Cantor Confusion
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From: mueckenh on 5 Jan 2007 08:05 Dik T. Winter schrieb: > 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. Name a level or a node or and edge which are in one of the trees only. Do not argue that paths can differ without any different node or edge. That is too obvious a nonsense. > > > > 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. But, according to your point of view, N exists as the union of all finite sequences {1, 2, 3, ..., n}. Why do infinite paths not exist as the union of all finite paths? Precisely here is the point where it becomes obvious that an infinite number of finite numbers does not actually exist. Regards, WM
From: mueckenh on 5 Jan 2007 08:06 Dik T. Winter schrieb: > 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? I thought I had lost several lines in the buried version. Regards, WM
From: mueckenh on 5 Jan 2007 08:43 Dik T. Winter schrieb: > 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. Infinite node numbers are not in any of the finite trees, if you cut them at a certain level (contrary to my initial definition with sequences of 111... and 000... ). Does the union of such "cut-off" trees not contain infinitely many nodes? > > > > 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. Similarly, for any level of the tree there are finite trees which contain this level. Similarly, for any node of a path, there is a finite path which contains it. You said that in the union of all finite trees there is no infinite path. Would you say that in the union of all finite trees there is a path which contains infinitely many nodes? And if so, why do you say that this path which contains infinitely many nodes, is not infinitely long? And if not, how can 2^finite be infinite? > 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. How can an infinite set N get into the union of all finite initial segments? > > > > > 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. I need no belief. I know (from your reaction on the tree, among other evidence) that there are no irrational numbers (except for a few names, if you wish to call them numbers), so I will not spend any effort to look at V=L. The result can only be an assertion that well-ordering is possible. If there really was a well-ordering accomplished, then we would no longer need the axiom but could use the well-ordering. (If you really have two parallels, then you do not need the axiom that parallels exist.) > > > 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. I have not to allow for anything. Every path which is possible does exist as soon as all its edges exist. > In the union of > finite trees you allow only finite path, when you want completion you also > do allow infinite paths. That is wrong. IF infinite path do exist, THEN they exist in in the union of all trees, because there is NOTHING to be completed if already ALL is together. Regards, WM
From: mueckenh on 5 Jan 2007 08:46 Virgil schrieb: > 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. "by introducing what does not exist." That is a wise sentence. Now you should try to find out who introduced what does not exist (like some paths in a tree). Regards, WM
From: mueckenh on 5 Jan 2007 08:54
Virgil schrieb: > > So the union U{n | n e N} = U{1,2,3,...,n} is not N? What is missing? > > In my book, U{n|n e N} is not equal to U{1,2,3,...,n}. What tells your book about the difference? As far as I am informed, two different sets can be distinguished by at least one element. > And, in addition to any common sense, it appears that an "h" is missing > from WM's question. Yes, *here* you are right. > > > (I do not ask for a member. I do not ask whether an infinite path is a > > member. I ask whether it is in the union of finite trees.) > > What is the "it" that you are asking about? An infinite path. > > U{n | n e N = N. Really? Isn't there a very important curly bracket missing??? but U{1,2,3,...,n} = n+1 in NBG. Wrong. n+1 = {0,1,2,3,...,n} Of course the union over all natural numbers has to be taken. Has the difference then disappeared? Regards, WM |