Cantor Confusion
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From: Franziska Neugebauer on 3 Jan 2007 17:09 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > >> > Correct. x is an idea. It is identical to the entity which you call >> > an irrational number, i.e., an idea y. >> > >> >> Your answer to question 3 states >> >> >> >> x is different from y (due to some reason) >> >> <-> there is a difference between x and y (Q) > > Here is your error: > x is different from y' > <-> there is a difference between x and y' (Q) > i.e., there is a difference between x and what you believe is y. The > error lies in your assertion that y = y'. Your answer to my second last post is too late. I already had prepared a much nicer explanation of your contradiction. Probably google cut it away so I will paste myself here: ,----[ <1167256258.863156.191060(a)73g2000cwn.googlegroups.com> ] | > 1. So you agree that there exist two entities x1, x2 e R which solve | > the equation x^2 = 2? | | Yes. [<- this is WM's answer] | > | > 2. Is it true that what "we" call an "irrational number" (x1, x2)) | > is identical to your "idea"? | | Yes. [<- this is WM's answer] `---- This answer to question 2 is (P) ,----[ continued ] | > 3. If so, I cannot see what a meaningful, substantial difference | > between your "irrational idea" and the common "irrational number" | > could be. What - besides pure terminology - is that difference? | | The difference is that an idea has no digits. [<- this is WM's | answer] `---- This part of your answer to question 3 is (Q). Irrefutibly it states that there is a difference, hence Q <-> ~P. ,----[ continued ] | And the numerical approximation of an idea like sqrt(2), written in a | list, has not enough digits to determine whether it is different from | infinitely many other numbers or numerical approximations of ideas | contained in that list. [<- this is WM's answer continued] `---- This part of your answer to question 4 we may name (Q'). Your three answers together state P & Q & Q' According to the rules of logic which still are in effect a & b & c -> a & b regardless of the truth of c. Hence from stating P & Q & Q' it follows that you also state P & Q Since P <-> ~Q you are stating a contradiction: P & ~P F. N. -- xyz
From: Dik T. Winter on 3 Jan 2007 17:58 In article <1167859855.107241.239690(a)k21g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Virgil schrieb: .... > > it is quite similar in that one has in both cases an infinite set of all > > whose members are finite. > > That is precisely the infinite string of finite digit indexes of > irrational numbers and similarly the infinite string of finite nodes of > infinite paths. But one has no infinite paths in the tree? Not in the union of finite trees, but it is in the union of the full tree. The two are not the same. > > WM again conflates being a member of a set (the uniion) with being a > > subset of that set. N is a subset of the union of all initial segments ( > > actually equals the union) but is not a member of that union. > > So the union U{n | n e N} = U{1,2,3,...,n} is not N? Wat is missing? It is equal. What Virgil is stating is that N in U{n in N} {1, 2, 3, ..., n} but that N is not one of the subsets used in the union. -- 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 3 Jan 2007 18:21 In article <1167860661.713987.282460(a)k21g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: Why do you post this twice, once buried in another article and now again? -- 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 3 Jan 2007 18:20 In article <1167860527.440727.143820(a)i80g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > But you did not say which edges or nodes distinguish the complete tree > > > from the union of all rational trees. > > > Do you believe that there are such distinguishing edges but that one > > > cannot name them? > > > Or do you think that there are same edges in both trees but that some > > > paths can form in one tree which cannot from in the other tree? > > > > The last one. In the union there is *no* infinite path. > > > > > Wouldn't both answers point to some matheology? > > > > No, they are based on the definition of union and some elementary logic. > > (1) Each node is in one of the finite trees, so it is also in the union. > > So there is an infinite number of nodes in the union of finite numbers > of nodes. Right. > > (2) Each edge is in one of the finite trees, so it is also in the union. > > So there is an infinite number of edges in the union of finite numbers > of nodes. Right. > > (3) Infinite paths are in none of the finite trees, so they are also not > in the union. > > So the set of natural numbers is not in the union of all initial > segments of natural numbers? Wrong. Infinite