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

Sort of Damocles
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julia199827
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    Hi @Julia199827, usually on Math.SE it's important to ask a question proactively, including what you've tried already, your thoughts on the problem, and the specific point(s) where you're feeling stuck. People often look poorly on questions like this, and are less likely to give helpful answers. You should consider editing your question with this in mind. – Sort of Damocles Apr 29 '20 at 15:23

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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.

Ali Dursun
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