Obections to Cantor's Theory (Wikipedia article)
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From: Martin Shobe on 26 Jul 2005 20:21 On Tue, 26 Jul 2005 14:11:40 -0400, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: >imaginatorium(a)despammed.com said: >> >> >> Tony Orlow (aeo6) wrote: >> > Martin Shobe said: >> <big snip> >> >> > > BTW, there is a caveat on convergence. You have to assume the >> > > standard topology. In other topologies, that sequence can converge. >> >> > You mean with a ring? That's really not what we're talking about, unless you >> > agree that the number line is a circle, and even then it's not relevant. In >> > pure quantitative terms, a sum of infinite 1's is infinite. >> >> Tony, could you please clarify: when you use the word "ring", what do >> you mean? >> >> (a) The algebraic structure known by mathematicians as a ring >> (b) Something else (in which case please call it a T-ring) >> (c) You're sure it is (a), but cannot actually sketch the axioms for a >> ring (a) >> >> If you select (c), please confirm you really meant (a) by sketching the >> axioms. >> If you select (a), please suggest why you think the name "ring" is >> used. >> >> (Since I really have no idea, I'd be interested in informed comments on >> the last question.) >> >> Brian Chandler >> http://imaginatorium.org >> >> >> >> > >> > So, please make up your mind. Do we increment to get a successor an infinite >> > number of times, or only a finite number of times, to get N? >> > > >> > > Martin >> > > >> > > >> > >> > -- >> > Smiles, >> > >> > Tony >> >> >I am not an expert in rings, nor am I going to sketch the axioms that you know >better than I, nor does it help the conversation when you snip the original >statement was repsonding to, which had soemthing to do with numbers being their >own multiples of more than 1, or something. No. The rules of addition, multiplication, etc. don't change just becuase the topology changes. What changes is, roughly speaking, which parts of N are close to other parts. Martin
From: Martin Shobe on 26 Jul 2005 20:36 On Tue, 26 Jul 2005 16:51:05 +0200, David Kastrup <dak(a)gnu.org> wrote: >Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > >> David Kastrup wrote: >> >>> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: >>> >>>>And this is precisely what anti-Cantorians find unacceptable. IMHO >>>>an "infinite" set cannot consistently have a "size". >> >>> This is a _perfectly_ valid point of view (as opposed to the views >>> of those you sympathize with which are wildly inconsistent). >> >> Uh, uh. How do you know with which I symphathize? Do you keep a record >> of my responses to everybody? > >Well, you use the word "anti-Cantorians" above. If this was not >intended to mean mostly a particular set of some outspoken people in >this Usenet group, it would appear that I misinterpreted this. > >> [ .. rest deleted .. ] >> >> OK. It seems that we finally have arrived somewhere. I have one >> final question, though. Is it "legal" (according to mainstream >> mathematics) to call a set "countable" if it can be brought in 1:1 >> correspondence with the naturals? > >That's the usual usage of the word. I believe that that is the definition of "countably infinite". A set is "Countable" iff there exists a one-to-one correspondence to a subset of the naturals. >> And uncountable otherwise? > >Well, in one direction: {1} can't be brought into 1:1 correspondence >with the naturals, either. But if you can't map the naturals to cover >the set exhaustively, than the usual term would be "uncountable". > >It may be that there is a bit of excluded middle: I think finite sets >are usually classed neither as countable nor uncountable. Not sure >about that, though. I'm pretty sure that finite sets are usually considered countable. Martin
From: Randy Poe on 26 Jul 2005 22:03 Chris Menzel wrote: > On Tue, 26 Jul 2005 16:39:58 -0400, Tony Orlow <aeo6(a)cornell.edu> said: > > When the only way to form a bijection is with a mapping function, > > How else? TO has a very restricted meaning to the term "mapping function", which seems to be restricted to functions which can be expressed in a half-dozen or so symbols and which preserve order. - Randy
