Cantor Confusion
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From: Virgil on 6 Feb 2007 14:46 In article <1170754878.219127.327100(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 4 Feb., 19:25, Virgil <vir...(a)comcast.net> wrote: > > > > > > The pathlength is not a natural number but it is a number (by > > > > > definition), namely omega. You see that there is no infinite set N > > > > > without an infinte number in it. > > > > > > As omega is not member of N I don't see that. > > > > > The total pathlength is the union of all single pathlengths. > > > > If all those separate pathlengths are finite ordinal numbers and there > > are more than any finite number of them, then the union is the first > > limit ordinal. > > So if there are infinitely many pathes of lengths 1, then the union of > all of them is the first limit ordinal? Are there are more than finitely many path lengths if all path lengths are 1? > You should distinguish between "how many" and "how large". For path /lengths/, more than finitely many requires larger than any finite size. > > > > The total > > > pathlength cannot be omega without omega being a pathlengt (in the > > > union). > > > > False! It is quite possible for a union not to be any one of the sets > > being unioned. > > Interesting. There is some additional ghost? WM must be ghost ridden to see them where they are not. The union over {{2,3},{5,7}} is {2,3,5,7}, which is not a member of {{2,3},{5,7}}. > > > > But omega cannot be in the union of finite pathlengths. > > Therefore there cannot be an actually infinite pathlength. Maybe not in WMtopia, but there is everywhere else. > > > > It can be, and is, everywhere except possibly in WM's weird world. > > The pathlengths are corresponding to the lines in the EIT. If there is > omega as a pathlength then there is omega as a line in the EIT. The "union" of all WM's lines is the non-finite diagonal which has the property that for every line there is a member of the diagonal not in that line. > > > > > "Infinite pathlength" means > > > only that every finite pathlength is surpassed by another finite > > > pathlength. > > > > Which requires a path with no end. > > No. Then name that alleged end of the union of all paths! > In particular because there cannot be such a path. In mathematics there can be such a path, regardless of WM's misrepresentations. > But if you > assume its existence, then you see that the reals are countable or > that identical nodes yield different path systems, an idea which > presumably only you can utter. I do not assume it, I prove it, and I see that the reals are uncountable and WM's pseudo-unions of trees are mathematically corrupt and self-contradictory, and those are ideas that any sane person can utter regardless of WM's presumption.
From: Virgil on 6 Feb 2007 14:48 In article <1170754974.135681.22780(a)p10g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 4 Feb., 19:49, Franziska Neugebauer <Franziska- > Neugeba...(a)neugeb.dnsalias.net> wrote: > > mueck...(a)rz.fh-augsburg.de wrote: > > > On 4 Feb., 14:34, Franziska Neugebauer <Franziska- > > > Neugeba...(a)neugeb.dnsalias.net> wrote: > > >> The path-length of the infinite path is the supremum of the set of > > >> paths-lengths of all finite paths. The latter is the supremum of the > > >> set of natural numbers. This supremum exists. > > > > > No. What does it consist of? > > > > It "consists" of (i.e. contains as members) every and all natural > > numbers, each of which is finite. > > > > > A supremum exists in many places but not here. > > > > Not where? > > > > > The simplest reason is that omega - n = omega for all n in N. > > > > Where did you get that from? Reference? EB? > > You could even figure it out by yourself. If one needs to adopt WM's crazy assumptions in order to come to WM's crazy conclusions, we are all better off without them. Both his assumptions and his conclusions.
From: Virgil on 6 Feb 2007 14:58 In article <1170755634.350929.67010(a)s48g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > The existence of the empty set is not at all guaranteed. It is in ZFC. ZFC requires at least one set to exist, and also requires existence of a subset of that set which does not contain any of the members of that set. > There is an > axiom which requires the existsence of the non existing and seems to > make some people happy If there is such an axiom anywhere, it is only an axiom in WM's system, not in anyone else's. > (like the axiom which requires the finity of > the infinite). Any such axiom exists only in WM's system, and not in anyone else's. Wm seems to be misrepresenting the axiom called the axiom of infinity, which says there is a set S such that {} is a member of S and such that for every member x in S, ( x union {x} ) is also a member of S.
From: Virgil on 6 Feb 2007 15:15 In article <1170756166.580698.67100(a)p10g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 4 Feb., 20:28, Virgil <vir...(a)comcast.net> wrote: > > > > > 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? It is not numbers but sets of naturals of which I spoke. learn to read, WM. > > > > Who said anything like that? WM misreads so consistently that he must be > > doing it intentionally. > > You say: ... but that [Induction] does not, according to the axiom of > infinity, exhaust the possible sets of naturals. So what remains to be > exhausted wich is not exhausted by induction? Sets of naturals that are not defined or definable inductively. > > > > > > > > > > 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. > > If > "If P holds for n then it holds for n+1" > is satisfied, then the set is infinite. Otherwise it is finite. WM's "If P holds for n then it holds for n+1" is valid for all naturals when, for example, P(n) represents "n = m + 1/2 for some natural number m." But there is no first natural, n, for which P(n) holds, and one can see that the set of naturals, n, for which P(n) holds is {}. > > > 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. > > In set theory infinity is always actual infinity, i.e., completed > infinity or esoteric infinity or finished infinity. WRONG! In set theory, there is only finite and not finite and no third option. > > > > 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. > > See above. WM again does not give his definition of finiteness, despite his "See above" remark. What is WM's test for the finiteness of a set? If he cannot give a clear exposition of this notion, perhaps it is because he has no clear notion of what he means by "finite".
From: Virgil on 6 Feb 2007 15:31
In article <1170757099.427483.83050(a)l53g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 4 Feb., 20:33, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > 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). > > You said: "E contains numbers which are larger than the *cardinal > number of E*." See below. > > > > > 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. > > Yes, because there is no cardinal number of E. Does WM claim that there are no ordinals which can be bijected with E? The cardinal of E is, in NBG, merely the smallest such ordinal, and the set of naturals is, as an ordinal set, that very minimal ordinal. > > > > > namely because there is no cardinal number for potentially > > > infinite sets. If one defines cardinals as in NBG, every well ordered set, and E is well ordered, has a cardinality. > > > > > > 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. > > Statement ii is false. every statement beginning with "As statement ii > does " therefore is irrelevant. It is only false in WM's philosophy, which abjures reason and logic to the extent that it is unacceptable in mathematics. > > > 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. > > Wong. There is no element in E which is outside of every F(e). Hence > there are two possibilities: > Either all elements of E exist and there is necessarily one F(E) > or not all elements of E exist, then here is no F(E). WM's quantifier dyslexia misleads him again. TRUE: for every x in E there is a y in E with y > x FALSE: there is a y in E such that for every x in E, y > x. > > > > 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. But that does not require, as WM tries to insist, that there are no ordinals order-isomoprphic to E. > > Potential infinity is in this case the sequence of sets F(e) - there > is no further set E other than this sequence. The union of that sequence is necessarily a set, as are all unions of sequences of sets, and that union /is/ E. > > > The counterexample is the > > potentially infinite set E. > > You seem to be unable to understand what potential infinity means. > Therefore we should stop here. WM seems to be unable to understand anything having to do with mathematics or logic, therefore he should stop here. |