Uncountability of the Rationals?
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From: Virgil on 11 Jun 2010 16:39 In article <3b31805b-e246-476e-b8a2-331d00a6e47e(a)g19g2000yqc.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 11, 11:45�am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > Tony Orlow <t...(a)lightlink.com> writes: > > > On Jun 10, 5:09�pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > >> Tony Orlow <t...(a)lightlink.com> writes: > > >> > On Jun 10, 3:05�pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > > >> >> Transfer Principle <lwal...(a)lausd.net> writes: > > >> >> > To me, the difference between ZFC and non-ZFC theories > > >> >> > is more analogous to the difference between a flight > > >> >> > on American Airlines from JFK to LAX, and a flight on > > >> >> > Virgin America from JFK to LAX. In this analogy, > > >> >> > MoeBlee is forcing everyone to fly only on American > > >> >> > Airlines and not Virgin or any other airline. > > > > >> >> No, he's not. �Moe has *never* said that ZFC is the only acceptable > > >> >> theory. > > > > >> >> > In another thread, he claims that he doesn't consider ZFC to be the > > >> >> > best theory, but the fact that he compares ZFC to an airliner and > > >> >> > other theories to rubber balls speaks for itself. > > > > >> >> The fact is that Tony, AP, etc., have *not* offered any coherent > > >> >> mathematical theory at all *and you know it*. �Thus, if Moe > > >> >> criticizes > > >> >> their blatherings, then it is extraordinarily disingenuous to claim > > >> >> this > > >> >> as evidence that he accepts only ZFC. > > > > >> >> -- > > >> >> Jesse F. Hughes > > > > >> >> "Really, I'm not out to destroy Microsoft. That will just be a > > >> >> completely unintentional side effect." -- Linus Torvalds > > > > >> > "Disingenuous" means "lying". I believe Transfer's comment falls into > > >> > the category of a best-guess interpretation of Moe's motives. You, > > >> > Moe, Virgie, The Tribble and others seem completely closed to the > > >> > concept of any improvement on the standard obfuscation. For, "none > > >> > shall drive us from the Garden which Cantor has created for us". If it > > >> > doesn't produce fruit, it's time to plant a new bed, or at least > > >> > fertilize. > > > > >> I have nothing against alternative definitions of set size, but you have > > >> offered no clear definition at all. > > > > >> Nor do I have anything against alternative theories. �Why should I? > > >> Indeed, in a previous life, I did a bit of work in ZFA, which is an > > >> anti-well-founded variant of ZFC. � > > > > >> The criticisms that you receive are based primarily on the fact that you > > >> have never offered a clear and complete definition of set size that > > >> really competes with cardinality at all. �Instead, you have a definition > > >> that applies to *some* (not all) sets and --- according to your own > > >> description --- gives different answers depending on one's perspective. > > > > >> That's not a promising mathematical definition. > > > > > Then I guess natural density is not very promising either. Huh! > > > > Not as a general notion of set size, no. �But it is useful in certain > > contexts. > > > > Let E be the set of even numbers and S the set of squares. > > > > It's perfectly sensible to say that, while E, S and N have the same > > cardinality, the natural density of E in N is 1/2, while the natural > > density of S in N is 0. � > > > > Who could criticize such a claim? � > > Who can criticize Bigulosity for claiming to have different measures > for, not only these three sets, but the sets of cubes, square roots > and logs of natural numbers? Who could possibly reject a method that > can accomplish that? I can't imagine.... It is the ambiguity/vagueness of "Bigulosity" which offends. For example, can TO prove trichotomy or transitivity for his Bigulosities? > > > > > Now, if I said instead that cardinality is foolish and we should use > > natural density as a measure of set size, then of course I am being > > silly. �Natural density is no replacement for cardinality. �It just > > doesn't do the same job. > > It does a slightly better job, at least distinguishing infinities with > a finite difference or ratio relative to omega. It misses greater > distinctions which IFR detects. What's the complaint? For one thing, TO has yet to show that his 'Bigulosity' satisfies the properties of an order relationship > You really have a JSH obsession, don't you? > > TOny
