Re: Herc and Russell's Paradox
- From: Balthasar <nomail@invalid>
- Date: Thu, 07 Aug 2008 02:45:21 +0200
On Thu, 07 Aug 2008 02:20:19 +0200, Balthasar <nomail@invalid> wrote:
In fact, there's a variant of the Russell's "Barber paradox" which quiteRight. There's a similarity in the FORM of the argument. (I already
T = {s in S : s not in f(s)}
this becomes
T = {s in N : s not in P(N)_s}
which becomes
T = {s in N : s not in the sth subset of P(N)}
which is:
the subsets that don't contain their own index
this logic is exactly the reasoning that determines Russel's
set doesn't exist.
wrote in this thread that Russell got the idea for his paradox from an
analyses of Cantor's proof.) BUT in contrast to Russell's paradox here's
nothing "paradoxical". :-) T is just a subset as any other. No
SELF-REFERENCE, or CIRCULARITY, or whatever is involved here.
closely mirrors the form of the argument (involved here):
"Barber pseudoparadox. The council of a certain village is said to have
given orders that the village's barber (supposedly unique and a man) was
to shave all the men in the village who did not shave themselves, and
only those men. Who shaved the barber?"
Well there's a simple solution: the barber does not live in the village;
and hence he might or might not shave himself.
Concerning the set T, we have: T is not indexed.
And hence f is not surjective. qed.
B.
--
"For every line of Cantor's list it is true that this line does not
contain the diagonal number. Nevertheless the diagonal number may
be in the infinite list." (WM, sci.logic)
.
- References:
- Herc and Russell's Paradox
- From: Balthasar
- Re: Herc and Russell's Paradox
- From: Balthasar
- Herc and Russell's Paradox
- Prev by Date: Re: Herc and Russell's Paradox
- Next by Date: Re: Herc and Russell's Paradox
- Previous by thread: Re: Herc and Russell's Paradox
- Next by thread: Re: Herc and Russell's Paradox
- Index(es):
Relevant Pages
|
Loading
