(a,b)}$ does not equal f element of $f_1 f_2 f_3 f_4$ what am I misunderstanding?
Let $A={(a,b)}$, now create the power set of A. And let $2^A$ be the set of functions:
2 Answers
$\begingroup$For any $f_i$, the corresponding $S$ is defined by:
$S_i = \{ x \in A : f_i(x)=1 \}$.
Thus:
$S_1 = \emptyset, S_2 = \{ a \}, S_3 = \{ b \}, S_4 = \{ a, b \}$.
Seeing it in the "other direction" for each of the four elements of $\mathcal P(A)$ - i.e. for each subset of $A$ - we have a different characteristic function $f_i$.
$\endgroup$ $\begingroup$$S$ is not "a set with elements $\{a,b\}$", rather it is a very specific subset of $A$ that is defined in terms of the (given) map$ f\colon S\to\{0,1\}$, namely $S$ is the set of precisely those elements that $f$ maps to $1$. In your example, if $f=f_2$ then $D=\{a\}$, if $f=f_1$ then $S=\emptyset$, etc. At any rate, the map $\mathcal P(A)\to 2^A$ defined previously sends $S$ to a map $f_S$, for whish we want to show $f_S=f$, that is: for all $x\in A$, we have $f_S(x)=f(x)$. By definition of $f_S$ (in terms of $S$), we have $f_S(x)=1\iff x\in S$. By definition of $S$ (in terms of $f$), we have $x\in S\iff f(x)=1$. Thus $f_S(x)=1\iff f(x)=1$.
$\endgroup$
