I know that $|\mathbb{R}| = 2^{|\mathbb{N}|}$, but is there a set $S$ such that $|\mathbb{N}| = 2^{|S|}$?
What is N the power set of?
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elementary-set-theory
2 Answers
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No, there isn't. If $S$ is finite, then $2^S$ is also finite. If $S$ is infinite, then $2^S$ has size strictly larger than $S$ and hence is uncountable. But this means that $2^S$ can never be countable, so there is no $S$ for which $2^{|S|}=\mathbb{N}$.
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No, we know by Cantor's theorem there is no injection of a power set onto a set, but any infinite set, by definition has an injection of the set into itself. So if $\Bbb N=2^S$ then $|S|\ge\aleph_0$ a contradiction.