Clarification on pairs of sets that make up a power set From the Wikipedia article on power sets, 

... one can select any element x of a set, S, and gather up the subsets of S itself into pairs (A-, A+) that are identical excepting that A- does not include x and A+ does.

So if S = {1,2,3}, and we choose x to be 3, I get pairs
{1},{1,3}
{2},{2,3}
{1,2},{1,2,3}
But there should be 23 = 8 sets. Which am I missing?
 A: Let us write the power set of $S=\{1,2,3\}$:
$$\mathcal P(S)=\Bigg\{\varnothing,\underbrace{\{1\},\{2\},\{3\}}_{\text{singletons}},\overbrace{\{1,2\},\{1,3\},\{2,3\}}^{\text{pairs}},\{1,2,3\}
\Bigg\}$$
You have chosen $3$, so we essentially going to write $S$ as $S\setminus\{3\}\cup\{3\}$. Note that this is to write a set of three elements as $2+1$, taking a power set, in terms of size it to raise by a power of $2$, so we have $2^3=2^{2+1}=2^2\times 2$. 
This means that indeed we can write $\mathcal P(S)$ as four pairs of the form: $\langle A,B\rangle$ where $A\subseteq S\setminus\{3\}$ and $B=A\cup\{3\}$. Of course we can replace $S$ by an arbitrary set (even infinite!) and $3$ by an arbitrary element of this set. The calculation pans out the same way:
First we write a table of all the subsets of $\{1,2\}$, and fill the $B$ coloumn:
$$
\begin{array}{lcr}
\begin{array}{|c|c|}
\hline
A & B\\\hline 
\varnothing & \\\hline
\{1\} & \\\hline
\{2\} & \\\hline
\{1,2\} & \\
\hline
\end{array} &
\text{We add the }3:
\begin{array}{|c|c|}
\hline
A & B\\\hline 
\varnothing & \varnothing\cup\{3\}\\\hline
\{1\} & \{1\}\cup\{3\}\\\hline
\{2\} & \{2\}\cup\{3\}\\\hline
\{1,2\} & \{1,2\}\cup\{3\}\\
\hline
\end{array} &
\text{and finally:}
\begin{array}{|c|c|}
\hline
A & B\\\hline 
\varnothing & \{3\}\\\hline
\{1\} & \{1,3\}\\\hline
\{2\} & \{2,3\}\\\hline
\{1,2\} & \{1,2,3\}\\
\hline
\end{array}
\end{array}$$
Indeed four rows, eight elements. We wrote all $\mathcal P(S)$ as a list of pairs as wanted. 
