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Does this proof work?

By definition:

$$[A \cap B = A] \wedge [A \cup B = B] \implies [A \subseteq B]$$

Therefore:

$$[\emptyset \cap B = \emptyset] \wedge [\emptyset \cup B = B] \implies [\emptyset \subseteq B]$$

Thanks in advance.

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  • $\begingroup$ meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Timbuc Feb 7 '15 at 13:44
  • $\begingroup$ What about: for any set $\;A\;$ we have that for all $\;x\in\emptyset\implies x\in A\;$ and thus $\;\emptyset\subset A\;$ ? $\endgroup$ – Timbuc Feb 7 '15 at 13:45
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    $\begingroup$ understanding $\emptyset \subseteq B$ is easier than understanding $[A \cap B = A] \wedge [A \cup B = B] \implies [A \subseteq B]$ $\endgroup$ – user 1 Feb 7 '15 at 13:50
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    $\begingroup$ If you take that relation as a definition, then this is valid, as long as you can show that $\emptyset\cap B=\emptyset,\emptyset\cup B=B$. $\endgroup$ – Wojowu Feb 7 '15 at 13:56
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    $\begingroup$ @G Ch Wojowu is right. So the question is how do you say $\varnothing \cap B = \varnothing$ and $\varnothing \cup B = B$ $\endgroup$ – R_D Feb 7 '15 at 14:11
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If you use this as your definition of subset, you need to justify that $\varnothing\cap B=\varnothing$ and $\varnothing\cup B=B$. I will do these here.

Note that $\varnothing\cap B=\left\{x:x\in \varnothing\land x\in B\right\}$. Since the condition $x\in\varnothing$ is never satisfied, the set is defined by an always-false condition, and so we have that $\varnothing\cap B=\varnothing$.

Now, note that $\varnothing\cup B=\left\{x:x\in\varnothing\lor x\in B\right\}$. Since, again, the condition $x\in\varnothing$ is never satisfied, the defining condition for this set is $x\in B$, and so we have that $\varnothing\cup B=B$.

Therefore, since $\varnothing\cap B=\varnothing$ and $\varnothing\cup B=B$, we have that $\varnothing\subseteq B$. Since $B$ was no set in particular, this makes $\varnothing$ a subset of every set, as required.

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