Let S be a set. If $S$ has $n$ elements, then S has $2^n$ subsets proof

Question

I'm reading a book and this I've been stuck on this proof for hours:

I cant understand what set is $V$ and what set is $W.$ $V$ is the set of all the subsets of $S$ (the power set) "and $a_{n+1}\notin S$" what does that "and .." mean? I don't really get it, is it $S$ without $a_{n+1}$?

Then, another question, I don't really understand that function, if $s_n=\{1,2\}$, for example, then its power set is $\mathcal{P}(S_n)=\{\phi, \{1\}, \{2\}, \{1, 2\}\}$, now this is the domain of that function, and the range is $\{\{3\}, \{1, 3\}, \{2, 3\}, \{1, 2, 3\}\}$ but how is this the same as "a mapping from $V$ onto $W$"?? maybe I don't understand this because I don't understand how $V$ and $W$ are defined.

Answer 1

What characterizes a subset of $S$? It is a set $ T$ of elements each of which belons to $S.$ So how do you build such a $T$ from $S?$ You look at every member of $S$ and ask "Does this member belong to $T$? The answer is "Yes" or "No." Since $S$ contains $n$ members the number of different possibilities for "yes"or "No","yes"or "No",...,"yes"or "No" is $2\times2\times\dots\times2=2^n$ because there are $n$ $2$'s in the product. So the number of subsets of S is $2^n.$

Answer 2

There is a typographical mistake:

$V=\{X\subseteq S:a_{n+1}\notin X\}$ and
$W=\{X\subseteq S:a_{n+1}\in X\}$

Next, if we put $n=2$ then $S=\{1,2,3\}$ and $S_2=\{1,2\}$. Then
$V=\mathcal{P}(S_2)=\{\emptyset, \{1\},\{2\},\{1,2\}\}$ and $W=\{\{3\},\{1,3\},\{2,3\},\{1,2,3\}\}$.
So the $f$ maps following:

$f:\emptyset\mapsto \{3\}$
$f:\{1\}\mapsto \{1,3\}$ and so on.

Answer 3

As @AVISEKSHARMA notes, $\notin S$ and $\in S$ should respectively read $\notin X$ and $\in X$. So $V$ is the set of subsets of $S$ that don't have $a_{n+1}$ as an element, and $W$ is the set of subsets of $S$ that do have $a_{n+1}$ as an element. We biject $V$ with $W$ using $f(A)=A\cup\{a_{n+1}\}$ for each $A\in V$. But the inductive step also uses the fact that $V$ has $2^n$ elements, because it happens to also be the powerset of a set with $n$ elements: that set is $S\setminus\{a_{n+1}\}$.