# $x$ is in $B$ and $C$, the same as $x$ is in B or $x$ is in $C$?

I am having trouble understanding the statement that says If $x \in B \cap C$, then $x \in B$ or $x \in C$. Shouldn't this read: If $x \in B \cap C$, then $x \in B$ and $x \in C$? • @StVincent Your modification makes the question impossible to understand. Please be more careful. – Did Sep 22 '13 at 9:01
• It is just a typo in the reference. The "or" at the end of the first line is supposed to be an "and". The rest of the argument shows that this is what the author had in mind. – Andrés E. Caicedo Sep 22 '13 at 19:03

Yes, you're correct. The definition of set intersection requires that $x \in B \cap C$ if and only if $x \in B$ and $x \in C$.
Suppose that $x \in A \cup (B \cap C)$. If $x \in A$, then $x \in A \cup B$ and $x \in A \cup C$ by definition of union, so $x \in (A \cup B) \cap (A \cup C)$. If $x \in B \cap C$, then $x \in B$ and $x \in C$, so by a similar argument, we find that $x$ lies in the correct set.
If $x \in (A \cup B) \cap (A \cup C)$, then $x \in A \cup B$ and $x \in A \cup C$. If $x \in A$, we're done; if $x \notin A$, then $x \in B$ and $x \in C$, so that $x \in B \cap C$ and we're again done.