Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: William Hughes on 29 Oct 2006 15:33 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > > There is no computable list > > > > of computable numbers. > > > > > > Why do you insist on this obvious fact? It does not support your > > > position. There is a list of all finite constructions or definitions > > > (encoded by numbers, Gödel). This is the definition of countablity. > > > The diagonal number of this list shows the listed numbers are > > > uncountable. Contradiction. > > > > > > There is indeed a list of all possible finite constructions or > > definitions. Call this list A. However, this list must contain things > > that look > > like definitions but are not because the method given to produce > > a number does not halt. We cannot get a diagonal number from A, > > because some of the members of A do not give numbers. > > So there is no contradiction. > > A contains all finite words (construction fromulas, theorems). The > diagonal (cannot be longer than the lines and hence) is a finite word > too. That is enough to obtain a contradiction. A also contains a number of things that looka like contruction formulas but are not (because they don't halt). Thus there is no diagonal. Thus there is no contradiction. You need a diagaonal before you can get a contradiction, therefore you need set B. > > > > So let's take list B. We get list B from A by taking only the true > > definitions > > from A (that is we throw away anything that will does not halt). > > So now we have the list B, so by definition B is countable. > > > Now we can get a diagonal number, d. This number cannot be a member > > of B. > > > > But note. d cannot have a finite definition. A finite definition of d > > would > > include not only the diagonal procedure (finite description) but also > > the list B (infinite description) or a finite description of a way to > > get the list B from the list A (i.e. a solution to the halting > > problem). > > > > So d is not a member of B. If we do not take > > a contrucivist viewpoint this is not a contradiction, there are lots > > of things that are not a member of B. > > > > But let's say we take a constructivist viewpoint. d is not computable, > > so d does not exist. (Nothing that is not a member of B > > exists.) We still do not have a contradiction because > > if we take a constructivist viewpoint the list B does not exist so we > > cannot use it to form the diagonal d. > > Funny, but not sufficient to confuse a clear thinking mind > sufficiently. > No matter how funny you find it, it is also true. B is not computable, so if you claim that only computable objects exists, then B does not exist. Whether or not you claim that B exists, there is still no contradiction. -William Hughes. > Regards, WM
From: mueckenh on 29 Oct 2006 16:10 Lester Zick schrieb: > Now without debating the issue of mathematical definition in general I > will say that if mathematikers insist on definitions drawn exclusively > in particular terms then WM may have some basis for his conclusions in > that the infinity of mathematical expressions cannot be accommodated > in a physically limited universe. (Of course I don't agree that the > universe is physically limited but that's another issue.) Please drop in: http://www.fh-augsburg.de/~mueckenh/MR.mht Among others you will find links to interesting papers by Edward Nelson, Princeton, and by David Isles: 2What evidence is there that 2^65536 is a natural number?" Further by L.M. Kraus and G.D. Starkman on Universal limits on computation > In other > words if mathematics insists its definitions can only be valid if cast > in particular terms "Don't you see that the whole aim of Newspeak is to narrow the range of thought? In the end we shall make 'contradiction' literally impossible, because there will be no words in which to express it." (after George Orwell in "1984") This and more quotes can be found on http://www.fh-augsburg.de/~mueckenh/MR/Zitate.mht Regards, WM
From: mueckenh on 29 Oct 2006 16:18 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > And once again WM deliberately confuses, "all positions are > > > finite", with "there are a finite number of positions". > > > > There is no confusing! Every finite position belongs to a finite > > segment of positions (indexes). > > Correct. > > But > every position belongs fo a finite set of positions, > does not mean > there are only a finite set of positions. > > The first means: > > for every position N, there exists a finite set > of positions A(N) such that N belongs to A(N) > > The second means: > > there exists a finite set of positions B, such that > for every position N, N belongs to B No it means only potential infinity: Every set B of positions is finite, although there is no largest set B but every set B has a larger superset B'. Actual infinity would yield: There is a completed, finished infinite set. That is nonsense, whther you call it finite set B or infinite set |N. Finished infinity is a contradiction in itself, even worse than quantifier changing. > You cannot simply exchange the quantifiers. It has nothing to do with that. > Why. Why does "there are an infinite number of positions" imply > that "at least one position is infinite"? Try to find an example for the opposite assertion: Try to construct an actually infinite set with only finite numbers (which differ by a constant value at least). Then you will know it. Regards, WM
From: mueckenh on 29 Oct 2006 16:19 William Hughes schrieb: > If there is a limit to how many numbers we can define there > is also a limit to how large a number we can define. Why. Why does "there is a limited number of numbers" imply that "every number is limited"? Regards, WM
From: mueckenh on 29 Oct 2006 16:22
Virgil schrieb: > In article <1162068521.341383.99300(a)k70g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > And once again WM deliberately confuses, "all positions are > > > finite", with "there are a finite number of positions". > > > > There is no confusing! Every finite position belongs to a finite > > segment of positions (indexes). If you don't believe that and assert > > the contrary, then try to find a finite position which dos not belong > > to a finite segment (of indexes). > > It is equally true that for every finite segment of indexes there are > positions beyond that segment. Correct. And those positions have finite indexes too. Every set of natural numbers has a superset of natural numbers which is finite. Every! > > > > It is simply purest nonsense, to believe that "all positions are > > finite" if "there are an infinite number of positions". > > There are certainly more than any finite number of positions. > And it is even purer nonsense to believe that in such a situation there > are only finitely many of them. Try to draw the graph of the function f(n) = n. The points lay on the diagonal of the first quadrant. As long as n is finite, f(n) is finite too and vice versa. It would be nonsense to claim that f(n) gets infinite while n remains finite. Regards, WM |