Cantor Confusion
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From: Dik T. Winter on 9 Jan 2007 21:33 In article <1168350148.472024.53880(a)s80g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1168291077.860958.62580(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: .... > > > 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. > > > The union of all finite trees (= union of all levels) of the tree > automatically contains all paths, because there is no path or part of a > path outside of this union. And again, in that case the set of paths in the union is not the union of the sets of paths. I did show that with a finite graph. So I wonder why you think it holds for an infinite graph (which a tree is)? > > > 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 all finite trees (= union of all levels) of the tree > automatically contains all paths, because there is no path or part of a > path outside of this union. Yes, but that does *not* make the *set* of paths the union of the *sets* of paths in the finite trees. > > 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. > > Correct. As the union of all finite paths turning always right is > 0.111... . Pray, define the union of paths. But I am talking about *sets* of paths. > > 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. > > But {0.111...} is {{0.1}, {0.11}, {0.111}, ...} (see below). Oh. > > 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. > > > And what is the set {0.1, 0.11, 0.111, 0.1111, ...}, or, translated to > digits, the set {1,2,3,...}? Is it not {0.111...} or, translated to > digits, N (or omega)? Perhaps. But what is the relevance? How about the set {0.1, 0.10, 0.101, 0.1010, ...}? How do you define the *union* of paths? In a level three tree I can encounter the paths 0.111 and 9.101. What is their union? > > So we go a bit abstract already. > > So consider the circles below. Their meaning should be unique. For you, apparently. > > > > > 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. > > In above circles there is no abstraction. You think so. I see (in my non-abstracted view) a line with circles, beneath it another line of circles, a dash and a third line of circles. I have no idea what the meaning of that is. A picture, apparently. Looks nice. But what is the meaning? -- 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 9 Jan 2007 21:47 In article <1168351276.094403.80470(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > What is a ghost path? > > A ghost path is a path which does not exist in a tree which contains > the paths of all real numbers the bits of which can be indexed by > natural numbers. That is something I do not understand. Please elaborate. > > > 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. > > > > Ah, again about that diagonal. No, you are wrong. > > All path the bits of which can be indexed by natural numbers are in the > union of all finite trees. But *not* in the union of the sets of paths in the finite tree. > You know it. You cannot point to a bit of any number of [0, 1] which is > missing. But you claim there were further numbers. > > Try to complete the union of all finite trees to the complete infinite > tree. Which level, node or edge would you add? Pray, properly define the union of trees. In my opinion a tree consists of three sets. First the set E of edges, next the set N of nodes and finally the set P of paths. What is the union of two trees? > > Why was it not a proof? > > Because you used crippled plants which are not under discussion in a > forest of well grown trees. My proof was about the set of paths in the union of finite trees not being the union of the sets of paths in the finite trees. What was wrong about that proof? Please, once come up with a proper definition of the union of two trees. Before you come up with such a definition it is impossible to even entertain a discussion. > > I am talking about the union of *sets* of paths, not the union of paths. > > Do you not see the difference? How do you define the union of paths? > > And how do you define the union of the *sets* of paths that are in the > > finite trees? > > Which edge of any infinite path is missing? Which bit could be added to > one of the numbers represented? What is the relevance? If you cna state that there is some infinite path in the union of the *sets* of paths in finite trees, you should also be able to point to a finite tree that contains that infinite path. There is no edge missing, it is only that your infinite path is not a path in any of the finite trees, so it is not in the union of the sets of paths in finite trees. > > Ok, so you do *not* use the union of the sets of paths but something else. > > The material from which the paths are constructed. Makes no sense. > > Indeed, you actually do not use those sets at all in forming your union. > > Nevertheless you want to draw conclusions about the cardinality of the > > *set* of paths in that union from the cardinalities of the *sets* of > > paths in the finite trees. > > The completeness of U{n e N} {1,2,3,...n} = N is the same as the > completeness of the union of all levels of the infinite tree, and that > is the same as the union of all finite trees. And all these unions are > counatble. As you still fail to provide proper definitions of unions of trees, I wonder. > > How do you *define* the union of two trees? > > I saw a tree as consisting of three sets: E, N and P for edges, nodes > > and paths. > > You can also unify the levels. Eh. Pray expand. > > And I defined the union of two trees T1 and T2 as: > > [ {E1, E2}, {N1, N2}, {P1, P2} ] > > (where those unions are sets, so duplicated elements can be elided). > > It is sufficient to build the union of all finite trees, or of all > levels, or of