On Jul 1, 6:40 pm, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > In article <9dfc93db-dfea-48b1-9f2b-04f257f37...@s9g2000yqd.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes: > > On 30 Jun., 16:12, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > > > > I do not argue that there is a bijection with all real numbers. > > > > > > You did assert that there was a bijection with the set of all paths in the > > > tree. You have not proven that. > > > > I do not argue that there is a bijection with what you dream of as > > "all real numbers". > > Sorry, no dreaming involved. You did show that you could cover each node by > countably many paths. You did *not* show that the set of all paths was > exhausted by that process. > > > I show that the number of paths that can be identified by sequences of > > nodes or bits is countable. > > Not at all, because you did not show a bijection. > > > > > The > > > > reason is that real numbers with infinite bit sequences do not exist > > > > *as bit sequences". > > > > > > Oh, well, you are apparently a non-standard form of "exist". By the > > > axiom of infinite, the infinite set N exists, and so there does exist > > > a function: > > > f: N -> {0, 1} > > > f(n) = 0 if n is odd > > > = 1 if n is even. > > > I would think that would be an appropriate infinite bit sequence of 1/3. > > > > And I have shown that this sequence is covered by finite sequences to > > the satisfaction of everybody who is using bits to identify a number. > > Ah, you are easily satisfied. So 0.010101 identified 1/3? > > > > > pi exists in form of many formulas, Vielta, > > > > Wallis, Gregory-Leibniz, Euler, for instance. But it does not exist > > > > asa a subject to Cantor's proof. All existing sequences are in my > > > > tree. And obviously there are only countably many. > > > > > > I would argue that the infinite bit sequence .0101010101 exists and is in > > > your tree. Your bijection above does not include that bit sequence. > > > > The tree will not differ in any detail from the construced one, if we > > use all finite paths as before but add the sequence 010101... instead > > of 000.... > > So your bijection did *not* include that path. Why do you not admit that there > is a path that was not included in your bijection? And when you add the > sequence 01010101... instead of 000... the bijection does not include 000... > which is also a path in your tree. > > > > > > What do you mean with "covering a path"? Until now you were talking > > > > > about "covering nodes". And whatever it may mean, in what way does > > > > > "every node is covered" show that there remains "no path to be > > > > > covered"? > > > > > > > > Try to find a path p that contains a node not yet covered by a path q > > > > used to construct the tree. > > > > > > Is *that* a definition of "covering a path"? But if your paths cover all > > > nodes, there is no such path. > > > > That' swhy you cannot find it. > > Yes, but not yet given any definition about "covering a path"... And there > are paths present in the tree where all nodes in the path are also covered > by other paths. Consider the tree as you presented (starting with a list > of nodes, of which only finitely many are labeled with 1, and, yes, I do > admit that that list is countable). You created a mapping of paths to nodes > such that each node was covered by a path. Also you did admit that the path > 01010101... was also in the tree, but your mapping did *not* include that path. > So you did exhaust all nodes but not all paths. > > Now come up with a *bijection* between the set of nodes and the set of paths > in your tree, and be sure that there is *no* path in the tree that is not part > of that bijection. Only when that is done I will believe in your so-called > bijection. > > We start with the following definitions: > a node is a finite sequence of 0's and 1's > a path is an infinite sequence of 0's and 1's > If you do not agree with these definitions, provide your own definitions. > > > > > It contains > > > > already all paths that can be distinguished by nodes, i.e., all reals > > > > that can be distinguished by digits. And As my construction B shows, > > > > this set is countable. > > > > > > And the path .0101010101... is not in that bijection, nevertheless there > > > is no node in the tree that is *not* covered by that path. So