From: imaginatorium on 26 Jul 2005 22:15 Tony Orlow (aeo6) wrote: > imaginatorium(a)despammed.com said: > > Tony Orlow (aeo6) wrote: > > > > > I don't see where you pointed out any specific flaw, except to rant about your > > > largest finite number again. > > > > No, well, I give up. Just for my curiosity, though, I still cannot > > understand your point when you complain about "ranting about my[sic] > > largest finite number". It has been pointed out to you so many times - > > with absolutely no effect - that the Peano axioms (or any similar more > > informal notion of pofnats) imply that there cannot be a largest > > pofnat. Just tell me: do you claim... > > > (sigh) > > (1) There _is_ a largest pofnat. > no > > (2) There is no largest pofnat (but the contradictions with your ideas > > escape you) > yes, please explain the contradiction, without the mantra. I have heard Virgil > claim that I think there is one, or that I MUST produce one, if I am to claim > there are infinite whole numbers. I see no such need. I ahve agreed that one > cannot count finitely from the finite to the infinite, and it has been agreed > that one cannot count down from the infinite to the finite. The first fact does > not mean the infinite whole don't exist, any more than the second means that > finite wholes cannot exist. So, where is the contradiction? You claim, simultaneously that: (a) There is no largest pofnat. (b) There are only a finite number of pofnats. You suddenly became very explicit (elsewhere in the thread) and appear to agree that to say a set is *finite* means that it can be counted against a ditty, and the ditty stops. OK, so arrange the pofnats in normal ascending order, and count them against a ditty. When the ditty stops, you have reached the last pofnat, which is therefore the largest. This is a consequence of (b), and contradicts (a). > > (3) The answer to "Is there a largest pofnat?" is somehow neither 'Yes' > > nor 'No'. > No, the answer is no, just like the answer to "is there a smallest infinite > number?" There is no distinct line between the finite and infinite. That line > is infinitely wide, and requires an infinite difference to cross. I wouldn't call it a "line", personally, but roughly speaking this appears to be a description of the set of surreals {0, 1, 2, 3, ...} U {..., w-3, w-2, w-1, w} (using union notation to prevent dotty confusion). Brian Chandler > > http://imaginatorium.org > > > > > > -- > Smiles, > > Tony
From: Virgil on 26 Jul 2005 22:45
In article <dc6fn8$51j$1(a)news.msu.edu>, stephen(a)nomail.com wrote: > In sci.math Daryl McCullough <stevendaryl3016(a)yahoo.com> wrote: > > Tony Orlow (aeo6) wrote: > >> > >>Daryl McCullough said: > > >>> But for the set we are talking about, there *is* no L. We're talking > >>> about the set of *all* finite strings. That's an infinite union: If > >>> A_n = the set of all strings of length n, then the set of all possible > >>> finite strings is the set > >>> > >>> A = union of all A_n > >>> = { s | for some natural number n, s is in A_n } > >>> > >>> This set has strings of all possible lengths. So there is no L > >>> such that size(A) = S^L. > > >>If those lengths cannot be infinite, then the set cannot be either. > > > Why do you believe that? > > >>Either you have an upper bound or you do not, and if there is no > >>upper bound on the values of the members, then they may be infinite. > > > Why do you believe that? > > >>> You are assuming that every set of strings has a natural number L > >>> such that every string has length L or less. That's false. > >> > >>I am saying that if L CANNOT be infinite > > > I'm saying that there *is* no L. So don't talk about the case > > where L is infinite or the case where L is finite. I'm talking > > about the case where there *is* no maximum size L. > > > Why do you think that there is a maximum size L? > > I doubt you will get any rational response. The idea > that a set of finite objects must be finite is so engrained > in some people's mind that they cannot see past it, despite > all the obvious contradictions. > > For example, presumably there is some maximum length > to the finite binary strings, which we will call L. > How many binary strings are there then? 1 + 2 + 4 + ... + 2^L, > which we all know is 2^(L+1)-1, which is finite, and is clearly > larger than L (assuming L > 0). Now why someone would believe that > there can exists 2^(L+1)-1 binary strings, but there cannot exist > binary strings with length 2^(L+1)-1 is quite beyond me. They > are both finite numbers. Why is the limit on finite string lengths > smaller than the limit on finite sets of finite strings? > > Stephen Nice point! 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? |