Cantor Confusion
in [Math]
|
From: William Hughes on 8 May 2007 19:10 On May 8, 5:29 pm, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 8 Mai, 17:11, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 8, 11:00 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > On 8 Mai, 14:41, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > On May 8, 8:20 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > > On 7 Mai, 14:10, William Hughes <wpihug...(a)hotmail.com> wrote: > > > > > > > > > Every node, n, of path p is shared with another path > > > > > > > > p'(n). Note that p'(n) can be different for every n. > > > > > > > > Note that p'(n) need not diffrent for every n. > > > > > > > However, it cannot be the same for all n. > > > > > > For which n must it change? > > > > > > Could you explain, please, how you define a set of at least two > > > > > different paths, p' and p'' (or even more) which are required to > > > > > accompany p for all n? > > > > > Here is a very simple, countable example. > > > > > Let p be the path 000... > > > > > Let p'(1) be the path 0111... > > > > Let p'(2) be the path 00111... > > > > Let p'(3) be the path 000111... > > > > ... > > > > > let P' be the set of all the p'(n) > > > > > In general p'(n) has a n 0 nodes, followed by only 1 nodes. > > > > (Note that there is no last p'(n)) > > > > > A: for any node k, of p, there are an infinite number of paths > > > > in P' which contain node k > > > > But why do you write down these paths: > > > > Let p'(1) be the path 0111... > > > Let p'(2) be the path 00111... > > > Let p'(3) be the path 000111... > > > > ? > > > Any set p(1) up to p(n) can be substitued by a single path. > > In fact? Why then did you write down so many useless paths? > > > > > Look! Over there! A pink elephant! > > > The set P' can be substitued by a single path. > > Oh, does the set P' consist of any paths which are not paths p(n) with > a finite argument n? i: Any set of paths {p(1) ... p(n)) can be replaced by a single path. ii: The set P' is the union of sets {p(1) ... p(n)} Look! Over there! A pink elephant! iii: The set P' can be replaced by a single path. -William Hughes
From: William Hughes on 8 May 2007 19:21 On May 8, WM <mueck...(a)rz.fh-augsburg.de> wrote: time does not play a role in set theory and To branch off is a temporal act. This is not a contradiction in Wolkenmuekenheim, where temporal does not refer to time when it is inconvenient for Muekenheim. - William Hughes
From: Dik T. Winter on 8 May 2007 21:52 In article <1178459226.093794.184990(a)e65g2000hsc.googlegroups.com> WM <mueckenh(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. -- 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 8 May 2007 22:09 In article <1178461785.380972.277030(a)e65g2000hsc.googlegroups.com> WM <mueckenh(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. > 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. > > > 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). > > > Bunches terminate by splitting into other bunches. If they would not > > > split, then they would not terminate. > > > > This is nonsense. Bunches *all* start at the root by your definition. > > If a bunch splits in two bunches at a node that would mean that those > > two new bunches start at the node, and they do not. > > All start at the root node, but they terminate by splitting. How can they split if there are no bunches starting at the splitting point? Into *what* do they split? If you think that at a node a bunch splits in the many bunches that have all nodes in common from the root, I would say that at each node the number of bunches is diminished by one, and does not increase. > > > They terminate because they split. > > > > You mean edges, not bunches. > > No. Look at the discussion about that: > > http://groups.google.nl/group/AOTI/browse_frm/thread/1f7faf620bb90808 No thanks, much more of the same. Moreover, there is no discussion there, at least, not yet. > > What are separated bunches? > > Let us continue this discussion there. No. > > > It is the number of separations which counts! > > > > Before you say something like this you should first define your terms. > > When are two bunches separated? When I analise carefully, I find the > > following: > > (1) A path bunch is the maximal set of paths that have some set of nodes > > in common. > > yes. > > > (2) Two path bunches are separated if they have no path in common. > > yes. > > > (3) From (1) we can biject the path bunches with (ordered) sets of nodes > > that start at the root and where nodes are connected by edges. So a > > path bunch is either a finite path or an infinite path. > yes. > > (4) Two bunches are separated if there is a node in one of them (after the > > bijection) that is not in the other and the other way around. > yes. > > (5) A bunch comes in at a node when that node is (after the bijection) in > > the set of nodes, and it is the last node. > > (6) A bunch comes out at a node when that node is (after the bijection) in > > the set of nodes, and it is *not* the last node. > > When I use this I find that I can biject the bunches that are finite paths > > with the edges in the tree (and they are countable), because for all such > > bunches there is a terminating edge. I am still at a loss with those > > bunches that are an infinite path (actually a single path). > > Do they occupy more nodes than all the nodes occupied by finite > bunches. Look at a formalizaton by Franziska Neugebauer which I > replied to in: > > http://groups.google.nl/group/AOTI/browse_frm/thread/1f7faf620bb90808 Sorry, if you want to go there, quit here. -- 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 8 May 2007 22:20
In article <1178463181.745792.46180(a)y80g2000hsf.googlegroups.com> WM <mueckenh(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. 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. This does *not* mean that each real number is finitely definable. > > 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 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |