Cantor Confusion
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From: Virgil on 21 Dec 2006 20:34 In article <1166749334.180879.244840(a)73g2000cwn.googlegroups.com>, cbrown(a)cbrownsystems.com wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > cbrown(a)cbrownsystems.com schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > cbrown(a)cbrownsystems.com schrieb: > > > > > > > > > > > > > > No. But I need no proof in order to see that > > > > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > > > > > > > > > > > > > Yes, it appears that you need no proof to make any of your assertions. > > > > > You simply "see" them; or "feel" them. That's not what I would call > > > > > "really correct mathematics". > > > > > > > > Call it whatever you want. > > > > > > I call it "mathiness". > > > > > > > I can calculate 1 + 1/2. > > > > > > Good for you! > > > > Bad for you, if you can't. > > > > > > > But considering the > > > > fact that you would call the following result "correct mathematics", I > > > > prefer to stay with my version. > > > > > > > > Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > > > > different paths and E(n) = 2*2^n - 2 different edges. So we find > > > > lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, > > > > E(n)/P(n) >= 1. Nevertheless "E(n)/P(n) < 1 in the actual infinity". > > > > > > It is /not/ a fact that I would call this "result" correct mathematics. > > > I would call it yet another example of your devotion to "mathiness" > > > whenever you claim to provide a "contradiction". > > > > > There can't be a contradiction in ZFC because everything looking like a > > contradiction is mathiness. > > Assertions don't "look" like contradictions, as much as you may "feel" > that they do. They must be /proved/ to be contradictions, which is to > say, they must be proved /in ZFC/ to be a contradiction /in ZFC/. > > > > While "E(n)/P(n)" is certainly well-defined when both E(n) and P(n) are > > > real numbers, it is not well-defined otherwise. You simply "feel" that > > > Cantor's result must imply "E(n)/P(n) < 1 in the actual infinity"; > > > without considering what that assertion /means/. > > > > > I know that 2^aleph0 > aleph0 is a result of Cantor's theory. > > Great! > > > If you > > don't know that, you should try to inform you. > > And I have proved that E(n) > P(n) for every n and for the limit n --> > > oo. > > > > Yes, you have. > > > > You apply "mathiness" to the problem. You "know" (need no proof) that > > > "2 < 3" implies that "2/3 < 1". > > > > I know that 2/3 < 1 is a convenient way to express 2 < 3. > > For heaven's sake. "2/3 < 1" is not just "a convenient way" of > expressing "2 < 3". One statement is /derived/ from the other: > /Because/ 2/3 < 1, it follows that 2 < 3; and /because/ 2 < 3, it > follows that 2/3 < 1. How can these statements be derived? Because we > have /defined/ what "x/y" means and what "x < y" means in this context. > > > But if you > > don't know that, then we can stay with 2 < 3 as well. > > > > > You "see" the assertion "|N| < |R|", > > > > This is a provable theorem in ZF > > > > Yay! > > > > and because it "looks" like "2 < 3", you "feel" that you can therefore > > > conclude "|N|/|R| < 1". > > > > > > Instead, the statements "|N|/|R| < 1" and "E(n)/P(n) < 1 in the actual > > > infinity" are simply mathematical nonsense: neither true nor false. > > > > Are you unable to see that the formal use of fractions like |N|/|R| < 1 > > is completely irrelevant but only a brief way to express |N| < |R| ? > > "The formal use of fractions like ...". What on earth is "formal" about > it? Why not simply /say what you mean/ instead of using misleading > language? > > Suppose you had simply said: > > Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > different paths and E(n) = 2*2^n - 2 different edges. So we find > lim{n-->oo} E(n)/P(n) = 2. Nevertheless, |E(n)| < |P(n)| in the actual > infinity. > > Then I would agree: the fact that E(n)/P(n) = 2 in the limit for > natural n indeed has no particular relevance to the provable fact that > |E(n)| < |P(n)| for infinite n; and there really is nothing surprising > about this. > > E(n)/P(n) -> 2 is a feature only of finite n. |E(n)| < |P(n)| is a > feature only of infinite n. /Many/ things that can be said in the > finite case are not true in the infinite case; for example E(n) is > finite for natural n, and not finite for infinite n. If WM were the least bit amenable to reason, he would see his logic requires that every infinite path in a infinite binary tree must have a terminal (leaf) node at its "end. If T(n) is the number of terminal nodes and P9N0 is the number of paths in a binary tree of order n, we see that T(n)/P(n) = 1 for all n, so that, by WM's argument, T(oo)/P(oo) = 1, and an infinite tree must have infinitely many terminal nodes, 1 for each path. But WM chooses to his own arguments when they are shown not conform to his ironclad prejudices. > > It is when you try to argue "lim n-> oo E(n)/P(n) = 2 contradicts > E(n)/P(n) < 1 in the infinite case" that your confusing notation > "looks" like a contradiction (to you; to me it looks like nonsense). > > Cheers - Chas
