(Substitute $ =, \subseteq, \supseteq $ to the place of $\bigcirc$) ,$\mathcal{P}(A \setminus B) \bigcirc \mathcal{P}(A) \setminus P(B)$

Question

$\mathcal{P}$- the power set

I take the sets $A= {\{1,2,3 \} }$ and $B = \{ {1,2 \} }$

For $A = \{ \emptyset, \{{ 1\},\{ 2\},\{ 3\},\{ 1,2\},\{ 1,3\},\{ 2,3\},\{ 1,2,3 \} }$

For $B = \{ \emptyset, \{{ 1\},\{ 2\},\{ 1,2\}}$

For $\mathcal{P}(A \setminus B) = \mathcal{P}{ \{ 3 \} } $ = $ \{ \emptyset, \{3 \} \}$

For $\mathcal{P}(A) \setminus \mathcal{P}(B) = \{ \emptyset ,{\{ 1,3\},\{ 2,3\},\{ 1,2,3 \} }$

I think that this is a good counter example to show that $\mathcal{P}(A \setminus B) \neq \mathcal{P}(A) \setminus \mathcal{P}(B)$

how can I continue it?

Answer 1

Should be $\subseteq$:

Let $x \in \mathcal{P}(A-B)$

$x \subseteq A-B$

$x \subseteq A$ & $x \nsubseteq B$

$x \in \mathcal{P}(A) - \mathcal{P}(B)$