proof on power set and complement Suppose A and B are two sets. Prove that ℘(A \ B) ⊆ (℘(A) \ ℘(B)) ∪ {∅}.
℘ - this symbol means power set, which is the set of all the subsets of a set P(S).
this is what I did 
let X ∈ ℘(A \ B)
then X ⊆ (A \ B)
implies X ⊆ (A) but X ⊄ ( B) 
, implies X  ∈ ℘(A) and X ∉ ℘(B), implies
X ∈ ℘(A)\℘(B) 
this is what I did till now
I don't know why {∅} comes into the picture
I know it will be there but I don't know where I went wrong...........
there is a second part to the problem
Show that it is false that ℘(A \ B) = ℘(A) \ ℘(B) ∪ {∅}.
can this be done by an example or do we need a formal proof for this too?
 A: The $\emptyset$ comes from for example:
Let $A=\{1,2\}$, $B=\{2,3\}$.
Then $\{\emptyset\} \in P(A/B)$ but $\{\emptyset\} \notin P(A)/P(B)$.
Therefore you need to unite $P(A)/P(B) \cup \{\emptyset\}$.
A: Suppose $X\in \mathscr{P}(A\backslash B)$, so that $X\subseteq A\backslash B$. Then $X\subseteq A$ and $X\cap B = \varnothing$. If $X=\varnothing$, then clearly $X\in \mathscr{P}(A)\backslash \mathscr{P}(B)\cup \{\varnothing\}$. Otherwise, suppose $X$ is nonempty. Since $X\subseteq A$, we have that $X\in \mathscr{P}(A)$. Further, since $X$ is nonempty and $X\cap B=\varnothing$, there exists $x\in X\backslash B$. It follows that $X\notin \mathscr{P}(B)$. Therefore, $X\in \mathscr{P}(A)\backslash \mathscr{P}(B)\cup \{\varnothing\}$. 
As we can see, the second line of reasoning doesn't work if $X=\varnothing$, for if it is, then $X\in\mathscr{P}(B)$. That's why we need to add the empty set into the set manually. 
A: You could translate this to mean "if $U \in 2^{A \backslash B}$ is nonempty then $U \in 2^A$ but $U \notin 2^B$" (I'm writing $2^X$ as being the power set of $X$). $U \in 2^{A\backslash B}$ means $U \subseteq A\backslash B$. This means that $U \subseteq A$ and $U \cap B = \emptyset$.
The next step is where whether or not $U$ is empty becomes important. If $U$ is nonempty $U \cap B = \emptyset$ implies $U \not\subseteq B$. But if $U$ is empty then $U \subseteq B$ (because the empty set is a subset of all sets). In the nonempty case we get that $U \in 2^A \backslash 2^B$. Because the empty set is in $2^{A\backslash B}$ we must add this set to the right hand handside, so that we can say for all sets $U \in 2^{A\backslash B}$ implies $U \in (2^A \backslash 2^B) \cup \{\emptyset\} $.
