An uncountable countable set
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From: Virgil on 11 Nov 2006 14:40 In article <4556122c(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > I've given a simple example many times. Starting with 0, your > > rules will never generate 3, 1/3, or any integer multiple or > > power of 3. You still have not responded to that little flaw, > > which would go a long way towards convincing others that > > you might be on to something. > > > > Why is that a flaw in the H-riffics, and not in the digital reals? You > cannot express 1/3 as a decimal fraction either. Not as a terminating decimal fraction, but every rational can be represented by either a terminating or an eventually repeating decimal fraction. > Do the decimal > fractions constitute a countable set? That's counter to Cantor's > Diagonal Argument for the uncountability of the reals. TO, as usual, puts his mind into neutral and his typing into high gear before posting. The terminating decimals, and even the terminating together with repeating decimals (which is no more than the rationals), form countable sets. If TO want a set of decimals which is uncountable, he must include non-terminating and nonrepeating ones. And the set of all such represents the reals but no countable subset of them does. The set of all > digital fractions is uncountable. The set of all H-riffics is > uncountable. Every digital fraction has successor and predecessor as > well, when the bits are mirrored to the other side of the digital point, > and ordered as naturals. Ah, you say, but 0.333... does not have a > natural mirror, since ...333 is not a natural. And yet, I say to you, > that is every digit in that string is in a finite position with respect > to the digital point, then there is no point in that string where it > achieves an infinite value, and so it represents some sort of finite but > unbounded quantity. Since any such unending strings also has a successor > and a predecessor (except perhaps for ...000 and ...999), it would seem > that in two significant ways such numbers are very like the terminated > finite strings you call naturals. You might be reassured by keeping in > mind that the set of such finite string representations of infinite > strings is still countable, since they depend on a repetition of a > finite string of digits. > > > > >> David - If every element in the reals has at least one successor, what > >> does that say to you? > > > > Not much, apart from the fact that some countable subset of the reals > > is covered by your generating rules. > > > > Your rules produce a countable set by their very nature. For any > > H-riffic H_n, I can produce H_2n and H_2n+1 from it. Which means > > that I can map every natural k to some H-riffic, and vice versa. > > Hence the H-riffics are a countable set. > > > > But there is no r in R for which you can say there is no successor. How > do you partition R into countable subsets? Which r in R does not have > two successors?
From: Virgil on 11 Nov 2006 15:38 In article <4556151c(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > Your BigUn notational is meaningless. I'm assuming that you're > > building on the existing positional digit notation we are all familiar > > with, which is: > > for all x in R, > > x = (sum{i=0 to oo} d_i x B^i) x B^p, > > for some base B and some integer p, and 0 <= d_i < B. > > First of all, you're summing digits only to the right of the digital > point, and why are you multiplying them by some power of the base after > that? I don't see any fractional component. 6/37 = 0.162162162... repeating 162 forever. But this only works for fractions less than 1 in absolute value. For improper fractions like 6000/37 it must be modified to allow representation of the integer part. 6000/37 = (162.1621621622...) = (sum{i=0 to oo} d_i / 10^i) x 10^3 where d_i = 1 for i ==1 mod 3 and d_i = 6 for i ==2 mod 3 and d_i = 2 for i == 0 mod 3 > > I'd say: > for all x in R > x = (sum{i=-oo to oo} d_i x B^i) > for some natural base B and some natural d_i such that 0<=d_i<B. > > That covers all digits, left and right. Except that it does not allow arbitrary arithmetic. Some numbers cannot be added, some cannot be multiplied, some cannot be subtracted and some cannot be divided, even if we a priori bar division by zero. > > > > > For naturals, we can simplify that a bit by eliminating fractional > > digits (negative powers of B): > > for all n in N, > > n = sum{i=0 to oo} d_i x B^i. Even here there is no arithmetical operation valid for arbitrary TOintegers. > > Sure. > > > > > Your notation looks to be, in more formal terms: > > for all n in T, > > n = (sum{i=0 to oo} d_i x B^i) + (sum{j=? to ?