Cantor Confusion
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From: mueckenh on 6 Jan 2007 13:21 Dik T. Winter schrieb: > 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. What is the difference, please? Is there a tree with more nodes than the union of all finite trees has? > > > 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. After which node does the first non-infinite = finite path of the union end? >Because in none of the finite trees there is a path that contains > infinitely many nodes. So the two questions following are irrelevant. None of the sequences {1,2,3,...,n} contains infinitely many numbers. Nevertheless, you assert that the union of all of them contains infinitely many numbers. > > > > 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. The same is valid for the infinite paths in the complete tree. The complete tree is the union of all finite trees. The infinite path is the union of all respective finite paths. > > > > 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. THEN GIVE IT PLEASE, HERE AND NOW ! Simply reproduce it. > If the axiom is false the well-ordering is not given. > How could it disappear after having existed? Who would force us to forget it, if it had been given? > > (If you really have two parallels, then you do not need the axiom that > > parallels exist.) > > How do you prove they are parallel? By measuring their distance in a plane. > 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. False. Consider your screen. _________________ x Draw more than one parallel to the line through the centre of the x. > > > > 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. What counts is only: There cannot be more unions than elements. > > > > 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. I have a union of all finite trees. The result is an infinite tree (which need not be an element of the union). > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Which is the postal address by which you would like to receive my book? Please write capitals where appropriate. Regards, WM
From: Virgil on 6 Jan 2007 14:15 In article <1168076831.751014.212600(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > I do not assume that 0.111... exists. I only assume that > > > > given that 0.111...1 exists you can show that 0.111...11 exists. > > > > > > > That position is correct. It is called potential infinity. > > > > > > > > > > That is the same with the lines. Why should the diagonal exist > > > > > actually > > > > > but the system of lines should not exist actually? (1/9 = 0.111...) > > > > > > > > Indeed. It is the same with lines. It is always possible to find > > > > another line. However, iii says nothing about whether the > > > > diagonal "actally exists" (or equivalently whether the > > > > system of lines "acutally exists"). > > > > > > But you always assert that the the complete diagonal exists, i.e., a > > > diagonal which could not be extended. Then you must also accept a line > > > system which cannot be extendend. > > > > No. At no time do I assume that the complete diagonal exists. > > > > > > > > > > > > > > > iv: Not all the elements of the diagonal exist > > > > > > > > > > Not all natural numbers in unary representation 0.1, 0.11, 0.111, ... > > > > > exist. If all elements of the diagonal exist (which are the last > > > > > digits > > > > > of the unary numbers) then these unary numbers must exist too, as far > > > > > as I understand existence. > > > > > > > > Any discussion of iv is completely irrelevant. iv is neither needed > > > > nor used. > > > > > > Then drop it. > > > > I have never used it ("neither needed nor used"). > > > > > > > > You used to use ~iv: all the elements of the diagonal exist > > > > > > > Only to point out that I neither need nor use the > > assumption ~iv. > > > > > > The contradiction is: > > > > > > > > A: L_D must have the property that given any set of elements > > > > that exist in L_D one can always find another element of > > > > L_D > > > > [This follows immediately from i ii and iii] > > > > > > It is not different for the lines. > > > > > > > > B: L_D has a largest element. [this follows from the fact the > > > > L_D > > > > is a line] > > > > > > This follows from the assumption (iv) of complete existence of L_D, > > > which implies complete existence of the system of lines. If you drop > > > it, then the contradiction vanishes. > > > > No, it follows from the fact that L_D is a line. A line has a largest > > element. > > The diagonal has not a largest element Since L_D does have a largest and the diagonal does not, the diagonal will contian elements larger than any in your L_D.
From: Virgil on 6 Jan 2007 14:43 In article <1168077250.963396.178270(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > 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! > > > > No one, except possibly WM, is claiming that an infinite binary tree is > > the same as the union of all binary rational, and therefore finite, > > trees. > > > What else could it be? It counld be, and is, the union of infinitely many finite trees. Since the union of two trees is not a tree, whyever should the union of more than two trees be a tree? > If it is something else: which nodes are distinguishing it from the > union of all trees? The fact that a union of two or more trees has more than one root node distinguishes that union from any single tree. > > > Those who have thinking brains are > > capable of distinguishing between the union of infinitely many finite > > trees and a single infinite tree. > > But they strictly refuse to tell what the difference is. Among other differences between individual trees and unions of trees, is that the union of more than one tree has more than one root node. > > According to set theory, most numbers have no names, no addresses and, > therefore, most probably, no existence. What axiom requires that a number have either a name or an address? > > ========================================== > > >> So where are the nodes and edges which make the infinite paths longer > >> than any finite path? It is the absence of any terminal edge or node in an infinite path that makes it different from a finite path. > ========================================== > > > I do not know what sort of trees WM is dreaming about, but in my > > infinite binary trees, at each node there are infinitely many paths > > branching left and infinitely many others branching right. > > Is in the union of all finite trees one node through which *not* > infinitely many paths > are branching left and infinitely many others are branching right? As each node in the union is a member of a particular finite tree and each node in a finite tree lies in only a finite number of paths, yes. > > ========================================== > > > And the only object that can be a member of that union of trees is an > > object which is a member of one of those trees. And every path in a > > finite tree is a finite path. > > What about the union of all finite path? When one forms the union of two trees one does not form a union of any path with any other path. To do that sort of "unioning" of paths requires a different operation than mere unioning of the trees, > > =========================================== > > > That infinite union of finite trees contains infinitely many nodes and > > infinitely many edges in infinitely many finite paths, but no infinite > > paths. > > No infinite path? Right! > Then every path is finite. Yes? In such a union, right! > The tree has oly a finite number of levels. Which tree? There are infinitely many finite trees in the union of all finite trees, each tree of which has only a finite number of levels, and the union contains only those original finite trees, nothing new. WM is imagining that when one forms the union of two or more trees one gets a new tree which includes all the separate trees as subtrees, but that new tree is not produced by mere union. If WM wants such a construction, he will have to call it something else besides "union" as that word already has a meaning which prohibits that sort of interpretation.
