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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.

Git Gud
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Mathman
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    Your proof does need neither fixing nor clarification. It is absolutely fine ;-) – Stefan Mesken Jun 16 '14 at 09:43
  • It is absolutely fine as it is. The only comment I could think of is of style: try not to use the words 'by definition' too often. In the end every proof comes down to a bunch of 'by definition's and it could get annoying to read those all the time. – HSN Jun 16 '14 at 11:46

2 Answers2

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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

Rudy Navarro
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$ \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.

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MarnixKlooster ReinstateMonica
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