An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Dik T. Winter on 15 Jul 2006 20:11 In article <1152979000.961294.290560(a)m79g2000cwm.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1152884145.033413.104250(a)75g2000cwc.googlegroups.com> muecken= > h(a)rz.fh-augsburg.de writes: .... Let's start with an English translation, makes it quite a bit easier for me (I have not to do two translations with three different languages): Works, p. 214, translated: The question, through which transformations the ordinal number of a well-ordered set is changed, and through which it is not changed, is easily answered, those, and only those transformations leave the ordinal number unchanged that can be rewritten as a finite or infinite set of transpositions, that is, of interchanges of two elements. Agree that this is a reasonable translation? > > No, but still, not *any* set of transpositions is permitted. Only those > > sets of transpositions that are the result of rewriting a transformation > > ("Umformungen"). So it remains interesting what Cantor is meaning when > > he writes "Umformungen". And when he is meaning with that word arbitrary > > re-orderings, he is wrong (because *all* re-orderings can be rewritten as > > an infinite set of transpositions). > > You do completely misunderstand what is defined and what is the > defintion. I think not. > Defined is Umformungen: "diejenigen und nur diejenigen Umformungen = > those and only those transformations" Yes, so that is not a definition of transformations, it restricts the transformations to a subset of them, namely only those transformations that have a particular property. Apparently there are transformations that do *not* have that property. So the question remains, what is the *definition* of "transformations"? (I can easily enough give a transformation that does not have that property at all: transform {1, 2} to {1} by mapping both elements of the first to 1. It can not be re-written as a sequence or set of transpositions. So while it is a transformation, it does not satisfy the requirements to leave the ordinal number unchanged.) So again, what is the definition of "transformation"? If it is unrestricted (and so any re-ordering is allowed), the statement is trivially false. If there are restrictions (and so not any re-ordering is allowed), the statement might be true. But whatever, when the first is true (any re-ordering is allowed as a transformation), the statement is false and you can not use it, because it is not proven. On the other hand, if there are restrictions, you have to show that your set of transpositions form a transformation before you can use the statement. > they are defined by: "welche sich zur=FCckf=FChren lassen auf eine > endliche oder unendliche Menge von Transpositionen, d. h. von > Vertauschungen je zweier Elemente =3D which can be traced back to a > finite or infinite set of transpositions, i.e., interchanges of two > elements." Yes, that is the definition of the property of the transformations that leave the ordinal number unchanged. But the question remains, given any sequence (or set) of transpositions. Is the result a "transformation"? I do not know, because I do not know the definition of "transformation". > > > Of course it is false, whether there is a sequence or not. My > > > transpositions form a sequence. One cannot do better. > > > > Yes, so what? Is yuor set of transpositions a rewriting of some > > "Umformung" in the sense Cantor implies? So there is no "of course" > > here. We need to know what Cantor means when he writes "Umformungen". > > > He says it clearly: those and only those transformations which are > realized by a finite or infinite set of transpositions, i.e., > interchanges of two elements, do not change the Anzahl =3D ordinal > number. Yes, that is a subset of the Umformungen. What is his definition of that term? What does he mean when he writes "Umformungen"? > > > Cantor's conclusion is correct, with no doubt, if infinite sets like > > > the set of my transpositions do exist. > > > > No, when you assume that "Umformungen" means *any* re-ordering, it is > > trivially false if infinite sets do exist. But pray *prove* Cantor's > > conclusion based on > > (1) infinite sets do exist > > (2) an "Umformung" is any re-ordering > > You state there is no doubt, so it should be easy for you to provide a > > proof. > > He says it: those and only those transformations which are realized by > a finite or infinite set of transpositions, i.e., interchanges of two > elements. I ask for a proof, you do not provide one. So I can conclude that you can not proved a proof, although you said "with no doubt". And, "he says it" is no proof. Either point to a place where he proves it or provide a proof yourself. > > > The implication that Cantor's statement could be correct (if special > > > constraints were obeyed by permutations) with the implication that > > > infinity could exist. > > > > Can you provide a proof of that statement? Because you state just above: > > "Cantor's conclusion is correct, with no doubt, if infinite sets like > > the set of my transpositions do exist." Which contradicts what you state > > here. > > No. Cantor obviously was correct, if infinite sets would exist. Again you state "obviously" when I ask for a proof. So you are apparently not able to provide a proof but think that what Cantor writes is always true. Who coined the term metheology? > But > they do not. So he was wrong. And you are wrong, as I have shown above > by Cantor's text which you misinterpreted grossly. See above where I have written my rebuttal. You apparently have no idea what a good definition entails. He does not define the term "Umformungen", but defines a subset of "Umformungen" that have some particular property. Until we know that the larger term "Umformungen" means, we have no idea what that subset contains. > > > Therefore it never becomes 0 but always remains > 0 like 0.11... never > > > becomes 1/9. > > > > But apparently you do not use the standard definition of 0.111... . Is it > > a number or something else? > > It is a number with aleph_0 places, which are all occupied by 1's. Eh, it is a number th