From: Virgil on 6 Jan 2007 14:45 In article <1168107015.838378.71940(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > > > > 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. > > There are no paths to find. A tree with no paths to be found is not a tree.
From: Dik T. Winter on 6 Jan 2007 20:21
In article <1168107015.838378.71940(a)51g2000cwl.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > 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. > > There are no paths to find. Paths do exist in the tree. > > > 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 > > Are you joking? We are talking about paths representing real numbers. > There is no real number which simutaneously has a 1 and a 0 at the same > place. Your example is totally useless. Are you joking? We were talking about unions of finite trees. > > 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. > > Only if you insist that there are real numbers which simultaneously > have different numerals at the same position. What nonsense. Can you point to an infinite path in any of the finite trees? So how does it come in the union of the finite trees? It is certainly *not* in the union of the sets of paths in the finite trees, by the definition of union. The example was just to show that when you look at the sets of paths in subgraphs, there union is not necessarily the full set of paths in the union of the subgraphs. > > 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. > > No? Who told you so? Or is this your own fantasy? Can't you follow a proof if presented? > There are infinite paths because the paths are not finite. You should > know from set theory: An infinite sequence is a sequence which is not > finite. But those infinite paths do *not* emerge from the union of the sets of paths in the finite tree. > > And you are arguing about the union of the sets of paths. > > Of course, I am arguing about the union of the set of all path which > really exist in a binary tree. You were arguing about the union of the sets of paths in the finite trees. > This union is an infinite set of paths (although the set of paths in > each finite tree is finite), and there is no hint on uncountability. I do not disagree that the set is infinite. I only disgree that that set contains an infinite element (path). That the set is infinite is trivial (just like U{n in N} {1, ..., n} = N. But there is no infinite path in the union of the sets of paths, just like there is no infinite number in the union of initial segments. And that is because *none* of the constituent sets in the union contain such a thing. > Take the most right path: It is the union of all numbers 0.111...1 with > n "1". This union is the number 0.111.... How can that path be the union if none of the constituents is that path? The union of the sets {0.1}, {0.11}, {0.111}, etc. would be the set {0.1, 0.11, 0.111, ...}, but 0.111... is not in it. If it is in it is must be in at least one of the constituent sets of the union. And I have no idea how you come to the "union" of numbers when they are rational. > Similarly the union of all > initial segments of a well-ordered countable set is this countable set. That is, eh, slightly different. > The union is not an element of the initial segments. Of course not. But each element of the union is an element of one of the initial segments. > But even in > infinity, there is only "one" union and not uncountably many unions. Indeed. And the union of sets of finite paths is a single set of finite paths. > > Right. And so is every notation of numbers. > > Not the notations below. You think so. > > Numbers are not concrete > > entities, they are abstract entities. > > No. Numbers are very concrete. Abstraction can be introduced but need > not, at least not for small numbers. When have you even seen five? It is only when you apply it to something that it becomes concrete. Five apples, five meters, five whatever. > > > 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. > > Do you need new spectacles? Yes, when driving. Not when reading. I have still no idea whether that is 8 or 7. If I count strokes I come to 8 strokes. If I have also to use length, I come to strokes with a total length of 7 units. So what is it? > > 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. > > They will understand, at least by experiment, > > oo > ooo > _____ > ooooo You apparently have not done the experiment. No, they will not understand, and such experiments have been done. > > In their > > culture only concrete entities are acceptable (and that also only if > > the person telling about it has personal experience with it). > > That is an advantage over people who see different identical trees. Perhaps, but you do not have history. Once every person that has personally experienced some event has died, the event fades away, because it has become too abstract. On the other hand, I see the difference between a union of finite graphs and the completed union, something you are not able to see. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |