Another AC anomaly?
in [Logic]
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From: WM on 13 Dec 2009 09:56 On 12 Dez., 22:44, Marshall <marshall.spi...(a)gmail.com> wrote: > On Dec 12, 9:45 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 12 Dez., 18:29, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > On Dec 12, 9:02 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > On 12 Dez., 17:19, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > On Dec 12, 1:32 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > On 12 Dez., 02:01, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > > > > Showing > > > > > > > a contradiction would qualify, but it's been well > > > > > > > established that you don't know how to do that. > > > > > > > Consider how a union of paths is counted (I copy > > > > > > from another posting, therefore the quotation symbols): > > > > > > Ascii diagrams don't qualify as a contradiction. > > > > > Why should pictures, diagrams, acoustic signals etc. qualify less than > > > > sequences of symbols? > > > > With pictures, diagrams, etc. the possibilities for tomfoolery > > > are endless. A formal proof is more resistant to human > > > error. Anyway, if your diagrams are sound, translating > > > them into formal proofs should not be out of reach. > > > > > Every bit of information, in what form ever, can > > > > be used in proofs. But here you are: > > > > > The union of all natural numbers is, according to set theory, omega.. > > > > If *actual* infinity is meant, this is plainly impossible, because the > > > > natural numbers count themselves. > > > > No natural number counts how many natural numbers there are. > > > > > This leads to the result that the same structure, namely the tree with > > > > all its nodes, contains only a countable set of paths and > > > > simultaneously it contains an uncountable set of paths. > > > > > And this is a contradiction. > > > > That actually would be a contradiction if it were true. > > > It is true as can be seern from my last posting. But those who try but > > cannot not understand these few sentences will not understand the > > proof in either form. > > Have you tried? > > Yes. It was riddled with obvious errors. Which error was obvious? Regards, WM
From: Virgil on 13 Dec 2009 14:49 In article <4934d93b-3b3f-4e04-a831-8fc6ee4465ab(a)m26g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 12 Dez., 22:44, Marshall <marshall.spi...(a)gmail.com> wrote: > > On Dec 12, 9:45�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 12 Dez., 18:29, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > > On Dec 12, 9:02�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 12 Dez., 17:19, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > > On Dec 12, 1:32�am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > > On 12 Dez., 02:01, Marshall <marshall.spi...(a)gmail.com> wrote: > > > > > > > > > > Showing > > > > > > > > a contradiction would qualify, but it's been well > > > > > > > > established that you don't know how to do that. > > > > > > > > > Consider how a union of paths is counted (I copy > > > > > > > from another posting, therefore the quotation symbols): > > > > > > > > Ascii diagrams don't qualify as a contradiction. > > > > > > > Why should pictures, diagrams, acoustic signals etc. qualify less than > > > > > sequences of symbols? > > > > > > With pictures, diagrams, etc. the possibilities for tomfoolery > > > > are endless. A formal proof is more resistant to human > > > > error. Anyway, if your diagrams are sound, translating > > > > them into formal proofs should not be out of reach. > > > > > > > Every bit of information, in what form ever, can > > > > > be used in proofs. But here you are: > > > > > > > The union of all natural numbers is, according to set theory, omega. > > > > > If *actual* infinity is meant, this is plainly impossible, because the > > > > > natural numbers count themselves. > > > > > > No natural number counts how many natural numbers there are. > > > > > > > This leads to the result that the same structure, namely the tree with > > > > > all its nodes, contains only a countable set of paths and > > > > > simultaneously it contains an uncountable set of paths. > > > > > > > And this is a contradiction. > > > > > > That actually would be a contradiction if it were true. > > > > > It is true as can be seern from my last posting. But those who try but > > > cannot not understand these few sentences will not understand the > > > proof in either form. > > > Have you tried? > > > > Yes. It was riddled with obvious errors. > > Which error was obvious? Among many others, your claim that in a complete infinite binary tree there are as many nodes as sets of nodes.
