Proving that the symmetric difference of sets is countable at most Let $A$, $B$, $C$ be sets. How do you prove that if $|A\Delta B|\leq \aleph_{0}$ and $|B\Delta C|\leq \aleph_{0}$ then $|A\Delta C|\leq \aleph_{0}$?

HINT: Remember that $A\triangle C=(A\triangle B)\triangle(B\triangle C)$, and that a subset of a countable set is countable.
