Let A = $\left\{1,\:2,\:\left\{1,\:2\right\}\right\}$, how many elements are in the set $P(A)\setminus A$?

Solution 1:

It looks like you have the right number of elements, but the wrong elements listed for $P(A)\setminus A$.

You correctly noted that the power set of $A$ is

$$P(A)=\left\{\quad \{\},\{1\},\{2\},\{\{1,2\}\},\{1,2\},\{1,\{1,2\}\},\{2,\{1,2\}\},\{1,2,\{1,2\}\}\quad\right\}$$

However, note that $\{1,2,\{1,2\}\}=A\not\in A$, so this element is not removed from the power set by taking the set difference of $A$.

Instead, note that $\{1,2\}\in A$ and also $\{1,2\}\in P(A)$. Thus this element is removed by taking the set difference.

As a result:

$$P(A)\setminus A=\left\{\quad \{\},\{1\},\{2\},\{\{1,2\}\},\{1,\{1,2\}\},\{2,\{1,2\}\},\{1,2,\{1,2\}\}\quad\right\}$$