Obections to Cantor's Theory (Wikipedia article)
in [Logic]
|
Prev: Derivations
Next: Simple yet Profound Metatheorem
From: Martin Shobe on 27 Jul 2005 21:57 On 26 Jul 2005 22:07:39 -0700, malbrain(a)yahoo.com wrote: >Martin Shobe wrote: >> On 26 Jul 2005 17:15:10 -0700, malbrain(a)yahoo.com wrote: >> >> >Chris Menzel wrote: >> >> On Tue, 26 Jul 2005 16:39:58 -0400, Tony Orlow <aeo6(a)cornell.edu> said: >> >> > ... >> >> > then that function needs to be taken into account. This nonsense >> >> > about an infinite set of finite whole numbers is pretty bad too, but >> >> > probably without any real consequences. >> >> >> >> You seem to agree that the set of whole numbers is infinite. But there >> >> was an inductive argument a few posts back that all the whole numbers >> >> are finite, and hence that the set of finite whole numbers is infinite. >> >> There was some real mathematics there. >> > >> >How does it follow that the count of finite whole numbers is infinite? >> >How is this established by the Peano axioms? >> >> A set, A, is infinite if, and only if, there exists a one-to-one >> function, f:A -> A, such that f(A) is a proper subset of A. >> >> Or equivalently, >> >> A set, A, is infinite if, and only if, there exists a function, f:A -> >> A, such that >> 1) for all x,y in A, f(y)=f(x) => x=y. >> 2) there exists an x in A such that for all y in A, f(y) =/= x. >> >> Since there are no sets, and we are interested only in the domain of >> PA, we have >> >> The domain of PA is infinite if, and only if, there exists a function, >> f, such that >> 1) for all x,y f(y)=f(x) => x=y. >> 2) there exists an x such that for all y, f(y) =/= x. >> >> The successor function meets those criteria. Therefore, the domain of >> PA is infinite. >> >> The only problem that I can see with this is that it's a theorem about >> PA instead of a theorem of PA. > >It's ABOUT PA because you back-filled a definition for infinite to PA? It's about PA because the question isn't even expressable in PA. There are no functionn in the domain of PA, just numbers. >> > You have Tony agreeing to >> >the axiom of infinity apriori, when this is not indicated. >> >> The axiom of infinity is not needed to prove that a set is infinite. >> The axiom of infinity is needed to prove that infinite sets exist. > >This KOAN is going to be a tough sell. karl m What do you think the axiom of infinity says? I'm not looking for the logical consequences, just what the axiom is. Martin
From: Virgil on 27 Jul 2005 22:18 In article <MPG.1d519905895f2783989fb5(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > Since TO will not abide by the axioms of any system containing the > > equivalent of the Peano axioms, all his claims must refer to what occurs > > in some other system on his own invention, for which he has not, and > > perhaps cannot, give a comprehensive axiom system. > > > You didn't like my extension of Peano's axioms? Since any such extention conflicts with many , if not all, standard models for the naturals, no.
From: Virgil on 27 Jul 2005 22:23 In article <MPG.1d51996d5a7522e5989fb6(a)newsstand.cit.cornell.edu>, Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > Virgil said: > > In article <MPG.1d5028071ff97742989f8f(a)newsstand.cit.cornell.edu>, > > Tony Orlow (aeo6) <aeo6(a)cornell.edu> wrote: > > > > > Actually there are two ways to look at it. In unsigned binary, yes, an > > > infinite > > > number of 1's is the largest number possible. Since we start with all 0's > > > representing 0, the size of the set, N, will be one more than 111...111. > > > It > > > will be 000...001:000...000, or one unit infinity. > > > > Until TO can produce unambiguous rules comparing sizes for all of these > > alleged pseudostrings of binary digits, it is all garbage. Unless TO > > wants to say that his "naturals" are not naturally ordered and not > > orderable. > > > > For example, given 1000...010...001 and 100...0110...001, where each > > ellipsis represents a psuedostring of infinitely many zeros, by what > > rule does one compare the composite psuedostrings to determine which is > > larger? > > > > Note that were the ellipses to each represent only finitely many zeros, > > the answer would be relatively trivial. > > > > And what is the effect of concatenation of two or more such allegedly > > infinite psuedostrings? > > > None of that affects the requirement that they be included in order to have > an > infinite set of whole numbers. None of it. Except that any such a requirement, at least in any standard system of naturals, is a defusinal figment of TO's imagination with no actual existence. Whether there is any system in which such a foolish requirement does not immediately make the system self-contradictory has yet to be demonstrated.
From: Daryl McCullough on 27 Jul 2005 22:13 >> >Robert Low said: >> > >> >> OK, so how many elements are there in the set of all finite >> >> natural numbers? Tony replied. >>>Some finite, indeterminate number. That is an out-and-out contradiction. Let FN be the collection of all finite natural numbers. You say that FN is finite. You say that that means that its size is equal to some finite natural number. So call that number L. If L is finite, then it must be an element of FN, because FN is the collection of *all* finite natural numbers. But that means that FN contains at least L+1 elements: 0, 1, 2, ..., L. That contradicts the claim that FN contains exactly L elements. Your theory is self-contradictory. Not that *you* would ever notice the contradiction, because you are just making things up as you go. You are just playing, not caring whether what you're saying makes sense or not. -- Daryl McCullough Ithaca, NY
From: Daryl McCullough on 27 Jul 2005 22:17
Tony Orlow (aeo6) wrote: >The absence of a largest finite does not in any way invalidate this proof. There is no proof. There is only you playing with words. >There is no finite length of string that can possibly produce an infinite set >of strings. That's right. If there is a bound on the lengths of strings in a set S, then S is finite. If there is no bound on the lengths of strings in S, then S is infinite. It's that simple. The collection of all finite strings has no bound. -- Daryl McCullough Ithaca, NY |