# What's wrong with my proof that $m^{\aleph_0} = n^{\aleph_0}$ for all $m,n \in \mathbb{N}$?

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

• Isn't $n=1$ a special case? Commented Oct 21, 2021 at 18:22
• This proof is pretty good, if overly wordy. Of course you must say $m,n \ge 2$ in your formulation. That bijection between $(0,1)$ and binary expansions is almost (but not completely) bijective. Same for other bases. Commented Oct 21, 2021 at 18:23
• A few small things: (1) You want to throw $||$ around those or something similar, because the sets aren't equal, but they're bijective. (2) You don't need the correspondence between $\mathbb{R}$ and $(0,1)$; as soon as you show that both $m^{\aleph_0}$ and $n^{\aleph_0}$ both biject to (0,1) you're done without even knowing the cardinality of the latter. And (3) as noted by GEdgar, you don't exactly have a bijection; there are some values $r\in(0,1)$ with multiple sequences from $n^{\aleph_0}$ mapping to them under your mapping. This is easily repairable, but it still requires repair. Commented Oct 21, 2021 at 18:27
• For any $m,n≤\aleph_0$ we have $$2^{\aleph_0}≤m^{\aleph_0}≤\left (2^{\aleph_0}\right)^{\aleph_0}≤2^{\aleph_0^2}=2^{\aleph_0}$$ Commented Oct 21, 2021 at 18:27
• There are two special cases $n = 1$ and $n = 0$ which you ignored. Other than that, this is pretty close to being a proof, except that you ignore that some numbers have multiple decimal representations. An easier proof is to note that for all $n \geq 1$, we have $(2^n)^{\aleph_0} = 2^{n \cdot \aleph_0} = 2^{\aleph_0}$. So given any $k \geq 2$, we have $2^{\aleph_0} \leq k^{\aleph_0} \leq (2^k)^{\aleph_0} = 2^{\aleph_0}$. Commented Oct 21, 2021 at 18:30

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 $$1.
(In fact we can go even further since $$2^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}$$, but that's a separate thing.)
• If you know the Schröder-Bernstein theorem, then I think it's clearer to forget real numbers by finding an $i \in \Bbb{N}$ such that $m$ embeds in $n^i$ and using that to embed $m^{\aleph_0}$ in $(n^i)^{\aleph_0} \simeq n^{\aleph_0}$ and similarly with $m$ and $n$ interchanged. Commented Oct 21, 2021 at 20:51