Cantor Confusion
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From: Virgil on 19 Dec 2006 23:39 In article <1166545513.802845.88170(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > [...] > > >> L_D is a line and is therefore bounded. > > >> > > >> Contradiction. > > >> Hence, the _assumed_ L_D does not exist. > > >> > > > Assumed is the diagonal. Premise: "If the diagonal is unbounded." > > > L_D is a line and is therefore bounded. > > > > Yes. And because L_D is bounded it cannot "contain" every element of the > > diagonal. Exactly this constitutes the contradiction to your claim P1. > > Wrong conclusion. Because every line (call it L_D or as you like) is > bounded the diagonal cannot be actually infinite. Wrong, and stupidly wrong! In any infinite binary tree, which WM has conceded one can have, it is easy to construct a model of WM's EIT which has infinitely many finite lines and an actually infinite diagonal.
From: G. Frege on 19 Dec 2006 23:59 On Tue, 19 Dec 2006 20:46:58 -0700, Virgil <virgil(a)comcast.net> wrote: >>> >>> Math, at least in ZFC or NBG, says there are infinitely many finite >>> values for each side of "1+1+1+...+1 = n". >>> >> Precisely that's a good reason why this insane theory should be >> abolished. (WM) >> >> > [...] I said that there are infinitely many equations of form "1+1+1+...+1 > = n" with equal finite values on each side, but made no claim of any such > equation with any infinite number on either side. > WM is not able to grasp the concept that there might be a set of infinitely many "finite" natural numbers: "If a set consists of different finite whole numbers only, then it has necessarily a finite number of members and, hence, is a finite set." (W. M�ckenheim) F. -- E-mail: info<at>simple-line<dot>de
From: Newberry on 20 Dec 2006 00:39 mueckenh(a)rz.fh-augsburg.de wrote: > Newberry schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Newberry schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Newberry schrieb: > > > > > > > > > > > Sorry that I joined a bit late. > > > > > Doesn't matter. > > > > > > > > > > >Are you saying that (in an infinite > > > > > > binary tree) the set of paths is uncountable but the set of edges is > > > > > > countable? > > > > > > > > > > The set of edges is obviously countable by, e.g., > > > > > > > > > > 1, 2, > > > > > 3,4,5,6, > > > > > 7,... > > > > > > > > > > As no path can separate from another one without the existence of two > > > > > more edges, the number of edges is an upper bound for the number of > > > > > paths. > > > > Is this formally provable in ZFC? > > I don't know. Up to now nobody has found a way. But also nobody has > found a contradiction in the tree. > > > > > > > > > > If it is so simple why do you still need the proof by inheritance of > > > > shares? > > > > > > I don't need it. But there are others who do not see the forest because > > > of too many trees blocking the view. > > > > Going back to the shares ... I can sort of see that each path will > > acquire enough shares to build two edges. What is supposed to follow > > from that? > > It follows that the cardinal number of paths is not larger than the > cardinal number of edges. How does it follow? >The edges have cardinal number aleph_0. Certainly. > A > subset of the paths can be shown to be in bijection with the real > numbers of the interval [0,1]. Which subset? Path as a set of intervals or path as a set of eges? How does the above follow from this? > > Regards, WM
From: Albrecht on 20 Dec 2006 03:39 William Hughes schrieb: > Albrecht wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > Franziska Neugebauer schrieb: > > > > > > > > > > > > > >> > > > All elements that can be shown to exist in the diagonal can be > > > > > >> > > > shown to exist in one single line. [(P1)] > > > > > > > > > > This proposition P1 has _not_ yet been proved (shown). > > > > > > > > This proposition is the definition of the diagonal. > > > > > > > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > > > > contained in the diagonal. > > > > > > > > > > The Diagonal is unbounded thus any _assumed_ L_D is not bounded, too. > > > > > > > > Correct is: > > > > If the diagonal is unbounded then any _assumed_ L_D is not bounded, > > > > too. > > > > > > > > > L_D is a line and is therefore bounded. > > > > > > Contradiction. > > > Hence, the _assumed_ L_D does not exist. > > > > > > - William Hughes > > > > > > - > > > William Hughes > > > > It's always the same game you play: you compare numbers and sets. > > Numbers are always complete, sets are complete only if they are finite. > > Now you claim that the set of the natural numbers contain all natural > > numbers and hence it is complete > > No, the claim is that it is potentially infinite (given > any finite set of natrual numbers one can allways find one more). That's okay. You can say: given *any* set X of natural numbers, one can allways find at least one natural number which isn't element of the set X. > No assumption of completeness is made or used. > (here "complete" is a property of a set or a potentially infinite > set. We say X is complete, if assuming X exists means that > we can assume that all elements of X exist). There are no potentially infinite sets. You may have a potentially infinite sequence of sets. You again mix up states (sets) and processes (algorithm) > > > > but in the same time |N isn't complete > > since |N is bijectable to the diagonal and the diagonal is never > > complete since there is no complete line (number) which covers the > > diagonal. > > No, the claim is that there is no line that covers the diagonal. > > - William Hughes We know, there are infinitely many natural numbers. We know also that there is no number amongst this numbers with the form ....11111. Why? It's easy to say: Since there is no number of the form ...11111 there is no quantity which elements has the quantity ...111111. Since there are only quantities with numbers and since numbers, in there easiest form, are also quantities (like 1, 11, 111, 1111, ...) there is no infinite number ...11111 with infinite many "1" since there is no quantity with the number ...111111 to number the quantity of the "1" in an infinite number. And so on. Best regards Albrecht S. Storz
From: Franziska Neugebauer on 20 Dec 2006 04:36
Albrecht wrote: > William Hughes schrieb: >> Albrecht wrote: >> > William Hughes schrieb: >> > > mueckenh(a)rz.fh-augsburg.de wrote: >> > > > Franziska Neugebauer schrieb: >> > > > >> > > > > >> > > > All elements that can be shown to exist in the >> > > > > >> > > > diagonal can be >> > > > > >> > > > shown to exist in one single line. [(P1)] >> > > > > >> > > > > This proposition P1 has _not_ yet been proved (shown). >> > > > >> > > > This proposition is the definition of the diagonal. >> > > > >> > > > > > You misinterpret L_D. L_D is that line which contains all >> > > > > > numbers contained in the diagonal. >> > > > > >> > > > > The Diagonal is unbounded thus any _assumed_ L_D is not >> > > > > bounded, too. >> > > > >> > > > Correct is: >> > > > If the diagonal is unbounded then any _assumed_ L_D is not >> > > > bounded, too. >> > > >> > > L_D is a line and is therefore bounded. >> > > >> > > Contradiction. >> > > Hence, the _assumed_ L_D does not exist. >> > > >> > > - William Hughes - >> > > William Hughes >> > >> > It's always the same game you play: you compare numbers and sets. >> > Numbers are always complete, sets are complete only if they are >> > finite. Now you claim that the set of the natural numbers contain >> > all natural numbers and hence it is complete >> >> No, the claim is that it is potentially infinite (given >> any finite set of natrual numbers one can allways find one more). > > That's okay. You can say: given *any* set X of natural numbers, one ^^^^^^^^^ *any* _finite_ set > can allways find at least one natural number which isn't element of > the set X. The set N is a set of natural numbers, but there is no natural number wich is not element of N. Hence your claim is wrong. >> No assumption of completeness is made or used. >> (here "complete" is a property of a set or a potentially infinite >> set. We say X is complete, if assuming X exists means that >> we can assume that all elements of X exist). > > There are no potentially infinite sets. > You may have a potentially infinite sequence of sets. You contradict yourself: Sequences "are" sets so it is valid that from the existence of potentially infinite _sequences_ the existence of potentially infinite _sets_ follows. > You again mix up states (sets) and processes (algorithm) There are no "processes" at all. >> > but in the same time |N isn't complete >> > since |N is bijectable to the diagonal and the diagonal is never >> > complete since there is no complete line (number) which covers the >> > diagonal. >> >> No, the claim is that there is no line that covers the diagonal. >> >> - William Hughes > > > We know, there are infinitely many natural numbers. Do "we"? > We know also that there is no number amongst this numbers with the > form ....11111. This is not a number but at most a representation. > Why? Because every n e N has a representation of the form x...x and not of the form ...x. > It's easy to say: Since there is no number of the form ...11111 there > is no quantity which elements has the quantity ...111111. You don't need metaphysical notions like "quantity" to simply state the fact that the "conversion" of any n e N into its representation never yields ...111. This is a purely formal issue. > Since there are only quantities with numbers and since numbers, in > there easiest form, are also quantities (like 1, 11, 111, 1111, ...) > [...] Circular reasoning may indicate ignorance. F. N. -- xyz |