Cantor Confusion
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From: mueckenh on 2 Feb 2007 05:55 On 2 Feb., 02:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170348637.768496.237...(a)a75g2000cwd.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > So the union of finite chains of nodes like the following > > 0 > > 12 > > 6543 > > does not contain an infinite chain of nodes of the form > > 0 > > 12 > > 6543 > > 789... > > ??? > > I did not state that. Pray try to read what I wrote, rather than giving > your own interpretation to it. > > > Frankly, I do not see much difference to: The union of finite paths > > contains an infinite path. > > But that is true. So you say: The union of finite trees contains an infinite tree. The union of finite chains contains an infinite chain. The union of finite paths contains an infinite path. > What I state again, again and again, but what you > misread each and every time is: "the union of sets of finite paths > does not contain an infinite path". So you say the union of finite paths p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... contains the infinite path p(oo) = {0,0,0,...}. But the union of finite sets of finite paths {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... does not contain the union of finite paths p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... I would say: The union of all finite sets of finite path contains all finite paths. The union of these finite paths can be consructed (it is a set of nodes) and is realized in the binary tree by its nodes. It contains all infinite paths of the infinite tree. The tree contains the union of all finite sets of finite path as well as the union of these paths. > > > > > How then does this infinite path manage to enter the union of the > > > > finite trees? > > > > > > Because the set of paths of the union of finite trees is *not* contained in > > > the union of the sets of finite paths. > > > > And it is not contained in any of the finite trees. Why is it said to > > be contained in the union of all finite trees? > > Because an infinite path is a union of paths. And when you construct > unions of *sets* of paths you are not constructing unions of paths. The union of a set of sets of elements is a set of elements, not a set of sets of elements. The union of one ot more sets of paths, as I defined it, can be a path. > > > > > If all natural numbers are finite, their union as defined for > > > > pathlengths cannot be infinite. > > > > > > Why not? > > > > Because there cannot be more than infinitely many natural numbers. All > > those infinitely many numbers, however, are allegedly finite. > > And there are infinitely many such. And they are, indeed, all finite. > The path lengths however do not map to natural numbers, which you assert. > That is valid for finite paths, but not for infinite paths. The finite natural numbers map on the finite pathlengths. The finite segments of natural numbers map on the finite pathlengths. If there is an infinite pathlength, then there must be an infinite number as well as an infinite segment of natural numbers. You say, the former does not exist in N while the latter does exist in the form of the whole set N. That is a contradiction. You see that there is no infinite pathlength possible without an infinite pathlength. > Let's, for > once, go to a proper definition (assuming all are ordered sets): > given a path {n_1, n_2, ..., n_k} then the path length is > |{n_1, n_2, ..., n_k}| = k. > Clearly for finite paths the path length is finite, and so a natural > number. However, when we construct the path: > p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... > we get as path: > {n_1, n_2, n_3, n_4, ...} > the path length in this case is *not* a natural number. It is the > cardinality of N. The pathlength is not a natural number but it is a number (by definition), namely omega. You see that there is no infinite set N without an infinte number in it. > > But you said that there is an infinite pathlength corresponding to an > > infinite set of paths. > > But that set of paths is *not* the union of the P(i). Above you said that the set of all paths p(n) contains the infinite path p(oo). > > > > > What was your argument distinguishing infinite unions of finite trees > > > > and finite paths, respectively? > > > > > > None. I distinguish unions of finite trees and unions of *sets* of > > > finite paths. Above you are using P(1) U P(2) U ..., where each > > > P(i) is a *set* of finite paths. > > > > What about unions of sets of finite trees? Would they work completely > > different from unions of sets of paths? In particular, would the union > > of sets of finite trees not contain or establish the inifnite tree? > > The union of sets of finite trees would neither contain, nor establish, > an infinite tree. It would not contain one, because none of the > constituent sets contains an infinite tree. It would not establish > one, because the union is a set of trees, not a tree. The union of a set of sets of elements is a set of elements, not a set of sets of elements. The union of a set of trees, as I defined it, is a tree. > A tree is a > set of nodes, a set of trees is a set of sets of nodes. And the union of two or more sets of nodes is a set of nodes. Therefore the union of a set of trees, as I defined it, is a tree. > > > > > > Wrong. Infinite paths do not have a path-length that is a > > > > > natural number. > > > > > > > > Pathlengths ARE natural numbers. > > > > > > For finite paths. > > > > In all cases, by definition. > > *What* definition? Definition: Map the pathlength x on the number x. Regards, WM
