Cantor Confusion
in [Math]
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From: mueckenh on 9 Mar 2007 03:55 On 8 Mrz., 16:25, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1173299798.650965.55...(a)30g2000cwc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 7 Mrz., 16:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Did you ask the referee? Or are you the arbiter yourself? C(oo) is a > > > > part of T(oo). > > > > > > What do you *mean* with "is a part of T(oo)". How do you *define* > > > C(oo). You continuously *fail* to give a proper definition of it. > > > And do not refer to "cross section" of T(oo), because that is undefined > > > either. Neither do refere to limits, unless you *define* those limits. > > > > I did define them by means of set theory: > > > > C(oo) = U(C(n)) where C(n) is the cardinal number |L(n)| of L(n) which > > is the last level of T(n) where T(n) is the binary tree with n levels > > where n is a natural number and the union has to be taken over all > > natural numbers. > > Ok, so C(oo) is the cardinal number aleph-0. > > > T(oo) = U(T(n)) according to the definiton I gave for the union of > > trees. This union can also be obtained by > > T(oo) = U(L(n)) with L(n) defined as above. > > So T(oo) is a set of nodes. > > > C(oo) c |T(oo)|. (Both are cardinal numbers, both are aleph_0.) > > And you mean *that* with C(oo) is part of T(oo). A strange formulation, > but correct. > > And how do you conclude from this that the number of paths is countable? > > > > > No, Dik. I know that the functions in set theory are not continuous > > > > and perform unpredictable jumps when leaving the finite and entering > > > > the infinite. One of these tricks is the set of all finite sets being > > > > countable and the set of all infinite sets being uncountable. Another > > > > trick is the famous observation: > > > > Forall n in N: |{2,4,6,...,2n}| < 2n > > > > while > > > > lim {n-->oo} |{2,4,6,...,2n}| > lim {n-->oo} 2n, and similar jokes. > > > > > > That is tricky of you. You have defined *neither* limit. > > > > > > I defined for several times: > > lim {n-->oo} {1,2,3,...,n} = N > > lim {n-->oo} |{1,2,3,...,n}| = aleph_0 > > > > lim {n-->oo} {2,4,6,...,2n} = set of all even numbers > > lim {n-->oo} |{2,4,6,...,2n}| = aleph_0 > > Ah, you need *four* definitions here? Only two would be sufficient, I > think. Because none of the definitions follows from any of the other > definitions. > > > The limit lim {n-->oo} n is a number which n comes as close as you > > like. This is not omega or any greater number. If it exists, then it > > can only be less. > > So, according to *your* definitions it is >. Not according to common > definitions. Would it not be possible that your definitions are not > consistent? It is your last paragraph which is inconsistent with the > other definitions. See how you *did* define: > > lim {n-->oo} |{1,2,3,...,n}| = aleph_0 > which means (as |{1,2,3,...,n}| = n: > lim {n-->oo} n = aleph_0? > > > > > Have you really been blinded for the fact that most of the subsets of > > > > nodes of the complete cannot serve as paths? > > > > > > That does not matter. Some subsets are paths. Of these some are infinite > > > and some are finite. The finite ones are a subset of a countable set, the > > > infinite ones are a subset of an uncountable set. But there *do* exist > > > uncountable subsets of uncountable sets. > > > > May be, but not in the tree --- as long as it stretches there is no > > uncountable set. > > Why not? > > > > At each finite level, the cross-section C(n) is the number of path-bundles. > > > This does not tell us anything about the infinite tree. > > > > Infinite trees have infinite cross section, however, the cardinal > > number is closely connected to the length n. Do you claim that there > > are paths of uncountable length (number of nodes)? If not, then you > > will obtain from the countability of the |2^omega| nodes the > > countability of the |2^omega| paths. > > Why? I see only that C(oo) is aleph-0, but I see no relation between > paths and aleph-0. Isn't it difficult to look from this perspectice? The cross sections *are* numbers of paths. 1) We know tha every digi of a real number stands on a finite place. (We know that here is no digit the last one, but that is here completely irrelevant.) 2) From (1) we obtain that every node of a path is at a finite place. That means, there is no part of a path which would jut out f the tree. 3) All paths which are in the tree cross each cross section (that is why it is named cross section). There is no path or even splitting of paths outside of every cross section. Therefore, the limit processes for paths-lengths and cross sections are identical. Regards, WM