From: Jesse F. Hughes on 11 Jun 2010 16:32 Brian Chandler <imaginatorium(a)despammed.com> writes: > Jesse F. Hughes wrote: >> Tony Orlow <tony(a)lightlink.com> writes: >> >> > On Jun 11, 12:21 am, David R Tribble <da...(a)tribble.com> wrote: >> >> I'm still waiting to hear what bigulosity/IFR says about the size >> >> of the set of natural squares. Any results yet? >> > >> > I already did that. According to IFR it's sqrt(omega), which according >> > to ICI is less than omega. >> >> So, it is a(n ordinal) number x such that x * x = omega, is that right? >> >> I'm not clever like you, but I'd wager that one can prove no such >> ordinal exists, when * stands for ordinal multiplication. Does that >> bother you? > > Come on! Of course it isn't an ordinal by any normal definition. Tony > uses our words to help us along, but only with his meanings. That's a funny use of the word "help". > I really think it is fairly clear how Tony really approaches all this: > Think of numbers, small (like 3) and large (like 672), but also very > very large (like 10^101). Now think that these numbers go on and on > and on and on, and well, really speaking they never stop, but when > they do, or at least when one gets in the general region of where they > would stop (if they didn't go on without stopping, that is), well, > these numbers are simply imponderably enormous. Let Bigun, or whatever > today's name is, be the representatively simply most totally > imponderably enormous of these numbers. Now the important thing is > that while imp. enor., Bigun is still a respectable number that can be > added divided, sqrted, and so on. Well, I've said this before, so I'll > stop. Maybe that's what he has in mind. I'm sure I wouldn't know. -- Jesse F. Hughes "If the above is not true, it could have been." -- Bart Goddard offers the perfect .sig.
From: Tony Orlow on 11 Jun 2010 17:58 On Jun 11, 12:00 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(a)lightlink.com> writes: > > On Jun 10, 5:30 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > >> David R Tribble <da...(a)tribble.com> writes: > > >> > Tony Orlow wrote: > >> >> No, That's IFR in particular which is used on countable sets of reals > >> >> mapped from some segment of N. > > >> > So then what is the *general* case for IFR? > > >> Could someone remind me what IFR stands for? > > >> -- > >> "I've noticed [...] I routinely have been putting up flawed equations > >> with my surrogate factoring work. My take on it is that I have some > >> deep fear that the work is too dangerous and am sabotaging myself." > >> -- James S. Harris > > > The inverse function rule. If a set of reals is bijected with a > > segment of the naturals using an invertible formula, then that inverse > > formula may be used to calculate the number of elements in the set > > within any given value range. > > What's an invertible formula? More specifically (sorry) a monotonically increasing real function, which necessarily has an inverse function, also monotonically increasing. > > > Where a<b, and where S is a set mapped from the naturals using f(n)=x, > > and where there exists an inverse function g(x)=n such that > > f(g(x))=g(f(x))=x, the number of elements in the set S within the > > range [a,b] is given by floor(g(b))-ceiling(g(a)) +1. > > f(g(x)) cannot equal g(f(x)), can it? The functions are f:N -> S and > g:S -> N, so f(g(x)) is in S, while g(f(x)) is in N. Yes, it most certainly can. If x is in S, then g maps to a member of N which f maps back to the same member of S. If x is in N, then f maps x to a member of S which g maps back to the same member of N. The domain of f is N, and the domain of g is S; the range of each is the domain of the other, given their own domains. Surely you can see that. > > > Try it on any finite set. Now imagine applying