Are these sets in the correct order according to cardinality?

Question

$$ \{3,7\} = P(\{1\}) < P(\{2,5\}) < P(\{\Bbb Q\}) = \Bbb R \setminus \{\pi, e\} = \Bbb R \setminus \Bbb Z = \Bbb R \setminus \Bbb N < P(\Bbb R\setminus \Bbb Q) $$ Where P is power set, R is the Reals, and N is the natural numbers

Answer 1

Yes, but there is a simple mistake. $P(\{\mathbb{Q}\}) = \left\{\emptyset, \left\{\mathbb{Q}\right\}\right\}$, and $P(\mathbb{Q})$ is the set of all subsets of $\mathbb{Q}$.

$\{3,7\}$ and $P(\{1\}) = \left\{\emptyset, \left\{1\right\}\right\}$ has 2 elements. $P(\{2,5\}) = \left\{\emptyset, \left\{2\right\}, \left\{5\right\}, \left\{2, 5\right\}\right\}$ has 4 elements.

For $P(\mathbb{Q})$, you can look Cardinality of the Continuum, and Cardinal Number Subtraction for the rest.

Since $\{\pi, e\}<\mathbb{R}$ and $\mathbb{R}$ is infinite and $\mathbb{Z}<\mathbb{R}$, $\mathbb{N}<\mathbb{R}$ and $\mathbb{Q}<\mathbb{R}$ according to the cardinality, $$\mathbb{R} \cong P(\Bbb Q) \cong \Bbb R \setminus \{\pi, e\} \cong \Bbb R \setminus \Bbb Z \cong \Bbb R \setminus \Bbb N \cong \mathbb{R}\setminus\mathbb{Q}$$

and finally, $P(\Bbb R\setminus \Bbb Q) \cong P(\mathbb{R}) > \mathbb{R}$ by the Cantor's Theorem.