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If $A \cap B = \emptyset$ then $A \subset B'$ and $B \subset A'$, where the prime symbol denotes the complement of each set.

Here are my thoughts:

Assume $A \cap B = \emptyset,$ since the intersection of $A$ and $B$ are empty, then an arbitrarily chosen element $x \notin A$ and $x \notin B.$ Thus $x \in A'$ and $x \in B'.$

How do I go about justifying that $A \subset B'$ and $B \subset A'?$

Maybe a direct proof is not the best way to do so? How about a proof by contradiction?

Thank you for any help or guidance!!!

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    $\begingroup$ You may draw a Venn diagram to guide yourself. $\endgroup$ – Salomo Apr 5 '15 at 19:09
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Hint:

Suppose for example that

$$a\in A\implies a\notin B\;,\;\;\text{lest}\;\;A\cap B\neq\emptyset\implies a\in B'\implies a\in A\cap B'$$

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  • $\begingroup$ is your suggestion a proof by contradiction? Sorry if my question seems dumb! $\endgroup$ – mathamphetamines Apr 5 '15 at 19:13
  • $\begingroup$ Not really, though the central part could be considered such. Is anything unclear ? $\endgroup$ – Timbuc Apr 5 '15 at 19:14
  • $\begingroup$ Take $a \in A$ thus $a \notin B$ since their intersection is empty. This makes sense, but how do you go to the next step: lest $A \cap B \ne \emptyset?$ $\endgroup$ – mathamphetamines Apr 5 '15 at 19:18
  • $\begingroup$ Well, $\;a\in B'\iff a\notin B\;$ , so we already have $\;a\in A\;$ and also $\;a\in B'\;$ , thus $\;a\in A\cap B'\;$ $\endgroup$ – Timbuc Apr 5 '15 at 19:20
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    $\begingroup$ Got it now. Thank you very much! $\endgroup$ – mathamphetamines Apr 5 '15 at 19:28
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For every $x\in A$ the assumption $x\in B$ leads to $x\in A\cap B$ contradicting that this set is empty.

So we conclude that for every $x\in A$ we have $x\notin B$ wich is the same as $x\in B'$.

This proves that $A\subseteq B'$.

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  • $\begingroup$ Thank you for your help!!!! $\endgroup$ – mathamphetamines Apr 5 '15 at 19:19
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You can argue this with logic pretty well, and directly, although I'm not sure if you were looking for this answer.

Suppose $$A \subset B'$$ then, take some element in A, called $x$, and by definition: $$(x \in A)\rightarrow(x \notin B)$$ So, using the logical equivalent to the implication statement, $$\neg(x \in A) \vee \neg(x\in B)$$ And with De Morgan's law, this is equivalent to $$\neg[(x\in A) \wedge (x \in B)]$$ which is equivalent to saying $$x \notin A\cap B \rightarrow A\cap B =\emptyset$$

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