An uncountable countable set
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From: Tony Orlow on 10 Nov 2006 15:10 imaginatorium(a)despammed.com wrote: > Tony Orlow wrote: >> imaginatorium(a)despammed.com wrote: > > <snip> > >>> Virgil was muddling everything up, as usual, but the difference between >>> a partial and total ordering is basically whether there are pairs of >>> elements for which the order is undetermined. The subset relation is a >>> very obvious example, where (for example) the set of reals in [0, 1] >>> and the set of prime integers cannot be compared, because neither is a >>> subset of the other. >> Okay, understood. > >>> "Bigulosity" has never been sufficiently clearly defined to tell, but >>> since you get very steamed up about subsets, and since the only known >>> coherent claim is that A proper subset of B -> b(A) < b(B), it's >>> extremely unlikely Bigulosity could be extended to become a total >>> ordering. >> Bigulosity is not based on the subset relation, but on formulaic >> mappings and infinite-case induction. > > Let me just point you to one of your problems here: do you understand > the normal set-theoretic definition of a "function" (or mapping)? (Go > look it up; I'm not going to type it here) > > The problem is that at school you learnt about "functions" - first > things like x+1, x^2+4x-7, which you later learnt are called > polynomials, then things like sin(x), and possibly sinh(x), Bessel > functions, and various other exotic varieties. I imagine these all fall > into the category of what you call "formulaic". But the general notion > of a mapping is unimaginably bigger than these very specific examples, > which were chosen (of course) because they are manageable. Any scheme > based on "formulaic" anything is not going to apply to almost all cases > (in some reasonable sense). > > A more mundane problem is that in practice your approach to a problem > is to intuit the desirable answer, wave your hands, produce a > "formula", then start arguing. But anyway... > >>> Please compare the Bigulosities of the set of polygons with vertices on >>> integral x-y coordinates and the set of topologically distinct >>> polyhedra. Show your working. (Of course you don't need to come up with >>> an "answer" like "ratio of 5 pi^2", but you need to show how such a >>> task would be approached. One of the things you still don't seem to >>> have realised is that before anything can be "maths" it has to be >>> teachable to other people. I don't think anyone but you has the >>> faintest idea what Bigulosity is really supposed to be, except in a >>> ragbag of specific cases.) >> It consists of a few ideas, and they cover a lot. > > Could you be slightly more specific: what is "a lot"? I grant you, > "formulaic" sequences of integers, and words in the language on n > symbols. Anything else? > >> I don't know where to >> start with your topologically distinct polyhedra. > > Hmm, not polyhedra, I see. > >> Why don't you give me a rundown of how YOU compare those two sets? > > OK; how would I think about "counting" the topologically distinct > polyhedra? First I'd observe that I could attack from the faces, the > edges, or the vertices (F, E, V; and I could remember that there's an > obvious vertex-face duality), so I might choose the faces. I'd assemble > a list of the first few, out of curiosity (which has to be the driving > force here: what are the conditions for a polyhedron to be its own V-F > dual, for example?) > > The minimum number of faces is obviously 4; not less than 3 faces must > meet each vertex, and there must be more than one vertex. I make a > little list of the number of possibilities: > > 4: 1 (tetrahedron) > 5: 2 (square pyramid, triangular prism) > 6: (pentagonal prism, cube, etc. at this point, cheat > http://www.research.att.com/~njas/sequences/table?a=944&fmt=4 ) > 7: and so on > > Now I would notice that by "topologically distinct polyhedron" I am > referring to a bounded geometrical object; not only bounded, but also > discrete, in the sense that if I have one in my hand I know I can count > the vertices. A bit of minor handwaving, and I would see that for any > number of vertices there will be a limited number of possibilities for > arranging that number of vertices into a polyhedron. So I know that I > can pick an ordering scheme, and put all of the polyhedra in it. So I > can "count" them, in the sense that I know that with my counting scheme > there will not be a polyhedron that escapes it. I also notice that this > counting sequence will never end, because there is no maximum to the > number of vertices. After all, if there were, then given a polyhedron > with that number of vertices I could simply pick any face, construct a > pyramid on that face, and get a polyhedron with more than the supposed > maximum number of vertices, which proves (by contradiction) that my > sequence of polyhedra never ends. (This is all incredibly obvious, but > I'm afraid I never know which incredibly obvious bits you still haven't > grokked.) > > So with the constraints of the present