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

Question

My thinking:

A = $\left\{1,\:2,\:\left\{1,\:2\right\}\right\} $ (which contains three elements)

The power set of $A$, $P(A)$ then contains $2^3 = 8$ elements which are:

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

Therefore, $P(A)\setminus A$ = $\left\{\right\},\:\left\{1\right\},\left\{2\right\},\left\{\left\{1,2\right\}\right\},\left\{1,2\right\},\left\{1,\left\{1,2\right\}\right\},\:\left\{2,\:\left\{1,2\right\}\right\}$ which has 7 elements.

I'm not too sure if this is right, but if anyone could help me out, that would be appreciated.

Answer 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:

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

Answer 2

$A$ has $3$ elements, so $P(A)$ has $2^3=8$ elements.

If $X$ and $Y$ are sets, $X\setminus Y=X\setminus(Y\cap X)$,
so $P(A)\setminus A$ is the same as $P(A)\setminus(A\cap P(A))$.

So how many elements does $A\cap P(A)$ have?

$A$ has $3$ elements, namely, $1$, $2$, $\{1,2\}$;
$1$ is not a subset of $A$, $2$ is not a subset of $A$, $\{1,2\}$ is a subset of $A$;
so $1\notin P(A)$, $2\notin P(A)$, $\{1,2\}\in P(A)$;
so $A\cap P(A)$ has just one element, namely $\{1,2\}$; so $P(A)\setminus A$ has $8-1=7$ elements.