An uncountable countable set
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From: mueckenh on 8 Oct 2006 08:24 Dik T. Winter schrieb: > In article <1160127267.123550.306050(a)b28g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > No reason at all, but when you invent a new notation, it would be better > > > if you did define the notation before use. > > > > Ok, but as we have agreement now, we can return to he main question: > > Why do you think that 0.111... with the index sequences 1,2,3,... or > > k+1,k+2,k+3 or -k, -k+1, -k+2, ... represents exactly *one* number > > only, as you asserted? > > Why do you still maintain that I think it represents a number? How many > times do I need to state that, without proper definition, it only is > a sequence of symbols that I on occasion call a "number". Because I have > not yet seen a definition of "number", and you have stated that you are > not able to give one... In the decimal system of current mathematics 0.111... = 1/9 and is a number without doubt. If omega does exist, then 0.111... has omega 1's. But whatever. As a sequence of symbols, > igoring the "0.", it is in bijection with N. It also is in bijection > with {k+1,k+2,...} for every k. As {k+1,k+2,...} is in bijection with N. A bijection with N does not define the indexes such that there was only a unique sequence. Therefore, there is not, as you asserted, one unique number 0.111... . Regards, WM
From: mueckenh on 8 Oct 2006 08:30 Dik T. Winter schrieb: > In article <1160126832.709755.186110(a)h48g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1160079778.871756.325200(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > > > The axiom of infinity does only state n+1 exists if n is given. > > > > > > It is realized by he numbers of my list. > > > > > > > > > > That is *not* what the axiom of infinity states. The axiom states > > > > > that there exists a set that contains all the successors of its > > > > > elements. > > > > > > > > Oh I must have read always wrong texts. I never came across the word > > > > "all" in connection with this axiom. > > > > > > What texts *have* you read where your version is stated? > > > > I read many texts. (Don't intermingle the quantity "all" with the > > quanifyer "forall".) > > Well, in another posting you did show that you do read texts, and in it > you were apparently aware that the formulation you gave above is a theorem > that follows from the axiom of infinity. That does *not* make the two > statements equivalent. A (nearly) equivalent statement would be (I dispense > of the Von Neumann notation of naturals, and just just numbers, the results > are equivalent): > There is a set N such that 0 in N and (forall x in N, x+1 in N) > (If you wish, you may replace 0 by 1, as 0 would lead to Bourbakianism, > which you detest. The results remain the same.) Fraenkel, Abraham A., Bar-Hillel, Yehoshua, Levy, Azriel: "Foundations of Set Theory", 2nd edn., North Holland, Amsterdam (1984), p. 46: AXIOM OF INFINITY Vla There exists at least one set Z with the following properties: (i) O eps Z (ii) if x eps Z, also {x} eps Z. There are several verbal formulations dispersed over the literature without any "all". In German: Unendlichkeitsaxiom: Es gibt eine Menge, die die leere Menge enthält, und wenn sie die Menge A enthält, so enthält sie auch die Menge A U {A} (oder die Menge {A}). > You question whether "all x in N" does exist, apparently. Based on what? Based on the impossibility to index the positions of our 0.111..., based on the vase, based on many other contradictions arising from "all x in N do exist". > > > > > > Yes, because you still do not understand that there is no "specific > > > > > position" such that "every position" is the same as "upto that > > > > > specific position". > > > > > > > > In my list there are all specific positions (if all natural numbers do > > > > exist). If the required number is not in the list then it is nowhere. > > > > > > Again, you imply that "every position" is the same as "upto some specific > > > position". There is no such "specific position". > > > > I imply that "every position" is equivalent with "up to evey position" > > in a linear set. > > No, that is *not* the implication. If the number is in the list, "every > position" is equivalent to "upto every position" is equivalent to "upto > some specific position" (namely the last position occupied by a digit). > If the "number" is not in the list, "every position" is equivalent > "upto every position" is *not* equivalent to "upto some specific position". > (Where "every position" uses indexing, and "upto every position" and > "upto some specific position" use covering.) As I have said time and time > again. > > > I imply that "every specific position" is equivalent with "up to evey > > specific position" in a linear set. > > Yes, and that is false and not provable. A set containing all positions "up to position x" is a superset of a set