audience, I would express this > by saying that I can see I can "count" the polyhedra, a process which > will account for every one of them, given enough time, but I can also > see that the process of counting will never end. > > It's much messier, but I can do a similar thing for the polygons with > vertices having integral x-y coordinates. > > How could I compare the two sets? I don't know - in both cases it's > true that I can find a way of counting them, and that the counting will > never end. Since it never ends, it's hard to see how I might think that > the counting of the polygons was going to be over before the counting > of the polyhedra, or vice versa. It's not even possible to say of a > process that never ends that after so many counts the processs is "an > appreciable way to completion", because there _is_ no completion. > > So my answer is rather limited: both sets are (ok, dammit, in normal > words) countably infinite; there's no way obvious to me that I could > regard either as "less" endlessly endless than the other. > > What's your answer? You appear to be the one claiming to have "numbers" > for counting things when the counting never ends: do you have any here? > Or are you happy to accept that these two sets, and the set of pofnats > can all be put in 1-1 correspondence, or (in normal words again) have > the same cardinality? Per
From: Tony Orlow on 11 Nov 2006 12:58 Michael Press wrote: > 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? > Hi Michael. Excellent questions. This arose in, I think, Calculus XOR Probability, but then I started a thread to discuss the topic itself, Infinite Induction and the Limits of Curves. Here's a google link: http://groups.google.com/group/sci.math/browse_thread/thread/244a0fcf9c0091a3/d4015d66efa494e4?lnk=st&q=diagonal+staircase+limit&rnum=2&hl=en The link to my paper in the post is expired. Here's a new link to my paper, which I just uploaded: http://www.lightlink.com/tony/Limits.htm In short, my contention is that inductive proof works, not only for finite n, but also for infinite n, when proving equalities between expressions, or inequalities based on differences which do not have a limit of 0 as x->oo. Chas' example was meant to be a counterexample to such use of infinite-case induction, by showing that such inductive arguments do not hold in the infinite case, since the infinite-case staircase "clearly" has a length of sqrt(2), not 2 as inductively proven. However, my immediate response to this was that the staircase in the limit is not the same object as the diagonal, because it still maintains those right angles, and is therefore not a straight line exactly, but rather a sort of fractal, whose length actually is 2. While the two objects seem to be the same from the point-set perspective, since points can only be distinguished if there is measurable distance between them, and the distance between corresponding points has a lower bound of 0, point-set is not the only perspective. Since points have no measure other than location, it is not surprising that measure would be lost in the translation of a curve into points. So, I reformulated the problem as a segment-sequence topology, which clearly shows a difference between the two objects. Then Chas asked for an example of another curve that WAS the same as the diagonal in the limit, so I concocted such a thing, a saw tooth that went from vertical to diagonal as n increased, in segment-sequence form, which gave an answer that differed from the original staircase by some infinitesimal amount, which could be discarded according to the notion that the square of an infinitesimal may be considered as nothing. Anyway, if you care to read the thing, it's short, and now you have the links. :) Tony
From: Tony Orlow on 11 Nov 2006 13:11 David R Tribble wrote: > 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. > 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. Do the decimal fractions constitute a countable set? That's counter to Cantor's Diagonal Argument for the uncountability of the reals. 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: Tony Orlow on 11 Nov 2006 13:23 David R Tribble wrote: > 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. 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. 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. > > 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. 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. :) > 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. > > 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. Can you index each of those points with a finite string? No, of course not, unless you have some infinite alphabet. It's uncountable. But, that doesn't mean it can't also be viewed as residing along the same linear sequence. > 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. I declare Big'un to be the number of reals per unit interval, and the length of the hyperreal line in unit intervals. Then I employ this value in formulas which can be compared using infinite-case induction. It's very straightforward.
