Cantor Confusion
in [Math]
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From: Dik T. Winter on 16 May 2007 20:23 In article <1179346082.257799.291260(a)k79g2000hse.googlegroups.com> WM <mueckenh(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: > > > On 10 Mai, 03:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > So in your opinion there are only finitely many paths in the > > > > > > infinite tree. .... > > > 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 MatheRealism. See the "in your opinion". And you also want to apply that rule to mathematics. > > > It is valid for those cases I did prove. I did not prove it (and one > > > cannot prove it) for "all natural numbers which obey this formula", > > > because all natural numbers which obey this formula do not obey this > > > formula. > > > > This sentence makes no sense. I do not ask it for "all natural numbers > > which obey this formula", I ask it for "all natural numbers". And what > > natural numbers that obey that formula do not obey that formula? > > Up to each one they obey it. *All* natural numbers obviously do not > obey the formula. 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. > > > My argument is: Why do you have to exclude the case 0.999... = > > > 1.000... if you remain always in the finite? For every finite index > > > both numbers are different. > > > > But when you use all finite indices they are the same. But the indices > > are still finite. And by the axiom of infinity we can use the set of > > all finite indices. > > 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. > > Sorry, that makes it unreadable to me. But from what I can decipher is > > that they define the infinite sum of cardinal numbers as the cardinality > > of the infinite union of disjoint sets. In common set theory the infinite > > union of sets is defined, and so this makes sense. But this has nothing > > to do with > > sum{i = 1..n} i = (n + 1) * n /2 > > where the sum is constrained to natural numbers n. Analysis and algebra > > are about real numbers, and in those fields infinite sums are not defined, > > except as a limit. Moreover, using their definition of limits of > > sequences of ordinal numbers we would get omega. It can make sense to > > make > > sum{i = 1..oo} i = aleph_0 > > but it does not follow in any way from the usual theorem, and requires > > some additional definitions. > > 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. 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. 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. 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. Where the last step requires the additional definition. But now I am using ordinal arithmetic... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 16 May 2007 20:24 In article <1179346353.430169.24280(a)y80g2000hsf.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 16 Mai, 04:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > The set is finitely defined. Not all lucky numbers can get finitely > > > defined. > > > > In general a recursive definition is not considered a finite definition. > > Every definition which ends after finitely many words is a finite > definition. Ah, so you disagree with common mathematical terminology. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 16 May 2007 20:49 In article <1179346617.111313.267350(a)o5g2000hsb.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 16 Mai, 04:13, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > Each of the numbers is finitely definable, because each is a halting Turing > > machine, and Turing machines are finitely definable. It is the set that is > > not finitely definable. > > If each number is finitely defined, then there are only finitely many > numbers, because there are only finitely many Turing machines. Of > course, this finite set then is finitely defined too. Back again to how you started. There are only finitely many natural numbers... That is not mathematics. > > > A well defined Cauchy sequence, for pi, for instance, makes pi well > > > defined - not as a number in the sense of MatheRealism, but as a > > > number in the sense of mathematics. > > > > Now you come up again with a new term. 'Well defined'. What is > > 'well defined'? > > A number is well defined if it is defined by a finite number of words > such that, in principle (i.e., given an infinite amount ressources) > the Cauchy epsilon can be made arbitrary small. So pi is well defined as the limit of the circumference of the inscribed n-gon? The sequence of those circumferences *is* a Cauchy sequence... > > > Please read: > > > I gave a bijection between the set of nodes and the set of branching- > > > offs of paths bunches. > > > > Yes. I do not understand. 0.1010101010... is a path bunch. So there is > > a branching-off of this path bunch. And you claim a bijection with the > > nodes. > > I showed that there are countable sets which cannot be bijected with > the set of natural numbers, for instance the set of finitely definable > numbers. But that set *can* be bijected with the set of natural numbers. It is indeed easy to show that there is an injection from that set to the set of natural numbers. Consider all finite sentences over some alphabet (let's say the 26 latin letters plus a space). Each such sentence can be considered as a base-27 number, so we have an injection. > > > What node is bijected with the branch-off of 0.101010101010...? > > I claim that there are no more branching-offs than odes and that there > are no more real numbers represented in the tree than are branching- > offs. There is no path ever finished, but it is only branching off > from other paths in infinity. But the number of paths separated from > other paths cannot surpass the number of branching-offs. The number of paths is the same from the root node, because every path starts at the root. Or are you suggesting that there are paths *not* starting at the root? > > Makes no sense. As all bunches start at the root, there cannot be more > > bunches that come out of a node than go in. By your own definitions a > > bunch that comes out of a node, also comes in it (when it arrives there > > from the root). > > One bunch goes in, because the two which come out have not yet been > separated when the paths which they consist of, went in. So it is your opinion that bunches that start of the root nevertheless do not come in at some node but only come out? Another question about chapter 10. Do you understand what a normal number is? I think not. Off-hand I do not know whether there are normal numbers that are normal with respect to all bases (although it is expected that pi is one). But if a number is normal with respect to some base that does *not* mean that the digits are unpredictable. Nor does unpredictability of digits mean that a number is normal. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 16 May 2007 21:03 In article <1179346754.744571.316980(a)k79g2000hse.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 16 Mai, 04:17, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1178808886.504395.245...(a)w5g2000hsg.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > On 10 Mai, 04:02, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > ... > > > > > I will answer there. > > > > > > > > I will not see it, so you clearly cut short the discussion. > > > > > > Please try, we could reduce the noise significantly and could kill any > > > polemics and insults. If you are not able to see it, I will repeat it > > > here. But I wonder why Google introduced this technique if it remains > > > inaccessible to most users and even to experts. > > > > 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. > > But I do not want to sign up with google > > just for the privilege of accessing it. And I do not want to use a web > > browser to access discussions. Why google introduced it beats me. I > > prefer my keyboard based access to newsgroups, no mouse for me. > > 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/5d9926c33ee6e40f?scoring=d&hl=de 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. > PS: What about the critique of ma chapter 10? See my first question about it in another article. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: WM on 17 May 2007 06:23
On 17 Mai, 01:42, Virgil <vir...(a)comcast.net> wrote: > In article <1179345320.393851.200...(a)u30g2000hsc.googlegroups.com>, > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > On 14 Mai, 22:09, William Hughes <wpihug...(a)hotmail.com> wrote: > > > On May 14, 3:17 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > Which of the following statements is not true > > > > Every element of R is different from p. > > > > Every node of p is in some element of R > > > Back to the facts > > Back FROM the fact, more like. > > > > > 1) No element r =/= p of R is required to cover any node of p. (I can > > show of every such element r =/= p which you may name that it is not > > required.) > > 2) No element of R is sufficient to cove all nodes of p. > > Hence your set R is not suitable to demonstrate your claims. > > Let R be the set of all finite initial subsets of N. > No member of R covers N yet R itself does. The first assertion is right, the second is wrong. > Regards, WM |