Prove that $|\mathcal X| = |\mathbb N^\mathbb N|$ Let $\mathcal X$ be the set of subsets of a set of natural numbers $X$, where $X$ includes all even numbers.
Prove, that $|\mathcal X| = |\mathbb N^\mathbb N|$
I have a serious problem with this example. Because if $X$ contains {2,4,6, ...}, then what is included in the set $\mathcal X$? Would it be something like {{2,4,6,...}}?   
I tried to translate instructions as well as possible, but if something is not clear I will do my best to correct my translation.  
 A: Assuming $\mathcal{X} = \{X \subset \mathbb{N}: 2 \mathbb{N} \subset X \}$ then we have $\vert \mathcal{X} \vert = \vert 2^\mathbb{N} \vert$. A standard way to see this kind of thing is to consider the function $\Phi: 2^\mathbb{N} \to \mathcal{X}$ defined by $\Phi(f) = 2\mathbb{N} \cup \{2n+1 : f(n) = 1 \}$ and verify that this is a bijection. 
Then it is a standard fact that $\vert \mathbb{N}^\mathbb{N} \vert = \vert 2^\mathbb{N} \vert$. You can prove this by noting $$\vert 2^\mathbb{N} \vert \leq \vert \mathbb{N}^\mathbb{N} \vert \leq \vert (2^\mathbb{N})^\mathbb{N} \vert = \vert 2^{(\mathbb{N} \times \mathbb{N})} \vert = \vert 2^\mathbb{N} \vert $$ and applying Cantor-Schroeder-Bernstein. Combining these gives $\vert \mathcal{X} \vert = \vert \mathbb{N}^\mathbb{N} \vert$
Edit: If instead we are meant to take $\mathcal{X} = P(X)$ for a fixed set $X$ such that $2\mathbb{N} \subset X \subset \mathbb{N}$ as in the edited form of the question then note that a simple application of Cantor-Schroeder-Bernstein gives $\vert X \vert = \vert \mathbb{N} \vert$ so let $g: X \to \mathbb{N}$ be a bijection. We can define a bijection $G: P(X) \to P(\mathbb{N})$ by setting $G(Y) = \{g(y) : y \in Y \}$ for $Y \subset X$ so $\vert \mathcal{X} \vert = \vert P(\mathbb{N}) \vert = \vert 2^\mathbb{N} \vert $. Now we can revert to the above.
A: There is a general result that subsumes this case:
If $\kappa,\ \lambda $ and $\mu$ are cardinals such that $\lambda\le \kappa\le \lambda^{\mu}$ and $\mu$ is infinite, then $\lambda^{\mu}=\kappa^{\mu}$. The proof is a simple calculation for we have $\lambda^{\mu}\le \kappa^{\mu}\le (\lambda^{\mu})^{\mu}=\lambda^{\mu\times \mu}=\lambda^{\mu}.$ 
*The second to last equality follows by a  $\textit{currying}$ procedure: for any $g:\mu\times \mu\to \lambda$ we have a $\phi:\mu\to \lambda^{\mu},\ $ defined by $\phi(x)(y)=g(x,y),\ $ and it is easy to verify that the map $g\mapsto \phi$ is a bijection.
