Cantor Confusion
in [Math]
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From: mueckenh on 16 Mar 2007 10:07 On 16 Mrz., 14:35, Carsten Schultz <cars...(a)codimi.de> wrote: > mueck...(a)rz.fh-augsburg.de schrieb: > > > > > > > On 16 Mrz., 01:31, Virgil <vir...(a)comcast.net> wrote: > >> In article <1173954799.919385.61...(a)y80g2000hsf.googlegroups.com>, > > >> For even binary trees ( where even here means all paths are of equal > >> length), > > > Only those are under discussion here. > > >> the number of paths increases exponentially with number of > >> levels (lengths of a path). Adding 1 to the number of levels doubles the > >> number of paths. > > >>> The tree is continuous because its nodes are connected by paths. > >> That is a distinctly non-standard meaning for "continuous" in > >> mathematics. > > > It shows, however, that the number of paths cannot jump from finite to > > uncountable. > > Using a word does not constitute proof. > > And indeed sup_{n<aleph_0} 2^n = aleph_0 < 2^aleph_0, > so in this sense the function kappa |-> 2^kappa is not continuous. If > you can prove (not claim!) by using your tree that it is, then you will > finally have succeeded in showing that ZF is inconsistent. > > Have fun, I had already quite a lot. The function of all cross sections, f: n |--> 2^n, is "continuous" in the sense that never a jump by more than a factor 2 can occur because the nodes of the tree are connected by an untearable network. The domain is the same as the range, namely N. That is fact, not by claim but by construction of the tree. That's why I constructed it. Regards, WM
From: Carsten Schultz on 16 Mar 2007 10:15 mueckenh(a)rz.fh-augsburg.de schrieb: > On 16 Mrz., 14:35, Carsten Schultz <cars...(a)codimi.de> wrote: >> mueck...(a)rz.fh-augsburg.de schrieb: >> >> >> >> >> >>> On 16 Mrz., 01:31, Virgil <vir...(a)comcast.net> wrote: >>>> In article <1173954799.919385.61...(a)y80g2000hsf.googlegroups.com>, >>>> For even binary trees ( where even here means all paths are of equal >>>> length), >>> Only those are under discussion here. >>>> the number of paths increases exponentially with number of >>>> levels (lengths of a path). Adding 1 to the number of levels doubles the >>>> number of paths. >>>>> The tree is continuous because its nodes are connected by paths. >>>> That is a distinctly non-standard meaning for "continuous" in >>>> mathematics. >>> It shows, however, that the number of paths cannot jump from finite to >>> uncountable. >> Using a word does not constitute proof. >> >> And indeed sup_{n<aleph_0} 2^n = aleph_0 < 2^aleph_0, >> so in this sense the function kappa |-> 2^kappa is not continuous. If >> you can prove (not claim!) by using your tree that it is, then you will >> finally have succeeded in showing that ZF is inconsistent. >> >> Have fun, > > I had already quite a lot. I can imagine. > The function of all cross sections, f: n |--> 2^n, is "continuous" in > the sense that never a jump by more than a factor 2 can occur because > the nodes of the tree are connected by an untearable network. The > domain is the same as the range, namely N. That is fact, not by claim > but by construction of the tree. That's why I constructed it. You constructed the tree to show that 2^{n+1} <= 2*2^n ? Well, that really must have been fun. Ok, I agree on this. Now we know a property of the function f : N -> N n |-> 2^n. This does not tell us anything about 2^aleph_0. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: mueckenh on 16 Mar 2007 10:23 On 16 Mrz., 15:15, Carsten Schultz <cars...(a)codimi.de> wrote: > mueck...(a)rz.fh-augsburg.de schrieb: > > > > > > > On 16 Mrz., 14:35, Carsten Schultz <cars...(a)codimi.de> wrote: > >> mueck...(a)rz.fh-augsburg.de schrieb: > > >>> On 16 Mrz., 01:31, Virgil <vir...(a)comcast.net> wrote: > >>>> In article <1173954799.919385.61...(a)y80g2000hsf.googlegroups.com>, > >>>> For even binary trees ( where even here means all paths are of equal > >>>> length), > >>> Only those are under discussion here. > >>>> the number of paths increases exponentially with number of > >>>> levels (lengths of a path). Adding 1 to the number of levels doubles the > >>>> number of paths. > >>>>> The tree is continuous because its nodes are connected by paths. > >>>> That is a distinctly non-standard meaning for "continuous" in > >>>> mathematics. > >>> It shows, however, that the number of paths cannot jump from finite to > >>> uncountable. > >> Using a word does not constitute proof. > > >> And indeed sup_{n<aleph_0} 2^n = aleph_0 < 2^aleph_0, > >> so in this sense the function kappa |-> 2^kappa is not continuous. If > >> you can prove (not claim!) by using your tree that it is, then you will > >> finally have succeeded in showing that ZF is inconsistent. > > >> Have fun, > > > I had already quite a lot. > > I can imagine. > > > The function of all cross sections, f: n |--> 2^n, is "continuous" in > > the sense that never a jump by more than a factor 2 can occur because > > the nodes of the tree are connected by an untearable network. The > > domain is the same as the range, namely N. That is fact, not by claim > > but by construction of the tree. That's why I constructed it. > > You constructed the tree to show that 2^{n+1} <= 2*2^n ? Well, that > really must have been fun. Ok, I agree on this. Now we know a property > of the function > > f : N -> N > n |-> 2^n. > > This does not tell us anything about 2^aleph_0. > aleph_0 is not a natural number. Don't mistake the infinite number of finite paths with infinite paths. In the union of all fiite trees every path has a finite length, given by a natural number of nodes. Presently we are considering the union of all such finite paths. (The union of all finite natural numbers is an infinite union - nevertheless this union cotains only finite numberrs.) Regards, WM
