Cantor Confusion
in [Math]
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From: Virgil on 20 May 2007 16:01 In article <1179655065.969901.59820(a)k79g2000hse.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 19 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > 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. Irrelevant, since it does not put 1/3 in the list. > It is the same case as lim[n-->oo] (1 - 0.999...9 with n 9's) = 0. Which is equally irrelevant to the Cantor theorems. The ONLY issue is whether there must be some number not in a arbitrary given list of numbers. Which has been proved satisfactorily. And WM's challenges to it are, so far, all based on irrelevancies. > > > > > 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. Even if the limit of the sequence of listed numbers converges to the diagonal, that is irrelevant, as it does not put the diagonal at any (finite) position within the list itself, and there is no infinite position IN the List..
From: Virgil on 20 May 2007 16:11 In article <1179663816.282116.232880(a)h2g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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. As each "column" can contain only w's or m's, there are no natural numbers in any column. The diagonal, of all w's for the given list, differs from the first string in place 1 and in the second in place 2, and so on, differing from the nth in place n, and thus differing enough to be different from EVERY member of the list. Thus it is not in the list. > 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). So that WM claims that if the nth listed string contains an "m" in position n, there must be a string listed with no "m"'s at all? What a strange world WM's MathUnRealism is. > > This argument shows All WM's "argument" shows is that WM's MathUnRealism world does not match any mathematically valid real world.
From: Virgil on 20 May 2007 16:21 In article <1179672671.380247.281320(a)p47g2000hsd.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > 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) is true if and only if all paths are in some finite tree. (2) is true in a CIBT if by "finite path q" one means an infinite path that coincides with p for only finitely many levels( one whose intersection, as a set of nodes, with p, as a set of nodes, is finite). > > (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? (1) is wrong in any CIBT, if one interprets the q in "finite paths q =/= p" as the intersection , as a set of nodes, of p with some path other than p.
From: WM on 20 May 2007 17:34 On 20 Mai, 17:57, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > 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. The diagonal number cannot have more 3's than every list number has 3's. If the diagonal number is complete in the sense that at every position indexed by a natural number there is a 3, then there must be a complete sequence of 3's among the list entries too. Otherwise the list is not complete, i.e., there is always a 6 at some natural index. If the list is not complete however, the diagonal number cannot be complete either. And the saying that at *every* position there is a digit b_n =/= a_n,n is nonsense. > > >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? Why only in this particular case? Is lim[i --> oo] (b_i - a_i) * 10^-i = 0 correct only in one special case? > > >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. No, that's correct. It is the only valid interpretation of the notion infinity. > 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 This criterion is false. In case lim [i --> oo] (10 - 9) * 10^-i = 0 we see it clearly. > > 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 Forall n e N we have 0 < 1 - 0.999...9 (with n 9's). > > So r does not appear on the list. So the problem of replacing 0 by 9 is not existing. But we have in the limit j --> oo: D_j = 0. Without this limit there are no real numbers existing but at most sequences of digits. Regards, WM
From: Dik T. Winter on 20 May 2007 22:01
In article <1179433414.192576.40870(a)n59g2000hsh.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 17 Mai, 02:23, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1179346082.257799.291...(a)k79g2000hse.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 16 Mai, 03:55, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > In article <1178795907.957410.94...(a)o5g2000hsb.