Cantor Confusion
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From: Fuckwit on 4 Feb 2007 13:47 On 4 Feb 2007 10:39:08 -0800, mueckenh(a)rz.fh-augsburg.de wrote: >>> >>> What is 4 in mathematics? >>> >> In axiomatic set theory it's (usually) the set >> >> {0, 1, 2, 3}. >> > What is 3, what is 2, what is 1, what is 0, what is > the empty set? > The empty set is the only set which does not have any elements. Its existence (in ZFC) is guaranteed by the axiom of subsets (Aussonder- ung). You know, modern math is based on _axioms_, not some sort of hand waving (you seem to prefer). Usually the empty set is denoted with "{}". Then we may define (in an entirely non-circular way): 0 := {} 1 := {0} 2 := {0,1} 3 := {0,1,2} 4 := {0,1,2,3} > > That is set theory, not mathematics. > Well, the last time I've checked it set theory was mathematics. F.
From: Franziska Neugebauer on 4 Feb 2007 13:53 David Marcus wrote: > Carsten Schultz wrote: >> Franziska Neugebauer schrieb: >> > mueckenh(a)rz.fh-augsburg.de wrote: >> >> If A is infinite, induction can show that A is infinite. >> >=20 >> > 1. Where did you read that? Reference? >> > 2. It reminds me of the M=C3=BCckenheim-Axiom >> > =20 >> > X is not finite -> there must be an x in X which is infinite >> >> This follows directly from the more fundamental M=C3=BCckenheim axiom >> "ev= ery >> set is finite". > > One might naively think that the second axiom makes the first > superfluous. Rather remarkable that they are both needed. The natural truth cannot be mentioned often enough. F. N. -- xyz
From: Virgil on 4 Feb 2007 14:07 In article <1170606249.100767.104120(a)j27g2000cwj.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > How can a union of finite elements contain an infinite element? WM again conflates the size of the sets being unioned with how many sets are being unioned. A union of infinitely many finite sets can be an infinite set just as easily as a finite union of infinite sets. > > > > > > > What I state again, again and again, but what you > > > > misread each and every time is: "the union of sets of finite paths > > > > does not contain an infinite path". > > > > > > So you say the union of finite paths > > > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > > > contains the infinite path p(oo) = {0,0,0,...}. WM deliberately misreads one more time. That is NOT what was said. If one regards a path as a set of nodes then one can form the union of a nested sequence of paths to form a path in the WM psuedo-union of the containing trees. But the union of two or more sets of paths is merely a set of paths and not itself a path, and more than the union of two sets of dogs is a dog. > > Right. As I have stated all the time. And so the set of paths in the > > complete tree is *not* a subset of the union of all finite sets of > > finite paths, something you have stated repeatedly. > > The union of all finite paths of a special kind contains the due > infinite path. The union of sets of dogs is not a dog. > I state that N is potentially infinite. If N represents a set, then it is either finite or not finite. If it is not finite, then it is actually not finite, not merely potentially not finite.
From: Virgil on 4 Feb 2007 14:28 In article <1170606676.336257.302350(a)m58g2000cwm.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 2 Feb., 21:01, Virgil <vir...(a)comcast.net> wrote: > > > > Why should it? But it can be able. For instance it can prove that > > > every set of natural numbers is finite while the size of sets of > > > natural numbers is unbounded from above. > > > > Not so. Induction can only prove every set of naturals which is bounded > > above by a natural is finite, but that does not, according to the axiom > > of infinity, exhaust the possible sets of naturals. > > There is no natural number in this set which is not covered by > induction. So, which number is missing to exhaust N? Who said anything like that? WM misreads so consistently that he must be doing it intentionally. > > > What Induction says is precisely: > > Given a set, S, of natural numbers such that > > (a) S contains the first (smallest) natural, and > > (b) whenever a particular natural is a member of S, > > the successor of that natural is also a member of S, > > then every natural is a member of S. > > > > And induction says nothing else. > > Induction says: If P holds for n then it holds for n+1. On the contrary, it says no such thing. It is entirely up to the person attempting to use induction to PROVE that if P holds for n then it holds for n+1. There is nothing in induction to require it. > This is the > main property of an infinite set as the set described by the axiom of > infinity originally was. > Only the dubious interpretation of esoteric > infinity removes this set from the realm of induction and from the > realm of mathematics too. The only esoteric version of infinite (not finite) is WM's. In standard set theory, a set is either finite or not finite with no third option. > > > > > Of course every set of natural numbers is finite, as proved > > > by induction