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.
Best Answer
Since there is no demand on the union of your sets you may consider the following disjoint sets to meet the cardinality conditions
$$A=\{2^K:k=1,2,3,...\}$$
$$B=\{3^k:k=1,2,3,...\}$$
$$C=\{5^k:k=1,2,3,...\}$$