Find the cardinality of $$ S = \{X \in \mathcal P(\mathbb N)\mid 2\in X \land 4 \notin X \}$$
I am not entirely sure how to solve this. I tried to use the Cantor - Bernstein theorem. To do this effectively, I need to have an upper and a lower bound of my set. I thought it might be easier to think about the power set in terms of the characteristic function of the set of natural numbers. Therefore, $S$ will be bounded from above by $\{0,1 \}^{\mathbb N}$ and from below by $\{0,1 \} \times \{0,1 \} \times \{ 1\} \times \{0,1\} \times \{ 0\} \times \{ 0,1\}^\mathbb N$ Therefore - the cardinality of this set is continuum.
I am pretty sure that my way of solving this problem is not natural - well, I don't even know if it's correct. Could you suggest an easier - elementary - method to solve this?
As $S \subseteq \mathcal P(\mathbb N)$ you have $\vert S \vert \le \vert \mathcal P(\mathbb N) \vert$.
Also
$$\begin{array}{l|rcl} f : & \mathcal P(\mathbb N \setminus \{1,2,3,4\}) & \longrightarrow & S \\ & X & \longmapsto & X \cup \{2\} \end{array}$$
is an injection and $\vert \mathcal P(\mathbb N \setminus \{1,2,3,4\}) \vert = \vert \mathcal P(\mathbb N) \vert$ . Hence $\vert \mathcal P(\mathbb N) \vert \le \vert S\vert$.
By Cantor-Bernstein theorem $\vert \mathcal P(\mathbb N) \vert = \vert S\vert$.