Cantor Confusion
in [Math]
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From: WM on 20 May 2007 05:57 On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1179398355.204677.123...(a)o5g2000hsb.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 17 Mai, 03:03, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > It is not inaccessible to me. > > > > > > > > So you saw my answer? > > > > > > No. I say it is not inaccessible to me. Not that I do access it. > > > > > > > This set F of functions can be bijected with the > > > > set R of reals. The reals form an intercession with the rationals Q. > > > > The rationals form a bijection with the naturals N. > > > > > > Sorry, makes no sense without context. > > > > The context was your due posting. Either you recall it or you access > > my group or you cannot get an answer on your question which uttered an > > unjustified doubt. > > So you refuse to post an answer in the forum where I asked the question, > giving proper references ... I have given you reasons *why* I do not access > that group. That you ignore those reasons just shows arrogance. We have > seen this same behaviour earlier by James Harris. Are you going to mimick > him? I have *no* idea what you mean with "this set F of functions". You defined the functions: (1) If two functions f and g are not equal, there is a smallest n such that f(n) != g(n). (2) We define f < g if f(n) < g(n), and f > g if f(n) > g(n). This is a complete ordering on that set of functions. > > > > > If you are interested in the discussion which we had about Cantor's > > > > mistake or not mistake concerning his second diagonal proof, you may > > > > enter > > > >http://groups.google.com/group/sci.math.research/browse_frm/thread/5d... > > > > > > If I look correctly, again, it is not the case that Cantor made an error. > > > It is my opinion that what I wrote about it was right, the interpretation > > > being that Cantor gave an additional prove that there are sets with > > > cardinality larger than the naturals. But in essence this is quite > > > irrelevant. Cantor may have erred at times, that is *not* related to the > > > current ways of set theory. > > > > It is. Set theory is simply biased. Consider the list > > > > 0.666... > > 0.3666... > > 0.33666... > > 0.333666... > > ... > > > > If the diagonal number is defined by "replace 6 by 3", then we have > > two answers none of which can be preferred by logic, but the second of > > which is suppressed by convention. > > But, again, that is *not* the diagonal proof of Cantor. And even with > that notation you write nonsense. "Replace 6 by 3" yields the sequence > 0.33333..., which is not in the list. For the entries E(n) of the list we find lim[n-->oo] (E(n) - 0.333...) = 0. It is the same case as lim[n-->oo] (1 - 0.999...9 with n 9's) = 0. > > > 1) Every entry of the list differs at some place from the diagonal > > number. > > 2) Every initial segment of the diagonal number is represented by an > > entry of the list. > > Both are true, if you replace (2) by: > 2) Every initial segment of the diagonal number is represented by the > initial segment of an entry of the list. > To wit: > 0.333666... > does *not* represent > 0.333, > or you have a very strange interpretation of the word "represent". If very initial segment of the diagonal number is represented by the initial segment of an entry of the list, then the full diagonal number is represented by an entry of the list. Regards, WM
From: WM on 20 May 2007 08:23 On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > Set theory is simply biased. Consider the list > > > > 0.666... > > 0.3666... > > 0.33666... > > 0.333666... > > ... > > > > If the diagonal number is defined by "replace 6 by 3", then we have > > two answers none of which can be preferred by logic, but the second of > > which is suppressed by convention. > > But, again, that is *not* the diagonal proof of Cantor. The following wm-proof certainly even in your opinion belongs to the diagonal proofs considered by Cantor: 0) mmm... 1) wmmm... 2) wwmmm... 3) wwwmmm... 4) wwwwmmm... .... .......... And if the list can be considered as a completed entity, then there must be all natural numbers in the first column. And there must be a line with all natural indexes mapped on w's, i.e., no w must be missing (as would be the case if one m was present). This argument shows that the list cannot be complete and that replacing the diagonal digit of every number of the list proves literally nothing. Regards, WM
From: WM on 20 May 2007 10:51 On 18 Mai, 01:13, William Hughes <wpihug...(a)hotmail.com> wrote: > Which of the following statments is wrong? [...] There are two: (1) The union of all finite paths q =/= p does not cover p. (2) The union of all finite paths q =/= p covers p. (1) ==> There is a node which belongs to p but not to any other path q =/= p. (2) ==> p is nothing but a union of finite paths. (1) is obviously wrong, whereas (2) shows the countability of all paths. Which is wrong? Regards, WM
From: Daryl McCullough on 20 May 2007 11:57 WM says... >> > It is. Set theory is simply biased. Consider the list >> > >> > 0.666... >> > 0.3666... >> > 0.33666... >> > 0.333666... >> > ... >> > >> > If the diagonal number is defined by "replace 6 by 3", then we have >> > two answers none of which can be preferred by logic, but the second of >> > which is suppressed by convention. The diagonal number is 0.333... which is not on the list. >For the entries E(n) of the list we find >lim[n-->oo] (E(n) - 0.333...) = 0. >It is the same case as lim[n-->oo] (1 - 0.999...9 with n 9's) = 0. That's true. In this particular case, the limit of the sequence is equal to the diagonal of the sequence. So what? >If every initial segment of the diagonal number is represented by the >initial segment of an entry of the list, then the full diagonal number >is represented by an entry of the list. That's false. To say that the number r appears on the list r_0, r_1, ... is to say that there is some natural number j such that r = r_j. If we let D_j = |r - r_j|, then the criterion for r appearing on the list is that exists j such that D_j = 0 In the case we are talking about, that is false. If r_0 = 0.6666...., r_1 = .3666..., etc and r = 0.333..., then |r-r_0| = .333... |r-r_1| = .0333... ... in general, |r - r_j| = 0.333... * 10^{-j} So we have forall j, D_j > 0 So r does not appear on the list. -- Daryl McCullough Ithaca, NY
From: Virgil on 20 May 2007 15:43
In article <1179654356.461792.242730(a)n59g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 19 Mai, 04:49, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > According to current mathematics, pi is well defined. Even according > > > to MatheRealism pi is well defined (as an idea). > > > > Your distinction between "number" and "idea" is just terminology, and not > > more than that. > > Its is more. You cannot answer the question whether the numbers P = > [pi*10^10^100] and P' = P with the last digit replaced by 3 nsatisfy P > < P'. There have been lots of questions put to WM which he has not, and presumably cannot, answer, but that does not seen to have slowed him down, so why should the existence of numbers which it is impractical to compare be of any import? > > > > > > What node is bijected with the branch-off of 0.101010101010...? > > > > > > For an injection you can choose whatever node you want. > > > > Wrong. For an injection it is needed that two paths do not map to the same > > node, so you have to be careful in your mapping. You simply refuse to give > > an injection because you are not able to give one. > > Map every node onto the path which leaves it to the left-hand side. Trivially NOT an injection as all infinitely many nodes on the path which always branches left are mapped onto that same path. While there are trivial injections, WM seems incapable of finding any of them. > > > > > > Every path starts at the root node. But in order to count the paths, > > > they must be distinguishable, i.e. separated. > > > > Makes no sense. > > How would you count inseparated paths? By bijecting them with a set which can be "counted" in other ways. The set of all paths in a CIBT bijects to the set of all subsets of N, which is known not to be injectable into N, and is therefore non-denumerable. > > You were talking about bunches going in and out of nodes. What you are > > doing is counting edges, not bunches, and the number of edges is countable. > > The number of paths cannot be larger than the number of edges. Only true in finite trees. The set of all paths in a CIBT bijects to the set of all subsets of N, which is known not to be injectable into N, and is therefore non-denumerable. |