you did > > > show only a bijection between a subset of the paths with the nodes. > > > > I showed hat all paths are in the tree that can be identified by > > nodes. > > Yes, so you did not show a bijection. That is more than what you do state now. > > > > > > > > > No. From that we can obtain that the number 1/3 is also in a list > > > > > > of terminating rationals. There does not exist a sequence > > > > > > 0.010101... > > > > > > that is longer than *every* finite sequence of that form. > > > > > > > > > > Ah, so now you contend that 1/3 is a terminating rational. > > > > > > > > No. It is a rational that has no decimal and no binary representation. > > > > It has a ternary representation though. > > > > > > Eh? You explicitly did state (as I did quote): > > > "From that we can obtain that the number 1/3 is also in a list of > > > terminating rationals." > > > I would state this to mean that 1/3 is a terminating rational. > > > > All that can be identified of 1/3 belongs to a terminating rational. > > Your "..." has no meaning in rigorous mathematics. > > So you would like to abolish limits from rigorous mathematics? Do you have > any idea about the meaning of the "..." there in rigorous mathematics? I > do not think so. > > Moreover, one left-over question you always snip and never answer: > > (1) FISON{n} = {1, 2, 3, ..., n} > union(k = 1...n) FISON(k) is a FISON > so > union(k = 1...oo) FISON(k) is a FISON > > (2) P(n) = 1/n! is rational > sum(k = 1...n) P(k) is rational > so > sum(K = 1...oo) P(k) is rational > > can you explaing why (1) is correct but (2) is wrong (that is what you state)? > -- > dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland;http://www.cwi.nl/~dik/
Because (2) is wrong and (1) is correct below: Results 1 - 10 for musatov bijection. (0.68 seconds)
1. Answer to Dik T. Winter - sci.math | Google Groups the bijection of the rationals > 0 and the naturals I have so often shown, .... Cosmic Lensing os very handy for Musatov's proof P=NP<<>> ... May 27 by ... groups.google.com/group/sci.math/msg/2f0140fa143ab15f 2. Math Forum Discussions that for every x in X there is a bijection between ... furnishes such a bijection. If E_x is "infinite to the ... Martin Michael Musatov. 6/11/09 ... mathforum.org/kb/thread.jspa?messageID=6749040&tstart=0 3. [PDF] <a href="http://arxiv.org/abs/0904.3116v1">arXiv:0904.3116v1 [cs ... File Format: PDF/Adobe Acrobat - View as HTML Apr 20, 2009 ... performs a bijection between L ...... D. Musatov, Extractors and an effective variant of Muchnik's theorem. Diplom (Master thesis). ... arxiv.org/pdf/0904.3116 by D Musatov 4. Math Forum Discussions On May 14, 12:07 pm, Martin Musatov <marty.musa...@gmail.com> wrote: ... forums ... write bijections between any infinite set S and S \F for any ... mathforum.org/kb/thread.jspa?messageID=6714416&tstart=0 5. Integer Format Sequence for linbox-use - linbox-use | Google Groups May 3, 2009 ... From: Martin Michael Musatov <marty.musa...@gmail.com> ..... D A000045 N.S.S.Gu,N.Y.Li and T.Mansour,2-Binary trees:bijections ... groups.google.com/group/linbox-use/browse.../7bc6976f1181638c 6. Variations on a theme of Debs and Saint Raymond remarkable theorem which includes that of Ivasev-Musatov as a particular case. Theorem 8. ...... be a bijection. Set N(0) = 0, ? ... jlms.oxfordjournals.org/cgi/reprint/79/1/33.pdf by TW Korner - 2009 - Related articles - All 5 versions 7. Result for query "keyword(s)=theorem author= title=" On a theorem of Ivasev-Musatov III, ...... by Theorem; bijection, we get that the theorem holds for all $\pi(\lambda)$ with; _ \chi(G_ 1)$, see \cite KR. ... nyjm.albany.edu:8000/cgi-bin/aglimpse/19/nyjm/Http/.../j?... 8. [PDF] ABOUT INTERPOLATION OF SUBSPACES OF REARRANGEMENT INVARIANT SPACES ... File Format: PDF/Adobe Acrobat - View as HTML and G is bijective. Theorem 1.5. Let r.i.s.'s X ...... "Mir",. Moscow, 1965 (Russian), translated from English by O. S. Ivasev Musatov. ... www.emis.de/journals/HOA/IJMMS/Volume25_7/465.pdf by SV ASTASHKIN - Cited by 9 - Related articles 9. Proving the Parallel Postulate of Geometry from the Number Systems ... May 28, 2009 ... From: Martin Musatov <marty.musa...@gmail.com>. Date: Fri, 29 May 2009 08:46:13 -0700 (PDT) ... C (* Binary expansions as a bijection ... groups.google.com/group/sci.logic/browse.../b01f8ea2c5255dbc? lnk... + + + Martin Musatov