From: imaginatorium on 11 Nov 2006 14:13
Tony Orlow wrote: > imaginatorium(a)despammed.com wrote: > > Tony Orlow wrote: > >> imaginatorium(a)despammed.com wrote: > > <snip bit about how comprehensive Bigulosity is...> > >> I don't know where to > >> start with your topologically distinct polyhedra. > > > > Hmm, not polyhedra, I see. > > > >> Why don't you give me a rundown of how YOU compare those two sets? > > > > OK; how would I think about "counting" the topologically distinct > > polyhedra? First I'd observe that I could attack from the faces, the > > edges, or the vertices (F, E, V; and I could remember that there's an > > obvious vertex-face duality), so I might choose the faces. I'd assemble > > a list of the first few, out of curiosity (which has to be the driving > > force here: what are the conditions for a polyhedron to be its own V-F > > dual, for example?) > > > > The minimum number of faces is obviously 4; not less than 3 faces must > > meet each vertex, and there must be more than one vertex. I make a > > little list of the number of possibilities: > > > > 4: 1 (tetrahedron) > > 5: 2 (square pyramid, triangular prism) > > 6: (pentagonal prism, cube, etc. at this point, cheat > > http://www.research.att.com/~njas/sequences/table?a=944&fmt=4 ) > > 7: and so on > > > > Now I would notice that by "topologically distinct polyhedron" I am > > referring to a bounded geometrical object; not only bounded, but also > > discrete, in the sense that if I have one in my hand I know I can count > > the vertices. A bit of minor handwaving, and I would see that for any > > number of vertices there will be a limited number of possibilities for > > arranging that number of vertices into a polyhedron. So I know that I > > can pick an ordering scheme, and put all of the polyhedra in it. So I > > can "count" them, in the sense that I know that with my counting scheme > > there will not be a polyhedron that escapes it. I also notice that this > > counting sequence will never end, because there is no maximum to the > > number of vertices. After all, if there were, then given a polyhedron > > with that number of vertices I could simply pick any face, construct a > > pyramid on that face, and get a polyhedron with more than the supposed > > maximum number of vertices, which proves (by contradiction) that my > > sequence of polyhedra never ends. (This is all incredibly obvious, but > > I'm afraid I never know which incredibly obvious bits you still haven't > > grokked.) > > > > So with the constraints of the present audience, I would express this > > by saying that I can see I can "count" the polyhedra, a process which > > will account for every one of them, given enough time, but I can also > > see that the process of counting will never end. > > > > It's much messier, but I can do a similar thing for the polygons with > > vertices having integral x-y coordinates. > > > > How could I compare the two sets? I don't know - in both cases it's > > true that I can find a way of counting them, and that the counting will > > never end. Since it never ends, it's hard to see how I might think that > > the counting of the polygons was going to be over before the counting > > of the polyhedra, or vice versa. It's not even possible to say of a > > process that never ends that after so many counts the processs is "an > > appreciable way to completion", because there _is_ no completion. > > > > So my answer is rather limited: both sets are (ok, dammit, in normal > > words) countably infinite; there's no way obvious to me that I could > > regard either as "less" endlessly endless than the other. > > > > What's your answer? You appear to be the one claiming to have "numbers" > > for counting things when the counting never ends: do you have any here? > > Or are you happy to accept that these two sets, and the set of pofnats > > can all be put in 1-1 correspondence, or (in normal words again) have > > the same cardinality? Perhaps this particular fact raises no objections > > from your intuition module? > > > > Notice that of course there are other things one can say about these > > sets. For example, the number of polyhedrons with n edges [E(n)] is > > always less than the number with n vertices [V(n)] (= the number with n > > faces); I can see that E(n) < V(n-2), but "on average" these are > > close(?). Does the ratio V(n)/E(n) approach a particular value? I don't > > know. > > > > Brian Chandler > > http://imaginatorium.org > > Thanks for your take on the matter. It seems like a general notion of > countability for the set, though that depends on there being only a > finite n umber of vertices, edges or faces for each polyhedron in your > set. Oh, you mean you agree that it would be possible to arrange all the different polyhedra in a (one-ended) line so as to count them, reaching any one eventually, but never getting to the nonexistent other end? Good. Of course my treatment of polyhedra depends on them being members of the class of objects normally referred to in mathematics as polyhedra. You still haven't grasped the utility of having standard terms for things so everyone knows what they are talking about. But you might ought to be aware that of course mathematicians have thought about unbounded, "infinite pseudopolyhedra" - for example in David Wells' Geometry "dictionary" (Penguin) there's a section under "honeycombs" with some nice pictures in. > I notice that the infinite topologically distinct polyhedra known > as tiling systems are not included in your set. I'll ignore that point > for now. No, that's right. I also excluded Borges 17 Chinese categories of animal, strawberry, vanilla, and pistacchio ice cream, the King of Peru, and a box of rotten carrots. (This is only a representative sample, to which the exclusions are not restricted. This is also not medical advice. (Hope this helps.)) > In order for me to formulate this question in terms of Bigulosity > Theory, I'm going to have to formulate the relationships between > vertices, edges and faces in general, and their relation to the number > of polyhedra possible. I've played with these numbers in relation to > vertex angle for regular polyhedrons with some interesting results, but > not in the general case for irregular polyhedra. So, let me mull this > over, and see if I can concoct some numbers for you, formulaic > expressions of the size of th |