Another AC anomaly?
in [Logic]
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From: Quaestor on 17 Dec 2009 16:52 In article <25016582-3715-4b47-a02f-5fe79b0a594b(a)m3g2000yqf.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 17 Dez., 22:02, YBM <ybm...(a)nooos.fr.invalid> wrote: > > WM a �crit : > > > > > > > > > > > > > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: > > >> In article > > >> <2cd3c08a-a072-4ef5-bc0b-f0aaa9126...(a)a21g2000yqc.googlegroups.com>, > > > > >> �WM <mueck...(a)rz.fh-augsburg.de> wrote: > > >>>> No. �Suppose we have the paths 0.000 and 0.100, what is their union? > > >>>> �And > > >>>> is it a path? > > >>> No. But the union contains two paths. And an infinite union of that > > >>> kind may contain the path 0.111... > > >> So according to WM a union of sets may contain an object not contained > > >> in any of the sets being unioned. > > > > > For paths and initial segments to contain and to be is the same. In > > > fact: > > > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > > > and also > > > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. > > > > As usual you confuse to be an element and to be a party of... > > > I am concerned with numbers. Your absurd distinctions are irrelevant. They are neither absurd nor irrelevant to mathematics. But then, comprehensible mathematics and WM seem to be incompatible. > Either N exists or not. If yes, then it is the union of all natural > numbers as well as the union of all initial segments of the ordered > set of all natural numbers as well as the union of all these and > itself, because > > {1, 2, 3, ...} U {1, 2, 3, ...} = {1, 2, 3, ...} > > If you see that this equation is true, then you acknowledge my > argument. On the contrary, that equation is true AND your argument is FALSE. > If you cannot see that, then either you are a matheologian > or you have too little intelligence to understand it or both. Actually, what we have is too much intelligence to "understand" that which is incapable of being understood, i.e., is nonsense..
From: K_h on 18 Dec 2009 01:25 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:Kusoo8.xz(a)cwi.nl... > In article <GLidnXhuo5t347TWnZ2dnUVZ_sWdnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > > news:KuqwG3.qv(a)cwi.nl... > ... > > > > > The definition you provided for a sequence of sets > > > > > A_n > > > > > depends on whether > > > > > each A_n is or is not a set containing a single > > > > > set as > > > > > an > > > > > element. > > > > > > > > > > Your definition leads to some strange > > > > > consequences. I > > > > > can > > > > > state the > > > > > following theorem: > > > > > > > > > > Let A_n and B_n be two sequences of sets. Let A_s > > > > > = > > > > > lim > > > > > sup A_n and > > > > > A_i = lim inf A_n, similar for B_s and B_i. Let > > > > > C_n > > > > > be > > > > > the sequence > > > > > defined as: > > > > > C_2n = A_n > > > > > C_(2n+1) = B_n > > > > > Theorem: > > > > > lim sup C_n = union (A_s, B_s) > > > > > lim inf C_n = intersect (A_i, B_i) > > > > > Proof: > > > > > easy. > ... > > > Well, the above theorem is still not valid with your > > > definition. > > > > What case did you have in mind? I found cases where it > > works fine. > > Let's have some arbitrary object 'a' and the natural > numbers. Create > the sequence A_n where A_n = {a} and the sequence B_n > where B_n = {n}. > According to your definition: > lim sup A_n = {a} > and > lim sup B_n = {N}. > Now create the sequence C_n: C_2n = A_n, C_2n+1 = B_n. > Again according > to your definition: > lim sup C_n = {a} > which is not equal to union (lim sup A_n, lim sup B_n). This is a good example, thanks. Your theorem only applies in special cases for the definition I have offered (although my definition satisfies some different but interesting theorems). k