From: Virgil on 21 Dec 2006 20:37 In article <1166749694.734273.263470(a)73g2000cwn.googlegroups.com>, cbrown(a)cbrownsystems.com wrote: > Virgil wrote: > > In article <1166737382.495151.299530(a)i12g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Newberry schrieb: > > > > > > > > but the question is > > > > whether you can generate a direct contradiction of the type P & ~P in > > > > ZFC. > > > > > > I do not think that this is of any interest after it can easily be > > > shown that the results of ZFC are inacceptable. > > > > Except that the only demonstration of such "inacceptability" would be > > the discovery of some internal contradiction within ZFC or NBG. > > > > You are discounting the fact that for WM, an assertion is unacceptable > iff it "feels" unacceptable. > > Cheers - Chas Damn right! Such emotionally inspired blindness to the logic of his own weird claims is itself unacceptable.
From: Dik T. Winter on 21 Dec 2006 20:40 In article <1166737132.026953.46380(a)48g2000cwx.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1166617425.221440.15570(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > We are talking about shares, but we do not need more than 3/2 per > > > path. > > > > 3/2 of what? But again you do not answer my question. In how many > > shares is the edge going left from the root divided, and how do you > > do that division? > > It is irrelevant for my proof. It is as irrelevant (and silly) as the > question: in how many shares is the unit divided in the last term of > the series 1/2^n. Nevertheless we can calculate the limit of this > series. It is very relevant for the proof. You assume you can split in shares and later on recombine those shares. When each edge is divided in finitely many shares that is allowed. It is not allowed if you divide in infinitely many shares, which you are doing. And the series 1/2^n is not relevant. We can calculate the limit of that series because for every eps there is an n0 such that for n > n0 the finite sum is on a distance not larger than eps from 2, there are no shares involved at all. > > Indeed, that can not be done, until at some moment there is a proof > > (without AC) that that axiom is true. So until such a moment we have > > to do with two forms of mathematics, one with AC and one without it. > > But when we assume AC, a well-ordering can be constructed, as I did > > show. > > No, you did no show it yet. You referred to some Hamel basis. A > construction either says: This is the first element and that is the > second etc. or it gives a prescription how this can be found. And I did. Construct a Hamel basis and do a lexicographic ordering of the elements R as a vector space over Q (with some caveats). That one is constructable, but I will not show it here (it takes too much space). > > But apparently you have problems when something can be shown to exist > > (in the mathematical sense) but can not be computed. That is close > > to the intuitionistic view. > > I stated that there is no possible way to construct a well-ordering. > You opposed. You cannot sustain your statement. That's all. Given AC, there is an Hamel basis and using that a well-ordering can be defined. > > > > But I explicitly stated how you could construct > > > > a well-ordering when AC is true. You can not do it in general, and > > > > nobody can, because AC is not necessarily true. > > > > > > Are there axioms which are necessarily true? Not according to your > > > point of view. > > > > Indeed. No axiom is necessarily true. > > How then should the truth of AC be proved? Not. Simply because it can not be proven (at least according to current mathematical thinking). But I also never stated that it can be proven. On the other hand, it is quite well possible that AC ultimately will be proven or disproven; you never know. Try to prove the parallel axiom using the other axioms from geometry. With the parallel axiom we get Euclidean geometry, without it we can get elliptic and hyperbolic geometry. The same with AC. With it we get mathematics with AC, without it we get mathematics without AC or with countable AC or