Another AC anomaly?
in [Logic]
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From: Dik T. Winter on 10 Dec 2009 21:50 In article <fec95b83-39c5-4537-8cf7-b426b1779f84(a)k17g2000yqh.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 10 Dez., 16:35, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Before 1908 there was quite a lot of mathematics possible. > > > > Yes, and since than quite a lot of newer mathematics has been made > > available. > > Most of it being rubbish. Nothing more than opinion while you have no idea what has been done in mathematics since 1908. Algebraic number theory is rubbish? > > Moreover, before 1908 mathematicians did use concepts without actually > > defining them, which is not so very good in my opinion. > > Cantor gave a definition of set. What is the present definition? Something that satisfies the axioms of ZF (when you are working within ZF). It is similar to the concepts of group, ring and field. Something that satisfies those axioms is such a thing. But I think you find all those things rubbish. > > > N need not exist as a set. If n is a natural number, then n + 1 is a > > > natural numbers too. Why should sets be needed? > > > > Ok, so N is not a set. What is it? > > N is a sequence of natural numbers. Within ZF a sequence is an ordered set. But as you refuse to distinguish beteen an ordered set and a non-ordered set, I think this goes beyond you. > > > There is not even one single infinite path! > > > > Eh? So there are no infinite paths in that tree? > > In fact no, but every path that you believe in is also in the tree, > i.e., you will not be able to miss a path in the tree. I believe in infinite paths, you state they are not in the tree. So we have a direct contradiction to your assertion. > > > But there is every path > > > which you believe to be an infinite path!! Which one is missing in > > > your opinion? Do you see that 1/3 is there? > > > > If there are no infinite paths in that tree, 1/3 is not in that tree. > > 1/3 does not exist as a path. But everything you can ask for will be > found in the tree. > Everything of that kind is in the tree. This makes no sense. Every path in the tree (if all paths are finite) is a rational with a power of 2 as the denominator. So 1/3 does not exist as a path. In what way does it exist in the tree? > > Otherwise 1/3 would be a rational with a denominator that is a power of > > 2 (each finite path defines such a number). > > > > > What node of pi is missing in the tree constructed by a countable > > > number of finite paths (not even as a limit but by the axiom of > > > infinity)? > > > > By the axiom of infinity there *are* infinite paths in that tree. So your > > statement that there are none is a direct contradiction of the axiom of > > infinity. > > Try to find something that exists in your opinion but that does not > exist in the tree that I constructed. In what way do numbers like 1/3 exist in your tree? Not as a path, apparently, but as something else. Similar for 'pi' and 'e'. So when you state that the number of paths is countable that does not mean that the number of real numbers is countable because there are apparently real numbers in your tree without being a path. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 10 Dec 2009 21:40 In article <5333fb9a-1670-4fcc-85d3-25e75fb5bd1d(a)f16g2000yqm.googlegroups.com> WM <mueckenh(a)rz.fh-augsburg.de> writes: > On 10 Dez., 16:29, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Without the axiom of infinity omega would not be immediately existing. > > So apparently there is a definition of omega without the axiom of > > infinity. > > Can you state that definition? > > Look into Cantor's papers. Look into my book. I have never seen there a proper definition of omega. > > There are no concepts of mathematics without definitions. > > So? What is a set? Something that satisfies the axioms of ZF for instance. > > > An infinite union *is* not at all. But if it were, it was a limit. > > > > It *is* according to one of the axioms of ZF, and as such it is not a > > limit. > > It *was* according to Cantor, without any axioms. Yes, so what? You are arguing against current set theory, in the time of Cantor it was still being developed. > > Where? Why do you think taking a limit and taking cardinality should > > commute? Should also the limit of te sequence of integral of functions > > be equal to the integral of the limit of a sequence of fuctions? > > If an infinite set exists as a limit, then it has gotten from the > finite to the infinite one by one element. During this process there > is no chance for any divergence between this set-function and its > cardinality. If a function exists as a limit, then it has gotten from the finite to the infinite one by one step. During this process there is no chance of any divergence between the function and the integral. Now, what is wrong with that reasoning? Stronger: lim(n -> oo) 1/n = 0 1/n > 0 If a number exists as a limit, then it has gotten from the finite to the infinite one by one step. During this process there is no chance of any divergence between the