Re: Cantor Confusion

MoeBlee wrote:

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

Oops, that should be:

S(n+1) = the unique finite sequence f such that:
length(f) = length(S(n))+1 and
f(0)= S(n)(max(dom(S(n)))) + 1 and
f is monotone per the standard ordering of omega

In other words, I forget to mention that f is monotone. But that was
obvious from my illustration anyway.

So, here's the proof again for reference:

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
f(0)= S(n)(max(dom(S(n)))) + 1 and
f is monotone per the standard ordering of omega

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.

MoeBlee

.

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