An uncountable countable set
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From: mueckenh on 29 Aug 2006 05:08 Dik T. Winter schrieb: > Let us remove omega from that set. What is the resulting cardinality? Not an actually infinite one, because without omega there are only finite natural numbers. > As there is a bijection between that set and the set including omega, their > cardinalities are the same. No. Without omega there is no actual infinity. > > > Your example shows very clear that either both LUBs have to be included > > - or none. In my opinion we choose none. But whatever you choose, you > > cannot avoid to see the symmetry, can you? > > I see the symmetry, I also choose none. And I think I have stated that > already. Fine. That means there is no actual infinity. > > > > > As long as we are in the naturals we know: Each natural is the sum of > > > > all preceding differences. Infinitely many differences require an > > > > infinite sum. > > > > > > Wrong. In mathematics the concept of "infinite sum" is not defined, > > > > But if there are infinitely many naturals then the infinite sum is > > necessary. > > Not at all. Why do we need it? You should know that the n-th natural number is defined by the sum of the number n-1 plus 1 (Peano). This can be retraced to the sum of n 1's. What is there to be defined? If there are infinitely many naturals (and if infinity is a number), then there is at least one infinite sum of 1's (one from each natural). > > OK, I kind of understand. I do not state that infinity is a number aleph_0. > There exists a host of infinities. And aleph_0 is the infinity that gives > the equivalence class of sets that are in bijection with the set of > natural numbers; the canonical set with cardinality infinity. If aleph_0 > n for n e N, and if there are even numbers x > aleph_0, then aleph_0 is a number. Cantor stated that. If you do not do so, then we are in agreement in that point. > > > > The only connection I know of is where it is used in connection > > > with real or complex numbers. And it can also denote some norm > > > (in general Euclidean norm) in vector spaces. And as the finite > > > cardinals equate to the non-negative integers, that equate to their > > > magnitude, you might consider a cardinal number to represent its own > > > magnitude. Do you mean that? > > > > Exactly. And in addition, as long as the numbers are finite, ordinal > > and cardinal is the same size. That is just why the cardinal cannot be > > infinite as long as the ordinal is finite (as expressed by "the > > infinite number of finite numbers"). > > But indeed. If the ordinality of a set if infinite (w or larger), then > so is its cardinality (aleph-0 or larger). On the other hand, if the > cardinality of a set is infinite (aleph-0 or larger), and if it can be > well-ordered, so is its ordinality after well-ordering (w or larger). As long as the set of natural numbers includes only finite numbers, its cardinal number is also finite. Regards, WM
From: mueckenh on 29 Aug 2006 05:10 Dik T. Winter schrieb: > In article <1156585631.775313.153510(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > 1/3 is that path which turns left, right, left, right, left, right, ... > > > > and never ceases to change its direction. > > > > Which edge do you miss? Which dge could not be enumerated? > > > > > > As in your tree, by definition of the tree, each edge terminates at a node, > > > what is the number of the edge that terminates at 1/3? What is the > > > number of the edge that terminates at 1/5? > > > > I don't understand these questions. 1/3 and 1/5 are infinite paths > > consisting of infinitely many edges. > > You state that you can number the edges. Each edge terminates at a node, > and if 1/3 and 1/5 are in your tree, there must be edges that terminate at > those nodes. What do you mean with "those nodes"? 1/3 and 1/5 are not nodes but they are paths. >In that case you should be able to tell me what the numbers > of those edges are. Or do you now claim that there is *no* edge that > terminates at those nodes? If so, where do the nodes come from? Out of > thin air? I did already tell you which edges belong to the path 1/3. The nodes can be enumerated like the edges. > > > > As the set of all edges is countable, you > > > should be able to state that. The problem with your counting method of > > > edges (count first level 1, next count level 2, etc.) you will already > > > have exhausted all natural numbers by counting all edges that are at the > > > end of finite paths. > > > > There are no finite paths and there are no ends at all in my tree. > > During your construction process all paths are finite, There is no construction process. The tree exists like the real numbers do exist. > and there is an > edge that terminates at a node terminating the path. I meant the paths > during the intermediate process of construction. There is no process of construction. If you have difficulties to comprehend that: it is the same as with Cantor's list. The tree is defined once and for all. That's it. > > > If > > you think that 0.1 would be represented by a finite path, what should > > then be represented by the path 0.1000...? > > The same. > > > By your arguing you would never be able to count the digits of 1/3 > > because "you will already have exhausted all natural numbers by counting > > all edges that are at the end of finite" sequences. > > Wrong. To count digits we need only one number for each level. A set of countably many countable sets is countable. > > > > > Do you believe that the set of edges of 1/3 is uncountable? > > > > > > That is not the set of all edges in the tree. > > > > No. But would you say that the set of all edges of 1/3 and of 1/5 and > > of 1/7 were uncountable? > > No. Assuming of course that 1/3, 1/5 and 1/7 are in your tree. All real numbers of the interval [0, 1] are there, some of them even twice. > > > How many paths are required to obtain an > > uncountable set of edges, according to your opinion? > > Depends quite a bit on a host of questions. In your tree with terminated > edges, also the paths are terminated, so neither becomes uncountable. > But also 1/3 is not in that tree. You seem to misunderstand the tree. If the diagonal is in Cantor's list, then 1/3 is in my tree. Can you give a reason why it should not (other than that then set theory is inconsistent)? You recently mentioned an interesting aspect: The algebraic numbers, i.e., the polynoms are countable, because they are finite. And, therefore, you wanted only to count the finite segments of paths in my tree. I oppose, unless you agree that the 1's in 0.111... also are uncountable. You see, the problem is the same: We can count finite segment but not infinity. Either you agree that all the edges of my tree are countable or you agree that Cantor's diagonal is uncountable. Or you state at least how many infinite path in my tree you would consider to have completely countable edges. Regards, WM
