Let $k$ be an infinite cardinal (i.e. not in bijection with any natural number), show that
$\mathbb{N} \leq 2^{2^k}$
This is very easy with choice, without it I don't even know where to start.
2026-04-05 14:54:09.1775400849
Cardinal Inequality without using Choice
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HINT:
If $k$ is infinite, then $k$ contains subsets of every finite size.
Of course, there will be many subsets of $k$ of each finite size.
Do you see how to identify a given natural number $n$ with a certain family of subsets of $k$?