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > > That is correct. But that does not mean that I said that there are > > > > > only finitely many paths in the tree, as you implied. > > > > > > > > Why not? I can ask only for finitely many paths, and (if I > > > > interprete your thinking correctly) path which I can not ask for do > > > > not exist. > > > > > > That is MatheRealism. > > > > See the "in your opinion". And you also want to apply that rule to > > mathematics. > > In general I do not apply MatheRealism when refuting set theory > (because on that basis there are no infinite sets to refute). But you do. Or why else did you ask me whether I could ask uncountably many questions? And I have seen quite a few of such occurences later in this thread... > > What do you mean with the "upto"? The formula holds for each natural > > number n, and so it holds for all natural numbers n. Consider the > > formula again: > > sum{i = 1..n} i = (n + 1) * n / 2. > > I state it holds for each natural number n (there is no upto involved in > > the statement), and so it holds for all natural numbers n. That is *not* > > a statement about: > > sum{i = 1..oo} > > because in that case we have not substituted a natural number for n. > > If something is stated to hold for all natural numbers, that means that > > it holds for each and every natural number, *not* that it holds for the > > set of natural numbers. > > > If a sum is stated to hold for every subset of natural numbers then it > has to hold for the whole set too. But that is not stated. There is *nothing* in the statement about every subset of natural numbers. When you want to see the statement as a statement about a set of numbers on the left hand side, the statement is about particular subsets, namely the finite subsets that start at 1 and continue until a finite natural number n. I see nothing in the statement about the set of even numbers, nor of the set of primes. > The same is true for initial > segments. The statement is about initial *finite* segments, namely those that terminate at a natural number n. > Therefore we can put the sum over all natural numbers on the > left hand side. And what natural number is 'n' in that case? > > > But not in sum{i = 1..n} i = (n + 1) * n / 2 ? > > > > That formula is about a single natural number n, it is *not* about a set > > of natural numbers. > > The left hand side is obviously about a set of natural numbers. The left hand side contains a set of natural numbers. The statement is about a particular natural number. > > > No. Then sum{i = 1 to oo} i could also yield another result. But > > > that is certaily false. It can neither be finite nor can it be > > > uncountable. The same holds for sum{i = 1 to oo} 1 = aleph_0. > > > > Why? That depends entirely on how you define it. I have looked, what you > > are missing is that in the book cardinal arithmetic is defined, and that is > > not the same as standard arithmetic, and with cardinal arithmetic the use > > of 'oo' is extremely misleading, and actually also your notation. Better > > is (and I think the writers of the book use that notation): > > sum{i in N} i = aleph_0, > > which is quite different because it does not suggest sequencing. > > The writers sum the *sequence* 1, 2, 3, ... with a definite and unique > result (no definition could lead to another result. Your claim that > this was not done or was impossible is wrong. No. They do not state it is a sequence. But definitions *can* lead to another result. It depends on the definitions. The sum of the set of all natural numbers is defined *by them*. Other definitions can give other results, because the result is not defined without their definitions. > > And indeed > > also sum{i in S} i is defined when S is an arbitrary set of cardinal > > numbers. But indeed, the book *gives* some additional definitions to > > get such results. In addition, it is also stated that those additional > > definitions require the axiom of choice. Did you really read it? In > > standard arithmetic sum{i = 1..oo} i is not defined. > > Here the sum *is* defined. You said it was not, but it is. It is not defined in standard arithmetic. It is defined in that book using some set theoretical definitions, that are not common. > > In cardinal arithmetic division (and subtraction, I think) are not defined. > > And I clearly stated that the formula was about natural numbers, so I think > > I implied sufficiently that I meant ordinary arithmetic. > > I said that the sequence of natural numbers can be summed up. This is > true. Depends. Did you miss the point that the axiom of choice was required? But *their* sum is not a sum of natural numbers, but of cardinal numbers, and it is not a sequence. > > But whatever the > > case when I use their definition of the limit of sequences of ordinal > > numbers, I can define: > > sum{i = 1..omega} i = lim{n -> omega} sum{1 = 1..n} i = > > lim{n -> omega} (n + 1) * n / 2 = omega. > > omega and aleph are not distinct in this case. Ordinal and cardinal > summation are not different in this case. Hrbacek and Jech show that 1 > + 2 + 3 + ... = aleph_0 (you can also call it omega). They do only show it when you use *their* definitions (and when you consider cardinal arithmetic). > And in fact it > cannot be else. But the formula which is true for all natural numbers > and for the sum of every initial segment of natural numbers is not > true for every initial segment of natural numbers. Oh. I can define division on cardinals such that (aleph_0 + 1) * aleph_0 / 2 = aleph_0, which would perfectly fit. Or do you have some other ideas why such a definition would not be possible? The only problem is that such definitions are not common. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |