Set difference between power sets Show that if $A$ and $B$ are sets then $\varnothing \notin  
\mathcal P(A) − \mathcal P(B)$ .
My Attempt:
If $A$ and $B$ are sets, then $ \varnothing  \subseteq  A $ since the empty set is subset of all sets. 
Similarly $ \varnothing  \subseteq  B$
Then by definition $ \varnothing  \in  \mathcal P(A)$ and $ \varnothing  \in   \mathcal P(B) $
Then by definition of set difference $\varnothing \notin \mathcal P(A) - \mathcal P(B) $
I don't think this is the most eloquent and concise proof. I would appreciate help in making the above more so. 
 A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Your proof is fine.
Just for fun, here is the same proof in an alternative style that personally I like better: for all sets $\;A,B\;$,
$$\calc
\varnothing \not\in \mathcal P(A) - \mathcal P(B)
\calcop{\equiv}{definition of $\;-\;$; logic: DeMorgan}
\varnothing \not\in \mathcal P(A) \;\lor\; \varnothing \in \mathcal P(B)
\calcop{\equiv}{definition of $\;\mathcal P\;$, twice}
\varnothing \not\subseteq A \;\lor\; \varnothing \subseteq B
\calcop{\equiv}{set theory: $\;\varnothing\;$ is a subset of every set}
\varnothing \not\subseteq A \;\lor\; \text{true}
\calcop{\equiv}{logic: simplify}
\text{true}
\endcalc$$
For some more references about this calculational style, see https://math.stackexchange.com/a/332186/11994.
A: For the sake of contradiction, assume A and B are sets and $\varnothing \in \mathcal P(A)- \mathcal P(B)$.
Then, since $\varnothing$ is a subset of every set, we have that $\varnothing \subseteq A$ and $\varnothing \subseteq B$.
Notice, $\varnothing \in \mathcal P(A)- \mathcal P(B)$ implies $\varnothing \in \mathcal P(A)$ and $\varnothing \notin \mathcal P(B)$ by definition of set difference.
So, we have that $\varnothing \subseteq A$ and $\varnothing \not\subseteq B$, by definition of power sets. 
Thus, we have $\varnothing \subseteq A$ and $\varnothing \subseteq B$, and $\varnothing \subseteq A$ and $\varnothing \not\subseteq B$. 
Or, equivalently, $\varnothing \subseteq A$ and $(\varnothing \subseteq B$ and $\varnothing \not\subseteq B$).
Notice, $\varnothing \subseteq B$ and $\varnothing \not\subseteq B$ raises a contradictio, so it must be true that $\varnothing \notin \mathcal P(A) - \mathcal P(B)$.
Q.E.D
