Cantor Confusion
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From: Virgil on 25 Nov 2006 15:18 In article <1164446864.918062.126880(a)j44g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1164307852.961046.140060(a)f16g2000cwb.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Virgil schrieb: > > > > > > > > > I will accept that it can be connected if the "right" uncountable > > > > > > set is > > > > > > removed, but I can think of uncountable sets of points whose > > > > > > removal at > > > > > > least appears to totally disconnect the set of those remaining. > > > > > > > > > > Of course, for example if you remove all points. > > > > > > > > The remaining empty set is trivially connected. > > > > > > That is nonsense. If there is nothing to be connected then there is > > > nothing connected! > > > > Topologically, a set is connected unless it is a union of two or more > > non-empty disjoint open sets. > > > > See http://en.wikipedia.org/wiki/Connected_space > > > > According to this definition, an empty set is connected. > > I prefer to think myself. If WM prefers to substitute his own definitions in place of standard mathematical definitions, his version of "mathematics" will necessarily and obviously be irrelevant to any standard mathematics. The point of having a standard mathematics is that everyone must agree to use the standard definitions in order to discuss it reasonably. Those who insist on using non-standard definitions are required to give mathematically adequate definitions of what THEY mean if they expect their definitions to be understood or acceptable. > > What definition of connectedness does WM use which disconnects the empty > > set? > > A set is connected, if it has at least two elements and if its elements > are all connected. What constitutes a connection versus a disconnection? Topology has a clear definition. You do not. Ergo, your "definition" is irrelevant. >A set with only one element cannot be connected or > disconnected. Why? Topology has a precise answer, Do you? > A set with less than one element is, strictly speaking, > not a set at all. It is in ZF and NBG. Does WM then wish to claim that the intersection of two sets is sometimes "not a set at all"? > So it is neither connected nor > disconnected. WM, on the other hand, is totally disconnected on such issues. > > > With other words, you did not understand them. But our question > > > concerned the appendix only. The proof given there is very simple, so > > > simple that even Mr. Bader asserted to have understood it (it seemed so > > > to him, at least). You cannot confirm that my proof is correct? > > > > In reading the beginning of that paper, I found enough errors to > > dissuade me from bothering. > > So it should be possible to name one, unless you prefer to join the > slanderer, Mr. Bader. It is asserted in that paper that the so called first Cantor proof of the uncountability of the reals applies equally well to the set of rationals. Cantor's proof relies on the fact that if one has a strictly increasing sequence of reals with each term less than all terms of a strictly decreasing sequence of reals, there is at least one real strictly between the two sequences, not being a member of either but larger than every member of the increasing sequence and smaller that every member of the decreasing sequence. This is not the case for rationals. For example, there are strictly increasing sequences of rationals whose squares converge to 2 and strictly decreasing sequences of rationals whose squares converge to 2. But there is no rational number between the sequences. For example a_1 = a, a_{n+1} = 4*a_n/(2 + a_n ^ 2) is an increasing sequence of rationals and b_1 = 2, b_{n+1} = (2 + b_n ^ 2) / (2 * b_n) is a decreasing sequence of rationals with only sqrt(2) between all the a_n and all the b_n. > > > > > > Regards, WM > > > > > > PS: You really did not see that Cantor's arguing in his first proof is > > > valid for rational numbers as well as for transfinite numbers? Not so, as the example above proves. There is an increasing sequence of rationals converging to sqrt(2) and a decreasing sequence of rationals converging to sqrt(2) such that there is no rational at all caught between the two sequences. > > > > If WM thinks this, he has not understood the first proof. > > > > There is an increasing sequence of rationals whose LUB is sqrt(2), > > e.g., f(n) = floor( 2^n * sqrt(2) ) / 2^n. > > > > And a decreasing sequence of rationals whose GLB is sqrt(2), > > e.g., g(n) = Ceiling( 2^n * sqrt(2) ) / 2^n > > > > If Cantor's first proof held for the rationals, as WM claims, then WM > > would have proved that sqrt(2) is rational. > > You misunderstood completely. No, it is you who misunderstand Cantor's proof. > I claim that Cantor's proof is > "symmetric". It can be applied to the algebraic numbers, showing that > the limit is not algebraic. It can be but need not be. It can be applied to any proper subset of the reals showing that there is a sequence of values in that subset which does not converge to a value in that subset. So that anything short of the full set of reals is not closed under Cauchy convergence. One definition of the real number system is the closure of the rationals under Cauchy convergence. > It can as well be applied to the > transfinite numbers, showing that the limit is not transfinite. This I would like to see. A sequence of non-finite numbers having a finite limit? By what definition of limit does this sequence even have a limit? For sequnces of real numbers one can use a delta-epsilon definition or a topological definition of limit, but what >This > has nothing to do with rational numbers and sqrt(2) because that would > not prove any uncountability at all (sqrt(2) is algebraic and as such > belongs to a countable set). The point is that for every sequence of reals, there are reals not in that sequence. For a set to be countable, it is necessary and sufficient that there be a sequence containing every member of that set. This can be done, and has been done, with the set of rationals and with the set of algebraics, but not with the set of reals. It CANNOT be done for the reals. Cantor's first proof shows that ANY sequence of reals omits at least one real, thus any attempt to ennumerate the reals, in the sense of surjecting the naturals onto them, fails.
