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.
Best Answer
$ \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.