containing "position x". A set containing "up to every position" defines a superset of set containing "every position". But "every position" cannot be a proper subset. Hence both sets are equivalent. > I will re-iterate. Let L be your list (tacitly assumed to be unary > representations of natural numbers), define a sequence of symbols such > that it starts with 0., and for each n in N the n-th symbol is 1. I > state that: > (1) That sequence can be indexed, because the n-th symbol is indexed > by the n-th item in your list. > (2) For each n that sequence can be covered by the n-th item in your > list upto the n-th symbol. > (3) There is no n in your list that covers the whole of that sequence. > Because if there were one such n, the (n+1)-st symbol is not covered. > You claim (at various times) that either the definition is false, or that > (1) is false. Well, clearly (1) is not false because of the definition. This definition "covers" only the naturals. If 0.111.. had only such positions, ... Well, let us stop here. Regards, WM
From: Ross A. Finlayson on 8 Oct 2006 08:51 Virgil wrote: > In article <d01ff$4524e557$82a1e228$25988(a)news2.tudelft.nl>, > Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > > > > Not liked by Carl Friedrich Gauss as well. I'm in fairly good company. > > > > Han de Bruijn > > Gauss is long since dead. From the neck up, HdB emulates him. Some have Gauss inventing the limit. Ross
From: Ross A. Finlayson on 8 Oct 2006 08:57 Virgil wrote: > In article <45251b3e(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > > > Virgil wrote: > > > In article <4523c954$1(a)news2.lightlink.com>, > > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > >> David R Tribble wrote: > > >>> Tony Orlow wrote: > > >>>>> On the other hand > > >>>>> I don't know why I said "neither can the reals". In any case, the only > > >>>>> way the ordinals manage to be "well ordered" is because they're defined > > >>>>> with predecessor discontinuities at the limit ordinals, including 0. > > >>>>> That doesn't seem "real" > > >>> Virgil wrote: > > >>>>> In what sense of "real". There are subsets of the reals which are order > > >>>>> isomorphic to every countable ordinal, including those with limit > > >>>>> ordinals, so until one posits uncountable ordinals there are no > > >>>>> problems. > > >>> Tony Orlow wrote: > > >>>> The real line is a line, with > > >>>> each point touching two others. > > >>> That's a neat trick, considering that between any two points there is > > >>> always another point. An infinite number of points between any two, > > >>> in fact. So how do you choose two points in the real number line > > >>> that "touch"? > > >>> > > >> They have to be infinitely close, so actually, they have an > > >> infinitesimal segment between them. :) > > > > > > But any "infinitesimal segment" within the reals is bisectable. > > > > Within the standard reals, it's one number, if it's closer than any > > finite distance of a that number. > > In Standard reals,"infinitesimal", if it means anything, merely means > very small but not zero. > In The Robinson, or similar, non-standard models, infinitesimals are > different from standard numbers but still non-zero. > In both, they are bisectable, and between two distinct numbers, even > when only infinitesimally different, there is always another. "... for finitely many bisections." Do you think you can bisect the number more than finitely many times and be able to tell the difference? Ross
From: David Marcus on 8 Oct 2006 12:29
Han de Bruijn wrote: > Randy Poe wrote: > > OK, if there's anyone out there reading this who knows > > what Han means by "non-disciplinary mathematics", > > could you please explain since Han is unable to? > > > > If *I* were to characterize undisciplined approaches to > > mathematics, I would include something like "introduction > > of terms which the author is unable to define but > > nevertheless says 'the meaning is obvious'" > > Precisely! An example is Mike Kelly's function A(n,t) for t >= 0: > > > Let A(n,t) be 1 if the ball n is in the vase at time t, 0 if it is not > > in the vase at time t. > > > > Let B(n) be the time that the nth ball is added to the vase and C(n) be > > the time that it is removed. > > > > B(n) = -1/(2^(floor((n-1)/10))) > > C(n) = -1/(2^(n-1)) > > > > Note that B(n) and C(n) are strictly less than 0. > > > > Now A(n,t) = { 1 if B(n) <= t < C(n) > > 0 otherwise } > > > > Note that A(n,0) = 0. > > Sure. Simply "define" something that's undefined. And create the self > fulfilling prophecy that suits you best. Consider this situation: At time 5 one ball is added to a vase. At time 6, the ball is removed. Is the following a valid translation into mathematics? Let the value 1 denote that that the ball is in the vase and the value 0 that the ball is not in the vase. Let A(t) be the location of the ball at time t. Let A(t) = { 1 if 5 < t < 6; 0 if t < 5 or t > 6 }. -- David Marcus |