From: Franziska Neugebauer on 2 Feb 2007 06:10 mueckenh(a)rz.fh-augsburg.de wrote: > On 2 Feb., 02:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: [...] >> However, when we construct the path: >> p = {n_1} U {n_1, n_2} U {n_1, n_2, n_3}, ... >> we get as path: >> {n_1, n_2, n_3, n_4, ...} >> the path length in this case is *not* a natural number. It is the >> cardinality of N. > > The pathlength is not a natural number but it is a number (by > definition), namely omega. You see that there is no infinite set N > without an infinte number in it. As omega is not member of N I don't see that. F. N. -- xyz
From: Dave Seaman on 2 Feb 2007 07:46 On Fri, 02 Feb 2007 10:20:00 GMT, Andy Smith wrote: > Dave Seaman <dseaman(a)no.such.host> writes >>On Thu, 01 Feb 2007 21:52:26 GMT, Andy Smith wrote: >>> Dave Seaman <dseaman(a)no.such.host> writes >>> (snip) >> >>>>> I had previously imagined that the "Continuum Hypothesis" related to >>>>> whether the set of reals (as 0 dimensional points, each with no >>>>> neighbouring point) could cover the line, with dimension 1 (Which seemed >>>>> like a good question). >> >>>>You are asking whether R and "the real line" are the same set. In order >>>>for that question to have a nontrivial answer, you must have some meaning >>>>in mind for "the real line" other than R. What is your definition? >> >>> Well until recently my naive view was that points and lines were >>> different entities, and you might as well ask how many metres in an acre >>> as how many points in a line. I would have said (in your terminology) >>> that there is clearly an injection from the reals into the line, but >>> that you can never cover it. But I doubt now if that is true, or if the >>> question is a sensible one to start with. >> >>I don't follow. Is it your claim that: >> >> (1) A line does not have any points as members at all, or >> (2) A line has points as members, but perhaps there aren't >> enough points to fill it? >> >>Your metres-in-an-acre analogy seems to suggest (1), but otherwise you >>seem to be saying (2). >> >>If (2) is your claim, then can you show me a point on the line that is >>not represented by a real number? If you can do this, then evidently >>"the line" to you must be something other than simply "the set of real >>numbers", as it is to most people. Can you elaborate on this? >> >> > Well I meant 2) - (and I can't find any location in the acre that is not > covered by a metre line either). And I can't answer your question . Yes, > of course, the moment that you specify a real number you have defined a > location on the line but I am suffering from dimensional anxiety, > particularly when you observe that any point on the line has no nearest > neighbour. If I was a point sized entity wanting to walk from 0 to 1 > without falling down a crack, how would I do it? (I know that is not a > particularly sensible question, just illustrating my mis-perspective). Zeno was bothered by this when he talked about an arrow in flight being instantaneously "at rest" at each moment. Thanks to calculus, we now know that the arrow is never actually "at rest" during flight. -- Dave Seaman U.S. Court of Appeals to review three issues concerning case of Mumia Abu-Jamal. <http://www.mumia2000.org/>
From: William Hughes on 2 Feb 2007 08:29 On Feb 2, 2:14 am, mueck...(a)rz.fh-augsburg.de wrote: > On 1 Feb., 18:44, "William Hughes" <wpihug...(a)hotmail.com> wrote: > > > On Feb 1, 11:53 am, mueck...(a)rz.fh-augsburg.de wrote: > > > > Yes. That is not in question. What I dispute is that the difference > > > depends on the cardinal number of the set. > > > It may or may not. For instance the question > > > "Is the cardinal number of the set a (finite) natural number > > (if the set does not have a cardinal number then the > > answer is no)?" > > > certainly depends on the cardinal number of the set. Sets of type > > I have a cardinal which is a (finite) natural number. Sets of type > > II do not. > > My questions are: > > What numbers are subject to complete induction? > What is the set the elements of which are subject to complete > induction? > > > > Yes. That is not in question. What I dispute is that the difference > > > (induction is possible or is not) depends on the cardinal number of > > > the set. > > > Induction on the elements of A > > is always possible given any set A of > > natural numbers. However, induction may not > > be able to prove that something is true of A > > Why should it? But it can be able. For instance it can prove that > every set of natural numbers is finite while the size of sets of > natural numbers is unbounded from above. > > > > > For example, every one of the elements > > of A is finite. This can be shown by induction. > > However, this does not show that the set > > A is not (potentially) infinite. > > It does. No. Let P be the set of prime numbers. Every prime number is finite. P is potentially infinite. So