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How to prove $A\cap B =\varnothing \Rightarrow B\subseteq \overline{A}$? If I going by definitions, there is no $x$ s.t $x\in A$ and $x\in B$. But, what do we can tell about $\overline{A}$? What i'm missing?

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    $\begingroup$ @Battani I think that $\overline{A}$ denotes the complement of $A$. $\endgroup$ – Crostul Aug 3 '15 at 10:07
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Let's assume $A\cap B =\varnothing$ (start hypothesis)

Let $x \in B$

Since $A$ and $B$ are disjoint (start hypothesis), then $x \notin A$

By definition of $\overline A$, since $x \notin A$ then $x \in \overline A ~~~~(= \Omega - A)$

Therefore $B\subseteq \overline A$, because for all $x \in B$, we have $x \in \overline A$


Note that the reciprocal is also true.

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Let $x \in B$. We want to prove $x \in \overline{A}$, i.e. that $x \notin A$.

So assume $x \in A$. But then, we have $x \in A$ and $x \in B$, which means $x \in A \cap B$. However, by assumption, $A \cap B$ is a empty. This is a contradiction.

This proves $x \notin A$.

Actually, the converse is also true: $A \cap B$ is empty if and only if $B$ is contained in $\overline{A}$.

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  • $\begingroup$ Dear downvoter. Could you please explain what is not clear in my answer? $\endgroup$ – Martin Brandenburg Aug 3 '15 at 10:19
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Suppose $B\nsubseteq \overline{A}$.

Therefore exists an element $x$ so that $x\in B$ and $x \notin \overline{A}$
Then $x\in A $ hence $x\in {A}\cap B $, that's a contradiction.

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