Cantor Confusion
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From: Virgil on 13 Jan 2007 22:37 In article <6284429.1168742543853.JavaMail.jakarta(a)nitrogen.mathforum.org>, Andy Smith <andy(a)phoenixsystems.co.uk> wrote: > David Marcus wrote: > > > > > > > 2) I was previously of the opinion that you > > couldn't > > > have a countable infinite monotonically increasing > > > sequence that had a greatest member. > > > > Even layman's language must be precise or you will > > only confuse > > yourself. You can have a countably infinite ordered > > set that has a > > largest element. You can't have a sequence with a > > last element. > > > > > 3) But you can, e.g. the infinite succession of > > zero crossings > > > in [-1,0] of sin(pi/x) which has a last member at > > 0.00.. > > > > What do you mean by "succession"? And, I don't think > > that is a good > > example, since sin(pi/0) isn't zero (it isn't > > defined). > > > > Isn't sin(pi/0) defined by the antisymmetry of the function? Surely it must > cross the x-axis at x=0 ? Except that it cannot exist AT X = 0. > And actually, that is rather the point about this > example that is different from others. > > You can have a countably infinite sequence converging > on some limit point e.g. > {1/2,3/4,7/8, ...} but you could not then say add the > value {2} as the last member of the sequence and claim > that it was infinite and with a maximum member. > > But isn't the zeo crossing at x=0 of sin(pi/x) subtly > different? even though the function is oscillating > infinately fast at x=0, it is continuous, and the zero > crossing at x=0 is attached by the function itself (like > a piece of string) to the preceeding zero crossings, > even if you can't identify the number of the last > crossing before x = 0.0...? If that sort of argument were legitimate, the function would also have to be equal to every other value between -1 and +1, inclusive, at x = 0.
From: David Marcus on 13 Jan 2007 23:14 Andy Smith wrote: > Isn't sin(pi/0) defined by the antisymmetry of the function? No. Division by zero is not defined. We can define f:R -> R by f(x) = sin(pi/x), if x <> 0, 0, if x = 0. However, f is not continuous at zero. > Surely it must cross the x-axis at x=0 ? Depends on how you define "cross". Although, it seems a bit odd to define "cross" for a function that isn't continuous. Maybe you would like to try g(x) = x sin(1/x), if x <> 0, 0, if x = 0. This function is continuous. > And actually, that is rather the point about this > example that is different from others. > > You can have a countably infinite sequence converging > on some limit point e.g. > {1/2,3/4,7/8, ...} but you could not then say add the > value {2} as the last member of the sequence and claim > that it was infinite and with a maximum member. Sorry, I don't know what you mean. Sequences don't have last members. 2 isn't infinite. The set {2,1/2,3/4,7/8,...} is countably infinite and has a maximum element (the maximum element is 2). > But isn't the zeo crossing at x=0 of sin(pi/x) subtly > different? even though the function is oscillating > infinately fast at x=0, it is continuous, f isn't continuous. g is continuous. > and the zero > crossing at x=0 is attached by the function itself (like > a piece of string) to the preceeding zero crossings, > even if you can't identify the number of the last > crossing before x = 0.0...? > > So the sequence of zero crossings up to and including the > point x=0 exist, and are ordered? Please define the word "sequence". > Incidentally, on this web Usenet interface, is there > some snappy way to scroll to the bottom of the thread? > Takes ages on my machine with this huge thread ... I'm not using a Web interface. Try using a non-Web interface. Feel free to start a new thread. We seem to have moved to a different topic than the main part of the thread (which is just people giving WM a hard time because he is a crank). -- David Marcus
From: mueckenh on 14 Jan 2007 07:55 Virgil schrieb: > Which ones of those specifically "rooted finite paths" does WM claim > contain infinitely many nodes? Which segment {1,2,3,...,n} of N contains infinitely many elements? Is N the union of all initial segments? > > Why do you call the set {{1}, {1,2}, {1,2,3}, ...} which is equivalent > > and even isomorphic to the set{1, 2, 3, ...