showing that every element of a set is finite does not show that the set is not potentially infinite. > Of course every set of natural numbers is finite, as proved > by induction over all initial segments. (Potential infinity.) If you define a potentially infinite set to be finite, and say that there are only sets with a fixed maximum and potentially infinite sets, then all set are finite. So what? Calling a potentially infinite set finite does not change its properties. > This > simple truth has only been veiled by the arbitrary assumption that > there is an actual infinity. No. Sets with a fixed maximum and potentially infinite sets have different properties, whether or not you assume an "actual infinity". > As this assertion cannot be proved, some > have argue that induction was not valid for the whole set. No. There are two different questions. Let A be a set of natural numbers. i: Can induction show something about the elements of A ii: Can induction show something about A. Note, for some sets A, the answer to i can be Yes and the answer to ii can be No. No one has claimed that the answer to i is anything but Yes. The converstion goes like this M: induction shows that A has property X O: induction is not valid for A [that is question ii is false for A] M: A is a set of natural numbers so induction is valid for A [that is question i is true for A] The statement "induction [is] not valid for the whole set" means that the answer to question ii is No, not that the answer to question i is No. - William Hughes
From: Andy Smith on 2 Feb 2007 08:52
In message <epvbqo$bjl$1(a)mailhub227.itcs.purdue.edu>, Dave Seaman <dseaman(a)no.such.host> writes >On Fri, 02 Feb 2007 10:20:00 GMT, Andy Smith wrote: >> Dave Seaman <dseaman(a)no.such.host> writes >>>On Thu, 01 Feb 2007 21:52:26 GMT, Andy Smith wrote: >>>> Dave Seaman <dseaman(a)no.such.host> writes >>>> (snip) >>> >>>>>> I had previously imagined that the "Continuum Hypothesis" related to >>>>>> whether the set of reals (as 0 dimensional points, each with no >>>>>> neighbouring point) could cover the line, with dimension 1 (Which seemed >>>>>> like a good question). >>> >>>>>You are asking whether R and "the real line" are the same set. In order >>>>>for that question to have a nontrivial answer, you must have some meaning >>>>>in mind for "the real line" other than R. What is your definition? >>> >>>> Well until recently my naive view was that points and lines were >>>> different entities, and you might as well ask how many metres in an acre >>>> as how many points in a line. I would have said (in your terminology) >>>> that there is clearly an injection from the reals into the line, but >>>> that you can never cover it. But I doubt now if that is true, or if the >>>> question is a sensible one to start with. >>> >>>I don't follow. Is it your claim that: >>> >>> (1) A line does not have any points as members at all, or >>> (2) A line has points as members, but perhaps there aren't >>> enough points to fill it? >>> >>>Your metres-in-an-acre analogy seems to suggest (1), but otherwise you >>>seem to be saying (2). >>> > >>>If (2) is your claim, then can you show me a point on the line that is >>>not represented by a real number? If you can do this, then evidently >>>"the line" to you must be something other than simply "the set of real >>>numbers", as it is to most people. Can you elaborate on this? >>> >>> >> Well I meant 2) - (and I can't find any location in the acre that is not >> covered by a metre line either). And I can't answer your question . Yes, >> of course, the moment that you specify a real number you have defined a >> location on the line but I am suffering from dimensional anxiety, >> particularly when you observe that any point on the line has no nearest >> neighbour. If I was a point sized entity wanting to walk from 0 to 1 >> without falling down a crack, how would I do it? (I know that is not a >> particularly sensible question, just illustrating my mis-perspective). > >Zeno was bothered by this when he talked about an arrow in flight being >instantaneously "at rest" at each moment. Thanks to calculus, we now >know that the arrow is never actually "at rest" during flight. > > But the positions are reversed? I would be very happy to consider distance along a line as a continuous function x and rate of change of position dx/dt etc. But by asserting (or, as I understand it, defining) the line is an infinite collection of points one reintroduces the conceptual granularity that bothered Zeno? You just start with points and define the real line as the set of all of them. I have an image of a line as a continuous thing of point width, and it is trying to marry up the perception of continuity with the set of real points that is difficult for me. I mentioned fractals because they are a clear example of e.g. how you can get infinite volume from a finite surface area, etc. so I thought that the essentially recursive structure of the reals (you can map the real line into the space between any two points, indefinitely) was some sort of fractal explanation of how the line could be simultaneously pointlike and continuous. But doubtless, as with everything else, my problem is getting my head to a point where I see things in the correct perspective. -- Andy Smith |