From: K_h on 18 Dec 2009 03:26 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:Kusoty.176(a)cwi.nl... > In article <yrydnX_FwfczULTWnZ2dnUVZ_vmdnZ2d(a)giganews.com> > "K_h" <KHolmes(a)SX729.com> writes: > ... > > lim(n ->oo) n = N is true for the standard definition of > > natural numbers using just the wikipedia definitions for > > the > > limit of a sequence of sets. > > But not with Zermelo's definition of the natural numbers. > Nor when we take > the natural numbers as embedded in the rational numbers. Yes, and that was never denied. I just point out that any limit definition for a sequence of sets will give answers that violate the intuitive notion of a limit for certain constructions. There are two ways one can deal with that. First, for a given class of constructions, have another definition that does not violate the limit notion. The second way is to look at the meaning that the wikipedia definitions have for the constructions. I now think that the latter approach is superior to the first since the first caused so much misunderstanding. With the second option we can restrict ourselves just to the wikipedia definitions and define the sets n by: n = 0 = {} n = 1 = {{}} n = 2 = {{{}}} .... Wikipedia gives liminf(n-->oo) n = 0. What this means is that nothing is `accumulated' in a limit set since each set does not persist past its introduction. So the notion of n growing bigger, as one tends to the limit, is not embodied here since |n| is 0,1,1,1,1... as one proceeds and is 0 in the limiting case. Under the standard construction each natural is `accumulated' in the limit set, since each set persists beyond its introduction, and this preserves the notion of n growing bigger as one proceeds: |n| is 0,1,2,3,... and is N in the limiting case. k
From: K_h on 18 Dec 2009 03:39 "K_h" <KHolmes(a)SX729.com> wrote in message news:SqSdnSUF1q8FobbWnZ2dnUVZ_uCdnZ2d(a)giganews.com... > > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > news:Kusoty.176(a)cwi.nl... >> In article >> <yrydnX_FwfczULTWnZ2dnUVZ_vmdnZ2d(a)giganews.com> "K_h" >> <KHolmes(a)SX729.com> writes: >> ... >> > lim(n ->oo) n = N is true for the standard definition >> > of >> > natural numbers using just the wikipedia definitions >> > for the >> > limit of a sequence of sets. >> >> But not with Zermelo's definition of the natural numbers. >> Nor when we take >> the natural numbers as embedded in the rational numbers. > > Yes, and that was never denied. I just point out that any > limit definition for a sequence of sets will give answers > that violate the intuitive notion of a limit for certain > constructions. There are two ways one can deal with that. > First, for a given class of constructions, have another > definition that does not violate the limit notion. The > second way is to look at the meaning that the wikipedia > definitions have for the constructions. I now think that > the latter approach is superior to the first since the > first caused so much misunderstanding. With the second > option we can restrict ourselves just to the wikipedia > definitions and define the sets n by: > > n = 0 = {} > n = 1 = {{}} > n = 2 = {{{}}} > ... > > Wikipedia gives liminf(n-->oo) n = 0. What this means is > that nothing is `accumulated' in a limit set since each > set does not persist past its introduction. So the notion > of n growing bigger, as one tends to the limit, is not > embodied here since |n| is 0,1,1,1,1... as one proceeds > and is 0 in the limiting case. Under the standard > construction each natural is `accumulated' in the limit > set, since each set persists beyond its introduction, and > this preserves the notion of n growing bigger as one > proceeds: |n| is 0,1,2,3,... and is N in the limiting > case. Yo, it should be |N| in the limiting case! k
From: WM on 18 Dec 2009 04:21
On 17 Dez., 22:47, Quaestor <quaes...(a)qqq.com> wrote: > > Please prove your assertion by showing which path I did not produce. > > But prove it by knowing only all nodes of all paths. > > Until you give some explicit way of telling which paths you do produce, > ther is no way of telling which ones you miss. I produce every path. There is no node in the complete binary tree that is not member of an infinitude of paths. > > > Information exceeding the infinite sequence of nodes of each path is > > not acceptable in mathematics. > > I do not accept your fiats on what is or is not acceptable in > mathematics, In mathematics, numbers are defined by digits. Additional information is not part of mathematics. If you do not accept that, then there is no common basis for discussion. My claim is: The set of numbers which can unambiguously be identified by digit sequences is countable. > > > Which path do you see in the tree that has not that property? > > In my tree, none. But in there are sets of nodes satisfying my condition > of pathhood which do not occur in your tree. Pure assertion. Give an example, please. Regards, WM |