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).
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!