Powerset of {x, {y}, {x,y}} I'm trying to get the powerset of {x, {y}, {x,y}}, but i'm not shure if I have 4^2 = 16 subsets or, like other people sugested, replace a = {y} and b = {x,y} and thus have 3^2 = 9 subsets?
Thanks
 A: Explicitly:
$\emptyset\\
\{x\}\\
\{\{y\}\}\\
\{\{x,y\}\}\\
\{x,\{y\}\}\\
\{x,\{x,y\}\}\\
\{\{y\},\{x,y\}\}\\
\{x,\{y\},\{x,y\}\}.$
The first one is the empty set, which is always a subset of any set. The next three are singleton sets, consisting of a single element of the given set (the fact that some of these elements are themselves sets is more than likely designed to confuse you). The next three after that are enumerations of all the possible two-element subsets (and there are:
$\binom{3}{2} = \frac{3!}{(3-2)!2!} = \frac{6}{2} = 3$ ("three choose two")
of these), and finally the complete set itself, being the only possible three-element subset.
A: Build a table, with column names $x, y, \{x,y\}$.
Each row has values 1 or 0.
0 - means the element is not in the subset, 1 means it is.
you have 1 row with $(0,0,0)$ which means no elements on that set - $\emptyset$.
Since you have exactly 2 values possible for each element you have $2\times 2\times 2=2^3=8$ subsets.
A: Your set has $3$ elements. So, you'll have $2^3=8$ subsets. If you're confused about the third element $\{x, y\}$, it's just like the other elements. It being a set doesn't affect the number of subsets. 
