Cantor's diagonal argument demonstrates that the set of all infinite sequences of 0's and 1's cannot be countable. By constructing a new sequence that differs from each sequence in a countable list, the argument shows that this new sequence is not included in the original countable set. Consequently, this leads to the conclusion that the set of all infinite sequences of 0's and 1's is uncountable.

1. Why can't you use Cantor's diagonal argument to prove that the positive irrational numbers are
not countable?
Solution
Cantor's original proof considers an infinite sequence S of the form (s1, s2, s3, ...)
where each element si is an infinite sequence of 1's or 0's. This sequence is countable, as to
every natural number n we associate one and only one element of the sequence. We might write
such a sequence as a numbered list: s1 = (0, 0, 0, 0, 0, 0, 0, ...) s2 = (1, 1, 1, 1, 1, 1, 1, ...) s3 = (0,
1, 0, 1, 0, 1, 0, ...) s4 = (1, 0, 1, 0, 1, 0, 1, ...) s5 = (1, 1, 0, 1, 0, 1, 1, ...) s6 = (0, 0, 1, 1, 0, 1, 1, ...)
s7 = (1, 0, 0, 0, 1, 0, 0, ...) ... For each m and n let sn,m be the mth element of the nth sequence
on the list. So, for instance, s2,1 is the first element of the second sequence. It is possible to build
a sequence s0 in such a way that its first element is different from the first element of the first
sequence in the list, its second element is different from the second element of the second
sequence in the list, and, in general, its nth element is different from the nth element of the nth
sequence in the list. That is to say, if sn,n is 1, then s0,n is 0, otherwise s0,n is 1. For instance: s1
= (0, 0, 0, 0, 0, 0, 0, ...) s2 = (1, 1, 1, 1, 1, 1, 1, ...) s3 = (0, 1, 0, 1, 0, 1, 0, ...) s4 = (1, 0, 1, 0, 1, 0,
1, ...) s5 = (1, 1, 0, 1, 0, 1, 1, ...) s6 = (0, 0, 1, 1, 0, 1, 1, ...) s7 = (1, 0, 0, 0, 1, 0, 0, ...) ... s0 = (1,
0, 1, 1, 1, 0, 1, ...) The elements s1,1, s2,2, s3,3, and so on, are here highlighted, showing the
origin of the name "diagonal argument". Each element in s0 is, by definition, different from the
highlighted element in the corresponding column of the table above it. In short, s0,n ? sn,n.
Therefore this new sequence s0 is distinct from all the sequences in the list. This follows from
the fact that if it were identical to, say, the 10th sequence in the list, then we would have s0,10 =
s10,10. In general, we would have s0,n = sn,n, which, due to the construction of s0, is
impossible. In short, by its definition s0 is not contained in the countable sequence S. Let T be a
set consisting of all infinite sequences of 0s and 1s. By its definition, this set must contain not
only the sequences included in S, but also s0, which is just another sequence of 0s and 1s.
However, s0 does not appear anywhere in S. Hence, T cannot coincide with S. Because this
argument applies to any countable set S of sequences of 0s and 1s, it follows that T cannot be
equal to any such set. Thus T is uncountable: it cannot be placed in one-to-one correspondence
with the set of natural numbers .