Obections to Cantor's Theory (Wikipedia article)
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From: Chris Menzel on 30 Jul 2005 23:36 On 27 Jul 2005 13:18:08 -0700, Daryl McCullough <stevendaryl3016(a)yahoo.com> said: > Tony Orlow (aeo6) wrote: >> >>Virgil said: > >>> Okay, TO! Why is the limit on the size of finite string lengths smaller >>> than the limit on size of finite sets of finite strings? >> >>Maybe you should ask Stephen, since I never made any such claim. > > That's because you never explore the *consequences* of the things > you do claim. That's the difference between you and a competent > mathematician. > >>He just made my point anyway, when he said "How many binary strings >>are there then? 1 + 2 + 4 + ... + 2^L, which we all know is >>2^(L+1)-1, which is finite". If you have >>finite lengths only, then you have a finite set. Thanks, Stephen. > > Stephen was pointing out how nonsensical your claim was. You > are saying that there is a maximum string length L. The number > of strings of that length or smaller is M (where M = 2^(L+1)-1). > M is bigger than L. But it's still finite, right? So why can't > you strings of length M? This is at least the fifth clear and decisive refutation you've provided for Orlow to which he hasn't replied, Daryl. He doesn't deserve your time.
From: Jesse F. Hughes on 31 Jul 2005 03:38 "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: [...] > But the niggling point I don't get is the different cases for the > finite and infinite sets A. When A is finite, you view 2^A as a > functor from Set to BoolAlg and otherwise to CompBoolAlg, but this ad > hoc change of codomain seems to strongly undercut the naturalness of > your analysis, doesn't it? Without it, you lose the identity, but > with it, the whole "should" seems much weaker: it *should* only if > 2^(2^A) retains those features of 2^A which come from A's being > finite/infinite[2]. But why those features and not other features of > A? Or why *any* features of A per se? Why not just BoolAlg? I think this is just silliness on my part. In both cases, we could take the codomain to be the category of complete Boolean algebras, right? That would be a uniform treatment of the finite and infinite sets and I suppose it would support that 2^(2^A) = A. -- Jesse F. Hughes "I want to really eat myself, so then I'll be a coalgebra." -- Quincy P. Hughes, Age 3 1/2
From: Jesse F. Hughes on 31 Jul 2005 11:50 "Jiri Lebl" <jirka(a)5z.com> writes: [...] > There's lots of things one can do wih countable sets (such as sum over > them etc...) that one can't do otherwise. You can do induction on > countable sets. I don't get this comment. Countability doesn't have a lot to do with induction from my perspective. How do you "do induction" on an arbitrary countable set? -- "So, at this time, I'd like to assure you that I am not interested in making sure mathematicians worldwide get fired. I've rethought my desire to go to Congress and try to get funding for mathematicians cut." -- James Harris is a reasonable man. Whew!
From: Jesse F. Hughes on 31 Jul 2005 14:58 "MoeBlee" <jazzmobe(a)hotmail.com> writes: > Han de Bruijn wrote: > >> MoeBlee wrote: >> >> > Meanwhile, I'm still fascinated by your inconsistent theory, posted at >> > your web site, which is: >> > >> > Z, without axiom of infinity, bu with your axiom: Ax x = {x}. >> > >> > Would you say what we are to gain from this inconsistent theory? >> >> No. But first we repeat the mantra: >> >> A little bit of Physics would be NO Idleness in Mathematics >> >> It is physically correct that every member of a set is at the same time >> a _part_ of the set, meaning that a e A ==> a c A , where e stands for >> membership and c for being a subset. The above axiom that a member of a >> set cannot physically distinguished from a set which contains only the >> member - the envelope {} means nothing, physically - is only a weaker >> form of this idea. >> >> There has been a thread on this topic in 'sci.math' as well, called >> "Set inclusion and membership". It is noted that a "set theory without >> the membership" actually exists. Google it up as "mereology" and you >> will find quite some clues. Bullshit. Mereology does not support the claim that x = {x}. If we read x = {...} in the obvious way -- namely, that all of the parts of x are listed between { and } -- then x = {x} only if x has no proper parts. You've googled on mereology, but you seem to have missed any clues. >> Now I simply add this axiom to ZFC and see what happens. Nothing >> else. You get inconsistency, silly boy. It is trivial to prove so. First, prove that the empty set exists -- a trivial theorem in ZFC. Then we have by your axiom that {} = {{}} and hence {} is an element of {}. But by definition of {}, we also have {} is not an element of {}. In fact, there are plenty of other proofs that this is inconsistent, not depending on the empty set at all. As long as we have the pairing axiom and also have distinct sets x and y, we get contradiction. After all, by your axiom, {x,y} = {{x,y}}, and so {x,y} is a singleton and hence x = y. So, if you have a pairing axiom (pretty harmless, that), then you better be content with a one-set universe. What a remarkably stupid axiom you have. -- "It has been shown that no man can sit down to write without a very profound design. Thus to authors in general trouble is spared. A novelist, for example, need have no care of his moral. It is there -- that is to say, it is somewhere -- and the moral and the critics can take care of themselves." --E.A. Poe
From: Jesse F. Hughes on 31 Jul 2005 17:15
Robert Low <mtx014(a)coventry.ac.uk> writes: > Because adapting it by including the new rule a={a} really > does break it at a very fundamental level. Apart from anything > else, it means that > > = {} > > (by taking a to be nothing at all) > > and hence that everything in the cumulative hierarcy then > also collapses down to absolutely nothing at all. > > So presumably you need a bunch of urelements. And then the > whole universe is going to collapse down just to the set > of urelements, because, for example, the ordered pair > (a,b) is {a,{a,b}} = {a,a,b} = {a,b}, and similarly the > power set of any set is just the same set back again. Much worse than that. Let a, b be given. Then {a,b} = {{a,b}}, so {a,b} is a singleton and hence a = b. There is not a bunch of urelements. There is only one thing in his universe and it is a set, assuming that we can form unordered pairs. -- But in our enthusiasm, we could not resist a radical overhaul of the system, in which all of its major weaknesses have been exposed, analyzed, and replaced with new weaknesses. -- Bruce Leverett (presumably with apologies to Ambrose Bierce) |