An uncountable countable set
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From: Virgil on 15 Sep 2006 19:43 In article <49edf$450aacfc$82a1e228$14539(a)news2.tudelft.nl>, Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> wrote: > Mike Kelly wrote: > > > Han de Bruijn wrote: > > > > Plagiarism? I don't get it. Who is plagiarising what? > > "Your" would-be arguments against mine are not really yours. They are > just a _plagiary_ of well-known "arguments" employed by the mainstream > mathematics community. > > Han de Bruijn What is common knowledge can be used by anyone without plagiarizing. Otherwise only its original author could use "2 + 2 = 4".
From: Tony Orlow on 15 Sep 2006 20:00 MoeBlee wrote: > Tony Orlow wrote: >> David R Tribble wrote: >>> Tony Orlow wrote: >>>>> Yes, I am including infinite values on the number line, since it's >>>>> "infinitely long". >>> David R Tribble wrote: >>>>> Yet another thing you have to define or prove. Where do these >>>>> "infinite values" appear on your real number line? >>> Tony Orlow wrote: >>>> Further from 0 than any finite number. >>> Then how are they "on" the same "line"? >>> >> By trichotomy. For all x ad y on the line, either x>y, x=y or x<y. >> That's what makes a line in concept. > > That's question begging. Indeed, trichotomy is necessary for an > ordering to be linear. But you have not proven the existence of such an > ordering with the values you claim to be in its field. Just saying that > your claim follows from trichotomy is just assuming what you are being > asked to show, viz. that there is such a linear ordering with the > values you claim to be in its field. But more basically, it doesn't > matter, since you have no axiomatization nor rules of inference upon > which to prove anything whatsoever in a mathematical system. > > MoeBlee > Well, MoeBlee, I have been trying to get that axiom together so that it properly ties together count with measure, and trichotomy is one of the axioms that defines the values along the line. If such a rule is declared for all pairs of members of a set, then that can be considered the definition of a linear set, be it what you would call a sequence, or a dense linear set like a real interval. I guess when I finally publish a set of axioms, then the subject will suddenly "matter". Funny how that works in transfinitology. My axiom set must be some sort of limit ordinal or something. ToeKnee
From: Tony Orlow on 15 Sep 2006 20:02 MoeBlee wrote: > Tony Orlow wrote: >> Unbounded but finite may >> be considered potentially, but not actually, infinite. > > That will be jiffy, once you give axioms and/or our definitions for > 'potentially infinite' and 'actual infinite'. Until then, it's pure > handwaving. > > MoeBlee > As I said, a potentially infinite set is unbounded, but will all element indices finite. It's "countable" in standard parlance. An actually infinite set includes elements with infinite element indices, like 1/3 has in the decimal reals. :) ToeKnee
From: Virgil on 15 Sep 2006 20:02 In article <450b06ce(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <45085f4c(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> David R Tribble wrote: > >>> Tony Orlow wrote: > >>>>> If you remove an element, the proper subset should ALWAYS be smaller by > >>>>> 1. That is the case for me. For a theory to claim a proper subset is the > >>>>> same "size" as the proper superset is an immediate deal-breaker for me. > >>> David R Tribble wrote: > >>>>> If by "different size" you mean that you cannot pair up all the > >>>>> elements from both sets, then you're going to have a difficult > >>>>> time proving that for any infinite set. (You have never show this, > >>>>> BTW.) > >>>>> > >>>>> If by "different size" you mean something other than some way of > >>>>> denumerating (counting) the elements of the set (e.g., by assigning > >>>>> them different natural indices), then you should use a different term, > >>>>> because it's confusing. Oh, and you have to prove that it works > >>>>> (you have never shown this, either). > >>>>> > >>>>> Start with a simple proof: Take the set of naturals, N = {0,1,2,3,...} > >>>>> and remove one element to get set S = {1,2,3,...}. Now show that > >>>>> the "T-size" of N is exactly one less than the T-size of S. In other > >>>>> words, find a way to show that every counting of S versus every > >>>>> counting of N always leaves one element of N (0) left over. > >>> Tony Orlow wrote: > >>>> Use IFR. > >>> A.k.a. a bijection. You see where this is going? > >>> > >> Yes, bijection with measure. > >> > >>>> N maps to S using f(n)=n+1. The inverse of that function is > >>>> g(x)=x-1. > >>> Which proves that every n in N has an x in S. Where is that > >>> leftover element that was removed from N? If N has more > >>> elements than S, shouldn't N have a member that can't be > >>> mapped to any member of S? > >>> > >> One can map all sorts of sets. N contains every element of S - map those > >> first, to themselves. Now you have one left over. > >> > >>>> So, over the range of 0 to N, |S|=|N|-1. > >>> Funny how you don't define what |X| is. You're using standard > >>> symbols but obviously with a different meaning, since "|X|" means > >>> "cardinality of X" when X is a set. Your IFR bijection proves that > >>> |S| = |N|. > >>> > >> |X| means size of, like the absolute value of a real. > >> > >>>> Map N to the Evens E using f(n)=2n. The inverse function is > >>>> g(x)=x/2, so over the range of N, > >>>> the evens have |N|/2 elements. Isn't that intuitively satisfying? And > >>>> gee, it works for finite sets accurately too!! > >>> Same thing as above, it proves that every n has an x. Where are > >>> the leftover unmapped elements of N that make S a smaller proper > >>> subset of N? > >>> > >> All the odds. N contains all elements of E, plus the elements of O. IFR > >> works to the level of accuracy of detecting a change of 0ne element out > >> of an uncountable number. > > > > But in standard mathematics, it is trivial that any ordering of sets in > > which every proper subset of a set is 'smaller' than its superset, is no > > more that a partial ordering, and can never be a total ordering, so that > > there must be sets which cannot be compared. > > > > So that TO's requirements are self-contradictory. > > You dingbat. I never said that the subset relation was the only method > of ordering by any means. But anything that includes it must fail.
From: Virgil on 15 Sep 2006 20:08
In article <450b0826(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <45098084(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Dik T. Winter wrote: > >>> In article <4506d1ae(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > >>> writes: > >>> > David R Tribble wrote: > >>> ... > >>> > > Start with a simple proof: Take the set of naturals, N = > >>> > > {0,1,2,3,...} > >>> > > and remove one element to get set S = {1,2,3,...}. Now show that > >>> > > the "T-size" of N is exactly one less than the T-size of S. In > >>> > > other > >>> > > words, find a way to show that every counting of S versus every > >>> > > counting of N always leaves one element of N (0) left over. > >>> > > >>> > Use IFR. N maps to S using f(n)=n+1. The inverse of that function is > >>> > g(x)=x-1. So, over the range of 0 to N, |S|=|N|-1. Map N to the Evens > >>> > E > >>> > using f(n)=2n. The inverse function is g(x)=x/2, so over the range of > >>> > N, > >>> > the evens have |N|/2 elements. Isn't that intuitively satisfying? And > >>> > gee, it works for finite sets accurately too!! > >>> > >>> How many elements has the set of primes? > >> There is no well-known function that maps n to the nth prime, so IFR > >> doesn't apply. Do you have an inverse function that specifies the nth > >> prime for all neN? Didn't think so. > > > > Then there are sets whose size TO cannot measure, which makes his > > "measure" less useful that cardinality, at least cardinality in ZFC and > > NBG. > > Haha!! Good one. "There are the same number of primes as there are > naturals, a proper superset." Good answer. It's almost as bad as, "There > are an infinite number of rationals between any two naturals, but there > are the same number of each." What a "theory"! It's mathematical > creationism. It could hardly be worse than TO's anti-mathematical cretinism. |