Cantor Confusion
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From: mueckenh on 20 Dec 2006 07:31 Dik T. Winter schrieb: > In article <1166545226.008570.102300(a)n67g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > ... > > > > So for finte trees there are more edges than paths but for infinite > > > > trees there are more paths than edges? > > > > > > Indeed. > > > > LOL. > > > > > > So for finte trees there are more edges than paths but for infinite > > > > trees there are more paths than edges? > > > > > > Right. It is taking conclusions about the infinite from the finite cases > > > when you think it is otherwise. > > > > Perhaps we cannot determine the limit of the series 1 + 1/2 + 1/4 + ... > > unless we admit some sophisticated calculation of limits. But without > > any such sophistication we can be sure the sum of arbitrarily many > > terms of that series will be less than 28.50. And you assert that it is > > greater than 100! That is what you call mathematics. But it is simply a > > wrong application of human brain. > > The inequality holds for all finite paths. Why it also should hold for > infinite paths escapes me. Because we can prove that 2 - 1/2^n is always, i.e., for all finite natural n, less than 100. And because even infinity does not supply other than finite natural numbers. Regards, WM
From: William Hughes on 20 Dec 2006 07:32 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > > > > > Element n+1 can be shown to exist in L_D (which is obviously a line > > > > > > > > > containing n+1). > > > > > > > > > > > > > > > > No. L_D is bounded. The largest element of L_D is n. > > > > > > > > L_D does not contain n+1. > > > > > > > > > > > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > > > > > contained in the diagonal. If your L_D does not contain them, then you > > > > > > > have the wrong L_D. > > > > > > > > > > > > Assume that there exists an L_D which contains all the numbers > > > > > > contained in the diagonal. L_D is bounded, > > > > > > > > > > If an unbounded diagonal exists, then obviously an unbounded line must > > > > > exist. > > > > > The conclusion is false, so the antecedent cannot be true. > > > > > > > > The diagonal is the potentially infinite set of natural numbers. > > > > This exists and is unbounded. > > > > > > Yes, but not in the way you understand "to exist". Potentially infinite > > > means: always finite. The diagonal is always finiter though not > > > bounded > > > > The only property of the diagonal that is used is that > > it is unbounded (i.e. given a finite set of elements of the diagonal, > > it is possible to add one element). > > That property is correct. > > > The existence of all the elements > > of the diagonal is neither claimed nor used. The "infiniteness" > > of the diagonal is neither claimed nor used. > > Fine. > > > > The contradiction (a line must be both bounded and unbounded) > > No line must be unbounded in case of a potentially infinite (i.e. > always finite) diagonal. > > If you add one element to the diagonal then all its elements are in > another line than before, but they are in a finite line. "another line than before" Therefore what this proves is A: For every natural number n, there exists a line L(n) such that n is in L > Therefor all > finite elements of the diagonal are in one single line. > This is B: There exists a single line L_D, such that for every natural number n, n is in L_D. B does not follow from A, and there is a simple proof that if the diagonal is unbounded, then B is false. - William Hughes
From: Albrecht on 20 Dec 2006 07:39 Franziska Neugebauer schrieb: > Albrecht wrote: > > > William Hughes schrieb: > >> Albrecht wrote: > >> > William Hughes schrieb: > >> > > mueckenh(a)rz.fh-augsburg.de wrote: > >> > > > Franziska Neugebauer schrieb: > >> > > > > >> > > > > >> > > > All elements that can be shown to exist in the > >> > > > > >> > > > diagonal can be > >> > > > > >> > > > shown to exist in one single line. [(P1)] > >> > > > > > >> > > > > This proposition P1 has _not_ yet been proved (shown). > >> > > > > >> > > > This proposition is the definition of the diagonal. > >> > > > > >> > > > > > You misinterpret L_D. L_D is that line which contains all > >> > > > > > numbers contained in the diagonal. > >> > > > > > >> > > > > The Diagonal is unbounded thus any _assumed_ L_D is not > >> > > > > bounded, too. > >> > > > > >> > > > Correct is: > >> > > > If the diagonal is unbounded then any _assumed_ L_D is not > >> > > > bounded, too. > >> > > > >> > > L_D is a line and is therefore bounded. > >> > > > >> > > Contradiction. > >> > > Hence, the _assumed_ L_D does not exist. > >> > > > >> > > - William Hughes > - > >> > > William Hughes > >> > > >> > It's always the same game you play: you compare numbers and sets. > >> > Numbers are always complete, sets are complete only if they are > >> > finite. Now you claim that the set of the natural numbers contain > >> > all natural numbers and hence it is complete > >> > >> No, the claim is that it is potentially infinite (given > >> any finite set of natrual numbers one can allways find one more). > > > > That's okay. You can say: given *any* set X of natural numbers, one > ^^^^^^^^^ > *any* _finite_ set Since there are no other sets, we don't need this constriction. Sets are entities like numbers. They don't change there quantity. There is no potenitially infinite number and no potentially infinite set. There are only potentially infinite processes like sequences and so on. > > > can allways find at least one natural number which isn't element of > > the set X. > > The set N is a set of natural numbers, but there is no natural number > wich is not element of N. Hence your claim is wrong. There is not set N. Since if there is a set with infinite many objects (numbers) there must be an element-index "infinity" and there must be a number "infinity". Both doesn't exist since this objects are contradictory to the definition of natural numbers. Hence there is no set containing infinitely many natural numbers, hence ther is no set N. > > >> No assumption of completeness is made or used. > >> (here "complete" is a property of a set or a potentially infinite > >> set. We say X is complete, if assuming X exists means that > >> we can assume that all elements of X exist). > > > > There are no potentially infinite sets. > > You may have a potentially infinite sequence of sets. > > You contradict yourself: Sequences "are" sets so it is valid that from > the existence of potentially infinite _sequences_ the existence of > potentially infinite _sets_ follows. You mix up something. Infinite sequences aren't sets since there are no sets with infinitely many elements since that would need an infinite number. A set may be defined by sequences if the sequence is finite. Otherwise there is no set. > > > You again mix up states (sets) and processes (algorithm) > > There are no "processes" at all. Essential aspects of our world sre describable in terms of dualism like "sein" und "werden". Would you exclude this elementary aspects of our world out of math? Why shouldn't there be "states" and "processes" in math as in the world. > > >> > but in the same time |N isn't complete > >> > since |N is bijectable to the diagonal and the diagonal is never > >> > complete since there is no complete line (number) which covers the > >> > diagonal. > >> > >> No, the claim is that there is no line that covers the diagonal. > >> > >> - William Hughes > > > > > > We know, there are infinitely many natural numbers. > > Do "we"? I hope so. You don't? > > > We know also that there is no number amongst this numbers with the > > form ....11111. > > This is not a number but at most a representation. > > > Why? > > Because every n e N has a representation of the form x...x and not of > the form ...x. > > > It's easy to say: Since there is no number of the form ...11111 there > > is no quantity which elements has the quantity ...111111. > > You don't need metaphysical notions like "quantity" to simply state the > fact that the "conversion" of any n e N into its representation never > yields ...111. This is a purely formal issue. > > > Since there are only quantities with numbers and since numbers, in > > there easiest form, are also quantities (like 1, 11, 111, 1111, ...) > > [...] > > Circular reasoning may indicate ignorance. > > F. N. > -- > xyz Circulate reasoning may discover circulate reasons. There are only infinite sets if there are infinite numbers. Since there are no infinite numbers there is no infinite set. How come out? Best regards Albrecht S. Storz
