Cantor Confusion
in [Math]
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From: MoeBlee on 11 May 2007 15:39 On May 11, 1:00 am, WM <mueck...(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 22:10, Virgil <vir...(a)comcast.net> wrote: > H&J could have written: "it does not matter that the set of these > ordered pairs is described by a property" in your opinion? Properties don't describe. We (or our formulas) describe something by mentioning its properties; it's not the properties that are doing the describing. > > In dealing with sets > > of ordered pairs in general, one need not describe it as anything but a > > set of ordered pairs. > > Did I specify the rule, process, mapping, in short: the formula? When > dealing with the general notion of function, i.e., when not dealing > with a special function you need not specify the formula. But without > a formula or formal description determining by what property the set > of ordered pairs is described, there is no set of ordered pairs, i.e., > *there is no function*. You don't understand set theory from the very first word. Existence does not depend on description. We prove that there exist certain sets having certain properties, but we also prove that there exist sets having a certain property even as we might not prove of any PARTICULAR set that it has those properties. So, we define 'function' by 'f is a function <-> (f is a relation & Axyz((<x y>ef & <x z>ef) -> y=z). As far as the DEFINITION goes, that is IT. PERIOD. There is NOTHING else involved in that definition. There is no mention of 'formula' nor 'domain' nor 'range'. (From now on, let 'f is many-one' be an abbreviation for 'Axyz((<x y>ef & <x z>ef) -> y=z)'. as "f is many-one".) Then, to prove 'f is a function' ONLY requires proving that f is a many one relation. In a particular instance it might or might NOT be the case that we mention the domain and range of f or a particular defininig formula for f. Whatever we mention, all that is REQUIRED to prove that f is a funtion is that f is a many-one relation. And, of course we prove that every function as its domain and its range, but those are THEOREMS ABOUT functions (actually about relations in general, and actually about any set whatsoever in certain treatments though Hrbacek & Jech is not one of them) and not part of the DEFINITION of a function. I KNOW you can understand that. It's simple: there is the defintion of 'function' and then there are ALSO various things that that defintion entails. Right? Another poster gave you the example that every natural number has its square but 'square of' is not a part of the definition of 'is a natural number'. And every function has its domain but 'domain of' is not part of the definition of 'is a function'. I KNOW that you can see that clearly now. Right? > And without having a domain, the function is not defined at all. There is no function that doesn't its domain domain. Yes, we ALL AGREE about that. But (1) To define 'is a function', we don't require mentioning 'domain', since the fact that every function has its domain is a CONSEQUENCE of the definition and is not part of the definition ITSELF. You can see this now. Right? (2) There is a difference between defining 'is a function' and definining a PARTICULAR set that is a function. And to define a particular set that is a function (and possibly also to prove that it is a function) might or might not involve mentioning or proving something about its domain; but in any case we DO know that it is a CONSEQUENCE of the defintion of 'function' that every function has its domain. > Of > course you can name every element of the domain separately. But that > does not veil the fact, that a domain is required and is existing for > any well defined function f. Every function has its domain. That has NEVER been at issue. Further, even though Hrbacek & Jech themselves do not adopt this tack, we can define 'domain', 'range' and 'field' so that EVERY set S has its domain, range and field: dom(S) = {x | Ey <x y>eS} range(S) = {y | Ex <x y>eS} field(S) = dom(s) u range(S) And, yes, we've done that excercise in which we prove: E!zAx(xez <-> Ey <x y>eS) and E!zAy(yez <-> Ex <x y>eS) MoeBlee
From: Virgil on 11 May 2007 15:45 In article <1178866486.658963.253390(a)y5g2000hsa.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 21:21, Virgil <vir...(a)comcast.net> wrote: > > In article <1178795907.957410.94...(a)o5g2000hsb.googlegroups.com>, > > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > On 10 Mai, 03:09, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > > > So in your opinion there are only finitely many paths in the > > > > > > infinite > > > > > > tree. > > > > > > > I did not say that. > > > > > > Quote: "Only from those few you can ask for". I can ask only for > > > > finitely > > > > many paths. > > > > > That is correct. But that does not mean that I said that there are > > > only finitely many paths in the tree, as you implied. > > > > Since your other comments imply that what one cannot ask for does not > > exist, it is reasonable to conclude you meant only finitely many. > > My personal opinion is not the starting point here. Since mathematical > entities exist only in the mind, such entities, which cannot exist in > the mind, cannot exist at all. Since your own mind is the only one for which you can speak, what goes on it other minds is