What's wrong with my proof that $m^{\aleph_0} = n^{\aleph_0}$ for all $m,n \in \mathbb{N}$?
Where $\aleph_0$ is the cardinality of the natural numbers $\mathbb{N}$.
First, we observe that $f(x) = tan(x)$ defines a bijective correspondence between $(-\infty,\infty)$ and $(\frac{-\pi}{2},\frac{\pi}{2})$
Then, $g(x) = \pi x – \frac{pi}{2}x$ defines a bijective correspondence between $(\frac{-\pi}{2},\frac{\pi}{2})$ and $(0,1)$.
So $g(f(x))$ defines a bijective correspondence between $(-\infty,\infty)$ and $(0,1)$
Thus the cardinality of $\mathbb{R}$ is the same as the cardinality of $(0,1)$.
Now, think of the points in $(0,1)$ as decimal numbers reduced modulo $2$. Thus the set $(0,1)$ is in bijective correspondence with all sequences of $0$ and $1$. We can now see that the cardinality of $(0,1)$ is $2^{\aleph_0}$.
Now, think of the points in $(0,1)$ as decimal numbers reduced modulo $3$. Thus the set $(0,1)$ is in bijective correspondence with all sequences of $0,1$ and $2$. We can now see that the cardinality of $(0,1)$ is $3^{\aleph_0}$
Now, think of the points in $(0,1)$ as decimal numbers reduced modulo $n$. Thus the set $(0,1)$ is in bijective correspondence with all sequences of $0,1,2,….,$ and $n$.We can now see that the cardinality of $(0,1)$ is $n^{\aleph_0}$
My claim follows
Best Answer
First of all, you need to assume $n\not=1$. More substantively, there is a slight issue around passing from $(0,1)$ to $m^\mathbb{N}$ since in general a number in $(0,1)$ may have multiple base-$n$ expansions. (There's also the issue that you really want $[0,1]$ - think about the sequences $\overline{0}$ and $\overline{n-1}$ in base $n$, respectively - but meh.) Basically, what your argument does get is for each natural $n>1$ an at-most-two-to-one surjection $m^\mathbb{N}\rightarrow [0,1]$, and you need a further bit of argument to massage that to a bijection. (Cantor-Shroeder-Bernstein is useful here - note that it does not require the axiom of choice!)
However, once you take that into account you will have a correct proof: it is indeed the case that $m^{\aleph_0}=n^{\aleph_0}$ whenever $1<m,n<\aleph_0$.
(In fact we can go even further since $2^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}$, but that's a separate thing.)