whatever. Within an axiom system, an axiom is there for something that can not be shown from a system without that axiom. You can expand that smaller system either with the axiom given, or with its negation, or with variants. The results are different systems. It may happen that you can show that adding an axiom to an already given set of axioms leads to a contradiction, in that case the new axiom system defines nothing. But to prove inconsistency you have to show *within that new system* that you can prove both a proposition and its negation. However, if you do show that, you do not show the axiom is "false", you merely show that the set of axioms is inconsistent. As an example from algebra. Let's assume the axioms for a field (these are pretty basic and can be found in any book on algebra). Let's add to that set of axioms the axiom: (1) 1 + 1 = 0 which can not be proven from the basic axioms. And indeed, with that axiom a specific class of fields is defined (all fields of characteristic 2). We can rather add: (2) 1 + 1 + 1 - 0 and that also works (now fields have characteristic 3). But when we try to add (3) 1 + 1 + 1 + 1 = 0 it does not work anymore. In a field (by the basic axioms) all elements have an invers (i.e. for each a there is an a^(-1) such that a.a^(-1) = 1). But with this axiom we find that 2 does not have an inverse (there is no 2^(-1)). So there is an inconsistency. That does not make the axiom false, we can drop another axiom and most properly het a ring rather than a field. Also we can not add both (1) and (2) because that also would yield an inconsitency. But neither is false in itself. But all this is pretty basic algebra, taught (I think) in the first year at university. So I think you should know it. -- 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 21 Dec 2006 21:01 In article <1166737239.205667.205290(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1166617872.309270.165750(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > Dik T. Winter schrieb: > > ... > > > > The inequality holds for all finite paths. Why it also should hold for > > > > infinite paths escapes me. > > > > [ This was about the number of edges in finite case n: 2^(n+1) - 2, > > and the number of paths in finite case n: 2^n.] > > > > > > Because we can prove that 2 - 1/2^n is always, i.e., for all finite > > > natural n, less than 100. > > > > Yes, so I did state. > > > > > And because even infinity does not supply > > > other than finite natural numbers. > > > > That has no significance to the question. You fail to see that in > > the finite case each path has a last edge, but in the infinite case > > *no* path has a last edge. > > The sum 1 + 1/2 + 1/4 + ... has no last term. Nevertheless we know, > even in ZFC, that it is larger than 1 and less than 10. That has > significance to our question. The series exactly maps the infinity of > the paths. The sum as stated has *no* meaning in mathematics unless you are using limits. And that is why it fails to map exactly. You are using limits in a disguised form, and are using them wrongly. The limit of a quotient is the quotient of the limits only if all three limits do exist. It is not necessarily true if some of the limits do not exist (i.e. are not finite or the value alternates). In the previous messages (and this one) you state something akin to: lim{n->oo} ((# edges@n)/(# paths@n)) = lim{n->oo} (# edges@n) / lim{n->oo} (# paths@n). The first limit is indeed 2. But that does not mean that both limits on the right hand side can be compared with that. The two limits do not exist. So the theorem "the limit of a quotient is the quotient of the limits" can not be applied because that has only been proven for the finite case. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Newberry on 21 Dec 2006 21:48
Dik T. Winter wrote: > In article <1166737132.026953.46380(a)48g2000cwx.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > In article <1166617425.221440.15570(a)80g2000cwy.