From: WM on 13 Dec 2009 15:47 On 13 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: > > > Yes. It was riddled with obvious errors. > > > Which error was obvious? > > Among many others, your claim that in a complete infinite binary tree > there are as many nodes as sets of nodes. So it may seem to you. Your impression is wrong. The binary tree contains all real numbers between 0 and 1. Each one is represented by one single path, some are represented by two paths. If every node is covered, then every real number of [0, 1] has been constructed. Remember the ZF-law for unions of index sets: {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} This means: After having constructed all the paths ending with tails of zeros 0.1000... 0.11000... 0.111000... .... also the path 0.111... has been constructed. (The indexes of the 1' in above paths are the finite sets {1}, {1, 2}, {1, 2, 3}, ...) Then you might think you could choose as an alternative "tail" 010101... As you already constructed in the first run all paths of the form 0.000... 0.01000... 0.0001000... .... you have also constructed the path 0.010101010... because the indexes of the finite bit sequences in above paths are the finite sets {1}, {1, 2}, {1, 2, 3}, ... . And the infinite set {1, 2, 3, ...} of them includes already the bit sequence 0.010101 of the path 1/3. We have, if ZF is correct, {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} That means, you have constructed with the first approach using tail 000... already all paths that are possible in the tree. You could have started also with any other choice of the "tail", say that of 1/pi - with the same result. Therefore, the set of paths has been constructed by countably many steps. Once in a while, two paths have been constructed simultaneously, namely when an infinite sequence has been "finished". Then you have obtained 1 + aleph_0 paths instead of only aleph_0 paths. But is there a difference? The final result is: aleph_0 paths have been used to construct the complete infinite binary tree with all its paths such that no further path remains to be constructed, because the tree has been exhausted. Regards, WM
From: Virgil on 13 Dec 2009 17:34 In article <f9b8ae25-83eb-4831-af4a-3e2d45e85e17(a)9g2000yqa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 13 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: > > > > > Yes. It was riddled with obvious errors. > > > > > Which error was obvious? > > > > Among many others, your claim that in a complete infinite binary tree > > there are as many nodes as sets of nodes. > > So it may seem to you. Your impression is wrong. Then you are claiming that an ordered non-empty set with no last element has as many subsets with last elements as it has subsets altogether. Which is false. > > The binary tree contains all real numbers between 0 and 1. My binary trees do not contain any numbers at all, they only contain nodes. And have certain subset of those nodes called paths. Each one is > represented by one single path, some are represented by two paths. If > every node is covered, then every real number of [0, 1] has been > constructed. Not so. One can cover EVERY node using only the binary rationals, which omits "most" reals. So that WM is wrong again as usual. > > Remember the ZF-law for unions of index sets: > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} There is no such law in ZF. > > We have, if ZF is correct, > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} But that does not mean that {a,b} = {{a},{b}}, so you lose again. > The final result is: aleph_0 paths have been used to construct the > complete infinite binary tree with all its paths such that no further > path remains to be constructed, because the tree has been exhausted. WM's alleged construction does not work every infinite binary tree. Consider: N is the set of all naturals, {1,2,3,...}, the minimal inductive set. for any n in N, N_n is the subset of N of all elements of N not greater than n in successor order. In my tree each path is a PARTITION of N into an ordered pair of subsets, (L,R). A node of level n in that tree is a PARTITION of N_n into an ordered pair of subsets, (L_n, R_n). Node (L_n, R_n) is a node on the path (N,R) if and only if L_n is a subset of L and R_n is a subset of R. The tree itself consists of the set of all nodes, D, and the entirely sepraate set of all paths, P. And in my tree, no path is a union or sequence of nodes and there are only countably many nodes but uncountably many paths.
