Re: ping GhostInTheMachine

In sci.logic, ken quirici
<kquirici@xxxxxxxxx>
wrote
on 25 May 2005 07:41:35 -0700
<1117032095.445227.308150@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:
>
>
> The Ghost In The Machine wrote:
>> > This article does not mention the diagonalization issue.
>> If T = {s in S: s not in f(s)}, then one can define
>> g(s,t) : S x S -> boolean : g(s,t) = (s not in f(t)).
>> Then T = {s in S: !g(s,s)}. I am not familiar with
>> Quine's "New Foundations" set theory.
>>
>
> Sorry, but I don't understand the above. Where's the diagonal?
> How is it used? How do you derive T = {s in S: !g(s,s)}?
>
> Ken
>

If a mapping K : S x S -> U exists then one can generalize the concept
of diagonal by looking at K(s,s) exclusively during construction of
something (in this case, T). S x S is a square matrix (although if
S is uncountably infinite it might look a little weird).

If U = {T,F} then one can construct a unique antidiagonal from
K(s,s) by simply taking "not", which I've symbolized here as
!K(s,s). This has applications to e.g. the "infinite number of
coinflippers" issue that HERC occasionally trots out.

As for the derivation, that's simplicity itself.

If g(s,t) = (s in f(t)) [typo earlier; mea culpa]
then the expressions

s not in f(t)

and

!g(s,t)

are equivalent. Since T = {s in S: s not in f(s)} one can then
perform a simple substitution and get T = {s in S: !g(s,s)}.

HTH

--
#191, ewill3@xxxxxxxxxxxxx
It's still legal to go .sigless.
.

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