Cantor Confusion
in [Math]
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From: Dik T. Winter on 23 Mar 2007 12:09 In article <1174664147.264581.197260(a)n76g2000hsh.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 23 Mrz., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > As the geometric series contains only the smallest possible infinity > > > of terms, it should not cause problems to read it from behind. > > > > If there were a last one. As there is not a last one I have some > > difficulty with it, because I do not know where to start. > > You need not to start. Reverse all terms of the series simultaneously. Oh. How do I do that? What do I interchange the first one (1 with)? > > > Consider all of them simultaneously. You like to do so when Cantor's > > > diagonal proof is concerned. Every exchanged digit is followed by > > > infinitely many digits which have to be exchanged. Nevertheless you > > > say, it is possible to exchange all of them at one time. Consider the > > > paths of he tree - all at the same time. > > > > Yes, I do. What now? Every two paths separate from each other at some > > specific node, through all nodes go uncountably any paths, there is no > > level where all paths do separate. > > Because there is no "all paths". Pray, provide a proof (within set theory, where we are arguing). > > > At no point of the tree (and of the universe) more than countably many > > > path can be distinguished. > > > > I would state that as "at no node in the tree more than countably many > > groups of paths can be distinguished", > > That is where we agree. > > or, equivalently "at no node in > > the tree terminate more than countably many terminating paths". > > No path terminates. Yes, you want to confuse issues. > This > > still does not say anything about the number of non-terminating paths. > > That is where we disagree. All paths belong to groups of paths. Even > single paths belong to groups and are groups, if isolated. Yes, so what? > I think we have cleared our positions sufficiently: We agree in: At no > level in the tree more than countably many groups of paths (including > single paths) can be distinguished. And outside of he tree there are > no paths. > > You nevertheless believe in the uncountable while I do not. Well, within set theory it can be proven. But you do not believe in set theory. On the other hand, you have *not* proven that set theory is inconsistent. That you do not believe in the uncountable is *not* an argument against set theory. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 23 Mar 2007 12:11 In article <1174664333.103257.4920(a)y80g2000hsf.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 23 Mrz., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > Pray define "isolated". > > > > > > Two path are isolated at level k if they have digits m and n with m =/ > > > = n at a level i <= k. > > > > So each two non-terminating paths are isolated from each other. And now > > what? > > Now we see that their number is countable, because there are never > more than countably many paths-bundles including such with only one > path isolated from one another. The last part, again, makes no sense. If we exclude path-bundles that contain a single path, there are indeed countably many of such path-bundles, because they can be clearly identified with terminating paths. But that does say *nothing* about the number of single paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: mueckenh on 23 Mar 2007 13:26 On 23 Mrz., 17:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174664147.264581.197...(a)n76g2000hsh.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 23 Mrz., 15:58, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > As the geometric series contains only the smallest possible infinity > > > > of terms, it should not cause problems to read it from behind. > > > > > > If there were a last one. As there is not a last one I have some > > > difficulty with it, because I do not know where to start. > > > > You need not to start. Reverse all terms of the series simultaneously. > > Oh. How do I do that? What do I interchange the first one (1 with)? You are not so squeamish when exchanging every digit of Cantor's diagonal. Cantor, unless working every digit simultaneously, will never finish. In fact we need not know how to proceed, but it is sufficient to know hat the sum remains 2 what ever we do. However, you may proceed as follows (but only after having read my book): By transpositions (as we know them from Cantor's work) bring the n-th term (with n = 2, 3, 4, ...) into the first position and after that bring the n+1-th term into first position. If you are really fast, then you reverse the whole sequence of terms (because you can determine, for *every* term number n, the number of transpositions required to have it at the first position). > > > > > Consider all of them simultaneously. You like to do so when Cantor's > > > > diagonal proof is concerned. Every exchanged digit is followed by > > > > infinitely many digits which have to be exchanged. Nevertheless you > > > > say, it is possible to exchange all of them at one time. Consider the > > > > paths of he tree - all at the same time. > > > > > > Yes, I do. What now? Every two paths separate from each other at some > > > specific node, through all nodes go uncountably any paths, there is no > > > level where all paths do separate. > > > > Because there is no "all paths". > > Pray, provide a proof (within set theory, where we are arguing). I am arguing within the tree. There is never an uncountable number of separated paths. You say all paths were uncountable and separable. Conclusion. there is no "all paths" within the tree. > > > > This > > > still does not say anything about the number of non-terminating paths. > > > > That is where we disagree. All paths belong to groups of paths. Even > > single paths belong to groups and are groups, if isolated. > > Yes, so what? Therefore, if they existed isolated and were so many as you say, they would populate a level with uncountably many nodes. > > > I think we have cleared our positions sufficiently: We agree in: At no > > level in the tree more than countably many groups of paths (including > > single paths) can be distinguished. And outside of he tree there are > > no paths. > > > > You nevertheless believe in the uncountable while I do not. > > Well, within set theory it can be proven. Within the tree it can be disproven. > But you do not believe in > set theory. On the other hand, you have *not* proven that set theory > is inconsistent. That you do not believe in the uncountable is *not* > an argument against set theory. That you do believe in set theory is not a crime. But that you do believe in uncountaly many separations without uncountably many separations, that is hard to believe. Regards, WM
From: mueckenh on 23 Mar 2007 13:33 On 23 Mrz., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174655745.265882.25...(a)e65g2000hsc.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > The union of all *finite* paths gives the same infinite set of nodes > > as the union of all finite trees. But there is no infinite path > > involved / required. > > Not in the uniting, no. But the union *does* contain infinite paths. Just > like the union of all finite initial segments of the natural numbers does > have as a subset an infinite initial segment of the natural numbers. Just like the set of all negative unit fractions does cover zero? > > > > > > The path-length > > > > > is the cardinal number of the set that represents a path (actually > > > > > one less if the path is finite). > > > > > > > > No. You are in error. The cardinal number of an infinite set of finite > > > > paths is infinite. But that does not produce any path of infinite > > > > length. > > > > > > But I do not state that it does. The infinite paths are *unions* of finite > > > paths, where each finite path is an initial segment of the infinite path. > > > I am not talking about the cardinality of a set of path, but of the > > > cardinality of a path (the number of nodes in it). That does not mean > > > that an infinite set of paths contains an infinite path, that would be > > > false indeed. > > > > The union of all finite paths is an infinite union of finite paths, > > but not an infinite path. > > It is not even a path. It is the infinite tree. The union of all paths with only zeros 0., 0.0, 0.00, ... does not contain and not possess an infinite path. And it does not require an infinite path to pass every finite level. > > > The set of all negative unit fractions is infinite but neither any > > member nor the whole set does cover zero. > > What is the relevance? > To inform you about the difference between numbers and length of paths. Regards, WM
From: mueckenh on 23 Mar 2007 13:35
On 23 Mrz., 17:11, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1174664333.103257.4...(a)y80g2000hsf.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 23 Mrz., 16:10, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > Pray define "isolated". > > > > > > > > Two path are isolated at level k if they have digits m and n with m =/ > > > > = n at a level i <= k. > > > > > > So each two non-terminating paths are isolated from each other. And now > > > what? > > > > Now we see that their number is countable, because there are never > > more than countably many paths-bundles including such with only one > > path isolated from one another. > > The last part, again, makes no sense. If we exclude path-bundles that > contain a single path, there are indeed countably many of such path-bundles, > because they can be clearly identified with terminating paths. But that > does say *nothing* about the number of single paths. If there were any single path, then it had to exist within the tree. The tree with its infinitely many levels contains every path which can exist, including the infinite paths. Regards, WM |