Is this Cardinal arithmetic an error?

Question

In my uni notes I have written $n^{\aleph _0}$ equals the continuum. Now it is easy to show it is $\leq$ the continuum but not the other way other around since if $n=1$ surely we have $1^{\aleph _0} = 1$ And therefore not equal to the continuum?

Answer 1

You are correct that your notes are missing hypotheses: you must assume $n\gt 1$. If $n=1$, then as you note you get $1^{\kappa}=1$ for any cardinal, since there is a unique function from $\kappa$ to $1=\{\varnothing\}$. And if $n=0$, then $n^{\kappa}=0$ if $\kappa\neq 0$ (in particular, $0^{\aleph_0}=0$), and $0^0 = 1$ (because here, $0^0$ is the cardinality of the set of all functions from $\varnothing$ to $\varnothing$, and there is exactly one such function, namely the empty function).

Now, if we assume that $2\leq n\leq \mathfrak{c}=2^{\aleph_0}$, then the result follows easily: $$\mathfrak{c} = 2^{\aleph_0} \leq n^{\aleph_0} \leq \mathfrak{c}^{\aleph_0} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\aleph_0} = 2^{\aleph_0} = \mathfrak{c},$$ so equality follows. (This of course requires $\aleph_0\aleph_0=\aleph_0$, but this is known even in the absence of the Axiom of Choice).