Next Real Number

I'm hoping to understand why there cannot be a next real number. I have read a few other answers, but their proofs feel they are missing something.

Appeal to density of the reals

A common argument against the ability to define the next real goes something like this. Assume the next real after $x$ is $y$, with $x<y$. Then $\frac{x+y}{2}$ is a different real number that lies between $x$ and $y$. Hence, $x$ cannot be the number before $y$, because there is always a number closer. Answers claim that we can do this ad infinitum.

However, I think this must be false-- we cannot continue this process forever. Choose a number $r_1$ strictly between $x$ and $y$ (i.e. $r_1$ satisfies $x<r_1<y$). Next, we pick a new number $r_2$ strictly between $r_1$ and $y$ (i.e. $r_1 < r_2 < y$). Etc. At each step, we chose a unique real number, because $r_{n-1}<r_n$. If we could repeat this process forever, then we could repeat this process $2^{2^{|\mathbb{N}|}}$ times. But, this implies we can find $2^{2^{|\mathbb{N}|}}$ many different real numbers, which is certainly impossible! This seems to say that at a certain point, it's necessary we cannot find a closer real number to y.

What went awry with my argument? Any help would be appreciated!

Solution 1:

There is no ad infinitum. There is no process. It is, if you will, a scaffold to build a refutation for any single candidate anyone could possibly put forward, and that's it. You could claim to have a "next real number after $x$", and I can use your $\frac{x+y}2$ to prove that you're wrong. No matter which real number you pick, the one you pick cannot be the "next real number after $x$".

If there were a "next real number after $x$", then you could pick that and the refutation would have to fail somehow (consider, in contrast, how that exact refutation fails for integers, where there indeed always is a "next number after $x$"). You don't get to keep picking and picking and looking at some sort of limiting process. That's just not the kind of result this is.

Solution 2:

Suppose I start with $0$ and keep adding $1$. By your logic, if I can do this forever (which of course I can), then I can do this $2^{2^{|\mathbb{N}|}}$ many times, implying there are at least that many natural numbers. .... but of course there are not that many natural numbers. So there is a flaw in your logic... .