beyond your competence to decide. > > > > > > > > Not at all. I ask the single question: "are all other paths different > > > > from > > > > a particular path?" > > > > > That is circular, because by "other" you imply that path p can be > > > distinguished by a node from all the others. > > > > Not so. It only implies that any two distinct paths can be distinguished > > from each other at some node. > > > > And they can, for if they are not the same it is only because at some > > node they branch differently. > > Then all different paths should be distinguishable. This is > contradicted by the fact that every node of p which you can ask for is > shared by the same path p'. Only when p = p'. Otherwise one can ask for the first n at which they disagree, and as N is well ordered, either such an n must exist or the set of naturals at which they differ is empty. > > > > > But we know, that this > > > not possible. > > > > Is that a royal "WE"? > > No, it includes only people capable of logic thinking and refuting the > notion of infinite sets, in particular of infinite definitions. So far it seems to include only WM himself, as everyone else who posts on this issue disagrees with WM. > > > > It is certainly not a plural "WE" in this thread. > > > > > Hence there is another path with p for all the time. > > > > If two paths share all their nodes, how can you tell that there is more > > than one of them? > > That is impossible. > Therefore there are not all nodes. A much simpler answer is that they are not different. > > There is an antinomy. > Therefore there are not actually infinitely many nodes. Then there is not a CIBT either. Which is quite a different issue. Once existence is granted, then the properties that WM so abhors necessarily follow. Except that in ZF and NBG there ARE CIBTs. > > Definition of equality of paths in any tree: > > if every node of path p is a node of path p', and vice versa. then p = > > p'. > > > > > > > > > You always state that this was true, but you cannot prove it. Is that > > > mathematics? One does not need to "prove" definitions. They may turn out not to be useful, but they are not "claims" of anything, merely abbreviations. If WM should prefer a different definition of equality of paths, let him state it, and we will discuss its usefulness in comparison to mine. > > For every node in a CIBT, there are at least two paths through that node. > > > > If p' is "with" p at every node of p, then p' = p. > > Your last two sentences show the antinomy, because for any node we can > show that one and the same path ' accompanies p for all nodes > investigarted. Unless you can show it for every node whether investigated or not, there is no equality. > > > > > It is valid for those cases I did prove. I did not prove it (and one > > > cannot prove it) for "all natural numbers which obey this formula", > > > because all natural numbers which obey this formula do not obey this > > > formula. I fail to see any possible justification for "because all natural numbers which obey this formula do not obey this formula" > > > > So that. Apparently, one of WM's axioms is that finite induction does > > not hold. > > In mathematics it holds for the potentially infinite set of natural > numbers > MatheRealism shows that it holds only for few natural numbers. MathRealism is math fantasy. > > > > > My argument is: Why do you have to exclude the case 0.999... = > > > 1.000... if you remain always in the finite? For every finite index > > > both numbers are different. > > > > We, and Cantor, exclude them by having a diagonal rule that never uses > > either 0 or 9. > > Why is that rule required? To exclude the diagonal from having any alternate representation. > > > > > > It is obviously wrong for all n for which it holds, if such set of all > > > n exists. Thus WM argues "F(n) is true" implies "F(n) is false". No wonder he has so much trouble with both mathematics and logic! > > > > For which naturals n is WM saying Sum[x = 1..n, x] n*(n+1)//2 is false? > > For the actually infinite set of "all naturals" it is false. The set of all naturals is not a natural, so you answer is purposely non responsive. WM claims there are naturals for which the formula is false but refuses to name any. All > naturals are finite, so we can sum all naturals by this formula, but > as the result is > > 1+2+3+...= aleph_0 As the formula is for a particular natural, and the only sum is for the naturals not greater that the given one, WM's "formula" is deliberately irrelevant. > > there must be an infinite natural, which is a contradiction. Why must there be one? > > > > Only on the right-hand side. The left-hand side formula is a statement > > > about a set. > > > > So what? It is about a finite set as specified by a (finite) natural. > > Every natural is finite. So we can use this formula for every natural > (right) and for all naturals up to every natural, i.e., for all > naturals. Then it is ONLY WM who is assuming that there is a natural greater than all naturals, No one else is being so foolish. > > Or is "all up to every" not the same as all? All up to and including a given natural is not the same as all. The sum is for "all up to and including a given natural" The equation says for any naturals the sum up to that natural is given by a shorter form. Only someone as pixilated as WM need continually conflate the two senses.
