An uncountable countable set
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From: MoeBlee on 8 Nov 2006 17:58 Lester Zick wrote: > And you should seriously consider not doing mathematics. No, I'd rather not emulate you. MoeBlee
From: Lester Zick on 9 Nov 2006 13:37 On 8 Nov 2006 14:58:03 -0800, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: >Lester Zick wrote: > >> And you should seriously consider not doing mathematics. > >No, I'd rather not emulate you. Then you and Brian should continue with your poetry. ~v~~
From: Michael Press on 9 Nov 2006 16:55 In article <1162432710.713593.264080(a)h54g2000cwb.googlegroups.com> , cbrown(a)cbrownsystems.com wrote: > I went round and round and round and round with Tony regarding the > "staircase" problem, in which, to make the point that the lim f(g(n)) > is not always the same as f(lim g(n)), I gave an absurd "proof" that > the length of the diagonal of a unit square is 2. (A google of "chas > diagonal" in sci.math brings you to a random but representative section > of the discussion under the thread "Calculus XOR Probability", if > you're feeling interested or even just excessively bored). > > Part of the argument was an (increasingly painstakingly detailed) > definition of "the limit of a sequence of curves", which gave the limit > of a sequence of progressively finer staircases in R^2 traversing > opposing corners of the unit square as being the diagonal of the unit > square. Each staircase has length 2; but of course (I said, spreading > my arms and smiling encouragingly) the length of the diagonal of the > unit square obviously /cannot/ be 2, so therefore... I cannot resolve this problem, and have not located the thread you cite. If we define a curve in R^2 as a continuous map from [0,1] to R^2, how do we exclude the staircase sequence as a converging to the curve? Possibly by saying that a piecewise linear approximation must have the terminal points of the line segments on the curve. But how do we know that other pathologies do not lurk? What exactly is necessary for a sequence of curves to converge to a curve, and what curve is it? -- Michael Press
From: David R Tribble on 9 Nov 2006 19:40 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]: >> No, there are no successors in your tree for that number p, >> because there is no last L/R branch in the path defining that >> H2-number. So there can be no different or next branching for >> the "last" node for either successor of p, because there is no last >> node in the path. >> >> I'm talking over your head here, because you obviously do not >> follow what I'm saying. > Tony Orlow wrote: > Geeze, David, flatter yourself a little, why don't you? > > You are upset because what is clearly an uncountably long string might > have a successor. That's understandable. But, allow me to elaborate a > tad. I'll break into pieces so you can respond bit by bit: > > The function 2^x, as a real function, is continuous over R. It goes from > an asymptotic y of 0 as x->-oo to an asymptotic y of oo as x->oo, and no > discontinuities in between, monotonically increasing from -oo to oo. Right? > > Since the function is monotonically increasing, one can always find an > intermediate value between a and b, which is 2^((log(a)+log(b))/2), that > is, f((g(a)+g(b))/2), where f(x)=2^x, f(g(x))=g(f(x)), xeR. Right? > > Does that not make this set continuous over the range of reals? Sure. > So, we're left with a little conundrum. > > Every unique real x has a unique real y such that 2^x=y, and such that > -2^x=-y. So each real has two successors. We can't pick a point where > that's not true. We can start anywhere (of course, zero's a good spot to > start), but in any case, in enumerating this uncountable set, we're > going to get to uncountable strings, which have successors. 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. > 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.
From: David R Tribble on 9 Nov 2006 19:56
Tony Orlow wrote: >> It holds for all finite naturals, but if there are an infinite number of >> naturals generating using increment, then there are naturals which are >> the result of infinite increments, which must have infinite value. > David R Tribble wrote: >> Can you show us one of those infinite naturals? >> >> And while you're at it, show us the finite natural that is its >> predecessor. Of course, you first have to define what you mean >> by "infinite increments". > Tony Orlow wrote: > I meant an infinite number of increments, each being a successive > difference of +1 in measure. > > Here is an infinite natural, Big'un: 100...000. That's in binary, and > there are log2(Big'un) 0's after the 1, so the most significant bit is > at log2(Big'un). Its predecessor, 0111...111, with the 0 also at > log2(Big'un), is also infinite, as are countably, or evenly uncountably > many, in the chain of predecession. 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. 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. 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) 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. 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. 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. |