Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: mueckenh on 30 Dec 2006 16:37 Dik T. Winter schrieb: > > > But whatever, you stated that a definable well-ordering of the reals would > > > lead to a contradiction in ZFC. But you have nowhere shown any proof of > > > that, and it is false. It has been proven that a definable well-ordering > > > of the reals is consistent with ZFC. > > > > Why don't you simply give the formula for the definable well ordering? > > The formula could be evaluated, and, step by step, we would get a list > > of all reals. > > Wrong. It would not give a list. It would give a well-ordering. Do you > not know the difference? Every non empty set of a well-ordering has to have a first element. If all reals are well-ordered, then all reals are first elements in some sets. > And I doubt whether you could apply the formula > one by one to each real. I doubt that too, in fact it is impossible in ZFC. Therefore, there is no well-ordering of all the reals. ========================= > > Because non-distinguishable sets are not different sets. You just > > proved that there is no element of a path which distinguished it from > > every other path. >Right. But that does *not* mean that it can not be distinguished from every > other path. If every edge of a path is shared by another path, how would you distinguish them? ======================== > Well, I understand why he [Leibniz] is your favourite. He is not my special favourite. But he was very wise and he said: The arithmetic of the infinite has to be based upon the arithmetics of the finite. Nach LEIBNIZ sollte das Unendliche dieselben Rechenregeln wie das Endliche befol¬gen: Die Regeln des Endlichen behalten im Unendlichen Geltung, wie wenn es Ato¬me (Elemente der Natur von angebbarer fester Größe) gäbe. (Kontinuitätsprinzip). All you know of before encountering mathematics is finite. Therefore you cannot start with the infinite. ============================ > Again, a finitistic view. There are algorithms that do calculate that > number. [pi*10^10^100] Is it even or odd? > That they can not be implemented in a finite world is of no > concern to the mathematician. On the other hand, there *is* a natural > number that is equal to that value. Is it even or odd? > But apparently you also eschew the proof that li(x) and pi(x) cross > each other infinitely often, because it is impossible to even calculate > the first cross-over. I do not "eschew" the proof for li(x) and pi(x). These "numbers" seem to eschew existence. Regards, WM
From: mueckenh on 30 Dec 2006 16:42 hagman schrieb: > mueckenh(a)rz.fh-augsburg.de schrieb: > > > cbrown(a)cbrownsystems.com schrieb: > > > > > Have you given up on your "rational relation" proof that |N| = |R|? > > > > No, why should I? It is correct. (A series with a first but no last > > term can be reversed to have a last but no first term - without oosing > > its value.) > > > > But it is easier to see, and should be visible even for such as you, > > that the tree built from the union of all finite trees (with > > representations of rational paths) is the same as the complete infinite > > tree. > > > > Regards, WM > > and while a finite binary tree with n levels has > 2^n-1 vertices > 2^n-2 edges > 2^(n-1) leaves > 2^(n-1) paths starting at the root and ending in a leaf, > the union of these finite trees is an infinite tree with > |N| vertices > |N| edges > 0 leaves > |R| paths starting at the root and dashing off to nirwana. > > What shall this demonstrate? > a) Because #edges >= #paths for all finite subtrees, the same relation > holds for the union (i.e. |N|>=|R|)? > b) Because #paths=#leaves for all finite subtrees, the same relation > holds for the union > (i.e. |R|=0)? > c) Because #vertices<2*#leaves for all finite subtrees, the same > relation holds for the union > (i.e. |N|=0)? Do you think that the union of all finite subtrees is identical with the complete infinite tree? If not, what is the difference (in terms of edges or nodes)? If yes, where are the paths representing irrational numbers? Regards, WM
