Cantor’s diagonal proof revisited In his diagonal argument (although I believe he originally presented another proof to the same end) Cantor allows himself to manipulate the number he is checking for (as opposed to check for a fixed number such as $\pi$), and I wonder if that involves some meta-mathematical issues. 
Let me similarly check whether a number I define is among the natural numbers. The number is $n+1$ and it is clear that $1$, $2$, $3,\ldots,\ n$ are not among these numbers. This “proves” that $n+1$ is not a natural number.
I have here, just like Cantor, a formula for a number, rather than a given number. What is the difference between our proofs?
It seems to me that the answer is that Cantor’s number (as opposed to mine) is being successively better bounded; the process of moving forward in his enumeration describes a converging series, such as we use to define the real numbers, using Cauchy limits. You don’t hear this added comment in the proof. Don’t you think it belongs to the proof? 
 A: Cantor's argument is roughly the following:
Let $s:\ \Bbb{N}\ \longrightarrow\ \Bbb{R}$ be a sequence of real numbers. We show that it is not surjective, and hence that $\Bbb{R}$ is not enumerable. Identify each real number $s(n)$ in the sequence with a decimal expansion $s(n):\ \Bbb{N}\ \longrightarrow\ \{0,\ldots,9\}$. Then for example the sequence
$$r:\ \Bbb{N}\ \longrightarrow\ \{0,\ldots,9\}:\ n\ \longmapsto\ 9-(s(n))(n),$$
where $(s(n))(n)$ is the $n$-th digit of the $n$-th real number in the sequence $s$. By construction $r$ differs from $s(n)$ in the $n$-th place for every $n\in\Bbb{N}$, so it is not in the sequence $s$. Hence the real numbers are not enumerable.
The point of the proof the 'messing-up-function'. Given any sequence $s$ of real numbers, it returns a real number $r$ not in the sequence. To be explicit, the messing-up-function used as an example is
$$m:\ \Bbb{N}^{\Bbb{N}^{\{0,\ldots,9\}}}\ \longrightarrow\ \Bbb{N}^{\{0,\ldots,9\}}:\ s\ \longmapsto\ (n\ \mapsto\ 9-(s(n))(n)).$$
A: No
For your suggestion on the natural numbers to work, it would need to be something like: 


*

*List all the natural numbers 

*Call the largest $n$ (if such a thing is possible)

*Add $1$ to $n$ to make $n+1$

*Show $n+1$ is not in the list, since it is larger than any natural number in the list.
But this fails, because there is no largest natural number, and so no $n$ to add $1$ to. So there is an unlimited number of natural numbers.
Cantor's diagonal proof says list all the reals in any countably infinite list (if such a thing is possible) and then construct from the particular list a real number which is not in the list.  This leads to the conclusion that it is impossible to list the reals in a countably infinite list.
A: In the diagonal argument, a function $f$ from the set of sequences of real numbers to $\mathbb{R}$ is defined. We start from any sequence $S$ of real numebrs. Then it is shown that $f(S)$ is not an element in $S$. The formula is not "changing during the process"; the number we are searching for, $f(S)$, is well-defined if $S$ is given. Now we have shown that for any sequence $S$ of real numbers, there is a real number which is an element of $S$.
The $n+1$ in your proof is not a definition of a number you are searching.
A: Cantor's proof is way more general than just showing that $\mathbb R$ is uncountable. The diagonal argument actually shows that no set can have the same cardinality as its power set (i.e. the set of all its subsets). 
Regarding the concrete case of the real numbers, already when you state that real numbers can be written using infinite decimal expansions, you are using that they are characterized via a convergent series. No need to keep repeating that. 
As for your argument, I cannot see what you mean when you say "not among those numbers". 
A: 
Let me similarly check whether a number I define is among the natural numbers. The number is n+1 and it is clear that 1,2,3 …n are not among these numbers. This “proofs” that n+1 is not a natural number.

No, all you have proven is that there is a positive integer ($n+1$) that is not in the list  $1, 2, \ldots, n$.
Such a fact does make for a perfectly good proof that $1, 2, \ldots, 73$ is not a list of all of the positive integers. It is even a proof that no list of the form $1, 2, \ldots, n$ (where $n$ is a positive integer) can be a list of all of the positive integers.
If the set of positive integers were finite, it would be of the form $1, 2, \ldots, n$. So it can even be used as a proof that the set of all positive integers is infinite.

All of that is analogous to the proof you reference. The diagonal argument (constructively) proves that every list of real numbers is missing a real number.
Thus, we conclude that there does not exist a list of all real numbers.
