I've been asked to prove that: $P(A\setminus B)\subseteq P(A)\setminus P(B) \cup \{\emptyset\}$.
By the laws of group subtraction, I've concluded that:
$\alpha =P(A\setminus B) = P(A \cap B^c) $
$\beta = P(A)\setminus P(B) \cup \{\emptyset\} = P(A)\cap P(B)^c \cup \{\emptyset \}$
But I'm not sure where to go on from here.
It can be said that both sets have a member in the form of the empty set. But how can I prove that each member of set $\alpha$ is also a member in $\beta$?
If $X\in\mathcal P(A\setminus B)$, and if $X\ne\emptyset$, then $X\subset A$; therefore, $X\in\mathcal P(A)$. But $X$ contains no element of $B$; in particular, and since $X\ne\emptyset$, $X\notin\mathcal P(B)$. So, $X\in\mathcal P(A)\setminus\mathcal P(B)$.