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > ... > > > > We are talking about shares, but we do not need more than 3/2 per > > > > path. > > > > > > 3/2 of what? But again you do not answer my question. In how many > > > shares is the edge going left from the root divided, and how do you > > > do that division? > > > > It is irrelevant for my proof. It is as irrelevant (and silly) as the > > question: in how many shares is the unit divided in the last term of > > the series 1/2^n. Nevertheless we can calculate the limit of this > > series. > > It is very relevant for the proof. You assume you can split in shares and > later on recombine those shares. When each edge is divided in finitely > many shares that is allowed. It is not allowed if you divide in infinitely > many shares, which you are doing. And the series 1/2^n is not relevant. > We can calculate the limit of that series because for every eps there is > an n0 such that for n > n0 the finite sum is on a distance not larger > than eps from 2, there are no shares involved at all. The issue here is whether the paths can be mapped onto the edges. More precisely can each path be mapped on a set of edge segments such that all the sets are disjoint. It seems to me that it can. > > > > Indeed, that can not be done, until at some moment there is a proof > > > (without AC) that that axiom is true. So until such a moment we have > > > to do with two forms of mathematics, one with AC and one without it. > > > But when we assume AC, a well-ordering can be constructed, as I did > > > show. > > > > No, you did no show it yet. You referred to some Hamel basis. A > > construction either says: This is the first element and that is the > > second etc. or it gives a prescription how this can be found. > > And I did. Construct a Hamel basis and do a lexicographic ordering > of the elements R as a vector space over Q (with some caveats). > That one is constructable, but I will not show it here (it takes > too much space). > > > > But apparently you have problems when something can be shown to exist > > > (in the mathematical sense) but can not be computed. That is close > > > to the intuitionistic view. > > > > I stated that there is no possible way to construct a well-ordering. > > You opposed. You cannot sustain your statement. That's all. > > Given AC, there is an Hamel basis and using that a well-ordering can > be defined. > > > > > > But I explicitly stated how you could construct > > > > > a well-ordering when AC is true. You can not do it in general, and > > > > > nobody can, because AC is not necessarily true. > > > > > > > > Are there axioms which are necessarily true? Not according to your > > > > point of view. > > > > > > Indeed. No axiom is necessarily true. > > > > How then should the truth of AC be proved? > > Not. Simply because it can not be proven (at least according to current > mathematical thinking). But I also never stated that it can be proven. > On the other hand, it is quite well possible that AC ultimately will > be proven or disproven; you never know. > > Try to prove the parallel axiom using the other axioms from geometry. > > With the parallel axiom we get Euclidean geometry, without it we can get > elliptic and hyperbolic geometry. The same with AC. With it we get > mathematics with AC, without it we get mathematics without AC or with > countable AC or whatever. > > Within an axiom system, an axiom is there for something that can not be > shown from a system without that axiom. You can expand that smaller > system either with the axiom given, or with its negation, or with > variants. The results are different systems. It may happen that you > can show that adding an axiom to an already given set of axioms > leads to a contradiction, in that case the new axiom system defines > nothing. But to prove inconsistency you have to show *within that > new system* that you can prove both a proposition and its negation. > However, if you do show that, you do not show the axiom is "false", > you merely show that the set of axioms is inconsistent. As an example > from algebra. Let's assume the axioms for a field (these are pretty > basic and can be found in any book on algebra). Let's add to that > set of axioms the axiom: > (1) 1 + 1 = 0 > which can not be proven from the basic axioms. And indeed, with that > axiom a specific class of fields is defined (all fields of characteristic > 2). We can rather add: > (2) 1 + 1 + 1 - 0 > and that also works (now fields have characteristic 3). But when > we try to add > (3) 1 + 1 + 1 + 1 = 0 > it does not work anymore. In a field (by the basic axioms) all > elements have an invers (i.e. for each a there is an a^(-1) such > that a.a^(-1) = 1). But with this axiom we find that 2 does not > have an inverse (there is no 2^(-1)). So there is an inconsistency. > That does not make the axiom false, we can drop another axiom and > most properly het a ring rather than a field. > > Also we can not add both (1) and (2) because that also would yield an > inconsitency. But neither is false in itself. > > But all this is pretty basic algebra, taught (I think) in the first > year at university. So I think you should know it. > -- > dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 > home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |