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Topic: Answer to Dik T. Winter
Replies: 441   Last Post: Feb 5, 2013 6:25 AM

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Re: Answer to Dik T. Winter
Posted: Jul 1, 2009 9:48 PM

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.
>
> 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.
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 ...
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 ...
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 ...
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 ...
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 ...
lnk...
+ + +
Martin Musatov

Date Subject Author
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Dik T. Winter
5/27/09 mueckenh@rz.fh-augsburg.de
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5/30/09 G. Frege
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