Misconception of Cantor's Theorem(no seqeuence can contain all real numbers.) When reading the proof of Cantor's Theorem(the one that says no sequence can contain all reals), I feel unsure. The Cantor's Theorem are proved by contradicting the fact that there are some real number, call it x, in the nested intersection. However, the assumption that all reals are in the sequence leads us to the fact that no element can be in the nested intersection and thus a contradiction. I am wondering what if we let this number, x, put this x in the original sequence. Then what happens?
 I thought about we can apply the same method and just prove it again. So that some number y in real number is still not in the sequence. But we shouldn't have to prove this again. What part of my thoughts went wrong?
The proof I am talking about is to use the nested interval theorem to show it. Not the argument Cantor proposed. This proof show the connection with the completeness axiom.
Proof: The proof is by contradiction. Suppose that such a sequence A exists. Let $I_0$ be the interval $[0, 1]$.
At least one of the thirds of I0 does not contain A(1). Let I1 be such a third.
At least one of the thirds of I1 does not contain A(2). Let I2 be such a third.
At least one of the thirds of I2 does not contain A(3). Let I3 be such a third.
At least one of the thirds of I3 does not contain A(3). Let I4 be such a third.
Etc.
This way we have constructed a nested sequence
I0 ⊃I1 ⊃I2 ⊃I3 ⊃I4... 4
of compact intervals. We see that the intersection of them is a singleton since the length of the interval is going to zero.
tervals Theorem.
Since α∈I1 andA(1)∈/I1,it must be that α=A(1).
Since α∈I2 andA(2)∈/I2,it must be that α=A(2).
Since α∈I3 andA(3)∈/I3,it must be that α=A(3).
Since α∈I4 andA(4)∈/I4,it must be that α=A(4).
Etc.
This shows that this real number α is not a term in A. Yet we have as
sumed that all real numbers appear as terms in A. A contradiction. The proof is complete! 
 A: The proof isn’t really a proof by contradiction, even though it’s often presented in that language. It’s really a demonstration that there is an algorithm that takes as input any infinite sequence of real numbers in $[0,1]$ and constructs as output a specific real number that is not in that sequence. This shows that no sequence can list all of the real numbers in $[0,1]$. There is no need to assume that the sequence enumerates $[0,1]$.
In other words, the argument simply shows directly that every possible list is incomplete.
A: Your thought is not wrong. Suppose we have an alleged "list of all real numbers" (indexed by natural numbers). Then we may apply diagonalization to obtain $x_0$, a real number not on the list. Suppose that we add $x_0$ back to the list (at position $0$, for example). Then we may apply diagonalization to the new list to obtain $x_1$. You will notice that, no matter how many (finite) times we repeat this process, we will always end up with an incomplete list of real numbers, for there is always a new real number not on the list that diagonalization can come up with.
Now this fact doesn't really mean that we need to apply the diagonal argument infinitely often. You may be under the impression that the new list we obtain after adding $x_0$ is somehow "larger" than the original list, and hence necessitates a new proof (in the manner of induction). But in fact the new list has exactly the same size as the original list! The "Hilbert's hotel" thought experiment demonstrates that, when you add one number to a countably infinite list, you still end up with a countably infinite list. So the original proof applies perfectly well to this situation as well, and you are not really proving anything other than that which you've already proven when you apply diagonalization to the new list and obtain another real number.
A: You are not proving this by contradiction. Instead you show that whenever you are given a countable set of real numbers, there is a real number which is not in that set.
When you add such real number, you have a new countable set, and you can find a new one. And so on and so forth.
It's not that after repeating twice, three times, or even countably many times, you can suddenly claim that you have collected all the real numbers. 