From: mueckenh on 30 Dec 2006 16:49 Virgil schrieb: > In article <1167492220.138771.111750(a)h40g2000cwb.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > cbrown(a)cbrownsystems.com schrieb: > > > > > Have you given up on your "rational relation" proof that |N| = |R|? > > > > No, why should I? It is correct. (A series with a first but no last > > term can be reversed to have a last but no first term - without oosing > > its value.) > > How does one "sum" a series with no first term? There is no place to > start! There is a place to start, namely at the end, because there is a last term. Did you ever hear of an order omega^star (w*). Why should it be impossible to reflect a sequence or the terms of a series and nevertheless maintain the value? > > > > But it is easier to see, and should be visible even for such as you, > > that the tree built from the union of all finite trees (with > > representations of rational paths) is the same as the complete infinite > > tree. > > In the union of all finite trees, there are no infinite paths at all, No? Where do they end? You should be able to answer this question and to prove your answer. Hint: In the union of all finite initial segments of N there is an infinite initial segment, namely N (so some people claim at least). > since to be represented in the union, a path would have to be in one of > the finite trees, ergo a finite path. > > In the same way, the union of any family of sets of finite naturals does > not contain anything but finite naturals. The question is not whether the edges have a finite distance from the root, but if there are infintely many of them. (Compare the irrational numbers with only finite indexes but, as you claim, infinitely many of them.) Regards, WM
From: William Hughes on 30 Dec 2006 17:12 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > > As not all natural numbers do exist, the set is potentially infinite, > > > > > i.e., it is finite. It has a maximum. L_D. Taking this maximum and > > > > > adding 1 or takin L_D ^ L_D or so yields another maximum. In any cae > > > > > there is a maximum. > > > > > > > > No. At any time there is a maximum of the numbers that have been > > > > shown to exist. > > > > > > So it is. > > > > > > > However, there is never a maximum of the numbers > > > > that it is possible to show exist. > > > > > > It is possible to show: > > > When these numbers will exist, the due maximum will exist. > > > That's enough. > > > > > > > L_D does not only have to include numbers that have been shown > > > > to exist, it must include all numbers that it is possible to show > > > > exist. > > > > > > Why? How should we know which can possibly exist? > > > > L_D must contain the diagonal. > > Thus, L_D must contain any element that > > can be shown to exist in the diagonal. > > It does, after the existence has been shown. However, before existence has been shown it does not. If L_D contains the diagonal then it contains any element that can be shown to be in the diagonal whether or not existence has already been shown. It must contain the element before existence has been shown. We cannot know about all the elements that can be shown to exist. However, a single example suffices. There is an element of the diagonal, that has not yet been shown to exist, but which can be shown to exist which is not an element of L_D. Therefore, L_D does not contain the diagonal. - William Hughes
From: Virgil on 30 Dec 2006 18:20
In article <1167513952.467758.82640(a)48g2000cwx.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > > I did not calculate |N|/|R| but edges per path. > > > > Wonderful. But what relevance does the calculation of "edges per path" > > have to do with the /actual question/: does there exist a surjective > > function from the set N to the set R? > > If there are not less edges than paths, then a surjection is possible That this is the case for every finite tree does not imply that it is the case for any infinite tree. And it has been proven not the case for infinite binary trees. > does there exist a > surjective function from the set N to the set R? NO! AS has been amply demonstrated. If WM disagrees. Let him demonstrate by constructing such a surjection. > > > > In the latter case, we are not asking "is there a surjective function > > from 1 + 1/2 + ... onto 1?", because that would be simply nonsense. > > Only if one does not know the meaning of the arithmetic of rational > numbers and limits. Sums of rational numbers are nothing else but sums > of integers, given in the unit of the greatest denominator. What about sum_{n=1..oo} 1/n^2, which converges but for which there cannot be any common denomiator for all terms. Similarly for sum_{n=1..oo} 1/n!, where n! means t=the factorial of n. Similarly for sum_{n=1..oo} (-1)^n/p_n, where p_n indicates the nth prime. > > Similarly, to try to show that |N| >= |R| by calculating the limit of > > edges per path is equally nonsense - it requires misreading the meaning > > of ">=". |