} = N finite (finite path)? > > Each of the sets {1}, {1,2}, {1,2,3}, etc., is both a member and a > subset of N, but N is not. And, in addition, it is not green. Yes. Relevance? > > Even the diagonal of the EIT is no longer infinite, if it is > > inconvenient for set theory? > > As properly described, that diagonal is actually infinite however > inconvenient that may be for WM's version of set theory. This diagonal (infinite path) is the union of all digits (nodes) of the initial segments and the limit of the sequence of the initial segments. > > You cannot "unary represent" every real in [0, 1]. You can unary > > represent every natural number, and you can unary represent real > > numbers which contain only numerals a_n = const for every n in N > > (behind the decimal point), like 0.111 or 0.1111111. In addition you > > can unary represent omega or N = {1, 2, 3, ...}, namely as 0.111... = > > {0.1, 0.11, 0.111, ...}. Notice that this set does not contain infinite > > strings of 1. > > Then why does WM keep claiming that it does for trees? Because it either holds in both cases or in none. > Then WM's infinite path IS the "infinite" tree, not a member of it. Of course. The set of trees is the forest. The set of infinite paths is the infinite tree, the union of the finite trees. >> The union is the set of infinite paths. > No nodes? No edges? Just paths that were not in any of the original > trees? Paths contain nodes and edges. >> The union is the infinite ordinal. > Which is not one of the objects being unioned, The union is not a finite ordinal but it is an ordinal. > so there is no reason in > that for the union of finite trees to be a tree itself. The union is not a finite tree but it is a tree. Regards, WM
From: mueckenh on 14 Jan 2007 07:58 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> >> > This and your further questions are answerd by the following > >> >> > texts. > >> >> > >> >> I did not pose any questions. I have informed you about the fact > >> >> that there is no time and hence no temporal process _in_ math. > >> > > >> > Be informed then that mathematics is in time. > >> > >> From that it does not follow that there is a notion of time in > >> mathematics. > > > > A failure > > Thanks again for confirming that there is no time im mathematics _now_. There is no time in what you may erronesously understand by mathematics. > > > need not and will not persist for ever in mathematics. (From > > this you can see that time *is* in mathematics.) > >> > [Brouwer] maintains [...] > >> > >> Please tell us, whether you want to discuss views of > >> > >> [ ] Cantor > >> [ ] Brouwer > >> [ ] Mueckenheim > >> [ ] Cantor interpreted by Mueckenheim > >> [ ] Brouwer interpreted by Mueckenheim. > >> > >> If more than one applies please tag each sentence of yours > >> accordingly. > >> > > I do not remember having posted interpretations, but only original > > quotes. > > I do not discuss _with_ or _instead_ _of_ or _against_ Cantor or Brouwer > both of which are dead. If _you_ want to argue _you_ must rephrase them > in your own words. I should create new words? Why that, if there are good old well-known words? ==================================================== >> But an infinite ordinal is that union! > There is no infinite number (ordinal) _in_ that union. Rephrased: There > is no infinite _member_ in that union. The infinite ordinal is that union. The infinite diagonal of Cantor's list is the unioin of all its finite digits. >> An infinite union of finite sets is an infinite object. > But that object does *not* have an infinite _member_. Why should it? Regards, WM
From: mueckenh on 14 Jan 2007 08:03
Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > Franziska Neugebauer schrieb: > [...] > >> > The set of paths is countable. > >> > >> 1. The set of all rooted _finite_ paths in an infinite tree is > >> countable. The union of all finite paths is exactly this set of > >> all rooted finite paths. > > > > Yes, of infinitely many paths, each of which has infinitely many > > nodes. > > _finitely_ many nodes. Your trees to be united do only have finitely > many nodes. There are only finitely deep trees. Hence the union does > not contain any infinite tree and hence not any infinite path. > > Recall: Do get it out you must have put it in. Recall Cantor's diagonal. It consists only of finite initial segments. Either it does not exist - or there are infinite paths in the union of finite trees. > > [...] > >> 2. To "unary represent" every real in [0, 1] > > > > You cannot "unary represent" every real in [0, 1]. > > It's rather unimportant whether I "_can_" do so. It is important that it is in principle impossible. > > But for the sake of argument let us assume, you "could" *not* represent > every real in [0, 1] in terms of representations. Then your proof is > meaningless already before writing it down. Obviuosly yoou intermingle "unary" and "binary". > "mutable sets" is a notifiable disease. "actually infinite sets" is a disease from which maths will recover. Regards, WM |