Re: Cantor Confusion

MoeBlee wrote:
mueckenh@xxxxxxxxxxxxxxxxx wrote:
MoeBlee schrieb:

For reference, here is my proof:

A sequence is a function such that the domain of the function is an
ordinal. A finite sequence has a natural number as its domain. A
denumerable (countably infinite) sequence has omega as its domain. An
uncountable sequence has an uncountable ordinal as its domain.

The entries in a sequence are members of the range of the sequence.
Each entry is indexed by a member of the domain. The elements of the
sequence are ordered pairs of the form <x y> where x is a member of the

domain (which is an ordinal) and y is a member of the range of the
sequence (so the y's are the entries).

dom(S) = the domain of S.

length(S) = dom(S).

In my example S is the denumerable sequence recursively defined as
follows:

S(0) = {<0 0>}

S(n+1) = the unique finite sequence f such that length(f) =
length(S(n))+1 and such that f(0)= S(n)(max(dom(S(n)))) + 1

So S is a denumerable sequence such that each entry of S is a finite
sequence.

The diagonal of S = the unique denumerable sequence D such that, for
all n in omega, D(n) = S(n)(n).

The diagonal of S is a denumerable sequence.

Therefore, there exists a denumerable sequence S of finite sequences
such that the diagonal of S is denumerable.

Now map it on a line. It is longer than any line entry.

The real line?

Obviously, I could be wrong, but I think WM means map it on a line of
the list. He seems to think that because we construct the diagonal from
the list, the diagonal must be one of the lines in the list. Why he
thinks this, I have no clue.

--
David Marcus
.

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