over all initial segments. Actually every natural is finite by definition, but there is no inductive proof that every set of naturals is finite because it isn't treu, inductively or otherwise, at least for any standard definition of finiteness of a set. What is WM's definition of a set being finite anyway. He has often been asked for it, but I do not recall his ever giving it. > > > > That only proves that such initial segments as have a last element are > > finite. It says nothing at all about any initial segment which does not > > have a last element, and there is one. > > Every finite natural defines a finite segment. An infinite segment can > be defined by an infinite number only, and there is none. Omega is an infinite ordinal which defines an infinite set of ordinals, those preceding it. Sp WM is wrong again. > > ================================== > > > The set of all sets with exactly 4 elements is not a subset of the set > > of all sets with exactly 5 elements since the set of all sets with > > exactly 5 elements does not contain any sets with less that five > > elements. > > 4 is also every set with 4 elements. Every set with 4 elements is a > subset of that set where one further element has been added. WM is conflating incompatible definitions of cardinality here. The cardinal of a set can be a representative set which bijects with the given set, as in NBG, or it can be the class of all sets bijectable with the given set, as in NF, but it can't be both simultaneously, as the two theories are incomaptible. > > > WM must have completely lost touch with reality to think otherwise. > > Contrary. Just because I have not lost touch with reality. When one conflates equivalence classes with representative elements, as WM has done above, wherever he is, it is not reality. > > > The "mess" is only in the minds of those who imagine it into existence. > > > For those of us who perceive no mess there is no mess. > > That is an excellent description of your position. The alleged "mess" is only perceptible to those who assume as axioms things that create such a mess. To those who are more careful of their axioms, there is no such mess. > > > In Peano, you do not have to know what any of the things are, all you > > have to know is that they work according to the rules stated. > > You do not have to know (or to be able to get to know at least) > whether or not one natural number is larger than the other? Unless "largeness" is defined in the rules, no! > > > If you don't know at all what you're looking for, you can't ever tell > > whether you've found it or not. > > Therefore I accept and reported in my book the definitions by Peano > and Dedekind. Who knew what they were looking for, and found it. If WM's book at all represents the ideas he tries to fob off here, he should retract it.
From: William Hughes on 4 Feb 2007 14:33
On Feb 4, 1:01 pm, mueck...(a)rz.fh-augsburg.de wrote: > On 4 Feb., 15:31, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > > Similarly one can prove by induction: Every set of even positive > > > numbers contains numbers which are larger than the cardinal number of > > > the set. > > > No. You make your usual mistake. > > No, you make your usual mistake by thinking that a potentially > infinite set can have an infinite cardinal number. I make no assumptions whatsoever about the cardinal number of a potentially infinite set. (I do not even assume that one exists). > > > Let E be the potentailly infinite set of even numbers. > > Two statements > > > i: For every element e of E, the set > > F(e) = {2,4,6,...,e} contains cardinal numbers which are > > larger > > than the cardinal number of F(e) > > > ii E contains numbers which are larger than the cardinal > > number of E > > > Statement i is true and can be shown by induction. Statement > > ii is false. > > Of course, My claim is that statment ii is false. Your reply is "of course". We both agree that statement ii is false. > namely because there is no cardinal number for potentially > infinite sets. > > > The fact that statment i is true does not mean > > that statement ii is true. > > The fact that statement I is true means that there are no elements of > the set which could increase its "number of elements" without > increasing the sizes of elements in the set. As statement ii does not say or imply anything about increasing the "number of elements" in E, totally irrelevent. There is no element that can be shown to be in E that is not in some F(e), but you need more than one F(e) . There is no single F(e) that contains every element that can be shown to be in E. The fact that there is no element in E that is not in some set F(e), does not mean that E has the same properties as the sets F(e). The fact that statement i is true does not imply that statement ii is true. Statement i is true, statement ii is false (note your "of course" above). The statement "Every set of even positive numbers contains numbers which are larger than the cardinal number of the set", is false. The counterexample is the potentially infinite set E. - William Hughes |