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Question

Is every free ultrafilter on the natural numbers identical modulo permutation of the numbers?

My thoughts

I just thought this was interesting because knowing how much structure all the free ultrafilters on the natural numbers share is kind of a prerequisite to understanding them, for me. I know there are $2^{2^k}$ free ultrafilters on a set of cardinal $k\geq \aleph_0$, so a prerequisite I guess would be that there need to be $2^{2^ {\aleph_0}}$ permutations of the natural numbers. I couldn't even verify if that was the case.

There are $n!$ permutations of $n$ numbers.

$\displaystyle\lim_{n\to\infty} \dfrac{n!}{2^n}$ is obviously going to diverge pretty quickly.

$\displaystyle\lim_{n\to\infty} \dfrac{n!}{2^{2^n}}$ I'm not so sure, so I'm unable to rule out the hypothesis on that basis.

EDIT

I found in this answer that the number of permutations of the natural numbers is at most the number of functions from $\Bbb N$ to $\Bbb N$ and therefore is cardinality $\aleph_0^{\aleph_0}\leq2^ {\aleph_0}$ and therefore $<2^{2^ {\aleph_0}}$ therefore it must follow that all the free ultrafilters on the natural numbers are not identical, modulo permutation of the numbers.

it's a hire car baby
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  • I'll just add the comment (answer) here that the linked duplicates show that the set of permutations of the natural numbers is too small to account for all the free ultrafilters on them, and one of the answers gives examples of two ultrafilters which differ structurally by a difference which is fixed by permutation of the natural numbers. – it's a hire car baby Sep 25 '23 at 08:23

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