It is well known that $2^{\aleph_0}>\aleph_0$. This seems to imply that there is an $n \in \mathbb{N}$ such that $2^{|n|} = \aleph_0$. If so, what is known about the least permitted value of $n$ such that this statement is true?
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If you undelete your question I will be able to show you the most elementary way https://math.stackexchange.com/questions/4118937/must-the-analytic-continuation-of-zetas-for-s-in-mathbbc-backslash – reuns Apr 27 '21 at 21:31
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No, $2^n$ is finite for all $n\in \mathbb N$ while $\aleph_0$ is an infinite cardinal.
Christoph
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Does this not make the function $2^{|n|}$ discontinuous, jumping from a finite cardinal to $2^{\aleph_0}$ as $n$ transitions from finite to infinite? It seems as though one "skips" over a cardinality of $\aleph_0$. – Master Drifter Mar 29 '20 at 13:48
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2Well, it skips lots of finite values as well e.g. $3, 5, 6, 7, 9$. All except powers of $2$. – badjohn Mar 29 '20 at 14:12
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It is known that $2^\kappa = \aleph_0$ has no solution $\kappa$ in cardinal numbers.
As you say, $2^{\aleph_0} \gt \aleph_0$ by Cantor's Thm., and exponentiation of cardinals is monotonic. So the only possibility is that $\kappa$ is less than $\aleph_0$.
But such $\kappa$ would be finite (since $\aleph_0$ is the least infinite cardinal), and then $2^\kappa$ would also be finite (not $\aleph_0$).
hardmath
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If $n$ is finite then $2^n$ will be as well.
See logarithms of cardinal numbers at Wikipedia for more details.
Is this unintuitive? Maybe, maybe not, intuition does not work well with infinite cardinals. I don't see that it is unintuitive but opinions may vary. $2^n$ for $n \in \mathbb{N}$ skips many finite values as well e.g. $2, 3, 5, 6, 7, 9$.
badjohn
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