element and the inequality. What is wrong with that reasoning? You are assuming that taking a limit is a final step in a sequence of steps. In the definition I gave for the limit of a sequence of sets there is no final step. -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: K_h on 11 Dec 2009 02:03 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:KuFy3L.Cxt(a)cwi.nl... > In article <QP-dnV0EIYPt2b3WnZ2dnUVZ_h2dnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > > "Virgil" <Virgil(a)home.esc> wrote in message > > news:Virgil-1E8B09.00355309122009(a)newsfarm.iad.highwinds-media.com... > ... > > > The issue between Dik and WM is whether the limit of a > > > sequence of sets > > > according to Dik's definition of such limits is > > > necessarily the same as > > > the limit of the sequence of cardinalities for those > > > sets. > > > > > > And Dik quire successfully gave an example in which > > > the > > > limits differ. > > > > I suspect those definitions are not valid. The > > definition I > > used is the one on wikipedia and is generally > > `standard' -- > > as I've seen it in numerous places, including books and > > websites. > > Have a look at <http://en.wikipedia.org/wiki/Lim_inf> in > the section > titled "Special case: dicrete metric". An example is > given with the > sequence {0}, {1}, {0}, {1}, ... > where lim sup is {0, 1} and lim inf is {}. > > Moreover, in what way can a definition be invalid? It depends on the context. When it comes to supertasks, limsup={0,1} is basically useless. That is why those definitions are not good, and invalid, for evaluating supertasks -- in response to WM's supertask issues. Those definitions are useful only to the extent that they show the limit does not exist: i.e. limsup=/=liminf but that is about all. k
From: K_h on 11 Dec 2009 03:23 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:KuFvyG.8r4(a)cwi.nl... > In article <Hv2dnXQ7LtSxUIPWnZ2dnUVZ_hSdnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > > news:KuAGqH.FrI(a)cwi.nl... > ... > > > Not at all. When you define N as an infinite union > > > there > > > is no limit > > > involved, there is even no sequence involved. N > > > follows > > > immediately > > > from the axioms. > > > > I disagree. Please note that I am not endorsing many of > > WM's claims. There are many equivalent ways of defining > > N. > > I have seen the definition that Rucker uses, in his > > infinity > > and mind book, in a number of books on mathematics and > > set > > theory: On page 240 of his book he defines: > > > > a_(n+1) = a_n Union {a_n} > > > > and then: > > > > a = limit a_n. > > But here an infinite union is *not* involved, that is the > crucial > difference. As stated, you may define N as a limit or > not, and > when it is defined as an infinite union as in: > N = union {1, 2, ..., n} > a limit is not involved. I'm not sure we're on the same page here. The limit set, a, does involve an infinite number of unions; this follows from a_(n+1)=a_nU{a_n}. But an infinite number of unions is also involved in the way I defined N as a limit in my previous post. So I guess I'm unclear what you mean when you write that N is defined as an infinite union by N=union{1,2,...,n}. k
From: Virgil on 11 Dec 2009 04:14
In article <6cCdncLLYcT2nL_WnZ2dnUVZ_qmdnZ2d(a)giganews.com>, "K_h" <KHolmes(a)SX729.com> wrote: > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:KuFvyG.8r4(a)cwi.nl... > > In article <Hv2dnXQ7LtSxUIPWnZ2dnUVZ_hSdnZ2d(a)giganews.com> > > "K_h" <KHolmes(a)SX729.com> writes: > > > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > > > news:KuAGqH.FrI(a)cwi.nl... > > ... > > > > Not at all. When you define N as an infinite union > > > > there > > > > is no limit > > > > involved, there is even no sequence involved. N > > > > follows > > > > immediately > > > > from the axioms. > > > > > > I disagree. Please note that I am not endorsing many of > > > WM's claims. There are many equivalent ways of defining > > > N. > > > I have seen the definition that Rucker uses, in his > > > infinity > > > and mind book, in a number of books on mathematics and > > > set > > > theory: On page 240 of his book he defines: > > > > > > a_(n+1) = a_n Union {a_n} > > > > > > and then: > > > > > > a = limit a_n. > > > > But here an infinite union is *not* involved, that is the > > crucial > > difference. As stated, you may define N as a limit or > > not, and > > when it is defined as an infinite union as in: > > N = union {1, 2, ..., n} > > a limit is not involved. > > I'm not sure we're on the same page here. The limit set, a, > does involve an infinite number of unions; this follows from > a_(n+1)=a_nU{a_n}. But an infinite number of unions is also > involved in the way I defined N as a limit in my previous > post. So I guess I'm unclear what you mean when you write > that N is defined as an infinite union by > N=union{1,2,...,n}. In ZF, no union can be shown to exist unless there is already a set of all the sets being unioned. Thus where 0 = {} and n+1 = {n,{n}}, one cannot form the union of all naturals unless it is already known (by the axiom of infinity) to exist. |