Cantor Confusion
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From: Ross A. Finlayson on 28 Sep 2006 20:25 david petry wrote: > the_wign(a)yahoo.com wrote: > > Cantor's proof is one of the most popular topics on this NG. It > > seems that people are confused or uncomfortable with it, so > > I've tried to summarize it to the simplest terms: [...] > > I don't know if anyone in this newsgroup is confused by Cantor's proof, > but what you have written misses the whole point of the endless debate > that takes place here. > > The argument of the "anti-Cantorians" is that "real" mathematics is > computationally testable. That is, valid, meaningful mathematical > statements must make predictions about the outcome of computational > experiments. (And note that any mathematics that is potentially > applicable must satisfy that requirement) And hence, all objects that > exist in the world of "real" mathematics must be computable. And there > cannot be more than countably many such objects. > > If you start with a well defined list of well defined real numbers (so > that every digit of every real number can be computed), then the > diagonal method gives us a way of constructing a new real number not on > that list, but that certainly does not imply that the well defined real > numbers are uncountable in Cantor's intended sense of the word. That illustrates a differentiation between me and Dave. Where Dave considers himself an anti-Cantorian, and negates in denial for application of otherwise suitable incompatible results the powerset result, I'm a post-Cantorian, and obviate the powerset result. Is that fair, Dave? Dave, Dave, I suggest to staunch Cantorians that they print a list of the naturals, just, not here, because I prefer to write it as N. If you use a (the) universe in your set theory, it's not ZF, and the powerset result doesn't universally hold. Successor is power type. Universe: infinite; infinite sets: equivalent. Ross
From: Poker Joker on 28 Sep 2006 21:04 "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message news:efhkpt$3136$1(a)agate.berkeley.edu... > If one proves that for EVERY function f:N->R, there exists x in R (which > depends on f) such that x is not in f(N), then this proves that EVERY > list of real numbers is incomplete. This proves that NO list can be > complete, and that's what you want the proof for in the first place. Why the switch to functional notation? I think you are trying to make it more complicated than it is. The OP already pointed out the problem with your argument. The point is that you ->CAN'T<- prove that for ->EVERY<- function unless you assume that ->NONE<- of them have R as their image. And since you must assume that ->SOME MIGHT<- have R as their image, there exists no such real number because the definition of that real is self-referential in that case.
From: Virgil on 28 Sep 2006 21:31 In article <C2ZSg.1209$3E2.571(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "Peter Webb" <webbfamily-diespamdie(a)optusnet.com.au> wrote in message > news:451b5130$0$28952$afc38c87(a)news.optusnet.com.au... > > > You produce ANY list of all Reals, I can show you a missing real. > > Therefore I can do it for ALL lists, and hence there is no complete list > > of Reals. > > If he shows you a list of all reals you can't. The point is that if there is no such list he cannot show you one, and the proof show that there is no such list. > I thought that would be > obvious > to even the casual observer. My very words!
From: Virgil on 28 Sep 2006 21:33 In article <1159489165.729011.63370(a)m7g2000cwm.googlegroups.com>, cbrown(a)cbrownsystems.com wrote: > Virgil wrote: > > In article <1159417542.425540.214160(a)i3g2000cwc.googlegroups.com>, > > cbrown(a)cbrownsystems.com wrote: > > > > > the_wign(a)yahoo.com wrote: > > > > Cantor's proof is one of the most popular topics on this NG. It > > > > seems that people are confused or uncomfortable with it, so > > > > I've tried to summarize it to the simplest terms: > > > > > > > > 1. Assume there is a list containing all the reals. > > > > 2. Show that a real can be defined/constructed from that list. > > > > 3. Show why the real from step 2 is not on the list. > > > > 4. Conclude that the premise is wrong because of the contradiction. > > > > That is a proof by contradiction, which many constructionists object to. > > > > I assume you are replying to the OP here, not me. > > Cheers - Chas Right! Sorry, should have sniped out your ID to make it clear I as only re[plying to the_wign(a)yahoo.com
From: Virgil on 28 Sep 2006 21:39
In article <8i_Sg.25507$QT.3131(a)tornado.rdc-kc.rr.com>, "Poker Joker" <Poker(a)wi.rr.com> wrote: > "Arturo Magidin" <magidin(a)math.berkeley.edu> wrote in message > news:efhkpt$3136$1(a)agate.berkeley.edu... > > > If one proves that for EVERY function f:N->R, there exists x in R (which > > depends on f) such that x is not in f(N), then this proves that EVERY > > list of real numbers is incomplete. This proves that NO list can be > > complete, and that's what you want the proof for in the first place. > > Why the switch to functional notation? I think you are trying to make it > more complicated than it is. > > The OP already pointed out the problem with your argument. The point > is that you ->CAN'T<- prove that for ->EVERY<- function unless you > assume that ->NONE<- of them have R as their image. Maybe you can't but Arturo can, and Cantor could, as that's the way it was originally done. > And since you > must assume that ->SOME MIGHT<- have R as their image One need not assume any such thing. One proves directly that for every list there exists a number not listed in that list. |