Cantor Confusion
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From: David Marcus on 8 Jan 2007 18:40 mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > I cannot accept ghosts in mathematics, but, of course, I cannot prove > > > that they do not exist. So we are finshed with this topic. > > > > WE, on the other hand, can prove that what WM regards as ghosts do exist > > in several versions of set theory. > > That makes these versions of set theory vakueless. > > >As WM has no specific version of his > > own, he effectively has none at all. > > There is no version of infinity other than by belief in ghosts (lacking > mathematical spirit) I thought you said you were finished with this topic. Why post again on the same topic? -- David Marcus
From: David Marcus on 8 Jan 2007 18:53 Virgil wrote: > In article <1168290342.751050.317770(a)51g2000cwl.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > > 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 could 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? > > > > The union of two trees is a tree. > > Then it will have nodes which are simultaneoulsy terminal nodes and not > terminal nodes. One can embed one tree into a larger tree, but that > operation is not union. Obviously, it is impossible to have a sensible discussion with WM. However, I think the nonsense discussion will be easier for everyone to follow if you and Dik assume that when WM says "union", he really means to have all the finite trees be subsets of the infinite tree. If we do it this way, then all the nodes and edges in the sequence of finite trees are also in the infinite tree, but there are paths in the infinite tree that do not appear in the finite trees. -- David Marcus
From: Franziska Neugebauer on 8 Jan 2007 19:05 mueckenh(a)rz.fh-augsburg.de wrote: > The finite union of two or more finite trees is a finite tree. An > infinite union of finite trees is the infinite tree. What is an infinite union? (Please give a definition) F. N. -- xyz
From: David Marcus on 8 Jan 2007 19:12 mueckenh(a)rz.fh-augsburg.de wrote: > There is a union of paths for every path. > There is a union of trees yielding the infinite tree. > Every finite tree reaches from the root to a level n. > The union contains every level which is a natural number. > > Every node which is placed on a finite level (= every digit with a > finite index) is in the union of all finite trees. Therefore every path > containing nodes on a finite level (= every sequence of digits at > places with finite indexes) is in the union of finite trees. There is > nothing remaining! Consider the path in the infinite tree that always goes left. Which finite tree is this in? -- David Marcus
From: Dik T. Winter on 8 Jan 2007 19:18
In article <1168291077.860958.62580(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1168107015.838378.71940(a)51g2000cwl.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > > > 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. > > Ok? Every infinite tree, which contains all levels enumerated with > natural numbers, contains all possible paths and, therefore, contains > the representations of all real numbers of [0, 1]. In that case the set of paths in the infinite tree is not the union of the sets of paths in the finite trees. > > > > 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. > > But not such crippled plants with horizontal edges. It is only to show that the union of sets of paths is not necessarily the set of paths in the union. > > 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 union of all finite initial segments {1,2,3,...n} is the infinite > initial segment N. So the infinite tree is the union of all finite > trees. Yes, so what? I repeat that I am talking about the union of the sets of paths. > The union of the one-level tree > > 0. > / \ > 0 1 > > and the two-level tree > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > > is the two-level tree > > 0. > / \ > 0 1 > / \ / \ > 0 1 0 1 > > > similar to the union of the segment {0,1} and the segment {0,1,2} which > is the segment {0,1,2}. > > And so on by induction. Ok, so you are *not* talking about the union of sets of paths? > > > > 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? > > A proof well, but not such nonsense of crippled plants which are not > trees in our sense. I am talking about the union of sets of paths. And I am only stating that the union of the sets of paths is *not* the set of paths of the complete tree. What is your problem with that? Because there is *no* infinite path in any of the sets of paths in the finite trees, there is also *no* infinite path in their union. > > > 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. > > You are aware that you jmust try to disregard this idea, because it > kills set theory. I see. Why? > Infinite paths emerge from the union of all finite paths as the union > of all finite initial segments {1,2,3,...n} is the infinite initial > segment N. So the infinite tree is the union of all finite trees. I am talking about the union of *sets* of paths. Not about the union of individual paths (whatever that may be). > > > 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. > > So no infinite set N emerges from the union of all finite initial > segments {1,2,3,...n} ? I agree! But, I am afraid, you do not. No, because I do not state that. I only state (I repeat): there is no infinite number in the union of initial segments and that means that N does not contain an infinite number. It contains only finite numbers (but infinitely many). And the union of all finite initial segments *is* N. > > 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. > > Yeah. You are beginning to see why the union of all finite numbers n > canot yield an infinite set. U{n is natural} {n} = N. Because every k in N is in one of the sets used in the union, namely {k}. {0.111...} is *not* in U[ {0.1}, {0.11}, {0.111}, ...} ] because it is in *none* of the sets used in the union. > > > Similarly the union of all > > > initial segments of a well-ordered countable set is this countable set. > > > > That is, eh, slightly different. > > Why? The union of {0.1}, {0.11}, {0.111}, ..., is *not* {0.111...}, it is: {0.1, 0.11, 0.111, 0.1111, ...}. Remember: unions are defined for *sets*. I do not know any definition of union for numbers as these. > > > 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. > > An infinite set. Right. But no infinite paths. > > > > > 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? > > It must be 8, because 7 woud not be unique (it could then also be 9). > But a minimum amount of convention is of course necesary. So we go a bit abstract already. > > > 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. > > Depends on their intelligence, which I don't know. It has nothing to do with intelligence. In their culture they do not use *any* abstraction. They do everything with concrete things only. > > 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. > > Do you also see the difference between the union of all finite natural > numbers and the infinite set N? If you mean the union of the sets {n}, no. If you do mean the von Neumann model, no. If you mean something else, I do not know. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |