Cantor Confusion
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From: mueckenh on 20 Nov 2006 09:53 Virgil schrieb: > > Please learn what "definable well-order" means technically. > > We can easily define well-order: any ordering of a set in which every > non-empty subset has a first element. Please learn what "definable well-order" means technically. This is different from defining the expression "well-order". > > > > > > > > We have no reason to suppose the we could not recognize one if it were > > > presented for inspection. > > > > We cannot recognize it, because it cannot exist. > > In ZFC, it must exist. In ZF one cannot prove it impossible. its > non-existence in ZF can be no more than an added assumption. > > > 1) It cannot be defined by a formula. > > Perhaps. Do you have any proof that it cannot be defined by a formula to > present here? That has been proven in the sixties, I think by Cohen. > > > 2) It cannot be listed because of cardinality reasons. > > I am not quite sure of what you mean by "listed". Would the listing of a > countable set of general rules constitute a listing? A well-order of the real numbers cannot be accomplished by a countable set of general rules. > > > 3) It cannot be determined in any other way. (Or do you have an idea?) > > Hence, we cannot recognize it. > > > > > > > > It is true that no one has actually constructed any explicit well > > > ordering. > > > > > > > > > It is true that no one will ever have constructed any explicit well > > ordering. > > That is a statement of faith only. No. A well-order of the real numbers is not definable. A well-order of the real numbers cannot be given by a list. There is no other means which could well-order the real numbers. To believe that it exists (where should it exist?) is a certificate of distinguished madness at an advanced level. > > > > > > > The axiom of infinity does not say anything about ordinal or > > > > > cardinal numbers. However, given that the set N exists and > > > > > the defnition of ordinal and cardinal numbers, it is easy to > > > > > see that if N exists it must have both an ordinal and a cardinal > > > > > number. > > > > > > > > > No. You assume the possibility of a bijection of the set with itself. > > > > > > The identity function on any set bijects it with itself. And such > > > functions are guaranteed, via the axiom of replacement. Not without a wrong interpretation of the axiom of infinity. > > > > > > > > > > That is not proven from the mere existence of the set if we cannot > > > > recognize or treat all of its elements. > > > > > > It is proven in ZF, even without C. > > > > The recognizability of all elements is not proven. > > And is not needed if a rule for all cases can be provided, such as > x |--> x. I do not see that any infinite set is covered. Your x is always finite and the number of numbers x treated is always finite too. So we have potential infinity but never actual infinity. To speak of "all cases" is a gross overstatement, unless you agree that "all cases" constitute a set which is not actually infinite. > > In ZF, at least, any model of such a "triangle" has infinite diagonal. > > >(The diagonal is a subset of the line ends.) > > The diagonal contains ALL the infinitely many line ends, so its set of > positions a SUPERset of the set of line ends. The diagonal is a subset and a superset of the line ends. This is just the proof you denied above. Regards, WM
From: William Hughes on 20 Nov 2006 09:54 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > > The diagonal consists of line indexes, i.e., of the line ends. > > > Therefore it is a subset of the line indexes. > > > > > > > Yes, the diagonal is the set of line indexes (a subset, but > > not a proper subset). > > Every line is a subset of the diagonal. And the diagonal is a subset of > the line ends. > > > > We need more than this. Consider the set > > X={1,2} and the two finite sequences A and B > > > > A= {1,2,1,2} > > > > B={1,2} > > > > A and B both consist of elements of the set X, but A is > > longer than B. So to say that both the diagonal and the > > lines are subsets of the line indexes is not enough > > to show that A cannot be longer than B. > > > > So you must have something more in mind when > > you claim that the fact that every element of the diagonal > > is a line index means that the diagonal cannot > > be longer than every line. What is this? > > Of course there are more conditions. They follow from the EIT. > Therefore I chose it. > 1 > 12 > 123 > ... > > Every line has only different