all edges, or of all nodes. This union is an infinite > tree. An infinite tree cannot be surpassed (by another tree with > naturally indexed levels). Therefore the union is *the* infinite tree > and, hence, it contains all possible paths. Again, no definition, only assertions. -- 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 9 Jan 2007 21:53 In article <1168351968.885524.129360(a)s34g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > No, that does AC (although the statement is indirectly). V=L is the axiom > > of constructibility and states that every set is constructible. Most > > mathematicians think it is false, but it can not be disproven and is > > "consistent" with ZF. It *gives* a construction. > > But, alas, this construction cannot be communicated. It is and remains > a top secret. Otherwise most mathematicians could easily be proved > wrong by simply constructing a well-ordering of R. It can be communicated. But you are not even willing to look at the axiom and its consequences. > > > > Eh? As I understand it, there is only one union. But, what is the > > > > relevance? > > > > > > > There is a union of paths for every path. > > > > How do you define a union of paths? > > For example: the infinite union of {0.01} and {0.010} and {0.0101} ... > yields the path corresponding to 1/3. I asked for a definition, not an example. How do you define the union of paths? What (for instance) is the union of {0.1111} and {0.1011}? > > > 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! > > > > But that does *not* mean that the set of paths is the union of the sets of > > paths in the finite trees. > > If the tree is complete with respect to edges, then it cannot be > completed any further. Makes no sense. It does not even refute my statement. > > You are again, *not* talking about the union of sets of paths. So what is > > the relevance? > > If the tree is complete with respect to edges, then it cannot be > completed any further. And again, *not* talking about the union of sets of paths. So what is the relevance? > > > Nevertheless, the representations in the union of all finite trees are > > > countable as the countable union of finite sets. > > > > A new term again. What are "the representations"? > > The paths are representations of real numbers. Oh. Perhaps. Pray define. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Andy Smith on 9 Jan 2007 17:46 > > Tony asked whether given a rational and a real, there > is a rational in > between them. I don't think you covered the case > where the real is > irrational. > > > Doubtless there are more elegant ways of proving > that > > I'd use the fact that for all e > 0, there exists n > such that 1/n < e. > Then, I'd let e = r2 - r1. > Yes, of course. > > ( and that illustrate that there > > are as many numbers between any 2 points as between > > 0 and 1. ) > > Actually, this is trivial. If the two points are p > and q, then > f(x) = p + x (q - p) is a bijection. > Ditto. > > Why? You can't just stick the words "by definition" > in your sentence to > cover up the fact that you can't prove something. How > you do define > "lands on the tortoise"? The obvious way to define it > is that the flea > is at the same position as the tortoise. I don't see > anything in this > definition about there being a last jump. > Yes, all right, speaking too loosely. But anyway, the flea lands on the tortoise only once, and stops. There is a unique correspondence between landing on the tortoise and a jump. There are no further jumps, so landing on the tortoise is the last jump? > > What is the logical difference between this > argument > > and the proof that there is no greatset member in > the set > > of all natural numbers? > > In one case, we prove there is no largest natural > number. In the other, > we fail to prove that if the flea lands on the > tortoise, there must be a > last jump. > > > So maybe, if you allow counting "infinitely fast" > you can > > count up to an infinite number? > > What is an "infinite number"? > What if I set out on the real line, from 0 to 1, the set 1/2,1/4,1/8, ... and then set out its mirror from 2 to 1, moving backwards 1/2,1/4,1/8 .... Then the sequence {1/2+1/4+1/8 ...} + {... + 1/8 +1/4 +1/2) covers the interval [0,2] without any gaps. So the infinite series {1/2+1/4+1/8 ...} + {... + 1/8 +1/4+1/2) exists and has a finite last member = 1/2? There is undoubtedly a logical flaw here - you can't have (by Peano) an ordered infinite series with a last member, and doubtless you will tell me my error. But you see the point? Actually, at the risk of getting labelled as being in some dodgy company, isn't there something a bit asymmetric about Peano? What happens if we say; 1) All natural numbers form an ordered set bounded by alpha (aka 0) and omega (aka infinity). 2) With the exception of alpha and omega, all natural numbers have a successor and a predecessor; alpha has no predecessor, omega has no successor. 3) For any number n there exists (omega-n) which is > n. Then you can happily say that there is a last jump of the flea (but of course, the only thing you can say about the size of the jump is that is less than any e, where e is any real number <1.
From: Andy Smith on 9 Jan 2007 19:59
> > 1) All natural numbers form an ordered set bounded > by alpha (aka 0) and omega (aka infinity). > 2) With the exception of alpha and omega, all > natural numbers have a successor and a > d a predecessor; > alpha has no predecessor, omega has no successor. > 3) For any number n there exists (omega-n) which is > > n. > 3) obviously doesn't work (omega-n) is a number so that would give (omega-omega+n)=n>(omega-n). Not so easy! I still want to put a sensible label on the finite last terms of the sequence 1/2,1/4,... ...1/8,1/4,1/2 though. |