paths are not in the union of finite trees because none of the finite trees contains an infinite path. All natural numbers are in the union of the finite segments because for all natural number there are initial segments that contain it. > > With which of the above three statements do you disagree? > > If (1), name a node not in one of the finite trees. > > If (2), name an edge not in one of the finite trees. > > If (3), name an infinite path that is in a finite tree. > > Apparently you agree with all three, but at the same time states that > > the completed infinite tree (which has infinite paths) is the same as > > the union of the finite trees. > > That is wrong. If an infinite number of nodes is in any path of the > union, then the path is infinite too. How can an infinite path get into the union of finite trees if none of the finite trees contains one? By magic? All paths that are in the union of finite trees after a finite number of steps go only either to the left or to the right. > > Do you not see a contradiction here? > > Of course. It lies in the assumption of actual infinity. It lies in the > assumption of a complete tree. No, it lies in your misunderstanding. > > Well, have a look at ZFC+V=L. Apparently a well-ordering of the reals is > > automatic with V=L. > > Apparently it does not. I am sure otherwise you would already have > posted it. Have you *looked* at V=L? But you are unwilling to look at it because you are unwilling to believe it. > > That is easy: {x} is one of those ordered sets. Here it is the first and > > only element. Note that this was in reply to your: > > As your above example shows, one must be able to find for every x the > > set where x is the first element Now I may have misunderstood your question, of course. > But you cannot do it in > another way too. So maintain to assert that V=L is the magical formula > which well-orders the real numbers. As I have seen, you are ready to > believe in even more incredible items. You have apparently no idea what V=L is. It is an axiom, look it up. It is more restrictive than the axiom of choice, so you should be able to think that you can prove stronger assertions with V=L than with the axiom of choice. > > Wrong? I don't know. But I think a quantifier problem again. I state (P > > is set of paths, E is set of edges, and p1 and p2 are different): > > If EVERY edge of a path is shared by another path, then both paths > cannot be distinguished. Your quantifier magic is without any value to > veil this fact in an actually existing infinity, i.e., in an infinity Why? You still do not prove such. > > (1) forall{p1 and p2 in P} thereis{e in E} e in p1 but not in p2 > > or every two different paths differ in at least one edge. > > (2) forall{e in E} thereis{p1 and p2 in P} such that e in p1 and in p2 > > or for every edge there are two different paths that share that edge. > > For some reason you conclude from this: > > The reason is that I do not believe in different trees which are > identical > > > (3) forall{p1 in P} thereis{p2 in P} forall{e in p1} e in p2. > > or there are different paths that share all edges. > > The paths obviously are not different, if they share all edges. This > example shall only show you that your paragraph (2) is wrong in an > actual infinity. It is, or at least could be, correct in a potential > infinity. Alas, these both notions, actual and potential, are usually > intermingled in set theory. This paragraph looks a bit like nonsense. > > Now (3) is clearly in contradiction with (1), so I wonder how you come > > to this conclusion. > > I come to this from (2). The reason is that I do not believe in > different trees which are identical, a belief which is required by the > adorants of quantifier magic. The trees are not different with respect to edges and nodes, it is the set of paths that you allow in the trees that are different. In the union of finite trees you allow only finite path, when you want completion you also do allow infinite paths. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Virgil on 3 Jan 2007 19:25
In article <1167859668.778067.101400(a)i80g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > > > Correct. x is an idea. It is identical to the entity which you call an > > > irrational number, i.e., an idea y. > > > > > >> Your answer to question 3 states > > >> > > >> x is different from y (due to some reason) > > >> <-> there is a difference between x and y (Q) > > Here is your error: > x is different from y' > <-> there is a difference between x and y' (Q) > i.e., there is a difference between x and what you believe is y. The > error lies in your assertion that y = y'. As there is only a y and no y', the error is entirely WM's by introducing what does not exist. |