Cantor Confusion
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From: mueckenh on 14 Oct 2006 10:00 MoeBlee schrieb: > Han de Bruijn wrote: > > Set Theory is simply not very useful. The main problem being that finite > > sets in your axiom system are STATIC. They can not grow. > > Set theory provides for capturing the notion of mathematical growth. > Sets don't grow, but growth is expressible in set theory. If there is a > mathematical notion that set theory cannot express, then please say > what it is. Obviously the notion of "rational relation" as used in the binary tree cannot be expressed by mathematical notion: Consider the binary tree which has (no finite paths but only) infinite paths representing the real numbers between 0 and 1. The edges (like a, b, and c below) connect the nodes, i.e., the binary digits. The set of edges is countable, because we can enumerate them 0. /a \ 0 1 /b \c / \ 0 1 0 1 .............. Now we set up a relation between paths and edges. Relate edge a to all paths which begin with 0.0. Relate edge b to all paths which begin with 0.00 and relate edge c to all paths which begin with 0.01. Half of edge a is inherited by all paths which begin with 0.00, the other half of edge a is inherited by all paths which begin with 0.01. Continuing in this manner in infinity, we see that every single infinite path is related to 1 + 1/2 + 1/ 4 + ... = 2 edges, which are not related to any other path. The set of paths is uncountable, but as we have seen, it contains less elements than the set of edges. Cantor's diagonal argument does not apply in this case, because the tree contains all representations of real numbers of [0, 1], some of them even twice, like 1.000... and 0.111... . Therefore we have a contradiction: Card(R) >> Card(N) || || Card(paths) =< Card(edges) Regards, WM
From: William Hughes on 14 Oct 2006 10:03 mueckenh(a)rz.fh-augsburg.de wrote: > Alan Morgan schrieb: > > > In article <990aa$452e542e$82a1e228$16180(a)news1.tudelft.nl>, > > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > >Dik T. Winter wrote: > > > > > >> In article <1160646886.830639.308620(a)c28g2000cwb.googlegroups.com> > > >> mueckenh(a)rz.fh-augsburg.de writes: > > >> ... > > >> > If every digit position is well defined, then 0.111... is covered "up > > >> > to every position" by the list numbers, which are simply the natural > > >> > indizes. I claim that covering "up to every" implies covering "every". > > >> > > >> Yes, you claim. Without proof. You state it is true for each finite > > >> sequence, so it is also true for the infinite sequence. That conclusion > > >> is simply wrong. > > > > > >That conclusion is simply right. And yours is wrong. Completed infinity > > >does not exist. > > > > sqrt(-1) doesn't exist either. Frankly, I have a much harder time > > believing in "imaginary" numbers than I do believing in infinite > > sets. > > But sqrt(-1) does not yield contradictions, as far as I know. And as infinite sets do not yield contradicitons, this statement is completely beside the point. - William Hughes > > Regards, WM
From: mueckenh on 14 Oct 2006 10:04 William Hughes schrieb: COLLECTED ANSWERS > Once you have chosen your set of 100 elements it will have > a largest size. Questions as to the method of choice have > no bearing on this. > Correct. Once you have chosen how to use all the bits of the universe, then you will have a largest number. Infinty does only come into the play as long as this choice has not yet been fixed. > It is true that in some cases the quantifier exchange is possible. However, the fact that the quantifier exchange is impossible in general means that you cannot use quantifier exchange in a proof without explicit justification of the step. Despite you protestations, this is exactly what you do. Despite of your belief I have proved that exchange is possible in linear sets with only finite elements. If you state a counter example, I can reject it by disproving it. You cannot, but only claim that for an "infinite set" your position was correct - of course without being able to show any infinity other than by the three "...". > Your putative proof of this "fact" depends on a step in which quantifiers are switched without justification. Give a proof of that by a counter example using any concrete natural numbers. > This requires proof. Try to produce one without any unjustified quantifier exchange. I did never employ unjustified quntifier exchange, but if you dislike that example, then consider the binary tree which i just posted another time. Regards, WM
From: William Hughes on 14 Oct 2006 10:06 mueckenh(a)rz.fh-augsburg.de wrote: > Tony Orlow schrieb: > > > > > > > Why shouldn't it? If every digit position of 0.111... is a finite > > > position then exactly this is implied. Your reluctance to accept it > > > shows only that you do not understand how an infinite set can consist > > > of finite numbers. In fact, nobody can understand it, because it is > > > impossible. > > > > > > Regards, WM > > > > > > > But Wolfgang, surely that consideration does not impact, say, the set of > > reals in (0,1], which are all finite, yet whose number is infinite. It > > is not a requirement that a set of all finite values be finite. That > > conclusion follows from the combination of that fact with the fact there > > is a constant positive unit difference between consecutive elements. > > Of course, Tony, you are right! But only a finite number of real numbers will every be described in the lifetime of the universe. Surely by your reasoning there must be a finite number of real numbers? - William Hughes > > Regards, WM
From: mueckenh on 14 Oct 2006 10:06
MoeBlee schrieb: > If we're still talking about the diagonal argument for the > uncountability of the reals, then there's no "self-reference" anyway. > > The proof is good from an effectively decidable set of axioms using > effectively decidable rules of inference. So if one claims that there > is anything objectionable in the proof, then one should just say which > axioms and/or rules of inference one rejects. Any other dispute with > the mechanics or details of the proof is mindlessness. The rules to cover up with infinity are objectionable. In decimal representations of irrational numbers the infinite string of digits leads to an undefined result unless the factors 10^(-n) are applied. In Cantor's diagonal proof each of the elements of the infinite string is required with equal weight. Regards, WM |