Prove $A \cap B = \emptyset$ iff $A \subseteq B'$.

I can understand why this would be true if it was just $A \cap B =$ iff $A \cap B$ but I have trouble when it comes to complements and so that part of the question is really throwing me off.


2 Answers 2


Let $x \in A$ be arbitrary. Since $A \cap B = \emptyset$, we have that $x \notin B$. Thus, $x \in B^c$. Since we showed that every element of $A$ is in $B^c$, we conclude that $A \subseteq B^c$.

You can show the other direction similarly.

  • 1
    $\begingroup$ How do you know that A is nonempty? $\endgroup$
    – Jerry Qu
    Jan 15, 2017 at 22:55

$\Leftarrow$: If $A \subseteq B^c$ then for any $x \in A$, we have $x \in B^c$ or in other words, for any $x\in A$, we have $x \notin B$. Thus, $A \cap B = \emptyset$ because there's no $x$ which is an element of $A$ and also an element of $B$.

$\Rightarrow$: If $A \cap B = \emptyset$ then for any $x \in A$, we have $x \notin B$ which means for any $x\in A$ we have $x \in B^c$ which means $ A \subseteq B^c$.

There is one little thing left, which is to consider the fact of if $A= \emptyset$. In that case, clearly our statement holds.


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