Cantor Confusion
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From: Dik T. Winter on 12 Dec 2006 20:39 In article <1165923504.410525.226600(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > So you can not show a bijection (in your opinion), but nevertheless you > > state that you have given a surjection. Do you not think you are > > contradicting yourself a bit? > > I give a surjection on all existing paths. But not from the edges. But from parts of edges. > > > No? Even if there are enough parts to gather more than a whole edge to > > > be mapped on every path? > > > > No. Each part of an edge may map to a single path, that does *not* give > > a map from each edge to a single path. > > Since infinity = infinity became an allowed equation, 1/2 + 1/2 = 1 can > no longer be true? So if the first half of ab edge maps to, say, 1/2 and the other half to, say, 3/4. To what single path maps the edge itself? > > I can. And I see that each edge maps, when you do this, to a > > plethora of paths, so that is *not* a surjection from edges to paths. > > The crucial thing in a surjection (and indeed for a mapping) from A to B > > is that each element of A maps to a *single* element of B. So, let me > > ask a different question. To which single path does the edge that goes > > left from the root map. > > That one which goes left from the root is not engaged, because we need > only half of the set of edges. So your construction is not a surjection. > That one which goes right, could be > mapped on that real number (path) which is asked for most frequently, > namely 1/3. That edge leaving the node 1 on level 1 left is mapped on > the next frequently mentioned number, namely pi. You will not be able > to ask for more nmbers than I can name edges. And you will not be able > to construct a "diagonal" path. And you do not know what a mapping is either. > But considering this catalog for another time, I remember that we have > twice as much edges as paths. Therefore, we can waste the first > countable infinity of edges, and if not a whole infinity, we can at > least waste all the edges up to level n (for any finite number n) and > notwithstanding this big loss, there remain enough edges to attach one > to each path. That is so overwhelming, that I in fact cannot > understand how you may dare to defend your deplorably swaying position. Because you do not construct a mapping, let alone a surjection. > My sympathy and condolences. Indeed. -- 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 12 Dec 2006 20:55 In article <457ecfca(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: .... > Dik, consider this. If the number of levels in the tree is countable, > then every path in the tree is finite, and marked by a specific edge > where the value arises. Wolfgang Mueckenheim talks about infinite paths, and there are indeed infinite paths when the number of levels is countable. > Can you have infinite paths/strings with a countable > number of levels/bit positions? Yes. But that is already an old fruitless discussion with you. -- 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 12 Dec 2006 20:52 In article <457ece72(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: > Dik T. Winter wrote: > > In article <457d8cc0$1(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> writes: .... > > > Why does that matter? > > > > It does matter because if you do not properly define your problem, > > mathematics is not able to give an answer. > > It's sufficiently defined if one assumes that there is a uniform > probability distribution. You can assume as much as you want, that does not make it a definition. What *is* a uniform distribution of all natural numbers? > > > This is the same thing as your stupid ball and > > > vase trick. Why do you need to label anything, or know what you're > > > choosing from the infinite set? > > > > Because that is part of the problem setting. Giving that setten will > > allow mathematics to model the question and give an answer. > > That problem has a clear answer with or without the labels: the sum > diverges as f(n)=9n. The labels are confounding, not clarifying. Yes, but that does *not* indicate anything about the limit, as I did show below: > > And it is bad to think that because for a sequence of sets holds that > > lim{n -> oo} |S_n| = k > > with some particular value of k, that also > > | lim{n -> oo} S_n | = k > > because the latter statement contains something that has not been > > defined in mathematics. > > I'm not sure what that statement is supposed to say. Can uoi give an > example? You have: |S_n| = 9n and you think that | lim{n -> oo} S_n | = lim{n -> oo} | S_n | = lim{n -> oo} 9n. but first "lim{n -> oo} S_n" is not defined, and second, when you define it you have to prove that the first part is equal to the second part. > > But even when we define it, it is not certain > > that it holds. Given the following (I think reasonable) definition: > > lim{n -> oo} S_n = S > So, what, S_n is supposed to be an initial segment of the sequence? I do not understand. S_n are sets indexed by n and so form a sequence. > > if: > > (1) for every element a in S there is an n0 such that a is in each of > > the sets S_n with n > n0 > > (2) for every element a not in S there is an n0 such that a is not in > > each of the sets S_n with n > n0. > In (2), it sounds like a would not exist in ANY S_n if it's not in S. No. Any S_n with index larger than some n0. > > So from some particular point an element either remains in the sets in > > the sequence or remains out of the sets. > > You mean, at some point you can tell whether a given element a is in S, > because if it were, it would be there by then? and if it were not it would not be there by then. (Note that the point where that is the case can be different for each element.) > > With this definition (when we look at the rationals) we have that > > lim{n -> oo} [0, 1/n] = [0] > Okay that interval degenerates to 0.... > > > and so: > > lim{n -> oo} | [0, 1/n] | = aleph0 != 1 = | lim{n -> oo} [0, 1/n] | > > (I am talking standard mathematics here). > > Are the |'s supposed to denote set size? If so, how can you claim that > [0,0] contains aleph_0 elements? Where do I claim that? > > So taking cardinality and limits can not be interchanged except in some > > particular cases. But that is not unprecedented in mathematics. > > limits and integrals can also not be interchanged except in particular > > cases. And so can the interchange is not in general passoble if one > > of the things you interchange is a limit. Even interchanging limits > > is not in general possible. Consider: > > lim{x -> oo} lim{y -> oo} (2x + 3y)/xy > > True, but is it relevant? Yes, because the same holds for cardinality. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Han de Bruijn on 13 Dec 2006 02:55 William Hughes wrote: > Tony Orlow wrote: > >>Well, the proof is simple. Any finite number of subdivisions of any >>finite interval will only identify a finite number of real midpoints in >>that interval, between any two of which will remain more real midpoints. >>Therefore, there are more than any finite number of real points in the >>interval. > > This just shows that the number of real points is unbounded. > It does not show it is infinite (unless of course you use the > fact that any unbounded set of natural numbers is infinite). Isn't unbounded the same as infinite, i.e. = not finite = unlimited = without a limit? Han de Bruijn
From: mueckenh on 13 Dec 2006 05:55
Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > [...] > >> 1. You do not present a convincing definition of "number". (Most > >> likely you have none). > > > > Definitions are abbreviations like the following: > > [too long, too old, Impossible. Its from my new book to appear within few days. Can future be too old? > too German; it is impossible to be too German. > no definition at all] It is clear that you have not understood. > > >> 2. You do not present a convincing definition of "numbers" and "sets" > >> which are "not fixed" or "un-fixed". > >> > >> 3. You do again try to discuss issues of neuro sciences > >> (representation of abstract entities in mind (or in the brain?)) in > >> sci math. > > > > Of course, because math requires mind and brain. > > Mind and brain and representation of (abstract) entities therein is > still off topic in sci.math. This decision is 1) wrong 2) not yours. Regards, WM |