From: Virgil on 25 Nov 2006 15:25 In article <1164447079.521499.79580(a)l39g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > > For finite indexes it is correct. > > > > > > > > Infinite things are not constrained to behave in all respects like > > > > finite ones. That is because "infinite" means "not finite". > > > > > > But *the lines are all finite*. > > > > The *set* of lines in WM's so-called EIT is not finite. > > The number of elements of every line is finite. And only that is > relevant. WM is wrong again. In considering anything related to ALL lines, the finiteness or infiniteness of the set of all lines is critical. Any consideration of the "diagonal", which is built from the ends of ALL lines, requires us to consider ALL lines. > > > > > That's just why I chose the EIT as > > > example. > > > > > So your magic belief is easily > > > destroyed. > > > > Not by those who cannot keep their quantifiers straight. > > In dealing with linearly ordered finite sets there is no quantifier > magic. There is considerable quantifier logic involved, that WM habitually mucks up, in dealing with infinitely many finite lines.
From: Virgil on 25 Nov 2006 15:26 In article <1164447265.545922.145370(a)j44g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Why did it take so long to switch from "I don't need any advice from > > > you" and from "slightly misleading" to "erratum"? > > > > It took "so long" because you made my wording an issue and because you > > prefered to ride this dead horse insistently claiming that it was not > > dead. > > > > So I am guilty again! Still!
From: Ross A. Finlayson on 25 Nov 2006 15:50 Virgil wrote: > In article <1164446864.918062.126880(a)j44g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > In article <1164307852.961046.140060(a)f16g2000cwb.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Virgil schrieb: > > > > > > > > > > > I will accept that it can be connected if the "right" uncountable > > > > > > > set is > > > > > > > removed, but I can think of uncountable sets of points whose > > > > > > > removal at > > > > > > > least appears to totally disconnect the set of those remaining. > > > > > > > > > > > > Of course, for example if you remove all points. > > > > > > > > > > The remaining empty set is trivially connected. > > > > > > > > That is nonsense. If there is nothing to be connected then there is > > > > nothing connected! > > > > > > Topologically, a set is connected unless it is a union of two or more > > > non-empty disjoint open sets. > > > > > > See http://en.wikipedia.org/wiki/Connected_space > > > > > > According to this definition, an empty set is connected. > > > > I prefer to think myself. > > If WM prefers to substitute his own definitions in place of standard > mathematical definitions, his version of "mathematics" will necessarily > and obviously be irrelevant to any standard mathematics. > > The point of having a standard mathematics is that everyone must agree > to use the standard definitions in order to discuss it reasonably. > > Those who insist on using non-standard definitions are required to give > mathematically adequate definitions of what THEY mean if they expect > their definitions to be understood or acceptable. > > > > What definition of connectedness does WM use which disconnects the empty > > > set? > > > > A set is connected, if it has at least two elements and if its elements > > are all connected. > > What constitutes a connection versus a disconnection? Topology has a > clear definition. You do not. Ergo, your "definition" is irrelevant. > > > >A set with only one element cannot be connected or > > disconnected. > > Why? Topology has a precise answer, Do you? > > > > A set with less than one element is, strictly speaking, > > not a set at all. > > It is in ZF and NBG. Does WM then wish to claim that the intersection > of two sets is sometimes "not a set at all"? > > > > So it is neither connected nor > > disconnected. > > WM, on the other hand, is totally disconnected on such issues. > > > > > > > > With other words, you did not understand them. But our question > > > > concerned the appendix only. The proof given there is very simple, so > > > > simple that even Mr. Bader asserted to have understood it (it seemed so > > > > to him, at least). You cannot confirm that my proof is correct? > > > > > > In reading the beginning of that paper, I found enough errors to > > > dissuade me from bothering. > > > > So it should be possible to name one, unless you prefer to join the > > slanderer, Mr. Bader. > > It is asserted in that paper that the so called first Cantor proof of > the uncountability of the reals applies equally well to the set of > rationals. > > Cantor's proof relies on the fact that if one has a strictly increasing > sequence of reals with each term less than all terms of a strictly > decreasing sequence of reals, there is at least one real strictly > between the two sequences, not being a member of either but larger than > every member of the increasing sequence and smaller that every member of > the decreasing sequence. > > This is not the case for rationals. For example, there are strictly > increasing sequences of rationals whose squares converge to 2 and > strictly decreasing sequences of rationals whose squares converge to 2. > But there is no rational number between the sequences. > > For example a_1 = a, a_{n+1} = 4*a_n/(2 + a_n ^ 2) is an increasing > sequence of rationals and b_1 = 2, b_{n+1} = (2 + b_n ^ 2) / (2 * b_n) > is a decreasing sequence of rationals with only sqrt(2) between all the > a_n and all the b_n. > > > > > > > > > Regards, WM > > > > > > > > PS: You really did not see that