Cantor Confusion
in [Math]
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From: WM on 9 May 2007 10:34 On 9 Mai, 04:20, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178463181.745792.46...(a)y80g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 6 Mai, 04:40, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1178366406.361131.321...(a)y5g2000hsa.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > ... > > > > Theorem 2: The set of finitely defined numbers cannot be put into a > > > > list. > > > > > > > > Proof: Assume such a list exists. This means, such a list has been > > > > finitely defined. > > > > > > Wrong. Existence does *not* mean finitely definable. > > > > Marthematical existence does mean definable. What else should it mean? > > Definitions can only be finite. > > Right you are. Nice to hear. Here are some mathematicians who will even accept infinite definitions. > But that a set of entities is finitely definable does *not* > mean that each element of the set is finitely definable. Consider Cauchy > sequences. By (a finite) definition a Cauchy sequence denotes a real number. > This is a *finite* definition of the set of real numbers. Yes, of the set. But it is only a definition of those real numbers for which Cauchy sequences can be finitely defined. > This does *not* > mean that each real number is finitely definable. And it does not mean that every Cauchy sequence has a finite definition. > > > > Have a look at > > > Turings work. There *does* exist a (finitely definable) list of Turing > > > machines that purport to generate numbers. So the list is certainly > > > countable. However, the subset of that list that actually *does* > > > generate numbers is not finitely definable. > > > > Fine. That means, the set is countable but there is no bijection with > > N definable. Why then do you require a definable bijection between > > paths and nodes in the tree? > > Because with Turing machines the countable I think you meant computable > numbers form a subset of the > set of Turing machines, for which a bijection can be defined. So they > are a subset of a definable countable set. You want to show the set of > paths countable. As the set of nodes is countable, this either requires > that you give a definable bijection between a superset of the paths and > some countable set. You do not even need a bijection, an injection is > sufficient. So either you give an injection of the set of paths to a > countable set, or an injection of a superset of the set of paths to a > countable set. I gave a bijection between the set of nodes and the set of branching- offs of paths bunches. As there can be not more results of branching- offs than branching-offs, the task is done. Regards, WM
From: WM on 9 May 2007 10:46 On 9 Mai, 03:52, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178459226.093794.184...(a)e65g2000hsc.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 6 Mai, 03:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > There are no articles in this discussiongroup, they are either expired or > > > removed... > > > > Excuse me, here are the valid URLs: > > > > Binay Tree: > >http://groups.google.nl/group/AOTI/browse_frm/thread/1f7faf620bb90808/# > > Just more of the same. > > > Intercession > >http://groups.google.nl/group/AOTI/browse_frm/thread/6386cb0c4c23e619/# > > Intercession works only for ordered sets (as you note), and depends on the > ordering. But regarding your last question, consider the set of functions > from N to N. We order as follows: > (1) If two functions f and g are not equal, there is a smallest n such that > f(n) != g(n). > (2) We define f < g if f(n) < g(n), and f > g if f(n) > g(n). This is a > complete ordering on that set of functions. > Now try to find an intercession of this ordered set with the natural numbers. > You might also try to find an ordering of this set which *has* an intercession > with the natural numbers... > > > > And, I do not discuss through google.groups. > > > > Sometimes it is preferable to have a calm atmosphere. > > I have first to sign up, next I get probably an interface that can only be > handled with a mouse. No, thanks, I have already too many problems with > mouse use. I have got the same interface as here. I will answer there. This thread has become fairly long. Some credit for that goes to me. Some credit goes to MoeBlee for his understanding of function. Regards, WM
From: William Hughes on 9 May 2007 10:46 On May 9, 10:34 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > I gave a bijection between the set of nodes and the set of branching- > offs of paths bunches. As there can be not more results of branching- > offs than branching-offs, the task is done. There are only a countable number of results of branching-off. Whoops, an inifinite path is not a result of branching-off Look! Over there! A pink elephant. There are only a countable number of infinite paths - William Hughes.
From: WM on 9 May 2007 10:57 On 9 Mai, 04:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178461785.380972.277...(a)e65g2000hsc.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 6 Mai, 03:53, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > > But there are not two different sides. Both statements are not in > > > > > contradiction with each other. > > > > > > > > Wrong. If there is always a path p' with p, > > > > > > What does this statement mean? > > > > Even in the *infinite* tree a path cannot be distinguished from all > > other paths. > > Why not? For each other path there is a finite node where it goes a different > way. Obviously not. For each node there is a co-path p' which has been with p all the way long. > > > That means, a real number which cannot be described by a > > finite formula (and most of them cannot) does not exist. > > That is simply opion. Indeed, only countably many can be described by a > finite formula (as is clear from the work of Turing). In mathematics that > does *not* mean that the other numbers do not exist. How do you address, represent or use it in any other form "in mathematics"? > > > > > Hence Cantor's diagonal proof fails. > > > > > > Which proof? > > > > Cantor's diagonal proof fails, because the diagonal number is never > > distinguished from all other real numbers (if uncountably many real > > numbers exist). > > But that is not the proof. The proof is that the diagonal number is > distinguished from the real numbers in the list (which are countably many). That's your (and others') error. The proof in the tree is, that for every path p there is another path *existing in the tree* which is not different from p. And if you apply Cantor's proof in the tree, by forming a "diagonal" by switching a bit for every path, then it is undisputed that the constructed diagonal number is represented by a path in the tree. Why should this be different in Cantor's list? Regards, WM
From: WM on 9 May 2007 11:10
On 9 Mai, 15:50, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1178628554.931048.21...(a)y80g2000hsf.googlegroups.com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 7 Mai, 22:36, MoeBlee <jazzm...(a)hotmail.com> wrote: > > > > > (1) Who is the author of that book? I'd like to look it up to see just > > > what is written there. > > > > Karel Hrbacek and Thomas Jech: "Introduction to Set Theory" > > Marcel Dekker Inc., New York, 1984, 2nd edition. > > They do not *use* the word "formula". No, they don't, but I use it. Nevertheless it is synonymous to what they use. > But that is not the definition, their formal definition is: > "A binary relation F is called a function (or mapping, correspondence), if > aFb_1 and aFb_2 imply b_1 = b_2 for any a, b_1, and b_2." And how do you think the a's and b's are selected? And from what sets do you think are they selected? And why do you think H&J give that informal statement? Just in order to confuse the students? It is ridiculous to follow this discussion. I can assure you, if one of my students would not know that a function is a formula (or rule or whatever) together with a domain where it is defined and a range, then he or she would not pass the exame. And this is the same in the better math courses in Germany. Regards, WM |