From: Carsten Schultz on 16 Mar 2007 10:54 mueckenh(a)rz.fh-augsburg.de schrieb: > On 16 Mrz., 15:15, Carsten Schultz <cars...(a)codimi.de> wrote: >> mueck...(a)rz.fh-augsburg.de schrieb: >> >> >> >> >> >>> On 16 Mrz., 14:35, Carsten Schultz <cars...(a)codimi.de> wrote: >>>> mueck...(a)rz.fh-augsburg.de schrieb: >>>>> On 16 Mrz., 01:31, Virgil <vir...(a)comcast.net> wrote: >>>>>> In article <1173954799.919385.61...(a)y80g2000hsf.googlegroups.com>, >>>>>> For even binary trees ( where even here means all paths are of equal >>>>>> length), >>>>> Only those are under discussion here. >>>>>> the number of paths increases exponentially with number of >>>>>> levels (lengths of a path). Adding 1 to the number of levels doubles the >>>>>> number of paths. >>>>>>> The tree is continuous because its nodes are connected by paths. >>>>>> That is a distinctly non-standard meaning for "continuous" in >>>>>> mathematics. >>>>> It shows, however, that the number of paths cannot jump from finite to >>>>> uncountable. >>>> Using a word does not constitute proof. >>>> And indeed sup_{n<aleph_0} 2^n = aleph_0 < 2^aleph_0, >>>> so in this sense the function kappa |-> 2^kappa is not continuous. If >>>> you can prove (not claim!) by using your tree that it is, then you will >>>> finally have succeeded in showing that ZF is inconsistent. >>>> Have fun, >>> I had already quite a lot. >> I can imagine. >> >>> The function of all cross sections, f: n |--> 2^n, is "continuous" in >>> the sense that never a jump by more than a factor 2 can occur because >>> the nodes of the tree are connected by an untearable network. The >>> domain is the same as the range, namely N. That is fact, not by claim >>> but by construction of the tree. That's why I constructed it. >> You constructed the tree to show that 2^{n+1} <= 2*2^n ? Well, that >> really must have been fun. Ok, I agree on this. Now we know a property >> of the function >> >> f : N -> N >> n |-> 2^n. >> >> This does not tell us anything about 2^aleph_0. >> > > aleph_0 is not a natural number. Exactly. > Don't mistake the infinite number of finite paths with infinite paths. I don't. > In the union of all fiite trees every path has a finite length, given > by a natural number of nodes. Presently we are considering the union > of all such finite paths. (The union of all finite natural numbers is > an infinite union - nevertheless this union cotains only finite > numberrs.) The set of all finite subsets of N is countable, the set of all subsets of N is not, and this will not change, regardless of the number of reformulations that you give. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page.
From: Dik T. Winter on 16 Mar 2007 11:02
In article <1174039357.642800.20000(a)d57g2000hsg.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 15 Mrz., 16:39, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > In article <1173954799.919385.61...(a)y80g2000hsf.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: .... > > > A path bundle splits off into two bundles which pass said node. > > > Therefore every set of path-bundles in the tree has a finite cardinal > > > number > > > > As long as path-bundles are finite sets of nodes that is right. > > You should say: As long as path-bundles are sets of nodes that is > right. No, the finite is required here. > There are only countably many nodes. Unless the number of path-bundles > first gets countably infinite it cannot become uncountably infinite. > Reason: Continuity of the tree. Pray define "continuity". > > > This, in the "limit", may yield an infinite number, but certainly not > > > an uncountable number without having an intermediate countably > > > infinite number. > > > > I do not know. What do you understand under "limit"? > > In this case nothing else but: "going on without end". > But I will agree to any other definition which you might supply. > The limit of U[k<n]{1,2,3,...,k} is the union of all finite initial > segments for example. By what definition of limit? Pray, for once, provide a definition. > > > The tree is continuous because its nodes are connected by paths. There > > > is never more than the factor 2. There are no interruptions possible > > > and no jumps from "finite" to "uncountable". Your claim would require > > > that. > > > > My claim requires nothing of the sort. The number of finite paths is > > countable. > > The number of paths of every set of finite paths is finite. Eh? So there are not even infinitely many paths? Why not? > The number of nodes of every set of finite paths is finite > > > The number of infinite paths is uncountable. There is no > > jump from countable to uncountable, there is a jump from finite paths > > to infinite paths. > > You seem to imply a difference between the set of all finite paths > (which is countable) and the set of infinite paths. Yes, there is a difference. The difference being that in one set all paths are finite and in the other set all paths are infinite. > In the union of all finite trees there are only finite paths. Why? The axiom of infinity states different. > So you > agree that in the union of all finite trees the number of paths is > countable? No, because in the union there are also infinite paths. Or do you now think that 0.010101... is *not* a path in that union? > The set of all finite paths covers the whole tree with all its nodes. > What is the difference, in terms of nodes, between the set of all > finite paths and the set of infinite paths? Not in terms of nodes, why should there be a difference in terms of nodes? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |