# If $A$ is infinite then it has two infinite subsets $B, C$ which are pairwise disjoint.

I'm dealing with a question and I don't know how to formulate my answer. The question is: Let A be an infinite set. Prove that there exist two sets: B and C such that they are subsets of A, and also B and C are pairwise disjoint, then B and C are infinite sets as well.

-
This is not true in general. We need more conditions on $B$ and $C$. At the extreme, $B$ and $C$ could be empty! Even if we ask that the union of $B$ and $C$ be $A$, one of $B$ or $C$ could be tiny. –  André Nicolas Apr 9 '12 at 15:26
But if $A = \mathbb N, B = \{1\}, C = A \setminus B$ you have a counterexample. –  Rudy the Reindeer Apr 9 '12 at 15:26
Sorry I should edit the question, because I need to prove that there exist such sets like B and C –  JanosAudron Apr 9 '12 at 15:28
Like $A= \mathbb N$, $B=$ the even numbers and $C=$ the odd numbers? –  Rudy the Reindeer Apr 9 '12 at 15:31
Yes but more in general... –  JanosAudron Apr 9 '12 at 15:33

From your comments it seems you want to prove that there are disjoint subsets $B$ and $C$ of $A$, both of which are infinite. To do this, you could use the
Hint: Let $A'$ be a countably infinite subset of $A$. Let $f$ be a bijection from $\Bbb N$ to $A'$. Consider the sets $f(\Bbb N_{\rm e})$ and $f(\Bbb N_{\rm o})$, where $\Bbb N_{\rm e}$ is the set of even positive integers and $\Bbb N_{\rm o}$ is the set of odd positive integers.
This argument doesn't work if $A$ is Dedekind-finite. I understand the theorem is false in this case (i.e. you need countable choice to prove this)? –  Mario Carneiro Feb 5 '14 at 11:25