SCI.MATH POLL - uncountable infinity
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From: herbzet on 6 Jun 2010 23:34 |-|ercules wrote: > "herbzet" <herbzet(a)gmail.com> wrote > > > In any case, this thread appears to be closed, aside from tying > > up any loose ends that may remain. Since it has been established > > that proofs of the existence of higher orders of infinity do not > > in general rely on the proof that |S| < |P(S)| for any set S, there > > really is no way anyone is going to answer the poll question in the > > affirmative. Not even the OP, apparently, is interested in further > > discussion here. > > This is a joke. I thought I'd heard 10,000 reasons skeptics don't need > to investigate claims, but sci.math wants the King Con Crown. Actually, I don't subscribe to sci.math. You may avoid my replies by not cross-posting to sci.logic. > You refuse to acknowledge the powerset proof of higher infinity is silly > (at least looks silly) because there's other proofs of higher infinity! Sorry, that's not what you asked. I'm not a mind reader, you know. > Even if I meticulously worded my argument Say, that's a thought -- why don't you give it a try? > in your NULL HYPOTHESIS > anti anything lingo and traced down all your good for nothing ZFC religion > a - x - i - o - m - s you'd still not understand the claim and wiggle and worm > out of that too. You're foaming at the mouth a bit. -- hz
From: |-|ercules on 6 Jun 2010 23:47 "herbzet" <herbzet(a)gmail.com> wrote >> You refuse to acknowledge the powerset proof of higher infinity is silly >> (at least looks silly) because there's other proofs of higher infinity! > > Sorry, that's not what you asked. I'm not a mind reader, you know. > >> Even if I meticulously worded my argument You really are a literal nut. Reword: You refuse to acknowledge the silly rewording of the powerset proof of higher infinity. Despite it's provability in ZFC! (HA) Herc
From: herbzet on 6 Jun 2010 23:52 |-|ercules wrote: > "herbzet" wrote No, I didn't, you did. > >> You refuse to acknowledge the powerset proof of higher infinity is silly > >> (at least looks silly) because there's other proofs of higher infinity! > > > > Sorry, that's not what you asked. I'm not a mind reader, you know. > > > >> Even if I meticulously worded my argument > > You really are a literal nut. I didn't say that -- you did. Are you off you meds? -- hz
From: George Greene on 8 Jun 2010 01:25 On Jun 6, 2:31 am, Transfer Principle <lwal...(a)lausd.net> wrote: > Therefore, any poster who doesn't like Cantor's Theorem > ought to consider NFU instead of ZFC. Nobody ought to consider NFU period for any but the most theoretical of reasons. For example, in NFU, the set of all 1-element subsets of a set is NOT the same size as the set! You canNOT prove the existence of the function mapping x to {x} for every x in some domain-set S. What you really want to consider, far more generically than NFU, is set theory with a universal set. There are plenty of web-pages devoted.
From: Aatu Koskensilta on 8 Jun 2010 21:32
Transfer Principle <lwalke3(a)lausd.net> writes: > I am confident that ZF is consistent, because I believe that if ZF > were inconsistent, a proof of this would have been found by now. Why? This is not uncommon idea but on closer scrutiny there's not much to recommend it. > If it does turn out that ZF is inconsistent, then there would have > been some underlying reason that the proof wasn't discovered for over > a century after the axioms were first given. What's there to rule out the possibility that the simplest proof of a contradiction in ZF is inhumanely complex, utterly beyond our comprehension, invoking, say, an obscure instance of Pi-20^20^20^20^4546^3214532 + 4145624^7542 + 897412 replacement? -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |