SCI.MATH POLL - uncountable infinity
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From: herbzet on 5 Jun 2010 19:14 George Greene wrote: > herbzet wrote: > > I'm not sure that this proof is really a "proof of uncountable infinity" > > anyway. A finitist, for example, would reject the notion that the naturals > > constitute an infinite set in the first place, but I see no reason > > why she would reject the proof that for any set S, |S| < |P(S)|. > > Please don't say "a finitist". There is no such thing. > There is no finite number of the finite things. It appears that someone has shamefully neglected to send me the memo. > Therefore, even if you want to contemplate only finite things, > there will still NOT be only finitely many things to contemplate. Why do you say that? > Finitism is simply not a coherent position. -- hz
From: Transfer Principle on 6 Jun 2010 02:24 On Jun 4, 12:04 pm, MoeBlee <jazzm...(a)hotmail.com> wrote: > On Jun 4, 1:37 pm, Transfer Principle <lwal...(a)lausd.net> wrote: > > How about this: ZFC proves _neither_ or _both_ of > > CH and its negation (Goedel and Cohen)? > I see. So do you have any confidence that ZF is consistent? Here is my opinion on this issue. 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. The rapidity of how a proof of the inconsistency of naive set theory was found does suggest that a proof of ~Con(ZF) would've been found equally quickly, if such a proof were possible. 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. This is why I often refer to the mathematician Ed Nelson, who's currently working on a proof that PA (and hence ZF) is inconsistent. Nelson's work depends on very large natural numbers -- naturals so large that one has to come up with a new operation, called tetration, to describe them. Tetration is an operation that isn't considered by most mathematicians, and so a proof which relies on tetration can easily be overlooked for more than a century. The fact that the consistency strength of PA (Gentzen) is epsilon_0, the smallest ordinal which can't be constructed from omega with finitely many additions, multiplications, and exponentiations, but can be written with finitely many _tetrations_ -- indeed, it is omega^^omega (Rucker) -- is suspicious. It indicates to me that the key to a potential proof of ~Con(PA) (hence ~Con(ZF)) is the operation of tetration. But that being said, even though it's unlikely that we'll ever see a proof that ZF is inconsistent (even from Nelson), this doesn't mean that I must a priori agree that Herc is wrong. Although in Cooper's polls, I will vote for the choice which favors ZFC (and thus Herc opposes), the other choice isn't "wrong." There can be theories other than ZFC in which the other choice in the poll is right. Similarly, I'll vote for "0.999... = 1" in a poll asking for whether these two values are equal, but I'll still consider theories which prove that they are distinct. To conclude, just because I believe that ZF is consistent, it doesn't mean I believe that Cooper must be wrong.
From: Transfer Principle on 6 Jun 2010 02:31 On Jun 5, 6:14 am, herbzet <herb...(a)gmail.com> wrote: > |-|ercules wrote: > > Because the most widely used proof of uncountable infinity is the > > contradiction of a bijection from N to P(N), which is analagous to > > the missing box question. > I'm not sure that this proof is really a "proof of uncountable infinity" > anyway. A finitist, for example, would reject the notion that the naturals > constitute an infinite set in the first place, but I see no reason > why she would reject the proof that for any set S, |S| < |P(S)|. MoeBlee pointed this out too: "Sure, but without the power set axiom, we can still prove that for any S, if S has a power set, then there is no surjection from S onto its power set, which is the "essence" of Cantor's theorem." Of course, whenever posters mention this, I immediately point to the theory NFU. NFU proves the existence of non-Cantorian sets, and a non-Cantorian set is precisely a set S such that card(S) < card(P(S)). The simplest example of such a set is the set V of all possible sets -- a set whose existence is provable in NFU (but not ZFC, of course). It's easy to find a surjection from V to P(V) -- since P(V) = V, the identity _bijection_ suffices. Therefore, any poster who doesn't like Cantor's Theorem ought to consider NFU instead of ZFC.
From: Sylvia Else on 6 Jun 2010 06:45 On 5/06/2010 3:57 PM, Barb Knox wrote: > In article<86od4nFja0U1(a)mid.individual.net>, > "|-|ercules"<radgray123(a)yahoo.com> wrote: > > [SNIP] > > Ah, the Townsville looney returns. > > Mate, you just gotta stop licking those cane toads. > > I haven't heard of anyone licking them, but they boil up a treat. Sylvia.
From: |-|ercules on 6 Jun 2010 23:17
"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. You refuse to acknowledge the powerset proof of higher infinity is silly (at least looks silly) because there's other proofs of higher infinity! Fck me! ONE AT A TIME! Even if I meticulously worded my argument 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. Here's an idea. Try answering the question! Given a set of labeled boxes containing numbers inside them, can you possibly find a box containing all the label numbers of boxes that don't contain their own label number? Herc |