Least $n \in \mathbb{N}$ such that $2^{|n|} = \aleph_0$?

Question

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?

Answer 1

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 $\aleph_0$).

Answer 2

No, $2^n$ is finite for all $n\in \mathbb N$ while $\aleph_0$ is an infinite cardinal.

Answer 3

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. $2, 3, 5, 6, 7, 9$.