Cantor Confusion
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From: Dik T. Winter on 5 Jan 2007 20:05 In article <1167999736.624629.206820(a)s80g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > And my question is: Do these two trees, namely the complete tree and > > > the union of all rational trees differ such that the one has edges or > > > nodes which are missing in the other? > > > > No. > > > > But the distinction is apparently difficult (although about first year > > at University for mathematics). The set of terminating binary expansions > > is countable, the set of non-terminating binary expansions is no > > countable. (You may replace terminating binary expansions with binary > > expansions terminating with either a continuous stream of 0'z or of 1's.) > > But these streams are *not* present in all paths of the union of all > rational trees! That has been overlooked, as it appears, even in the > last years of university mathematics - in the last 130 years. Oh. > Can you really believe that a thinking brain will accept your assertion > that two absolutely identical systems of nodes and edges will supply > different systems of paths, i.e., strings of nodes and edges? Yes, it entirely depends on how you find your paths. Consider a graph consisting of three sets, (1) the edges, (2) the nodes and (3) the paths. Consider the following three graphs: x x x / \ / \ x x x---x x---x where we identify the nodes from the different graphs and also the common edges. It is clear that if we define a path in a graph as a sequence of nodes connected by edges, we see that in each complete graph there are three paths. So we may say that the left hand graph consistes of the three sets: {n1, n2, n3}, {e2, e3}, {n2-e3-n1, n1-e2-n3, n2-e3-n1-e2-n3} (e2 is opposite n2, etc.) When we take the union of the first two of them we get the following graph: {n1, n2, n3}, {e1, e2, e3}, {n2-r3-n1, n1-e2-n3, n3-e1-n2, n2-e3-n1-e2-n3, n3-e1-n2-e3-n1) And if we take the union of all three we get the following graph: {n1, n2, n3}, {e1, e2, e3}, {n2-e3-n1, n1-e2-n3, n3-e1-n2, n2-e3-n1-e2-n3, n1-e2-n3-e1-n2, n3-e1-n2-e3-n1} See, the same set of nodes and edges, but different sets of paths. Moreover, they all miss one path from the complete graph with three nodes and edges, the circular: n2-e3-n1-e2-n3-e1-n2 We clearly see that the union of a set of (complete) graphs does not necessarily give a complete graph. It is quite similar with your tree. The union of all finite trees gives a tree with all the nodes and edges, but not all the paths of the complete tree. While the union of the sets of edges and nodes behave normal, the union of the sets of paths is *not* the set of paths of the complete tree. And you are arguing about the union of the sets of paths. > > > > You are wrong. sqrt(2) has a pretty good representation: "sqrt(2)". > > > > > > That is a name. Name ist Schall und Rauch. > > > > In the same way '2' is a name, '10' is a name. '10' is clearly a name > > for a quantity that depends on the context. In the same way 'sqrt(2)' > > is the name for a quantity that depends on the context. For instance, > > the wedge that is used for the quantity 7 in Hyderabad Arabic is used > > for the quantity 8 in Devanagari and for the quantity 6 in Javanese. > > Therefore these wedges are not numbers but only names which can mean > anything we define. Right. And so is every notation of numbers. Numbers are not concrete entities, they are abstract entities. And we name them by symbols or strings of symbols according to particular convention. > But in Devanagari and Hyderabad Arabic and Javanese it is clear what we > mean by > > |||||| > ||||||| > |||||||| Well, even I do not know. I would say either 8 or 7. But try that in a culture that has no idea about abstract entities. Yes, there is one such culture, a tribe of Indians along the Amazone. They can not count, and do not count, and do not understand counting at all. In their culture only concrete entities are acceptable (and that also only if the person telling about it has personal experience with it). -- 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 5 Jan 2007 20:20 In article <1168002324.590103.255240(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1167859855.107241.239690(a)k21g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > That is precisely the infinite string of finite digit indexes of > > > irrational numbers and similarly the infinite string of finite nodes of > > > infinite paths. But one has no infinite paths in the tree? > > > > Not in the union of finite trees, but it is in the union of the full tree. > > The two are not the same. > > Name a level or a node or and edge which are in one of the trees only. Why should I do that? I can name a *path* that is in the complete infinite tree but not in the union of finite trees. See my previous article about that. > > It is equal. What Virgil is stating is that > > N in U{n in N} {1, 2, 3, ..., n} > > but that N is not one of the subsets used in the union. > > But, according to your point of view, N exists as the union of all > finite sequences {1, 2, 3, ..., n}. Why do infinite paths not exist as > the union of all finite paths? How many times do I have to explain that to you? Use the definition of the union of an infinite collection of sets. If an element is in one of the sets from the infinite collection it is in the union, if