From: William Hughes on 20 Dec 2006 07:46 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > You have correctly show that: > > > > A: For every natural number n, there is a line L(n), such > > that n is contained in L(n). > > > > You need to prove > > > > B: There is a single line, L_D, such that every natural > > number n is contained in L_D. > > > > A and B are not the same statement. You have not given > > a proof of B. (B follows from (A + linear ordering) iff there > > is a last line) > > Please give a counter example for a finite number n. No such counterexample exists. The claim is that despite the fact that given any n we can find a single line, L(n), that works for this n, there is no single line, L_D, that will work for all n. > Remember, even if there is not a last line, every n eps N is finite. Which means that for any n we can find a single line L(n) that works for this n. It does not mean that that we can find a single line L_D, that works for all n. > Therefore your claim is void, unless you can find a linear set of > finite elements which forbids exchange of quantifiers. Consider the (potentially infinite) set of natural numbers. This is linearly ordered. Every element is finite. For any natural number n, there exists a natural number m(n), such that m>n However it is not true that: There exists a natural number m, such that for any natural number n, m>n. - William Hughes
From: William Hughes on 20 Dec 2006 07:59
Albrecht wrote: > William Hughes schrieb: > > > Albrecht wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > Franziska Neugebauer schrieb: > > > > > > > > > > > > > > > > >> > > > All elements that can be shown to exist in the diagonal can be > > > > > > >> > > > shown to exist in one single line. [(P1)] > > > > > > > > > > > > This proposition P1 has _not_ yet been proved (shown). > > > > > > > > > > This proposition is the definition of the diagonal. > > > > > > > > > > > > You misinterpret L_D. L_D is that line which contains all numbers > > > > > > > contained in the diagonal. > > > > > > > > > > > > The Diagonal is unbounded thus any _assumed_ L_D is not bounded, too. > > > > > > > > > > Correct is: > > > > > If the diagonal is unbounded then any _assumed_ L_D is not bounded, > > > > > too. > > > > > > > > > > > > L_D is a line and is therefore bounded. > > > > > > > > Contradiction. > > > > Hence, the _assumed_ L_D does not exist. > > > > > > > > - William Hughes > > > > > > > > - > > > > William Hughes > > > > > > It's always the same game you play: you compare numbers and sets. > > > Numbers are always complete, sets are complete only if they are finite. > > > Now you claim that the set of the natural numbers contain all natural > > > numbers and hence it is complete > > > > No, the claim is that it is potentially infinite (given > > any finite set of natrual numbers one can allways find one more). > > That's okay. You can say: given *any* set X of natural numbers, one can > allways find at least one natural number which isn't element of the set > X. > > > > No assumption of completeness is made or used. > > (here "complete" is a property of a set or a potentially infinite > > set. We say X is complete, if assuming X exists means that > > we can assume that all elements of X exist). > > There are no potentially infinite sets. You just stated "given *any* set X of natural numbers, one can allways find at least one natural number which isn't element of the set X." Thus, by definition, the natural numbers form a potentially infinite set. > You may have a potentially > infinite sequence of sets. If the term set has quasi-religious significance to you, substitute "potentially infinite sequence", for "potentially infinite set". >You again mix up states (sets) and processes > (algorithm) > > > > > > > > but in the same time |N isn't complete > > > since |N is bijectable to the diagonal and the diagonal is never > > > complete since there is no complete line (number) which covers the > > > diagonal. > > > > No, the claim is that there is no line that covers the diagonal. > > > > - William Hughes > > > We know, there are infinitely many natural numbers. We know also that > there is no number amongst this numbers with the form ....11111. Why? > It's easy to say: Since there is no number of the form ...11111 there > is no quantity which elements has the quantity ...111111. Since there > are only quantities with numbers and since numbers, in there easiest > form, are also quantities (like 1, 11, 111, 1111, ...) there is no > infinite number ...11111 with infinite many "1" since there is no > quantity with the number ...111111 to number the quantity of the "1" in > an infinite number. And so on. > The question is: "Is there a line that covers the diagonal" The question is not "Is there a number with the form ....11111?" Please address the question. - William Hughes |