Cantor Confusion
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From: Han de Bruijn on 18 Dec 2006 10:42 William Hughes wrote: > Han de Bruijn wrote: > >>William Hughes wrote: >> >>>Han.deBruijn(a)DTO.TUDelft.NL wrote: >>> >>>>William Hughes schreef: >>>> >>>>>Does the "potentailly infinite set" of natural numbers >>>>>exist? A potentially infinite set is a function on sets >>>>>that takes on the values true and false. The >>>>>potentially infinite set of natural numbers >>>>>takes on the value true for a set containing only >>>>>natural numbers, and false for any other set. >>>>>(i.e. is there a way of recognizing a set of >>>>>natural numbers?) >>>> >>>>I don't understand this question. >>> >>>Stop me at the first line you don't understand. >>> >>>[Note if you do not like the term "potentially infinite >>>set" substitue "recognizer".] >>> >>>A "potentially infinite set" is a function on sets. >> >>It's already here: "function" and "set". > > You answered the question > > "Is there an unbounded set of natural numbers?" > > No. > > Now you claim you don't know what I mean > by "set". . > > Clearly you cannot answer simple questions. Whatever. It's the combination that confuses me: what are the domain and the range of your "potential infinite set" function? Han de Bruijn
From: Albrecht on 18 Dec 2006 10:48 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 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. Why should there be infinite sets but no infinite numbers? Easy: sets are infinite by declaration, numbers can't be infinite since they count and you could not count to infinity. But how should there be sets which are infinite without doing infinite things in? You do it in at once? Ah, real magic. :-) You mix up states (numbers) and processes (complete induction). It's an error of modern math to handle these two things as the same (in the form of sets). Best regards Albrecht S. Storz ***By the way: Complete is that what contains all what should be in.***
From: William Hughes on 18 Dec 2006 10:59 mueckenh(a)rz.fh-augsburg.de 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. > > > Assumed is the diagonal. The diagonal is the potentially infinite set of all natural numbers. The diagonal is not bounded. L_D is a line as therefore bounded. Conclusion. L_D does not contain every element that can be shown to be in the diagonal. - William Hughes
From: William Hughes on 18 Dec 2006 11:38 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > William Hughes schrieb: > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > William Hughes schrieb: > > > > > > > > > > > > > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > > Virgil schrieb: > > > > > > > > > > > > > > > > > > > > > > > > > (It is contained in the union of all lines, but the > > > > > > > > > > > > > > union of all lines is not a line) > > > > > > > > > > > > > > > > > > > > > > > > > > That is a void assertion unless you can prove it by showing that > > > > > > > > > > > > > element by which the union differes from all the lines. > > > > > > > > > > > > > > > > > > > > > > > > Not quite. In order to achieve that the diagoal is not in any linem all > > > > > > > > > > > > that is required is: > > > > > > > > > > > > Given any line there is an element of the diagonal not in THAT line. > > > > > > > > > > > > It is not requires that: > > > > > > > > > > > > There is an element of the diagonal that is not in any line. > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > For linear sets you cannot help yourself by stating that the diagonal > > > > > > > > > > > differs form line A by element b and from line B by element a, but a is > > > > > > > > > > > in A and b is in B. This outcome is wrong. > > > > > > > > > > > > > > > > > > > > > > Therefore your reasoning "there is an element of the diagonal not in > > > > > > > > > > > THAT line. It is not required that: There is an element of the diagonal > > > > > > > > > > > that is not in any line." is inapplicable for linear sets. You see it > > > > > > > > > > > best if you try to give an example using a finite element a or b. > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > In every finite example the line that contains > > > > > > > > > > the diagonal is the last line. > > > > > > > > > > > > > > > > > > Every example with natural numbers (finite lines) is a finite example. > > > > > > > > > > > > > > > > > > > Your claim is that there is a line which contains the diagonal. > > > > > > > > > > > > > > > > > > Because a diagonal longer than any line is not a diagonal. > > > > > > > > > > > > > > > > > > > Call it L_D. Question: "Is L_D the last line?" > > > > > > > > > > > > > > > > > > There is no last line > > > > > > > > > > > > > > > > Then, there is a line that comes after L_D. > > > > > > > > > > > > > > > > Therefore :L_D does not contain every element > > > > > > > > that can be shown to exist in the diagonal. > > > > > > > > > > > > > > All elements that can be shown to exist in the diagonal can be shown to > > > > > > > exist in one single line. > > > > > > > > > > > > > > > > > > > Call it L_D > > > > > > > > > > > > L_D contains a largest element. n. > > > > > > > > > > > > L_D is not the last line, so there is > > > > > > a line with element n+1, > > > > > > > > > > > > Element n+1 can be shown to exist in the diagonal. > > > > > > > > > > > > > > > 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. The antecedent is true therefore the conclusion is true. -William Hughes
From: William Hughes on 18 Dec 2006 11:55
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). 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). > 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 |