Construct a countably infinite subset of P(ℕ) with continuum cardinality

[somebd]

Prove that exist X ⊂ P(N)

P(N) is a power set of natural numbers
1 cardinality of X is continuum
2 for each A (element of X) in X, A is infinite
3 for each A and B in X, A is not equal to B, A intersection with B is finite

Please, I have no idea how to solve this

 

[micromass]

What you're searching for is called an almost disjoint family. A full treatise on this can be found in "Set theory, an introduction to independence proofs" by K. Kunen.

Here is a sketch of the proof:
If $$X\subseteq \mathbb{N}$$, define $$A_X=\{X\cap \{0,...,n\}~\vert~n\in \mathbb{N}\}$$. If X is countable, then $$A_X$$ is countable. If $$X\neq Y$$, then $$A_X\cap A_Y$$ is finite.
Let $$\mathcal{A}=\{A_X~\vert~X\subseteq \mathbb{N}~\text{and X infinite}\}$$. Then X has the cardinality of the continuum.
The problem is now, that it is not true that $$\mathcal{A}\subseteq P(\mathbb{N})$$. To solve this, let $$I=\{A\subseteq \mathbb{N}~\vert~A~\text{finite}\}$$. It is not that hard to see that I is countable. Let $$f$$ be a bijective function from I to $$\mathbb{N}$$. Then $$\{f(A)~\vert~A\in\mathcal{A}\}$$ is the set you want...

 

[somebd]

Wow, thanks you very much for the fast response and solution idea :)