Cantor Confusion
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From: Virgil on 5 Jan 2007 17:07 In article <1168005851.007466.284050(a)11g2000cwr.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1167860661.713987.282460(a)k21g2000cwa.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > If EVERY edge of a path is shared by another path, then both paths > > > cannot be distinguished. > > > > That is only true in finite trees. In an infinite tree, given any path > > and any finite n, there is anoth path sharing the first n edges but > > separating after n edges. > > We are looking for two paths sharing every edge. Why look for what cannot be found? > (Otherwise there must > be an edge not shared by two ore more paths.)> If one has two distinguishable paths in a given tree, then what distinguishes them is that there is at least one edge in one not in the other. But which edge or edges are not shared depends on which paths one is trying to distinguish between.
From: Virgil on 5 Jan 2007 17:16 In article <1168006304.299046.175620(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > > > It is without any value to follow your text. > > > > LOL. Head-in-the-sand? > > Lack of interest in useless sophisms. > > If you are interested in meaningful discussion, then try to find out, > for instance, the difference between the complete binary tree (with > irrational numbers) and the union of all finite binary trees (without > such representations). Your party is going down. > > Regards, WM One obvious difference is that the union of finite trees is not a tree, at least if union is taken in its usual sense. If the family of trees is ordered by having every path in a "smaller" tree be an initial segment of a path in a "larger" tree, then the family is well-ordered, and one can posit a "limit" tree, which, for a suitably infinite set of trees, would be an infinite tree in which each path in each finite tree is an initial segment of an infinite path in the limit tree, and every path in the limit tree is the limit of a sequence of such initial segments. Note that this "limit" construction is quite different from a mere union of finite trees.
From: Virgil on 5 Jan 2007 17:19 In article <1168006581.705686.85150(a)v33g2000cwv.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > I do not assume that 0.111... exists. I only assume that > > > > given that 0.111...1 exists you can show that 0.111...11 exists. > > > > > > > That position is correct. It is called potential infinity. > > > > > > > > > > That is the same with the lines. Why should the diagonal exist > > > > > actually > > > > > but the system of lines should not exist actually? (1/9 = 0.111...) > > > > > > > > Indeed. It is the same with lines. It is always possible to find > > > > another line. However, iii says nothing about whether the > > > > diagonal "actally exists" (or equivalently whether the > > > > system of lines "acutally exists"). > > > > > > But you always assert that the the complete diagonal exists, i.e., a > > > diagonal which could not be extended. Then you must also accept a line > > > system which cannot be extendend. > > > > No. At no time do I assume that the complete diagonal exists. > > > > > > > > > > > > > > > iv: Not all the elements of the diagonal exist > > > > > > > > > > Not all natural numbers in unary representation 0.1, 0.11, 0.111, ... > > > > > exist. If all elements of the diagonal exist (which are the last > > > > > digits > > > > > of the unary numbers) then these unary numbers must exist too, as far > > > > > as I understand existence. > > > > > > > > Any discussion of iv is completely irrelevant. iv is neither needed > > > > nor used. > > > > > > Then drop it. > > > > I have never used it ("neither needed nor used"). > > > > > > > > You used to use ~iv: all the elements of the diagonal exist > > > > > > > Only to point out that I neither need nor use the > > assumption ~iv. > > > > > > The contradiction is: > > > > > > > > A: L_D must have the property that given any set of elements > > > > that exist in L_D one can always find another element of > > > > L_D > > > > [This follows immediately from i ii and iii] > > > > > > It is not different for the lines. > > > > > > > > B: L_D has a largest element. [this follows from the fact the > > > > L_D > > > > is a line] > > > > > > This follows from the assumption (iv) of complete existence of L_D, > > > which implies complete existence of the system of lines. If you drop > > > it, then the contradiction vanishes. > > > > No, it follows from the fact that L_D is a line. A line has a largest > > element. > > L_D has a largest element? If it is to be a line, yes. > > > L_D is not the system of lines. > > L_D consists of line ends, i.e., just of these largest elements. Including an end to the ends. > > > No assumption about the complete existence of the > > system of lines is needed or used. > > You use the complete existence of L_D. But perhaps you do not recognize > it. WM defined L_D to be a line, therefore to have an end, but also to contain a next for each of its members, therefore not to have an end. And contradictions arise from WM's initial self-contradictory definitions, nothing else. > > Regards, WM
From: Andy Smith on 5 Jan 2007 07:12 Thanks for the reponse. >The Peano axioms are what define the natural numbers. If you don't >accept those axioms, then for you the natural numbers do not exist. I>f that's the case, then why are you arguing about the Cantor proof of >the uncountability of the reals? After all, neither of the sets N or R >exists for you and therefore the entire question is vacuous. Yes, I was talking about axioms of arithmetic rather than Peano - e.g. (according to Cantor) infinity + 1 is a new cardinal number but 1 + infinity is infinity. >Wrong. It's not a matter of whether the counting is complete. The >problem is that when the counting is complete, there is no integer that >maps to 1/3 in your scheme. By induction, each natural number maps to a >rational that can be represented in a finite number of digits. Unless you concede that an actually infinite natural number exists. >It seems that extrapolating without justification from the finite case to >the infinite case is perfectly acceptable, as long as you are the one who >is doing it. Extrapolating *with* justification (by appealing to the >axioms) is *not* justified, in your view, if it is someone else's >argument. Fair enough, I'm not trying to get into a fight - just trying to sort things out for myself. Back to the flipped set of {N}, the resulting finished set in [0,1] is bounded with one element at 0.000... and one < 1.000... If we consider an element with N successive binary 1's there will be an element with N+1 successive 1's, and this is true for any N, so by induction there must be a number .1111.... ? >> You didn't comment on the application of Cantor's diagonalisation argument >> to {N} - again, with the inverted system above, that >> would seem to suggest, if the argument is valid (and >> in this case, the infinite matrix is definitely diagonal) >> that there are integers not present in {N}. >I thought I did comment on that. The process produces an infinite digit >string, which does not represent a member of N. But {N} represents the set of ALL natural numbers, so there cannot be a member outside of it - so the Cantor diagonalisation argument is fallacious (which was rather the point) ? Thanks anyway for your time. Regards.
From: Virgil on 5 Jan 2007 17:24
In article <1168020763.180583.195440(a)51g2000cwl.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Perhaps there is a gap in your understanding of numbers and their > representation in set theory? And perhaps, as has been amply demonstrated of WM's profound misunderstanding of trees, that gap is in WM's understanding of numbers and their representation in set theory? |