Cantor Confusion
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From: David Marcus on 8 Jan 2007 13:37 Andy Smith wrote: > A slightly different question. > > A real point, such as pi or e has a genuine existance > independent of its e.g. binary representation. Also integers > e.g. 0 = 0.00... does not need to be defined as the limit point > between 1/2^n and -1/2^n. > > What about .11111.... ? That has a definite existance on > th line - can one define that as the point that has no > zeroes in its binary representation? > > If so then one can argue that .1111... is different from > 1.0000... and that there are no rationals between these > two pints (because any rational must have a terminating > string of 0000... or a repeating string which is other > than all ones? Some real numbers have more than one decimal representation. If a real number has a decimal representation that ends in all nines, then it has another representation that ends in all zeros. All other real numbers have only one decimal representation. -- David Marcus
From: David Marcus on 8 Jan 2007 13:38 Tony Orlow wrote: > Dave Seaman wrote: > > A number is not a digit string. The fact that there are no rationals > > between two reals constitutes proof that those reals are identical. > > See the reference above for an explanation of Dedekind cuts. > > What if one of those reals is a rational itself? Is there necessarily a > rational between every rational and every real, or are there irrational > reals "adjacent" to rational reals? Why don't you try to figure this out for yourself? If you post your proof, we'll check it for you. -- David Marcus
From: mueckenh on 8 Jan 2007 16:05 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. That is the same as the union of two initial segments of natural numbers which is an initial segment of natural numbers. 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}. > > > 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. The finite union of two or more finite trees is a finite tree. An infinite union of finite trees is the infinite tree. > > > > > > 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? No axiom requires this but the definition of a number as an imaginated item requires that an individual number needs an individual address to think about it. Even you have emphasized the correct idea, which I answered to. You asked: > Do unnamable numbers have any existence at all? What numbers are ever > referenced in serious mathematics except by being named? Obviously you had a lucid moment, when asking this rhetoric question. > > > > ========================================== > > > > >> 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. In the union of all finite paths there is no final node. In the unuion of all last levels of the set of finite trees there is no last level. > > ========================================== > > > > > 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. Each node lies in an infinite number of paths. Or which node does not? Regards, WM
From: mueckenh on 8 Jan 2007 16:17 Dik T. Winter schrieb: > 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. 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]. > > > > > 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. But not such crippled plants with horizontal edges. > > > > 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 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. 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. > > 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? A proof well, but not such nonsense of crippled plants which are not trees in our sense. > > > 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. 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. > > > > 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. 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. > > > 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. Yeah. You are beginning to see why the union of all finite numbers n canot yield an infinite set. > > > 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 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. 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? 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. > > > > 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. Depends on their intelligence, which I don't know. > > > > 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. Do you also see the difference between the union of all finite natural numbers and the infinite set N? Regards, WM
From: mueckenh on 8 Jan 2007 16:23
Dik T. Winter schrieb: > In article <1168107244.731296.9350(a)v33g2000cwv.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > Name a level or a node or and edge which are in one of the trees only. > > > > > > Why should I do that? > > > > In order to prove that there is evidence for at least one path being > > not in both trees. > > But for that I do not need to mention a node or edge, I only need to mention > a path. Unless you can mention a node or edge, your path may be a ghost. The union of all finite trees is also the union of all last levels. That is the union of all levels. If some path is missing, then it must stretch over more than all levels indexed by natural numbers. > > > > I can name a *path* that is in the complete infinite > > > tree but not in the union of finite trees. > > > > Without different nodes that is an empty assertion. > > Well, I have shown it by proof. So it is not an empty assertion. That was not a proof. > > > > 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. > > > > Do you claim that the union of all finite trees is not an infinite > > tree? > > It is. > > > Why do you deny the same for the paths? > > The union of the sets of paths is an infinite set of paths. But there is > no infinite element in there. The union of all finite initial segments (paths) is not an infinite segment (path)? > > > The infinite paths in the union of trees are the unions of the > > respective finite paths. So much should be clear to every reader. > > What would be clear to every reader is that you are confusing issues. > Unions are about sets. A tree has a set of nodes, a set of edges and > a set of paths. When we do a union of trees we can do a union of those > three sets for each tree. A tree is determined by its set of nodes as well as by its set of edges. Trees with same nodes and edges are identical with respect to paths. > Unless you define union completely different > (and if so, pray give a definition) we can not use unions of elements > of those sets. So you can not take the union of paths but only the > union of sets of paths. On the other hand, once you define your union > such that we also must consider unions of individual paths, I also want to > consider unions of individual edges. Can I? If not, why not? The union is easily defined: Every finite tree (up to level n) contains the union of all smaller trees (up to levels n - 1) and its own level n. Regards, WM |