From: mueckenh on 9 Mar 2007 05:57 On 7 Mrz., 22:57, Virgil <vir...(a)comcast.net> wrote: > > > C(oo) c |T(oo)|. (Both are cardinal numbers, both are aleph_0.) > > Then either represent the cardinality of the set of all paths in a > complete infinite binary tree. > C(n) is the cardinality of level L(n). There is no largest n. Therefore this holds throughout the whole infinte tree. At no finite position the infinite paths can be more than countably many. Either there is an infinite index. Or there are less than uncountably many paths. > > > The limit lim {n-->oo} n is a number which n comes as close as you > > like. > > Is this supposed to make sense? Otherwise limits do not make sense. > > > This is not omega or any greater number. > > Why not? > Because omega - n = omega > 1 > > > > > But you do *not* define the cross-section of the infinite tree, neither > > > do you define the limits you are using. > > > I defined the cross section for every existing tree. > > No! Only for finite trees. For EVERY LEVEL which a paths can populate (if it can populate only every finite level). > > > > May be, but not in the tree --- as long as it stretches there is no > > uncountable set. > > When one finally gets a complete infinite binary tree all "stretching" > is over and done with, and one has all those uncountably many paths > with nary a further stretch required. And after death you will enter the paradise and all pain will end and you will no longer need any logics. But as long as you are here in the vale of tears, there is logic dictating the facts. Regards, WM
From: mueckenh on 9 Mar 2007 06:02 On 7 Mrz., 23:07, Virgil <vir...(a)comcast.net> wrote: > In article <1173300252.453214.228...(a)t69g2000cwt.googlegroups.com>, > > > > As each infinite binary tree bifurcates infinitely many times > > > at as many nodes > > > > and each > > > path chooses one of those bifurcations infinitely many times, there are > > > at least as many possible such paths as there are infinite binary > > > strings, and there are, therefore at least an uncountable number of > > > infinite binary strings in each complete infinite binary tree. > > > The number of bifurcations is countable, isn't it? > > The number of binary digits in the numeral for a real number is > countable isn't it? > > The number of binary bits in an infinite binary string is countable > isn't it? > > Yet the set of all such paths, set of all such reals and set of all such > strings are all uncountable. Yes, I know, that's your wrong belief. Everything in the world looks yellow. > > > > > > One wonders which branchings WM denies in his incomplete trees. > > > None. Alas the number of branchings is countable. > > But the number of possible sequences of branchings is what we need to > count. And that is not countable. The number of paths which are generated by branchings is not counted by sequences of branchings but by branchings. Regards, WM
From: mueckenh on 9 Mar 2007 06:10 On 7 Mrz., 16:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1173215257.272668.121...(a)64g2000cwx.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > Neither do refer to limits, unless you *define* those limits. For the cross section we would not even need a limit. It is enough to know that at every level n (with finite n) the number of paths (- bundles) is countable. That is sufficient because we know that every path consists only of nodes (or edges) at finite places, i.e., enumerated with finite numbers n. How many of these levels are populated by paths is completely irrelevant. > > Have you really been blinded for the fact that most of the subsets of > > nodes of the complete cannot serve as paths? > > That does not matter. Some subsets are paths. Of these some are infinite > and some are finite. The finite ones are a subset of a countable set, the > infinite ones are a subset of an uncountable set. But there *do* exist > uncountable subsets of uncountable sets. Again and again the creed is muttered. > > Part of which argument? I cannot see any argument at all. You state > > (1) that Cantor's argument is valid (at this place I do no judge about > > it) and you state (2) that set theory is free of contradictions (I > > don't know who told you). From (1) and (2) you conclude that contrary > > to all mathematical evidence (I remind you of the ration of 2 nodes > > per path and the cross section of the tree which counts its paths) my > > result cannot be true. > > As you do not *define* the cross section of the infinite tree, your > argument is empty. There is no need to define it. It is sufficient to know that EVERY cross section is countable. Regards, WM
From: G. Frege on 9 Mar 2007 06:24
On 9 Mar 2007 03:10:56 -0800, mueckenh(a)rz.fh-augsburg.de wrote: >> >> As you do not *define* the cross section of the infinite tree, >> your argument is empty. >> > There is no need to define it. [...] > A very good example of M�ckenmathics! :-) F. -- E-mail: info<at>simple-line<dot>de |