indexes (there are not two equal indexes > in one line). Yes > Line n has all indexes from 1 to n. Yes > Every initial segment of line ends is a subset of a line. Yes > Every initial segment of line ends is a subset of the diagonal. Yes > Every initial segment of the diagonal is a line. No. There are two types of initial segments Initial segments with a largest element Initial segments without a largest element. There is an initial segment of the diagonal that does not have a largest element. (Recall: the diagonal has a largest element if and only if there is a last line. There is no last line.) Every line has a largest element. Thus there is an initial segment of the diagonal that is not a line. Assume that a sequence S is longer than every initial segement of S with a largest element. Then since S is an initial segment of itself, it follows that S does not have a largest element. You are trying to argue The diagonal cannot be longer than every initial segment with a largest element because there is no set of lines without a largest line. There is no set of lines without a largest line because the diagonal cannot be longer than every initial segment with a largest element. This is circular. - William Hughes
From: mueckenh on 20 Nov 2006 09:58 Virgil schrieb: > > For any set S, B := { <x,x> | x e S} defines a bijection from S to S. > > > > > But there are more than can be > > treated (= included in any proof) other than by induction. > > Not in ZF. > > > > > > > That is not proven from the mere existence of the set > > > > > > It _is_ proven using admissible techniques. > > > > Who admitted? > > ZF > > > It is proven by postulating that it is proven. > > It is provable from the axioms of ZF. Nothing further is needed. How then can countable models of ZF exist? Regards, WM
From: Sara M on 20 Nov 2006 10:15 > But the set of all lists is countable (as is any quantized or > discontinuous set), so is the set of all list entries. That's absurd. First of all, I'm not really clear what "discontinous" means when applied to a set, but if it means "connected" (or, path connected) then it's certainly false that any disconnected set is countable. The irrationals are no-where-connected (ie, they have no connected subset with more than one element) and yet they're not countable. Sara
From: Franziska Neugebauer on 20 Nov 2006 10:16
mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: >> mueckenh(a)rz.fh-augsburg.de wrote: >> >> > Franziska Neugebauer schrieb: >> >> Virgil wrote: >> >> >> >> > In article >> >> > <1163856355.707119.306700(a)m7g2000cwm.googlegroups.com>, >> >> > mueckenh(a)rz.fh-augsburg.de wrote: >> >> > >> >> >> Virgil schrieb: >> >> [...] >> >> >> And there are lines as long as the diagonal is. >> >> > >> >> > Name one. >> > >> > The elements of the diagonal are a subset of the line ends. >> >> Though Virgil posed this question: Name one single _line_ which is as >> long as the diagonal ist. > > Name one single element of the diagonal which is not contained in a > line (which contains this and all preceding elements). 1. Burden of proof weighs upon you who claimed "And there are lines as long as the diagonal is.". The question remains: Which lines? 2. Since neither Virgil nor I did posit any claim which needs support, not even by "naming one single element of the diagonal ...", we don't have to. > This is what I call the one-eyedness of set theory: Its proponents see > that for every line, there is a diagonal element not contained in this > and all preceding lines. This is your wording not mine and very likely not Virgils, too. As you know from my matrix-view of your list every diagonal element d_nn is necessarily located in row n and column n. Where else? Outside of this matrix there are no diagonal elements. > But they don't see, or at least dispel it, that there is no element of > the diagonal which is outside of any line. The sequence of diagonal elements is "embedded" in the matrix. There are neither n of the index set which have no d_nn nor are there d_mm which are not of the index set of the matrix. > The first observation leads to the theorem: The diagonal is superset > of all lines. Only sets have supersets. Define "all lines" please. > The second observation (together with the fact that > every line is a superset of all preceding lines) leads to the theorem: > There is at least one line which is superset of the diagonal. Please prove. > As there is no line with omega elements, there can be no diagonal with > omega elements. Non sequitur ("can be" replaced by "exists"). F. N. -- xyz |