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Let $A$ be a set. If $A\subset B$ for any set $B$, then $A=\emptyset$.

Here is my proof.

Suppose to the contrary that $A$ is not an empty set. Then there exists an element $a$ in $A$. Now choose $B$ as an empty set. Then for $a\in A$, $a\notin B$. So, $A\not\subset B$. This contradicts to the hypothesis. Thus $A=\emptyset$.

What do you think about this?

Daniel W. Farlow
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learner
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  • Looks fine to me –  Mar 17 '17 at 15:28
  • Looks good to me. – Arnaldo Mar 17 '17 at 15:29
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    This is fine, although, the use of a proof by contradiction seems unnecessary, as there is a direct proof (but your proof is fine) – Maxime Ramzi Mar 17 '17 at 15:40
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    Are you using $\subset$ as a proper subset? If so, then bear in mind that $\emptyset\not\subset\emptyset$. – Daniel W. Farlow Mar 17 '17 at 15:55
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    To expand on Max's comment the direct proof is: A $\subset \emptyset$ by hypothesis and $\emptyset \subset A$ (since the empty set is a subset of every set) hence $A = \emptyset$. – Kai Rüsch Mar 17 '17 at 16:28
  • by contrapositive, $A\neq \emptyset \to \exists B: A\nsubseteq B$, see http://math.stackexchange.com/questions/331908/proof-a-neq-emptyset-implies-a-nsubseteq-emptyset – mle Mar 17 '17 at 22:16

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