If $A\cup B = U$ and $A\cap B = \emptyset$ then $B = A^c$ — contradiction?

I somehow got a weird result when proving it.
Suppose $A \cup B = U$. Then by de Morgan's Law,
$A^c \cap B^c = \emptyset$.
Taking the intersection of both sides with $A$,
$A\cap A^c \cap B^c = A \cap \emptyset = \emptyset$.
$U \cap B^c = \emptyset$
$B^c = \emptyset$?

$U$ is the universal set
Where did I go wrong here?

• The assertion is fine. – Wuestenfux Jul 26 '17 at 13:05
• Why do you say $U\cap B^{\mathrm c}=\varnothing$? – Bernard Jul 26 '17 at 13:06

It seems to me that the flaw occurs in this step: $$A\cap A^C\cap B^C=\varnothing\implies U\cap B^C=\varnothing$$ Because it seems that you assumed $$A \cap A^C=U$$ when, in reality, $$A\cap A^C=\varnothing$$ because the intersection of any set with its complement is the empty set. You must have mistaken the $\cap$ for a $\cup$, because $$A\cup A^C=U$$ really is a true statement.

Your mistake is going from

$$A\cap A^C\cap B^C=\emptyset$$

(which is correct) to

$$U\cap B^C=\emptyset$$

which is not correct.

It looks like you replaced $A\cap A^C$ with $U$, which is a mistake since $A\cap A^C=\emptyset$, while $A\cup A^C=U$.

You need to take the union of $A^C\cap B^C=\emptyset$ with $A$ giving $A\cup(A^C\cap B^C)=A\cup \emptyset$. Thus $A\cup B^C=A$ and hence $B^C\subseteq A$. Similarly, $A\subseteq B^C$.