Cantor's arguing in his first proof is > > > > valid for rational numbers as well as for transfinite numbers? > > Not so, as the example above proves. There is an increasing sequence of > rationals converging to sqrt(2) and a decreasing sequence of rationals > converging to sqrt(2) such that there is no rational at all caught > between the two sequences. > > > > > > If WM thinks this, he has not understood the first proof. > > > > > > There is an increasing sequence of rationals whose LUB is sqrt(2), > > > e.g., f(n) = floor( 2^n * sqrt(2) ) / 2^n. > > > > > > And a decreasing sequence of rationals whose GLB is sqrt(2), > > > e.g., g(n) = Ceiling( 2^n * sqrt(2) ) / 2^n > > > > > > If Cantor's first proof held for the rationals, as WM claims, then WM > > > would have proved that sqrt(2) is rational. > > > > You misunderstood completely. > > No, it is you who misunderstand Cantor's proof. > > > I claim that Cantor's proof is > > "symmetric". It can be applied to the algebraic numbers, showing that > > the limit is not algebraic. > > It can be but need not be. > > It can be applied to any proper subset of the reals showing that there > is a sequence of values in that subset which does not converge to a > value in that subset. So that anything short of the full set of reals is > not closed under Cauchy convergence. One definition of the real number > system is the closure of the rationals under Cauchy convergence. > > > > > It can as well be applied to the > > transfinite numbers, showing that the limit is not transfinite. > > This I would like to see. A sequence of non-finite numbers having a > finite limit? By what definition of limit does this sequence even have > a limit? For sequnces of real numbers one can use a delta-epsilon > definition or a topological definition of limit, but what > > > > >This > > has nothing to do with rational numbers and sqrt(2) because that would > > not prove any uncountability at all (sqrt(2) is algebraic and as such > > belongs to a countable set). > > The point is that for every sequence of reals, there are reals not in > that sequence. > > For a set to be countable, it is necessary and sufficient that there be > a sequence containing every member of that set. > > This can be done, and has been done, with the set of rationals and with > the set of algebraics, but not with the set of reals. It CANNOT be done > for the reals. > > Cantor's first proof shows that ANY sequence of reals omits at least one > real, thus any attempt to ennumerate the reals, in the sense of > surjecting the naturals onto them, fails. That's not so where the reals, non-standardly or whatever, can be addressed as contiguous points on a line. Megill's proof checker found the rationals uncountable. Well-order the reals, biject them to an ordinal. Does Cantor's first in transfer apply? At some point in time there's a degenerate interval, and before that not, extend inductive reasoning to infinite ordinals, oops, now you're talking about the set of naturals as if it were an element, of itself, for inductive purposes. There is no universe in ZF. There isn't a set of sets in ZF, there are no sets in ZF, identity is too big. Cauchy/Dedekind definitions (acknowledgment, placement, in context) of real numbers are: not sufficient. Biject R[0,1]^x to N^N and sample towards an N-sequence, that appears to be a random natural integer. It is you, Virgii, who misunderstand, and so vociferously, and much. That's not to say others don't: you do. Ross
From: Dik T. Winter on 25 Nov 2006 21:25
In article <1164445719.457858.15120(a)l39g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1164281187.964067.115190(a)m7g2000cwm.googlegroups.com> muecken= > h(a)rz.fh-augsburg.de writes: > > ... > > > Cantor wrote: "It is remarkable that removing a countable set leaves > > > the plaine *being connected*." But it is not remarkable that removing > > > an uncountable set leaves the plaine being connected? What a foolish > > > assertion. > > > > But removing an uncountable set can leave the plane either connected or > > disconnected. What is remarkable about that? > > Nothing. In order to discuss Cantors paper one should know it. Of > course Cantor knew your trivial example and those proposed by Mr. > Bader, who is seems to be used to take trivialities for the main issue. > > > Remove the line x=0 from > > the plane. You remove an uncountable set and the result is clearly > > disconnected. > > Yes, but Cantor considered only point sets which are dense. You did not state that above. > "Betrachten > wir irgendeine Punktmenge (M), welche innerhalb eines n-dimensionalen > stetig zusammenh�ngenden Gebietes A �beralldicht verbreitet ist und > die Eigenschaft der Abz�hlbarkeit besitzt,..." But under *those* conditions there there are still uncountable sets that leave the plane connected and also sets that leave the plane disconnected. > Cantor wrote that the countability is an important condition for his > proof: "Es gen�gt, diesen Satz f�r den Fall n = 2 als richtig zu > erkennen; sein Beweis beruht wesentlich auf dem in Art. 1 bewiesenen > Satze, da�, wenn irgendeine gesetzm��ige Reihe reeller Gr��en ... > vorliegt, in jedem noch so kleinen willk�rlich gegebenen Intervalle > reelle Gr��en gefunden werden k�nnen, die in jener Reihe nicht > vorkommen. > > And this condition (of countability) is obviously not necessary. Obviously not. But it is sufficient, and that in itself is already remarkable enough. Remove all irrational points and in addition those points that have x = 0. You have a disconnected set. But I can not find in anything you quote that Cantor thought it was necessary. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |