An uncountable countable set
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From: Mike Kelly on 12 Sep 2006 14:56 Tony Orlow wrote: > David R Tribble wrote: > > Mike Kelly wrote: > >>> Better in what sense? What mathematics are you hoping to be able to do > >>> with your new foundation that cannot be done with ZFC? > > > > Tony Orlow wrote: > >> It accounts for changes of a single element between infinite sets and > >> therefore provides for a full spectrum of ordered infinite sets, thus > >> making the Continuum Hypothesis null and void. It relates the continuum > >> to the hypernaturals formulaically, and provides for an integration of > >> sets and measure in the infinite case. It rids us of anomalies like > >> omega-1=omega. > >> > >> 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. > > > > 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. > > > > 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? No. Now what? -- mike.
From: Virgil on 12 Sep 2006 14:57 In article <4506d4eb(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > Tony Orlow wrote: > >> Yes, I am including infinite values on the number line, since it's > >> "infinitely long". > > > > Yet another thing you have to define or prove. Where do these > > "infinite values" appear on your real number line? > > Further from 0 than any finite number. But as the finite numbers have no end, that means that TO's suppositional numbers must be beyond that non-existent end, which puts them beyond existence. > > > > > Are these infinite values connected (in the mathematical sense) to > > the finite values on the line? If not, how is it a "line"? > > > > > > > Tony Orlow wrote: > >>> You are mistakenly applying > >>> the standard cardinalistic fact that the number or reals in [0,1] is the > >>> same as the number of all reals. That is false in my system. What "system" is that? We have seen no system, only a miscegenated coglomeration of impossibilities. > > > > Unless you have a definition or axiom about denseness that > > is consistent with your theorem above? (The standard accepted > > definition of "dense" is not, of course.) > > > > I don't understand why you assume Big'un is finite, when it's an > infinite unit. I don't understand why TO assumes that his imaginings can form a consistent system when there is so much evidence to the contrary. >
From: Mike Kelly on 12 Sep 2006 15:05 Tony Orlow wrote: > David R Tribble wrote: > > Tony Orlow wrote: > >>> ....11111 binary (all bit positions finite) > > > > Mike Kelly wrote: > >>> That isn't a natural number, Tony. > > > > Tony Orlow wrote: > >> Are you sure? Pay close attention. > >> > >> For any finite bit position n, it and all predecessors can only sum to a > >> finite bit string value of 2^(n+1)-1. Since there are only finite bit > >> positions in the string, it can never achieve any infinite value at any > >> position in the unending string of bits. Therefore the value must be finite. > > > > Breathtaking. You have the annoying habit of ignoring the > > completely obvious when it suits you. If ...111 is a binary number > > with an infinite number of digits, why is it a finite value? > > Shouldn't an infinite value have an infinite number of digits? > > > > Yes, it should, and Wolfgang's and my position is that N is unbounded > but finite, given that it only contains finite values, and only one per > unit of value range. Since the value range is finite, given that no two > naturals are infinitely different, and given that only a finite number > of elements can be fit sparsely in a finite range, the set is finite, > even though it's unbounded and infinite by the Dedekind definition. > > At which bit position can the string achieve an infinite value? None > that exists in that string. Two statements : 1) A finite string of 1s represents a (finite) natural number. 2) An infinite string of 1s represents a (finite) natural number. 1) doesn't imply 2). -- mike.
From: Virgil on 12 Sep 2006 15:06 In article <4506d58e(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > You are obviously missing something. > > > > What I left out was that the power set includes countably infinite > subsets of N, and such countably infinite bit strings are not considered > standard naturals. However, they represent finite whole values For any standard meaning of finite, TO's claim is FALSE! In order for a binary string (function from N to {0,1}) to have a standard correspondence with any finite natural number value it must have only finitely many non-zero bits, and thus be equivalent to a finite string with leading bit 1 or with only one bit. Any string with infinitely many non-zero bits cannot correspond to any finite numerical value.
From: Virgil on 12 Sep 2006 15:14
In article <4506d7f6(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > Tony Orlow wrote: > >>> ....11111 binary (all bit positions finite) > > > > Mike Kelly wrote: > >>> Unless that string has only finitely many bit positions as well as only > >>> finite bit positions, it is not a natural at all, as it is then neither > >>> the first natural nor the successor of any natural, and every natural > >>> has to be one or the other. > > > > Tony Orlow wrote: > >>> It is the successor to ....11110. Duh. I've already proven that this is > >>> a finite value, given that all bit positions are finite, and that > >>> therefore no place in that string can achieve an infinite value, and > >>> that any such number has predecessor and successor. The cute thing is > >>> that the successor to ...1111 is 0, and that ...1111 is essentially -1. :) > > > > Mike Kelly wrote: > >>> Does it not bother you that nobody else agrees with, or even > >>> understands, your proof? > > > > Tony Orlow wrote: > >> I find it disappointing, but not surprising, that you don't understand > >> such a simple proof, since it's contradictory to your education. I do > >> find it annoying that you feel the right to disagree with it without > >> understanding it. If you feel there is a problem with the proof, please > >> state the logical error I made. > > > > As I posted earlier, it leads to the conclusion that 0 > 0. > > > > True, considering the number circle breaks the transitive nature of >, > except that it's only one way of looking at the number line. In computer > science, Computer science is not mathematics, and those whose minds are too feeble to tell the difference should restrict their postings to computer newsgroups. > > From the dense set of reals in the unit interval. That's my unit > infinity. Once we allow that value to exist AS the "number" of real numbers in the unit interval is precisely the same as in any other non-degenerate real interval, and all such intervals are order isomorphic to each other, TO's "unit" length is simultaneoulsy equal to every positive real number. |