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I proof by contradiction that: let $ S $ a set, and $ S \subseteq \emptyset $, then $ S= \emptyset $; if $ S \neq \emptyset $ then $\exists x \in S ( x \notin \emptyset ) $ but by hypothesis this is contraddictory, in fact if $\exists x \in S ( x \notin \emptyset ) $ then $S \nsubseteq \emptyset $, therefore $S = \emptyset $. Is it correct? Thank you all in advance.

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up vote 4 down vote accepted

You should write it more consistent, when $S\neq \varnothing$ then there are two cases. There is a $x$ that is in $S$ but not in $\varnothing$ or there is a $x$ that is $\varnothing$ but not in $S$. Or if you don't like words $$ S\neq \varnothing \iff ((\exists x \in S \wedge x\notin \varnothing)\vee (\exists x \in \varnothing\wedge x\notin S)) $$ As $x\in \varnothing$ is not true you only need to check $x\in S$ and $x\not \in \varnothing$. But than $S\not\subseteq \varnothing$.

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of course!! I understand thank you!! – mle Mar 14 '13 at 21:38

Suppose $S \subseteq \emptyset \rightarrow S= \emptyset$ is not correct. Then for some nonempty set $S$ we have $$S\subseteq \emptyset$$ Because $S$ is nonempty there's some $s\in S$. So $$s\in S\subseteq\emptyset$$ therefore $s\in \emptyset$. This is a contradiction.

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