From: mueckenh on 29 Aug 2006 05:12 Dik T. Winter schrieb: > In article <1156609759.075864.266590(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1156364007.547996.270100(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > > I do not think so. With blocks of height 1/2^n and width 1 - 2^n, > > > > > after k steps the total height is 1 - 2^k and the total width > > > > > 1 - 2^k. When we complete we get the limiting case. But there is > > > > > still no block at either height 1 or with width 1. > > > > > > > > > > And there is no smaller container that contains the complete stair. > > > > > > > > According to Cantor: The infinite set of finite numbers. aleph_0 is > > > > actually infinite, no natural number is actually infinite. The > > > > staircase has width [0, 1] and height [0, 1). It is a difference. > > > > > > I do not see a proof in your statements. The staircase is within > > > [0, 1) and [0, 1) after completion. How is that in contradiction with > > > what Cantor states? > > > > Cantor states (analogously) that *all stairs exist*, that the width of > > all ot them is L, but that none of them has hight L. Width [0, 1] and > > height [0, 1). > > A quote please. But indeed, the width of all of them is L, but there is > none of them with width L. And the heighth of all of them is L, but there > is none of them with heighth L. So width [0,1] and heighth [0,1]. That > is, if you define both as the smallest box containing the completed stair. > If you define both as the largest numbers that can be obtained, you get > width [0,1) and heighth [0,1). I agree with you. But Cantor said (Werke, p. 409): "So stellt uns beispielsweise eine veränderliche Grö�e x, die nacheinander die verschiedenen endlichen ganzen Zahlwerte 1, 2, 3, ..., �, ... anzunehmen hat, ein potentiales Unendliches vor, wogegen die durch ein Gesetz begrifflich durchaus bestimmte Menge (�) aller ganzen endlichen Zahlen � das einfachste Beispiel eines aktual-unendlichen Quantums darbietet. " And I am arguing against this actual understanding of infinity. Regards, WM
From: mueckenh on 29 Aug 2006 05:14 Dik T. Winter schrieb: > > > > > It is your lack of a proper proof that if each digit of a number can > > > > > be indexed that number can be covered. And such a proof does not > > > > > exist. > > > > > > > > I do not see how I could avoid my conclusion. But if you are so sure > > > > then give us at least one example how you completely index a number > > > > without covering all the preceding numbers. > > > > > > I have done so many times, and am doing it here again. > > > > You gave an example how a number of the form 0,111...1 with n digits > > indexes the n-th digit but does not cover all digits with m =<n ??? > > No, because that does not exist. How then can you believe and assert that the number 0.111... which has *only finitely indexable positions* could be completely indexed but not covered. Regards, WM
From: mueckenh on 29 Aug 2006 05:17
Dik T. Winter schrieb: > In article <1156689126.335154.133150(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > The first part of his first proof shows that a complete ordered field > > > > > has cardinality larger than the natural numbers. In his proof he did > > > > > not rely an any properties of the reals other than that they form a > > > > > complete ordered field (he uses reals to exemplify). > ... > > > Note what I wrote: "in his proof he did not rely on any properties of > > > the reals other than that they form a complete ordered field". What > > > other properties of the reals did he use? > > > > He had no field and he used no filed. What is a field without the > > axioms of the field? Nothing. Cantor disliked axioms if he did not hate > > them. His only concern were numbers, numbers, numbers and their > > *reality*. No fields and no axioms. > > Perhaps. I thought we were discussing current set theory. And not what > in some time long ago was et theory. He may not have liked them, but in > current terminology "the first proof shows that a complete ordered field > has cardinality larger than the natural numbers". In the terminology of the next century it may show even other things, more general perhaps, with even more insight. I was discussing what Cantor did. You were accusing me that I had not understood or misquoted him. > > > > > He wrote that it was a simpler proof for his first theorem, i.e. the > > > > theorem that the reals are uncountable. > > > > > > And he did *not* write that. > > > > Aus dem in =A7 2 Bewiesenen folgt n=E4mlich ohne weiteres, da=DF > > Quoting again only in part. I will quote the translation I gave, with > annotation of the complete paragraph, not only the part you like to quote: > In the article, titled: "Über eine Eigenschaft des Ombegriffs aller > reellen algebraischen Zahlen (Journ. Math. Bd. 77, S. 258) [hier > S. 115], can for the first time be found a proof of the theorem > that there are infinite sets that are not in bijection with the set of > natural numbers 1, 2, 3, ..., v, ..., or, as I say in general, do not > have the same cardinality as the natural numbers 1, 2, 3, ..., v, ... . > So apparently he is thinking that his first article proves the theorem > that there are sets that are not in bijection with the natural numbers. > Which is true. Of course: The set of the reals and the set of the transcendental numbers are such sets. So he has the right to speak generally about "sets". Which mathematician would not like to generalize his theorems? > From what has been proven in section 2 follows that e.g. the > set of real numbers in an arbitrary interval can not be put in > a sequence w_1, w_2, w_3, ..., w_v, ... . > So he is now thinking that that proof was just an example for such a set. > Which it is. The "beispielsweise" is telling. > It is however possible to construct a much simpler proof for > that theorem that is independent from the observation of irrational > numbers. Independent of the *necessary use* of irrational numbers. > And the "that theorem" (in German "jenem Satze") can, in my opinion only > refer to the theorem mentioned in the first sentence in this paragraph. That theorem the proof of which depends on irrational numbers. Regards, WM |