From: Dik T. Winter on 15 Jul 2006 20:33 In article <virgil-CDCEBE.15390415072006(a)news.usenetmonster.com> Virgil <virgil(a)comcast.net> writes: .... > > What is "|N"? I've never seen that before. What does the pipe > > ( "|") on the left side represent? > > I think it is sometimes used to indicate that the following character > should be printed in bold typeface. No. It is customary in many parts of Europe to state the sets N, Q, R, C, H and whatever (when they are to represent numbers) with a doubled stem on the left side. See <http://www.unicode.org/charts/PDF/U2100.pdf> what they look like (although some of them are quite different from what I did use at university). To start, C is as it should be, H I have seen with both stems doubled or only the left one, N I have only seen with a doubled left stem, not as is shown, P I have never seen, Q only doubling on the left, as R, and Z as it was. The prefix of the vertical bar is an invention by Mueckenheim. It may work in a German (or Dutch) newsgroup, but not in an English newsgroup. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 16 Jul 2006 04:37 Dik T. Winter schrieb: > In article <1152883753.767753.301720(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > This definiion is > > > > wrong. I show this by the fact that 0.111... is not in the set of unary > > > > reprsetations of all natural numbers. aleph_0, the number of 1's in > > > > 0.111... is *not* the cardinal number of |N. > > > > > > Proof, please. > > > > Define 0 *+ 0 = 0, 0 *+ 1 = 1 *+ 0 = 1 *+ 1 = 1. > > Do the following addition *+ > > > > 0.1 > > 0.11 > > 0.111 > > ... > > How can I do that? You have not defined your operators on those objects. > Yes, I know that you want to do it digit-wise on the digits. In that > case there is still no definition for 0.1 *+ 0.11. So let us extend > all numbers with zeros: > 0.10... > 0.110... > 0.1110... > so now we can go. But you have not defined how to do that "infinite > sum". You use a lot of things for which you provide no defition. And, > there is one natural number missing: 0.0... That is not a natural number. And in the unary representation I defined it would be expressed as "0.". But you may introduce it. The Addition *+ was defined by "if one 1 is present the result is 1, else it is 0". This covers also the infinite case. > > So let's see: > {0.0...} has cardinality {0.10...} > {0.0..., 0.10...} has cardinality {0.110...} {0.10...} is not a cardinality but a set. By indexing we see {0.10...} has cardinality 0.10... {0.10..., 0.110...} has cardinality 0.11... .... > > What is the result? 0.111... > > How do you get there? What rule are you using to calculate that > "infinite sum"? But even if we allow that sum, we know that that > is not the unary representation of a natural number. But it is a unary representation of a whole number, namely of the number of digits of a real number. Nevertheless, in decimal representation 0.111... does exist, but not in the list. The addition of the list numbers gives the largest number of the list. Whatever this may be, it is not 0.111... . Proof: *Every* list number ends up with at least one zero. Therefore there must be at least one column with only zeros. These summed up gives the result zero. 0.111... however has no zero. > > This number is not among the numbers to be > > added. But it must be, if it is different from any list number and if > > the result of the addition is 0.111... . > > I see no reason for the "must be". Please first provide a proper definition > of how that infinite operation should be performed. If a sum includes one or more 1's, it is 1. Therefore we have no problem with infinitely many 1's. If a sum includes only zeros, it is zero. The addition is done column by column, strating from the left hand side. If a sequence of numbers, each of which has trailing zeros, is summed up, the result must have trailing zeros (because there was no number without trailing zeros). 0.1 0.11 0.111 .... _________ 0.111... has no trailing zeros. So we have an equation which equates two different magnitudes: one with trailing zeros, the other without trailing zeros. > > What cannot be covered by naturals cannot be indexed by naturals. > > Makes no sense to me. I did state: > For all p there is an n such that An[p] = K[p]. > You said that was wrong. I asked you why that was wrong, and you come > up with some nonsense. I state: select n = p. Isn't it a fact that > For all n, An[n] = K[n]? All n which can be indexed are in the list. K is not in the list. All segments K[n] are in the list. K is not in the list. What distinguishes K from all its segments if not at least one more 1at a position which cannot be indexed? K has nothing else to offer. But you believe that there was something else. That makes no sense to me. Regards, WM
From: mueckenh on 16 Jul 2006 04:38 Virgil schrieb: > In article <1152882972.884911.319600(a)75g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > In the binary tree everyone can see that no path can split without two > > > > additional edges. So it is by no means possible that there were more > > > > paths than edges. > > > > > > As long as paths all end. > > > > And if thy do not end, they run without edges? Flying? > > If no path ends then more paths than edges is not only possible, it is > necessary. > > Does WM deny that for infinite binary trees there is a bijection > between the set of paths and the power set of the set of edges? No, not if infinity would exist. But therefore this kind of mathematics is self-contradictory. The result depends on the way you obtain it. Because you cannot reasonably argue (but only shut your eyes and believe with all your power) that a fractional surjecton proves the opposite. Regards, WM
From: mueckenh on 16 Jul 2006 04:40
Virgil schrieb: > In an infinite tree one can show that the endless paths are uncountable" We know that. Therefore we have the following circulus falsus: |N| << |P(N)| || || |{edges}| >= |{paths}| Regards, WM |