I don’t want to claim that I have ‘refuted Cantor’ or something here, I just want to understand it adequately.
I do understand that the proof works something like this:
You assume that you can map the naturals onto the real numbers like so, where each letter represents some arbitrary digit:
$$1 — 0.abcd\cdots$$
$$2 — 0.efgh\cdots$$
$$3 — 0.ijkl\cdots$$
And then you constrict the ‘mystery number’ $x$ which differs in at least one digit of each number that is assigned to a natural number. This mystery number, by definition, cannot be mapped onto any of the natural numbers. This is fine, but why can’t you just assign $0.abcd\cdots$ to $2, 0.efgh\cdots$ to $3$, and so on, then assign $x$ to $1$, which would now be open?
What is the problem with doing this?
Update: Sorry for the poor notation and formatting, I’m writing this on my phone.
Best Answer
Yes, if you are missing a real number, then you can always add it to the front of your list, but this not mean that the new list is now complete. (this is LordSharktheUnknown's point: if the mystery number was the only one that as missing, then yes, you'd get a complete list, but it may not be the only one missing).
In fact, we know the new list cannot be complete, because we can go through the same diagonalization to find a new mystery number that is not on this new list (this is spaceisdarkgreen's point)
And finally, remember that when you do the diagonalization on the original list, we did this within a proof by contradiction. That is, we assumed that there is a complete list ... but when we take any such supposedly complete list we can find a mystery number that is not on the list ... and so we reach a contradiction already. Coming up with a new list does not take away from the contradiction that you obtain when assuming there is a complete list at all, so you are doing nothing to refute the proof (and this is Cheerful Parsnip's point)