Another AC anomaly?
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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
From: WM on 18 Dec 2009 04:24 On 17 Dez., 22:52, Quaestor <quaes...(a)qqq.com> wrote: > > > > 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. Ok, then you have a different idea of mathematics. In common mathematics, numbers can be defined by digit sequences. > > > 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 Fine, Virgil, that is one big step to comprehension. Regards, WM
From: Dik T. Winter on 18 Dec 2009 07:22 In article <SqSdnSUF1q8FobbWnZ2dnUVZ_uCdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: .... > 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. Only when we use the definition of limit on sequences of sets, not when we use the definition of limit on sequences of numbers. You have to distinguish the two and clearly state which one you are using. > 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. Yes, as in many cases: lim | S_n | is not necessarily equal to |lim S_n|, whatever the definition of set limit. The same holds for limsup and liminf. -- 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 18 Dec 2009 07:25
In article <eaCdnfgmKeFaorbWnZ2dnUVZ_hmdnZ2d(a)giganews.com> "K_h" <KHolmes(a)SX729.com> writes: .... > > 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! Eh? With your definition, lim {n} = {N} (as you stated earlier) and so lim n = N. But now you actually state (as |n| = n), lim n = |N|. Are N and |N| equal or not? -- dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |