Number of elements of powerset If $A$ is a set with $k$ elements, so $A=\{ x_1, x_2, \dots, x_k\}$, then the number of elements of $\mathcal P(A)$ is $2^k$.
$$A(i_1, i_2, \dots, i_k), \ \ \ \ \text{ where } i_j \in \{0,1\},j=1, \dots, k$$

How can it be that $x_m \in A(i_1, i_2, \dots,i_m, \dots i_k )$ ?
 A: This is giving an alternate notation for the subsets of $A,$ largely to make it easier to prove that a set $A$ with $k$ elements has $2^k$ subsets. Let's consider an example to make things a bit clearer.
Say that $A=\{x_1,x_2,x_3\}.$ Then each subset of $A$ can be denoted by $A(i_1,i_2,i_3),$ where $i_1,i_2,i_3\in\{0,1\}.$ In particular, the value of $i_j$ will be $1$ if and only if $x_j$ is in the subset denoted this way. So: $$A(0,0,0)=\emptyset\\A(0,1,0)=\{x_2\}\\A(1,1,0)=\{x_1,x_2\}\\A(1,1,1)=A$$ Does this clear things up for you?
Added: To effect a proof, there are four things that you should show:


*

*There are $2^k$ different ordered $k$-tuples $(i_1,i_2,...,i_k)$ with each $i_j\in\{0,1\}.$

*Given an ordered $k$-tuple $(i_1,i_2,...,i_k)$ with each $i_j\in\{0,1\},$ we have that $A(i_1,i_2,...,i_k)\subseteq A.$

*If $B\subseteq A,$ then there exist $i_1,i_2,...,i_k\in\{0,1\}$ such
that $B=A(i_1,i_2,...,i_k).$

*$A(i_1,i_2,...,i_k)=A(i_1',i_2',...,i_k')$ if and only if $i_j=i_j'$ for all $j\in\{1,2,...,k\}.$


Finally, you must tie the above observations together to complete the proof. Let me know if you have trouble with any of the parts, with the final "tying together," or with the intuition behind any of this.
