Yes, it is extremely difficult to get the mathematical community to "own up to" a false statement. And what nylo is saying is simply false. The crucial point is
Yes, you can! We do it every day when we work with , , or even 1/3, each of whic has an infinite number of digits. We deal with infinite sequences and infinite series in basic Calculus courses. Have you never taken Calculus?
WRONG! You cannot claim such a thing! You cannot pretend that you can find a number which, in fact, you cannot find, because finding it requires an infinite number of operations and is, therefore, impossible!
What "approximation" are you talking about? There are no approximations involved here.
Notice that I don't reject that Cantor's s0 would be real. It certainly would. But to prove that it is not in the sequence, you need to do an infinite operation. What this means is that, no matter how good your approximation to s0 is, how close you get to s0, you will always find that your approximation DOES EXIST in the sequence.
Again, you are asserting that we cannot have numbers involving an infinite number of decimal places. That would remove all irrational numbers and most rational numbers (all those except fractions whose denominators involve only powers of 2 and 5. Do you really want to assert that neither 1/3 nor 1/7 can be written as decimals?
There are always more rows at the bottom. And given that you will never be able to know s0 exactly, you will never be able to claim having found a sequence which was not in the original sequence of sequences.
No, there is no "last element".
Look at this example. Imagine that I build my infinite sequence of sequences this way:
s1 = 0,0,0,0,0,0,0,0... (ad infinitum)
s2 = 1,0,0,0,0,0,0,0... (same)
s3 = 1,1,0,0,0,0,0,0... (same)
and each new sn just adds a new 1. The last element of that sequence would be, basically, all 1s, that would be the "limit".
That doesn't even make sense. The statement the "'infinites" are or are not "exacty the same number" makes no sense. "infinities" are NOT numbers in sense that you are using the word.
A limit that we never reach, as the list is endless, but that doesn't matter to Cantor, who insists on finding a diagonal which cannot be found, because an infinite sequence of sequences with infinite digits produces a Matrix that is not necessarily square. In fact, it is shapeless. It's like pretending that an infinite plane's shape is square, or finding what is the diagonal of an infinite plane. Cantor's method would produce an s0 that he would claim to be the diagonal, but that's only true if the matrix is square and that's something that makes no sense to claim. Cantor's s0 would look like this, given the above sequence, after he has finished:
s0= 1,1,1,1,1,1,1,1,1,1,1,1,1.... (ad infinitum) ...1,1,0. And he would say, as this is the oposite of the diagonal, it is none of the sequences of my sequence. But wait! What do you mean, that it is none of them? If I have built the sequence in the way that I said, s0 of course exists in my sequence. My sequence includes all possible sequences of 1s followed by 0s, including all 1s and all 0s. So s0 is a particular case, and it IS there. The problem is that in this case Cantor would not have found the diagonal, as my matrix is not square. He would think it is, but it isn't. This matrix I have created has strictly 1 more rows than columns, despite both the number of rows and columns being infinite. Indeed, for a sequence of any fixed number of digits n, the number of combinations of 1s followed by 0s is n+1.
Cantor's method can only work if, in the infinite sequence of infinite sequences, both infinites are exactly the same number, making the resulting matrix perfectly square. He has no basis to claim such a thing. Two infinites cannot be compared unless you reach them as limits of algebraic expresions that you can compare. And what is it that allows Cantor to claim that those two infinites are exactly the same number in any sequence that I may propose? I've just proposed one that wasn't.