From: K_h on 13 Dec 2009 21:02
"Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote in message news:878wd7lczh.fsf(a)phiwumbda.org... > "K_h" <KHolmes(a)SX729.com> writes: > >> "Jesse F. Hughes" <jesse(a)phiwumbda.org> wrote in message >> news:87vdgcksy2.fsf(a)phiwumbda.org... >>> "K_h" <KHolmes(a)SX729.com> writes: >>> >> >> So, you're claiming that he is not using { and } just to >> bracket the argument (i.e. the X_n to be limited) but >> {X_n} >> refers to a set containing the one set X_n. > > Er, yes. Though, something seems to be wrong with your > notation. At > issue is the set X_n = {n}, not the set {X_n}. > > In summary: > > |lim X_n| = |lim {n}| = |{}| = 0. > > lim |X_n| = lim |{n}| = lim 1 = 1. > >> Then it seems like the meaning has changed because in >> previous posts >> he writes that lim|S_n|=/=|limS_n| follows from a >> wikipedia >> definition applied to sequences of natural numbers n -- >> not to the >> non-naturals {n}. For example: >> >> > > I have explicitly defined the limit of a sequence of >> sets. With that >> > > definition (and the common definition of limits of >> sequences of natural >> > > numbers) I found that the cardinality of the limit >> is >> not necessarily >> > > equal to the limit of the cardinalities. > > And that's absolutely correct, as we see above. Only if the sequences were of the non-naturals {n} not sequences of the naturals n. >> Okay, if {X_n} refers to a set containing the single set >> X_n >> then lim(n-->oo){n} is not a limit of the natural numbers >> since the naturals are not the sets {n} but the sets n. > > Er, yes. Of course. > >> In this case my proof shows that lim(n-->oo)n=N. >> Applying the >> wikipedia definitions to n is sensible but applying them >> to {n} >> makes a mockery of the notion of a limit. > > You have some very odd notions yourself. It's a simple > application of > a perfectly sensible definition of limit. It violates the spirit of what a limit is in some cases. So, although it is sensible, it is not perfectly sensible. >> The basic idea behind a limit is that things in one state >> tend to >> some final state and a good definition and application of >> a limit >> should embody that. In looking at the sequence {n}, with >> 0={} and >> 1={0}, saying that it tends to 0={} is a betrayal of the >> core idea >> behind a limit: >> >> 1, {{0}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... --> 0 >> >> The basic idea of what a limit is suggests that an >> appropriate definition for lim(n-->oo){n} should yield >> lim(n-->oo){n}={N}: >> >> {}, {{0}}, {{0,1}}, {{0,1,2}}, {{0,1,2,3}}, ... --> >> {{0,1,2,3,4,...}} >> >> In other words, applying the wikipedia definitions to {n} >> is >> an abuse of those definitions. The definition that is >> used >> for a limit should make sense for the kind of object it >> is >> applied to. > > You're welcome to your own cockamamie opinions about > whether a > particular definition is sensible or not, but they're > utterly > irrelevant to the issue at hand. The fact is that with > this > *perfectly standard* definition of limits, we see that > > lim |X_n| != |lim X_n|. > > That's all there was at issue. The sensibility of a definition is the real issue. Applying the so-called standard definitions to {n} leads to a cockamamie limit which is at odds with the general notion of a limit. For a better definition, first choose one of the wikipedia definitions. If a sequence of sets, A_n, cannot be expressed as {X_n}, for some sequence of sets X_n, then lim(n-->oo)A_n is defined by the wikipedia limit. Otherwise let L=lim(n-->oo)X_n be the specified wikipedia limit for X_n. If L exists then: lim(n-->oo)A_n = lim(n-->oo){X_n} = {L} otherwise lim(n-->oo)A_n = lim(n-->oo){X_n} does not exist. Under this definition lim(n-->oo){n}={N} and |lim(n-->oo){n}|=lim(n-->oo)|{n}|=1. Even this definition can be improved. In the spirit of what a good definition of a limit should be, we should require that, for example, lim(n-->oo){n,n,n}={N,N,N}. This can be done by a simple generalization: if a sequence of sets A_n cannot be expressed as {X_n,Y_n,Z_n,...}, for one or more arguments, then lim(n-->oo)A_n is defined by the specified wikipedia limit. Otherwise let L=lim(n-->oo)X_n, K=lim(n-->oo)Y_n, J=lim(n-->oo)Z_n, ... be the specified wikipedia limits for X_n, Y_n, Z_n, ... . If L, K, J, .. all exist then: lim(n-->oo)A_n = lim(n-->oo){X_n,Y_n,Z_n,...} = {L,K,J,...} otherwise lim(n-->oo)A_n does not exist. Viewers of this thread may want to see if there is a way to generalize and/or improve this definition further. A good definition should always seek to capture the essence of the notion it is defining. k |