What is $n^{\aleph_0},n\in\mathbb N$

Question

Can I say that $$n^{\aleph_0}=2^{\aleph_0\log_2n}=2^{\aleph_0}=\aleph_1$$

Answer 1

No, there's no $\mathrm{log}_2$ in cardinal arithmetic, and you're assuming the continuum hypothesis to get from $2^{\aleph_0}$ to $\aleph_1$.

But note that:

Proposition. For all infinite cardinals $\kappa$, we have $\kappa^\kappa = 2^\kappa.$

So if $n \leq \aleph_0$, then:

$$n^{\aleph_0} \leq \aleph_0^{\aleph_0} = 2^{\aleph_0}$$

Hence if $n$ is a cardinal number in the interval $[2,\aleph_0]$, then $n^{\aleph_0} = 2^{\aleph_0}$.

Edit. Regarding logarithms, note that since the natural numbers aren't closed under the usual logarithm, hence we certainly cannot extend it to the cardinal numbers. What we might try instead is to define a "best over-approximation" variant. Explicitly, given cardinals $\kappa \geq 0$ and $\nu \geq 2$, define that $\mathrm{log}_\nu(\kappa)$ is the least cardinal $\lambda$ such that $\nu^\lambda \geq \kappa$. For example, ZFC proves that $$\mathrm{log}_2(2^{\aleph_0}) = \aleph_0.$$

Side remark. Order theoretically, all we've done is defined $\mathrm{log}_\nu\Box$ as the left-adjoint of $\nu^{\Box},$ which reads: $$\mathrm{log}_\nu (\kappa) \leq \lambda \iff \kappa \leq \nu^\lambda.$$

Now assuming the injective continuum function hypothesis, we have: $$\mathrm{log}_2(2^\kappa) = \kappa$$

Answer 2

This is set power. Recall that for cardinals $A$ $B$, $A^B$ is the cardinal of $$\{f: B \to A\}.$$ How many applications $$f : \aleph_0 \to n$$ do you find for fixed $n$ ?