Cantor Confusion
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From: Virgil on 27 Nov 2006 19:55 In article <1164636685.066274.319580(a)j72g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > 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, > > So the diagonal is not outside of any line? I didn't know that lines had insides. > So there is a finite line > as long as the diagonal? That may be WM's claim but is not mine. My claim is that every single line is exceeded by the diagonal. > > > > > > In dealing with linearly ordered finite sets there is no quantifier > > > magic. > > > > There is considerable quantifier logic involved, > > No, it is not logic but a mental disease. The mental disease in question is one that WM exhibits, and in sci.math is generally called quantifier dyslexia, an inability to keep one's quantifier meanings straight. Ax in S Ey in S ((x != y) --> (y > x)) is true in every ordered set,S, of 2 or more members. Ey in S Ax in S ((x != y) --> (y > x)) is false in every ordered set, S, which does not have a largest member, such as N or Q or R.
From: Dik T. Winter on 27 Nov 2006 20:00 In article <1164637800.628277.302340(a)14g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... You wrote: > > > > > 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. .... > > > Yes, but Cantor considered only point sets which are dense. > > > > You did not state that above. > > I wrote about algebraic and transcendental numbers. So being dense is > implied and need not be mentioned. Perhaps. But that was not clear from what you wrote. And you even let the examples where the inner portion of a circle were removed go, while the apparently do *not* satisfy what Cantor considers! > > But under *those* conditions there there are still uncountable sets that > > leave the plane connected and also sets that leave the plane disconnected. > > Of course, but Cantor did not recognize it. And as far as I know nobody > recognized it before I did. At least nobody mentioned it. Of course nobody mentions it, because it is highly unremarkable! Remove all irrational points and you still have a connected space. Remove in addition a line and you have a disconnected space. What is remarkable about that? What *is* remarkable is that whatever countable dense set you remove, the result is still connected. And the property of being dense is not even needed for both results. > Cantor tried > to apply his result to the continuity of the physical space: He closed > his paper by: "Daher liegt es nahe, den Versuch einer modifizierten, > f�r R�ume von der Beschaffenheit A g�ltigen Mechanik zu unternehmen, > um aus den Konsequenzen einer derartigen Untersuchung und aus ihrem > Vergleich mit Tatsachen m�glicherweise wirkliche St�tzpunkte f�r die > Hypothese der durchg�ngigen Stetigkeit des der Erfahrung > unterzulegenden Raumbegriffs zu gewinnen." He tried to explain the > matter by countably many particles, the ether by uncountably many > particles. Therefore the distinction between these both cardinals was > very important for him. There are many "scientist" that tried to prove something outside their field of expertise without real understanding. -- 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 27 Nov 2006 20:14 In article <1164637973.064418.67510(a)45g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > Nothing more than opinion. You think there are no such representations, > > so everything that states that there are such representations is wrong. > > I proved that there are no such representations. If they were (the > paths in the infinite tree, for instance), then we had a contradiction. You are, with your tree, wrong. In your tree (where the nodes represent bits), it can be shown that all nodes represent a number: the collection of bits found when going from the root to the current bit. So the nodes represent all rational numbers in [0,1) where the denominator is a power of two. You state that the paths represent numbers. Let us analyse that, and begin with the finite paths that terminate at some finite edge. When you consider such paths, you can assign numbers to each final edge: the same number as the node were it comes from (nodes represent bits, so when you come at an edge you can only add the bit of the last node you passed, you can not look in the future). So your paths (whether finite or infinite) all represent a number that is also represented by a node. -- 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 27 Nov 2006 20:25 In article <1164638600.073963.74180(a)h54g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > You do not require that one digit represents the number 1/3 in Cantor's > > > list. You do not require that any digit there represents an infinite > > > sequence. > > > > Of course not. > > > > > But you require that one digit (node) represents 1/3 in the tree? > > > But you require that one digit (node) in my tree represents an > > > infinite sequence? > > > > Of course. > > That is unfair. So be it. > > > There is no node in the tree which represents 0.000... and no node > > > which represents 0.1 But these real numbers are represented in the > > > tree by paths. > > > > I do not understand. You state that the bits are the nodes. > > Yes. And an infinte sequence of bits is the representation of a real > number. Yes. But as I did show in another article, when you assign bits to nodes, you can show that the nodes represent numbers, and also that when you want paths to represent numbers, each edge can be shown to represent the number the node represents where it comes from, and so that each path represents the edge where it terminates. So the paths represent no more numbers than the nodes do represent. In your infinite tree, all edges come fome a node at finite distance, and so all edges together only represent the numbers with a finite binary expansion. > > I have > > shown how you can extend that to the nodes representing numbers. > > I do no want not extend anything to nodes representing numbers. The > paths represent numbers. Look above. > > And > > I also have shown how the numbers represented by the nodes all have > > a finite sequence of bits. There is *no* node that represents 1/3. > > No. There is no digits in Cantor's list representing 1/3. Wrong. > > And when you switch to paths representing numbers you have something > > completely different. > > I did not switch to paths but defined from the beginning that the paths > represent the numbers. > > > Let's assign to a finite path the number of the > > node where the last edge terminates. > > There is no last edge, because the paths are infinite, like the decimal > representations used in Cantor's list. See above. > > Also in this case 1/3 is not in > > your paths. > > Then 1/3 is not in Cantor's list. It is. > > There are no numbers assigned to the non-terminating > > infinite paths. > > Then there are no numbers assigned to non-terminating decimal > representations. You are wrong. > > I think you wish that the edges represent the bits, > > not the nodes. > > No, I do not wish that. (I could do so, but it would again confuse you. > So we it as it is.) Again, see above. As there is no node at an infinite distance, there is also not an edge that comes from such a node, and so all edges represent a number with a finite binary expansion. -- 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 27 Nov 2006 20:32
In article <1164638966.304736.177910(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1164446537.991233.319900(a)j72g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > > I do not accept (3). It effectively states: > > > > forall{n in N}thereis{m in N} (index m is not in line n) -> > > > > thereis{m in N}forall{n in N} (index m is not in line n) > > > > > > For *finite, linearly ordered* sets, the quantifiers can be exchanged. > > > Every line is a finite, linearly ordered set. > > > > But the sets are not the lines. The set considered is N, which is > > linearly ordered but not finite. > > That N is the diagonal (and also the columns). Eh? N is the set of natural numbers. The quantifiers can be exchanged when N is finite. Not when something completely different is finite. > But I do not consider > them but the elements of *one* line (without fixing this line). The set > of elements of one line (of each line) are linearely ordered and > finitely many. Therefore I can state, for every line, that it is > finite. Yes, right. But that does *not* mean that you can exchange the quantifiers. The set considered in the quantifiers is the set of natural numbers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |