Cantor Confusion
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From: Virgil on 30 Oct 2006 17:13 In article <1162219026.697699.10850(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > It is unbounded but always finite. Unbounded implies not finite, as finite implies bounded.
From: Virgil on 30 Oct 2006 17:21 In article <1162219429.164644.314310(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > 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"? > > > > > > > Not by simply reversing the quantifier. > > > > The reason for believing that there are only a finite > > number of integers is that there are only a finite number > > of bits in the lifetime of the universe, and therefore only > > a finite number of different integers will be defined. > > Correct. > > > > But note, to define an integer, the definition must be > > communicated. We have only a finite number of bits > > to use for this communication (this includes comunicating > > the details of any compression scheme). So there are > > only a limited number of integers that it is possible to define. > > Among these is the largest integer that it is possible to > > define. > > Sorry, I can't see that conclusion. You can always define a new and > larger base., a new and more efficient way to use the bits. Unless this each time erases from existence some previously defined number(s), you are admitting that the number of numbers it is possible in infinite time to define is infinite. You cannot have it both ways.
From: Virgil on 30 Oct 2006 17:27 In article <1162219714.772929.189460(a)h48g2000cwc.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > > 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. > > > > I have in mind a list of infinitely many character strings with the nth > > string having n characters. As all I have to do to "have" it is to > > imagine it, I'm done. > > As long as your strings have a finite number of characters, you have a > finite number of strings. That may be how things work in in WM's world, but not in mine. Since I start, in ZF or equivalent, with at least one actually infinite set, |N. I can get as many actually infinite other sets as I wish. If WM claims that his world does not work that way, he is only making equally unprovable assumptions himself. But WM does not have the power to determine reality, only the power to assume that he knows it.
From: Virgil on 30 Oct 2006 17:30 In article <1162219925.966481.179840(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > Correct. And those positions have finite indexes too. > > > Every set of natural numbers has a superset of natural numbers which is > > > finite. Every! > > > > Not so. > > > > The set of even natural numbers has no finite superset. > > > > The set of prime natural numbers has no finite superset. > > Sorry, I meant "segment". > > > > > > > > > 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. > > > > So what? > > This shows that both are finite or both are not. So what? The set of even naturals and the set of odd naturals are both finite or both infinite, too. But that does not mean that either is finite.
From: Virgil on 30 Oct 2006 17:32
In article <1162220492.974057.66580(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > What kind of well ordering of the reals do you claim to exist? > > > > A well ordering in which each non-empty subset has a smallest member. > > That is the definition. Exactly so! > > > > > Defined? > > > Catalogue? > > > List? > > > Else? (Please specify) > > > > A well ordering in which each non-empty subset has a smallest member. > > > > That's what well orderings are all about. > > I know. But the definition does not guarantee existence. In particular > it does not say how the order in R differs from the order in N or Q. > > > > If you are asking for a rule for determining which objects come before > > others, you should know that no such explicit rule is possible, but that > > does not make their existence imposible. > > What kind of existence do you have in mind? In a system that has assumed the axiom of choice, the kind that that axiom insists on. These things do not exist > unless they exist in some mind. But you say that they do not exist in > any mind. So what *is* existence in this case? > > Regards, WM |