From: Virgil on 11 May 2007 15:52 In article <1178870415.724641.38810(a)n59g2000hsh.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 22:10, Virgil <vir...(a)comcast.net> wrote: > > In article <1178817355.571399.114...(a)y80g2000hsf.googlegroups.com>, > > > > A binary relation is, therefore, determined by giving all ordered > > > pairs of objects in that relation; it dos not matter by what property > > > the set of these ordered pairs is described. > > > > > !!!!!!!!!!!! But it obviously does matter *that* it is > > > described. !!!!!!!!!!!!!!! > > > > That depends on what one wants to do with the set. > > H&J could have written: "it does not matter that the set of these > ordered pairs is described by a property" in your opinion? They could have, with no loss of correctness, said that. > > > In dealing with sets > > of ordered pairs in general, one need not describe it as anything but a > > set of ordered pairs. > > Did I specify the rule, process, mapping You claimed the necessity of one without which no function can be defined. > > And without having a domain, the function is not defined at all. If one can deduce a domain, it is not necessary to specify one. The question is whether it needs to be specified before a function can defined or whether it can be determined afterwards, and only if needed. Of > course you can name every element of the domain separately. But that > does not veil the fact, that a domain is required and is existing for > any well defined function f. Only after the fact. Given a function, it is possible to show that a domain exists, but it is not always necessary to show it.
From: Virgil on 11 May 2007 15:59 In article <1178870843.168676.226300(a)e51g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 22:25, Virgil <vir...(a)comcast.net> wrote: > > > > 1) for every path p', there is a node where p' leaves p > > > and > > > 2) there is no node where every path p' has left p > > > then you must be able to name a node K(n) such that for nodes K(m < > > > n) > > > at least two paths p' and p'' are required to accompany p. > > > > This is true of EVERY path and for each of its nodes in an infinite tree. > > > > For every node K(n) of P, there is one path p' which accompanies it to > > the next node and then branches off, and another, p'', which does not > > branch off until the node after that, and so on, ad infinitum. > > And the Moon is round and January is the first month in the year. So > what? I did expect that. And what you said is true. But it is not an > argument. Then WM does not know what an argument looks like. > > Ad infinitum there is a path p' with p, because ad infinitum there is > no node where p is single. Therefore, ad infinitum, there is a path > which shares with p the node in question and all earlier nodes too. There is an ambiguity in your argument. While it is true that at every node there is more than one path through that node, it is also true that for any two paths there is a ladt node that they have in common. If two fixed paths, p and p', share ALL their nodes, then p = p'.
From: Virgil on 11 May 2007 16:03
In article <1178871714.942838.90580(a)w5g2000hsg.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 10 Mai, 22:25, Virgil <vir...(a)comcast.net> wrote: > > > > 1) for every path p', there is a node where p' leaves p > > > and > > > 2) there is no node where every path p' has left p > > > then you must be able to name a node K(n) such that for nodes K(m < > > > n) > > > at least two paths p' and p'' are required to accompany p. > > > > This is true of EVERY path and for each of its nodes in an infinite tree. > > > > For every node K(n) of P, there is one path p' which accompanies it to > > the next node and then branches off, and another, p'', which does not > > branch off until the node after that, and so on, ad infinitum. > > And the Moon is round and January is the first month in the year. So > what? I did expect that. And what you said is true. But it is not an > argument. > > Ad infinitum there is a path p' with p, because ad infinitum there is > no node where p is single. Therefore, ad infinitum, there is a path > which shares with p the node in question and all earlier nodes too. But not all subsequent nodes unless they are the same path. Ad infinitum there are differing paths p' "with" p, but there is no single path, p', always "with" p unless p' = p. |