it is in none of the individual sets it is also not in the union. You were talking about the union of sets of finite paths. Asking for an infinite path in that union is the same as asking for an infinite number in the (particular) union of sets of finite numbers. -- 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 5 Jan 2007 20:37 In article <1168004631.146333.67680(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: ..... > > > > (3) Infinite paths are in none of the finite trees, so they are > > > > also not in the union. > > Infinite node numbers are not in any of the finite trees, if you cut > them at a certain level (contrary to my initial definition with > sequences of 111... and 000... ). Does the union of such "cut-off" > trees not contain infinitely many nodes? Yes, but not infinite node numbers. > > > So the set of natural numbers is not in the union of all initial > > > segments of natural numbers? > > > > Wrong. Infinite paths are not in the union of finite trees because none > > of the finite trees contains an infinite path. All natural numbers are in > > the union of the finite segments because for all natural number there are > > initial segments that contain it. > > Similarly, for any level of the tree there are finite trees which > contain this level. > Similarly, for any node of a path, there is a finite path which > contains it. > > You said that in the union of all finite trees there is no infinite > path. > Would you say that in the union of all finite trees there is a path > which contains infinitely many nodes? No. Because in none of the finite trees there is a path that contains infinitely many nodes. So the two questions following are irrelevant. > > How can an infinite path get into the union of finite trees if none of the > > finite trees contains one? By magic? All paths that are in the union of > > finite trees after a finite number of steps go only either to the left or > > to the right. > > How can an infinite set N get into the union of all finite initial > segments? It *is* the union. It does not contain any infinite element because none of the constituents contains an infinite element. And if you think that N elementof N, you are wrong. > > Have you *looked* at V=L? But you are unwilling to look at it because > > you are unwilling to believe it. > > I need no belief. I know (from your reaction on the tree, among other > evidence) that there are no irrational numbers (except for a few names, > if you wish to call them numbers), so I will not spend any effort to > look at V=L. The result can only be an assertion that well-ordering is > possible. If there really was a well-ordering accomplished, then we > would no longer need the axiom but could use the well-ordering. You are dense. The well-ordering is given under the conditions of the axiom. If the axiom is false the well-ordering is not given. > (If you really have two parallels, then you do not need the axiom that > parallels exist.) How do you prove they are parallel? Note moreover that Euclid postulated that there was only a single parallel. So in a finitistic world Euclid's postulate is false. There is more than a single parallel. > > The trees are not different with respect to edges and nodes, it is the set > > of paths that you allow in the trees that are different. > > I have not to allow for anything. Every path which is possible does > exist as soon as all its edges exist. No, you are talking about the union of sets of paths. That union is different from the complete set of paths, as is easily shown. > > In the union of > > finite trees you allow only finite path, when you want completion you also > > do allow infinite paths. > > That is wrong. IF infinite path do exist, THEN they exist in in the > union of all trees, because there is NOTHING to be completed if already > ALL is together. In that case do *not* talk about unions of sets of 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: Dave Seaman on 5 Jan 2007 21:06 On Fri, 05 Jan 2007 19:28:19 EST, Andy Smith wrote: > OK, thanks. I will go back to physics. > I do now see what you mean, but it still strikes me as a > bit Humpty Dumpty - " aword means what I see it means, > nothing more and nothing less". So you can't have an > actually infinite natural number, because by definition > all natural numbers are finite, although the set of all > natural numbers is infinite. So numbers such as ..1111 are > not realisable, and the countability of the reals is > fundamentally different from that of the integers. The reals are uncountable. > I still wonder what the label attached to the point > exactly at 2.000... on the sum 1 + 1/2 + 1/4 ... is going > to read though. You haven't assigned it one. If I were to give it a label, it would probably be omega, the first transfinite ordinal number. Omega is not a natural number. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: MoeBlee on 5 Jan 2007 21:07
Andy Smith wrote: > I do now see what you mean, but it still strikes me as a > bit Humpty Dumpty - " aword means what I see it means, > nothing more and nothing less". Read the chapter on mathematical definitions in Patrick Suppes's 'Introduction To Logic'. This is the best general textbook introduction to the subject I've seen, and just technical enough without being too technical. It will cast a great deal of light for you on how definitions work in mathematical theories. MoeBlee |