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Topic: The Diagonal Argument
Replies: 28   Last Post: Dec 29, 2012 12:11 AM

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ross.finlayson@gmail.com

Posts: 921
Registered: 2/15/09
Re: The Diagonal Argument
Posted: Dec 28, 2012 2:09 AM
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On Dec 27, 9:35 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <c3b9462b-6826-46fd-bfe3-39c2d95ab...@pe9g2000pbc.googlegroups.com>,
>  Graham Cooper <grahamcoop...@gmail.com> wrote:
>

> > one must consider the audience Virgil!
>
> > SWAPPING DIGITS DOWN THE DIAGONAL
>
> > seems to be the only mathematics he can grasp!
>
> Actually, Cantor's original argument does not even use digits.
>
> Cantor considers the set, S, of functions from the set of naturals |N as
> domain, to the two-letter set of letters {m,w}, and shows that there
> cannot be any surjective mapping f: |N -> S by  constructing a member g
> of S not in Image(f)
>
> Since  f: |N -> S, each f(n) is a function from |N to {m,w}
> So that when  g(n) is a member of {m,w}\f(n)(n) for each n, then g is
> not a member of S.
> --


That's not "Cantor's original argument", for what he may have first
stated it.

http://en.wikipedia.org/wiki/Cantor's_theorem#History

For subsets M of N, the ordinal indices of S range from zero in alpha
through omega, let f_alpha(M) be onto {0} and f_omega(M) be onto {1},
then, G_alpha(M) = 1 - f_alpha(M) -> {1} = f_omega(M).

Here f satisfies the hypothesis of being a function from N at least
into S and doesn't see the contradiction. Here there are obviously
infinite ordinals between alpha and omega, between which there are
functions from N onto {0,1}. Basically this S has only one of the two
values on the ends, and two in the middle, with symmetry and
reflection, and the ordinal omega would look like 2^omega. Basically
for each member of S from zero, there is a corresponding bit-wise
complement, in the same order, in reverse, from omega, such that
G_alpha = f_omega-alpha. Thusly, G is not: not in S.

Then, that would get back into Cantor himself justifying counting
backward from "limit ordinals", or that omega is simply the next limit
ordinal.

For Russell's, let phi-x be or include that "not-phi-x is false",
i.e., truth.

If all the propositions in the language have truth values, and the
theorem is about their self-referential content, then admit their self-
referential statement, here that in the language one, or the other, of
the statement, and its negation, is a statement in the language.

Powerset is order type is successor, Hancher, you rainbow-puking
regurgitist.

Ah, then excuse me, Hancher's quite monochromatic, besides
melodramatic, as "puke parrot", or for that matter, "puke maggot".

Yeah Hancher, Cantor discovered this construction some hundred years
ago, I'm glad you've dug back a few more years, try and catch up.

Regards,

Ross Finlayson



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