} d_j x B^j) > > > > You mean the two ends of the string (and there may be extra limit points > within the string) each constitute a countable neighborhood of digits > within the string? That's correct. The uncountable portions oft he > string are filled in with a repeating pattern that equates to a rational > portion of an infinite quantity. It does end up needing to be > represented as a multi-part formula, as you suggest. That's the whole > idea, to represent such notions as oo^2-3*oo-log2(oo)+sqrt(oo), as a > digital string. :) Garbage. > > > I put '?' for the limits of the right sum, because I can't figure out > > what j is supposed to be once you've used up all the powers of B > > with B^i in the left sum. > > It is a declared formulaic infinity based on Big'un. it is a declared nonsense based only on TO's mental short circuits. > > > > > I mean it looks like you're saying that there are digits beyond > > an infinite number of digits, but that creates a problem for you, > > because you've run out of indexes for those extra digit positions. > > I don't see any more problem with that than saying you have traversed > and infinite number of points from the origin, but you are at a specific > point, with an infinite number of more points ahead if you continue. Except that WE make no such claims for discretely ordered index sets (in which every index but a first must have a unique predecessor and all have a unique successor) like N. WE never claim to have reached an index with infinitely many predecessors. WE leave that sort of nonsense to TO. > Can > you index each of those points with a finite string? We can index any one of them with a finite string, and we do. > No, of course not, > unless you have some infinite alphabet. It's uncountable. What is uncountable, TO's alphabet? > But, that > doesn't mean it can't also be viewed as residing along the same linear > sequence. Definition of ordinal limit points ( as distinct from topological limit points): P is a left limit point of an ordered set S if P is a member of S and for every Q in S with P < Q there is an R with P < R < Q. P is a right limit point of an ordered S if P is a member of S and for every Q in S with Q < P, there is an R in S with Q > R > P. In standard notation, index sets have no limit points of either type. In the standard set of natural numbers, N, there are no limit points of either type. In any standard ordinal which contains a limit ordinal as a member, those members which are limit ordinals are right limit points. but there are no left limit points in any ordinal. > > > Which puts you back at square one, because in order to define > > some meaning for those supra-infinite digit positions, you need > > some kind of supra-infinite indexing numbers, which is what > > you're trying to define the T-numbers to be in the first place. > > Which is just circular reasoning. > > > > Not circular at all. Then spiralling and into the oblivion of a central black hole. > I declare Big'un to be the number of reals per unit > interval, and the length of the hyperreal line in unit intervals. TO can declare anything he wants to, but declaring things does not make them so. Then I > employ this value in formulas which can be compared using infinite-case > induction. It's very straightforward.
From: David R Tribble on 14 Nov 2006 20:08
Tony Orlow wrote: >> What part of my definition says that? For the positives: >> 1 e H >> x e H -> 2^x e H >> x e H -> 2^-x e H > David R Tribble wrote [some paragraphs later]: >> You're confusing things here. You're going to get countably infinite >> long digits strings (I suspect that most of your H-riffic numbers are >> irrational). However, you will only be able to derive a countably >> infinite number of them. Most of the reals will not be derived by >> your rules, regardless of the starting point. >> >> I've given a simple example many times. Starting with 0, your >> rules will never generate 3, 1/3, or any integer multiple or >> power of 3. You still have not responded to that little flaw, >> which would go a long way towards convincing others that >> you might be on to something. > Tony Orlow wrote: > Why is that a flaw in the H-riffics, and not in the digital reals? Because the two are defined completely differently. The reals are not defined in a countable manner, but your H-riffics are. Every H-riffic can be assigned a unique natural. David R Tribble wrote: >> Your rules produce a countable set by their very nature. For any >> H-riffic H_n, I can produce H_2n and H_2n+1 from it. Which means >> that I can map every natural k to some H-riffic, and vice versa. >> Hence the H-riffics are a countable set. > Tony Orlow wrote: > But there is no r in R for which you can say there is no [H-riffic] successor. Sure there are, a lot of them in fact. For instance, 3 - what are its predecessors? Its successor? |