Another AC anomaly?
in [Logic]
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From: Aatu Koskensilta on 15 Dec 2009 22:11 "Jesse F. Hughes" <jesse(a)phiwumbda.org> writes: > So, a set is not something that satisfies the axioms of ZF. Rather, > it is an object in a structure that satisfies the axioms of ZF. So anything and everything is a set? We don't in fact find any formal or mathematical definition of set in mathematics -- instead we find various more or less informal explanations. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: K_h on 15 Dec 2009 22:23 "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message news:Kuq5DH.18H(a)cwi.nl... > In article <Kup2uH.JC7(a)cwi.nl> "Dik T. Winter" > <Dik.Winter(a)cwi.nl> writes: > > In article > > <r_6dnUXDv4pMfbvWnZ2dnUVZ_tWdnZ2d(a)giganews.com> "K_h" > > <KHolmes(a)SX729.com> writes: > > > "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote in message > > > news:KunF0v.9y9(a)cwi.nl... > > ... > > > > By what definition is: > > > > lim(n -> oo) n = N? > > > > > > Any of the wikipedia definitions. Here is the proof > > again. > > ... > > > Theorem: > > > lim(n ->oo) n = N. Consider the naturals: > > > > > > S_0 = 0 = {} > > > S_1 = 1 = {0} > > ... > > This presupposes a particular construction for the > > natural number. There are > > other constructions that are consistent with ZF. Is the > > limit valid for all > > those possible models? > > For starters, try it with > 0 = {} > n+1 = {n} > which is a valid construction of the naturals in ZF. > > Even with your definition > lim sup(n -> oo) {n} = {} Why is it so important to you to have a limit definition and a construction of the naturals such that lim(n->oo){n}={}? The general idea of a limit is that the limiting state is what you get when you go through all sequences. If one defines the naturals as you have done above then the general notion of a limit suggests that the limiting state should be something like: {...{{{{{{...{}...}}}}}}...} = limit We could construct a defintion of a limit so that this is the end result but it may be that a better definition for the limiting case of 0={} and n+1={n}is a defintion where lim(n -> oo)n does not exist. k
From: K_h on 15 Dec 2009 22:56 "WM" <mueckenh(a)rz.fh-augsburg.de> wrote in message news:9904ae9d-de2c-4d5f-86da-be165e8e9d7e(a)p30g2000vbt.googlegroups.com... On 15 Dez., 13:32, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > Look two lines above: > > Let {1} U {1, 2} U {1, 2, 3} U ...={1, 2, 3, ...} > hence > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...}={1, 2, 3, > ...} . All limit ordinals, X, satisfy UX=X. That means that the union of the members of a limit ordinal is equal to the ordinal itself. w=N is a limit ordinal so UN=N. So: {} U {0} U {0,1,} U {0,1 2,} U ... = {0,1,2,3, ...} is a true equation. However, w+1 is the ordinal wU{w} or NU{N}. So w+1 looks like: {0,1,2,3, ...}U{{0,1,2,3, ...}}={0,1,2,3, ...,{0,1,2,3, ....}} {0,1,2,3, ...}U{{0,1,2,3, ...}}={0,1,2,3, ...,N} {0,1,2,3, ...}U{{0,1,2,3, ...}}={0,1,2,3, ...,w} k
From: Virgil on 16 Dec 2009 00:07 In article <9904ae9d-de2c-4d5f-86da-be165e8e9d7e(a)p30g2000vbt.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 15 Dez., 13:32, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > �> It is nonsense to define a pink unicorn. > > > > That statement is nonsense. �Let pu be a pink unicorn is a proper > > definition. > > However there is nothing that satisfies that definition. > > That is a matter of taste. So WM is declaring that it is a matter of taste whether there is something that satisfies a definition of a pink unicorn? > > > > �> � � � � � � � � � � � � � � � � � � � � �The set N does not exist as > > �> the union of its finite initial segments. This is shown by the (not > > �> existing) path 0.000... in the binary tree. > > > > As you have defined your tree as only containing finite paths it is trivial > > to conclude that the infinite path 0.000... does not exist in your tree. > > However, this does *not* show that N does not exist. > > If N does exist, then the tree containing all finite paths contains > also infinite paths. In that case, you have proved that N "must" exist. > > > > �> Let {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...}. > > �> What then is > > �> {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} ? > > > > Ambiguous as stated. �What is "..." standing for? > > Look two lines above: > > Let {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > hence > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...} . An ellipsis with last term, as WM has indicated above, can only represent a finite sequence in HONEST mathematics. What WM may have meant, but was too dishonest or too dumb, or both, to express honestly is ( {1} U {1, 2} U {1, 2, 3} U ... ) U {1, 2, 3, ...} with the indicated parenthesis ABSOLUTELY necessary. > {1} U {1, 2} U {1, 2, 3} U ... = {1} U {1, 2} U {1, 2, 3} U ... U {1, > 2, 3, ...} {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...}, without any parentheses, is mathematically meaningless > > There is no space in the binary tree to contain infinite paths in > addition to the sequeneces of all finite paths. Neither finite paths nor sequences of finite paths are themselves paths in infinite trees. > The sequence 0.0, 0.00, 0.000, ... when being completely constructed, > is already the path 0.000... > You cannot add it separately. If you consider it only as a sequence of nodes then that set (or sequence) of all of those nodes can be identified with a path in the infinite tree. > > Therefore: When all finite paths have been constructed within aleph_0 > steps, then all paths have been constructed. There are absolutely no such things as finite paths in an infinite tree. No finite sequence can do more than identify a single node, or finite set of single nodes, in such a tree, and a node is not a path in any infinite tee. > There is no chance to > construct the path > 0.000... > after all finite initial segments have been constructed. There is no chance to even define, much less construct, an infinite binary tree by any of the false methods that WM proposes. Though in ZF it is trivial to construct such a tree. Example of an infinite binary tree, which exists in ZF: Starting with the infinite set N = {1,2,3,...}, certain infinite subsets are paths: a subset that contains 1 and for each n in N, contains one and only one of 2*n or 2*n+1 is a path, and no other subsets of N are paths. Each natural is a node but sets of naturals are never nodes and finite sets of naturals are never paths. WM has absolutely no model for an infinite binary tree that is even self-consistent, much less compatible with any acceptable set theory.
From: Jesse F. Hughes on 16 Dec 2009 07:23
"K_h" <KHolmes(a)SX729.com> writes: > Why do you ask? There are many ways a limit can be defined > in ZF but the definition should embody the general idea of > what a limit is. Well, it's not at all clear to me what this means, but I do have a question about general set convergence as presented in Wikipedia (http://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior#General_set_convergence). In general, the term limit is defined topologically: any topological space comes with its own natural definition of limit. But it's not at all clear to me whether "general set convergence" is a limit in this particular sense. Is there a topology on Set so that the "natural" definition of limit coincides with general set convergence? If not, then I guess K_h has a point about the naturalness of this definition of limit (and, conversely, if so, then K_h has no good point at all). -- "Eventually the truth will come out, and you know what I'll do then? Probably go to the beach. I'll also hang out in some bars. Yup, I'll definitely hang out in some bars, preferably near a beach." -- JSH on the rewards of winning a mathematical revolution |