Prove: there exists 3 sets: $A, B, C \subseteq \mathbb{N}$ such that: $A\cap B\cap C =\emptyset$ and $|A|=|B|=|C|=\aleph_0$? Prove: there exists 3 sets: $A, B, C \subseteq \mathbb{N}$ such that:  $A\cap B= B \cap C = A \cap C = A\cap B\cap C = \emptyset$ and $|A|=|B|=|C|=\aleph_0$?
also, the sets must exists: $$|\mathbb{N} \setminus {A}| = \aleph_0$$
$$|\mathbb{N} \setminus {B}| = \aleph_0$$
$$|\mathbb{N} \setminus {C}| = \aleph_0$$
in other words, I'm looking for a way to substitute $\mathbb{N}$ into 3 disjoint sets, such that the cardinality of each of them is equal to the cardinality of $\mathbb{N}$
EDIT:
the point for this question is to prove a lemma (finding $A, B, C$ as described), as given the set: 
$$ M = \{ A \in P(A) \vert \ \ |A| = \aleph_{0} \ \land \ |A^c| = \aleph_{0} \}$$
Prove that: $|M| = \aleph$
by finding 3 sets, such that $|A| = |B|=|C| =|\aleph_0$, then I could determine that: $(B\cup C) \in M$ as $|B \cup C| =\aleph_0$,
and:
 $(B\cup C)^c = \mathbb{N}\setminus(B \cup C) = A$, as $|A| = \aleph_0$. 
using the lemma, I'd argue that for every set $\beta \subseteq B: \ (\beta\cup C) \subseteq (B\cup C) \Longrightarrow \ \forall \beta: (\beta \cup C) \subseteq M$.
this is true because $|C| =\aleph_{0}, \ \forall \beta: |\beta \cup C| = \aleph_{0}$  and $ A \subseteq (\beta \cup C)^{c} \Longrightarrow |(\beta \cup C)^{c}| = \aleph_0$. 
finally, the collection of all $\beta$ sets is: $\{\beta | \beta \subseteq B \}$, Hence $\{\beta | \  \beta \subseteq B \} = P(B)$. 
Notice that $|P(B)| = 2^{\aleph_{0}} = \aleph$
this means that $\left(P(B) \cup C \right) \subseteq M \Longrightarrow \ 
 \aleph =|\left(P(B) \cup C \right)| \leq |M|$. 
because $M \subseteq P(\mathbb{N}) \Longrightarrow |M| \leq |P(\mathbb{N})| = \aleph$, hence, by Cantor Bernstein theorem we conclude that $|M| =\aleph$, as wished. 
 A: Since there is no demand on the union of your sets you may consider the following disjoint sets to meet the cardinality conditions
$$A=\{2^K:k=1,2,3,...\}$$
$$B=\{3^k:k=1,2,3,...\}$$
$$C=\{5^k:k=1,2,3,...\}$$
A: Split $\mathbb{N}$ by considering the remainder of the division by $3$ of each integer: $A=\{3k: k\in\mathbb{N}\}$, $B=\{3k+1: k\in\mathbb{N}\}$, and  $C=\{3k+2 : k\in\mathbb{N}\}$. 
A: Pick $A = \{0,3,6,...,3n,...\},\ B = \{1,4,7,...,3n+1,...\}$ and $C = \{2,5,8,...,3n+2,...\}$. Then, we can define a bijective function $f: A \to \mathbb{N}$ with $f(a) = \frac{a}{3}$ (I leave the verification of injectivity and surjectivity to you). Then, you can also define bijective functions $g: B \to \mathbb{N}$ and $h: C \to \mathbb{N}$ in a similar way.
A: Perhaps you simply need to be told that this is an easy question and that there isn't actually any trick to it whatsoever?
I suppose the first counter intuitive thing all students of mathematics face is how can it be that a subset can be the same "size" as its superset.  Intuition says that if, say, every even number is an whole there must be "fewer" even numbers than whole numbers.  
But once we get over that hurdle it becomes very easy to partition an infinite set into two disjoint sets equally infinite.  The odd integers and the even integers;  ; the primes and the non-primes; numbers that contain the string "$17$" in them and those that don't. etc. etc. etc.
So why would dividing it into three be any harder?  After all you could just divide them in two; say, the odd and the even integers, and then divide one of them further it two; say, the odd numbers, the even numbers divisible by $4$ and the even numbers not divisible by $4$..  Or the prime integers, the composite integers that are divisible by $34$ and the composites that are not divisible by $34$ or... pretty much anything you want.
One thing to point out is that there is no requirement that the union of these sets be the entirety of $\mathbb N$.  You could do.  All whole numbers whose first digit is $2$, all numbers whose first digit is $7$, and all number whose first two digits are $18$, for example.
