Cantor Confusion
in [Math]
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From: mueckenh on 7 Mar 2007 15:36 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: > > > On 6 Mrz., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1173176849.548735.27...(a)n33g2000cwc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > ... > > > > Of course. And we need not use the limit definition (of Hrbacek and > > > > Jech or else) in order to get this result, because the whole tree > > > > T(oo) has only a countable set of nodes. Therefore C(oo) cannot be > > > > larger than countable. > > > > > > The latter is an invalid conclusion. > > > > 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. 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. C(oo) c |T(oo)|. (Both are cardinal numbers, both are aleph_0.) > > > 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 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. > > Therefore I devised the tree. Its nodes are connected by continuous > > paths. They prevent any discontinuity. If we have a cross section less > > than the number of nodes for any finite tree, then the cross section > > cannot surpass the number of nodes of the infinite tree. The structure > > of the tree forces the function C(n)/|T(n)| to remain continuous under > > any circumstances. > > 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. > > > > > > Wrong. Each path is a set of nodes, hence a subset of the complete > > > > > set of nodes, hence an element of the powerset of this set of nodes. > > > > 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. > > 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. Regards, WM
From: mueckenh on 7 Mar 2007 15:44 On 7 Mrz., 19:48, Virgil <vir...(a)comcast.net> wrote: > What is "continuous" about pairs of nodes? The chance to get, by definition, from one path to another without leaving the system. > . > > What is "half" of an infinite number? What is whole an infinite number? > > 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? > > One wonders which branchings WM denies in his incomplete trees. None. Alas the number of branchings is countable. > > > As the cross sections show, these uncountably many are not in > > the tree, neither in U(T(n)) nor in T(oo). > > Then neither of them can contain all the paths in the complete infinite > binary tree, since THAT tree must have uncountably many paths, but less bifurcations? Or as mayn? > as has > been repeatedly proven and never refuted. Never. And will never be refuted. Regards, WM
From: Virgil on 7 Mar 2007 16:57 In article <1173299798.650965.55920(a)30g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 7 Mrz., 16:01, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > 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 Does WM mean his version of a cardinal as the class of all sets bijectable with a given set or as the more usual cardinal as the least ordinal bijectable with the given set? |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. > > 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. > > 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. > > > > > > 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 Then, lim {n --> oo} S_n = Union{n e N} S_n and lim{n --> oo} |S_n| = | Union{n e N} S_n | for S_n a set for each n in N, at least as far as one can tell. > > The limit lim {n-->oo} n is a number which n comes as close as you > like. Is this supposed to make sense? > This is not omega or any greater number. Why not? > > > > 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. > > > > > > > > Wrong. Each path is a set of nodes, hence a subset of the > > > > > > complete > > > > > > set of nodes, hence an element of the powerset of this set of > > > > > > nodes. > > > > > > 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. 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. It's a shame WM cannot manage to stretch that far.
From: Virgil on 7 Mar 2007 17:07 In article <1173300252.453214.228750(a)t69g2000cwt.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > On 7 Mrz., 19:48, Virgil <vir...(a)comcast.net> wrote: > > > What is "continuous" about pairs of nodes? > > The chance to get, by definition, from one path to another without > leaving the system. If one is going, as in the pairs I asked about, between parent node and child node of a path, one is strictly within one path, and not switching paths at all. > > . > > > > What is "half" of an infinite number? > > What is whole an infinite number? > > > > 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. > > > > 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. > > > > > As the cross sections show, these uncountably many are not in > > > the tree, neither in U(T(n)) nor in T(oo). > > > > Then neither of them can contain all the paths in the complete infinite > > binary tree, since THAT tree must have uncountably many paths, > > > > as has > > been repeatedly proven and never refuted. > > And will never be refuted. Agreed.
From: Dik T. Winter on 8 Mar 2007 10:25
In article <1173299798.650965.55920(a)30g2000cwc.googlegroups.com> mueckenh(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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |