An uncountable countable set
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From: mueckenh on 16 Jul 2006 12:20 Franziska Neugebauer schrieb: > You may recall that every sequence member has trailing zeros. Indeed. I recall. Therefore the linearity of the list numbers enforces a column with zeros. > > > As the first case, call it mixed number, is excluded by the definition > > of the list, > > there is only the > > second case remaining, call it linear number. > > This is also "excluded" by the definition of the list. So it's pointless > to discuss it either. Then the sum cannot be 0.111... Or "forall" must have a changed meaning (because, you know, there are no mixed numbers with zeros jumping around). > > > Hence (which means the > > same as "therefore, for this reason, thus, that is why, ...") there > > are only two logical alternatives: > > Either 0.111... is in the list, or the *+ sum of the list numbers is > > not 0.111... . > > 0,111... is not "in the list" by definition of the list _and_ > the *-sum of the list still is 0.111... > > This is neither inconsistent nor contradictory. That is your pretension, but this claim claims a gobbledygook. > > > You claim a third alternative which is prohibited by tertium non > > datur. > > My "list of claims" is finite and already finished. There is no third > alternative. That is what you pretend, but this pretension pretends a mumbo jumbo. > > > This can be proven in another way too: > > 1) If you remove all numbers with more than n 1's, then every line of > > the list contains at least one zero. Mixed numbers are not present, > > hence (which means the same as "therefore, for this reason, thus, that > > is why, ...") the zeros mount up at the right hand side. There must be > > a column with zeros. > > And there are only n columns which give the *+ sum 1. > > > > 2) If you reintroduce a number larger than n, then the number of > > columns with 1's grows. > > > > 3) If you reintroduce all numbers with a finite natural number of 1's, > > then the number of columns grows further, but there are only a finite > > natural number of columns with 1's. You said all these numbers carry > > trailing zeros. But now you refrain from this all-quantifiyer which > > according to he definition of *+ sum would yield a sum zero, but > > assert that the 1's of these numbers cover all columns. > > > > 4) If you at last introduce 0.111... in the list (as the last line or > > as the diagonal number, each of its 1's in a separate line), then the > > *+ sum remains he same. As 0.111... carries more 1's than all numbers > > in the list (i.e., not less than and not equal to any other) it is odd > > that its introduction should not have an effect. > > > > Concluding, you can assert further, that all these deductions are in > > vain because infinity could not be refuted as little as the existence > > of God can be refuted. And if you refrain from logical deductions you > > are right. But if infinity were a number and would behave like a > > number, it could be refuted. Hence your improvable infinity is > > completely outside of mathematics, it is not a number and is not good > > for anything else. > > I am not going to waste my precious time on scrutinizing this futile > mess. You could utter this fully justified with respect to set theory. Is it in particular point 4 which excites your fury? Regards, WM
From: mueckenh on 16 Jul 2006 12:22 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > >> Once again: There is no column without a 1. /Hence/ every column > >> *-sums up to 1. > > > > Contradiction to A n : n has trailing zeros. > > There is no contradiction. For all j e N there is an n e N > such that a_jn = 1 namely n = j, since a_jj = 1. Just that establishes the contradiction. forall n : n has trailing zeros. And the linearity of the list numbers enforces the column without zeros. Regards, WM
From: mueckenh on 16 Jul 2006 12:32 David Hartley schrieb: > In message <1153051856.174981.269510(a)35g2000cwc.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de writes > > > >David Hartley schrieb: > > > >> I cannot comment on the German text, but Dik's reading of the English is > >> clearly correct, WM's wrong. To put it even more clearly > >> > >> Question: Which transformations preserve ordinal number? > >> > >> Answer: Those which are realised by a finite or infinite set of > >> transpositions. > >> > >> "Transformation" is not defined here. > > > >But "such transformations which do not change the well-order" are > >defined here. And only those are interesting. > > They are not defined. They are characterised within the class of all > transformations, but transformations themselves are not described at > all. To use Cantor's proposition, to show that a particular > set-theoretic "object" is an order-preserving transformation, you would > need to show it can be written as a set of transpositions *and* that it > is a transformation. Nowhere - in the sections you've quoted - does he > say what a transformation is. He presumably knew what he meant; he may > well have written a definition somewhere else, but you do not seem to be > able to find it. Cantor was often redundant. Because he was very eager to get understood quite well. But he did never defined what "is" is or what "to write" means. Have you got me? In all of his collected works he does not define "Umformungen" because there is nothing do define. An "Umformung" is just a re-ordering. And such re-orderings which consist of a set of transpositions, were *defined* by him. Regards, WM >
From: Virgil on 16 Jul 2006 12:54 In article <1153039038.086809.295250(a)p79g2000cwp.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. It is to John von Neumann and he know more about it than you do. > > > > 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. In NBG cardinalities are sets. > > By indexing we see {0.10...} has cardinality 0.10... > {0.10..., 0.110...} has cardinality 0.11... Not in NBG. > ... > > > > 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. Unary representation not contain any radix points. > 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. If you mean concatenation, which is the algorithm of addition for unary, then 0.1 + 0.1 = 0.10.1
From: Virgil on 16 Jul 2006 12:59
In article <1153039117.024570.215850(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Not in my world. > The result depends on the way you obtain it. All results do. > Because you cannot reasonably argue (but only shut your eyes and > believe with all your power) that a fractional surjecton proves the > opposite. I can reasonably argue all sorts of things in NBG that "mueckenh" reflexively shuts his eyes to. And even with my eyes tightly open, I have no notion of what a self-contradictory notion like a fractional surjection would be. |