it to the range > > [0,omega]. > > > I hope that is clear enough for you. > > I'm not sure. Let's try. Let S = {1/4, 1/16, 1/64}. That's a finite > set. Stop. You have listed three elements. You have not defined how they are mapped from N. Given 1/2 to the -2, -4, and -6 powers, I will surmise that these are the initial three elements of {2^-2n: n in N} for N starting at 1. Good? Continue... Let g:S -> N be defined by > > g(x) = x^{-1/2} > > and f:N -> S be defined > by > f(n) = n^-2. Where on Mother Earth did these functions come from? How do they relate your set with N? f:N -> R is supposed to map the naturals to your proposed set. So, now I have two problems here now, both fine examples. First, I'll address your orignal set. The inverse function of the f:N -> R I proposed for your set (correct me if you meant something else) was f(n)=2^-2n. I'm probably screwing this up, but I think the inverse is: g(n)=-2/log2(n) No, I think that's right. (I screwed it up first, error omitted) Anyway, if so, then we would apply IFR and apply the function to get the number of elements within the range of the set. Within the stated range of the set, [1/4, 1/64] (actually monotonically decreasing, but as it turns out, no matter. Nice try thought (ntt)). We have that |S|: [1/4,1/64] (the number of elements in S in the range [1/4,1/64]), with g(n)=-2/log2(n) ... =floor(g(1/64))-ceiling(g(1/4))+1 =floor(-2/-6)-ceiling(-2/-2)+1 =3-1+1 =3 > > Then f(g(x)) = (x^{-1/2})^{-2} = x and similarly g(f(n)) = n. Okay, the following is copied from above and quoted, to address the second set/sequence suggested: (BTW, I especially like the alliteration in that statement, because the "slash" adds to it :)) " Let g:S -> N be defined by > > g(x) = x^{-1/2} > > and f:N -> S be defined > by > f(n) = n^-2. " Okay, now. We start N at 1, so this sequence may be enumerated as {1, 1/sqrt(2), 1/sqrt(3), 1/2, ... etc.}. Wait, no. You started with the inverse function. Your set is therefore (1, 1/4,1/9,1/16,...}, squares of the inverses of naturals. Okay. Back to the inverse function. So, what was our range? Call it [a,b]. We have that |S|:[a,b] (as above) = floor(g(b))-ceiling(g(a))+1. Let's try a few examples, shall we? Try, I dunno, [6,50]. floor(50^1/2)-ceiling(6^1/2)+1 = 7-2+1=6 No, we have no elements at all between 6 and 50. This is an excellent question. I welcome it. Thank, you, Jesse. Something has been brewing, a question about countably infinite sets withina finite range, and this question addresses it. So, let's go back, and instead consider the number of elements between 1/6 and 1/50, and go back to using floor(g(b)) and ceiling(g(a)). Then we have 7-2+1=6, which include: {1/4,1/9,1/16,1/25,1/36,1/49} So, there is definitely a rule I am missing, in the case of compactified countable sets. I'll print this out and consider it. Thanks again. I hope I made the first example easy anyway, and resolved this one at least to your tentative satisfaction that I may be able to resolve whatever issues remain unaddressed. > > Now, what do you mean by the number of elements in the set S within the > range [a,b]? I think you mean S n [a,b]. Is that right? (I'm confused > by your use of the term "range".) I'm confused by your use of the symbol 'n', actually. I mean, |(xeS ^ x>=a ^ x<=b)| for finite a and b. > > Let's try. > > S n [0,1] = S, of course. Is that given by floor(g(b))-ceiling(g(a)) > +1? Again, I don't get 'n'. Sorry. Is there some bijection or invertible function you're referring to? > > floor(g(1)) - ceiling(g(0)) + 1 = floor(1^-1/2) - ceiling(0^-1/2) + 1 > = 1 - 0 + 1 = 2. > > Well, that didn't work. How about looking at [1/64,1/4]? > > floor(g(1/4)) - ceiling(g(1/64)) + 1 = floor(2) - ceiling(8) + 1 > = -5 > > Okay, now I'm sure. I'm sorry, but I really *don't* have any idea what > you meant by the above. Can you be a bit clearer about it? If you can explain what you mean by 'n' I can probably respond, but for now, I have nothing to say about that failure except that it may have to do with the first one. The second one, however, was spectacular, whether you meant it or not. Thanks, again, TOny > > -- > "So now, The Hammer is here, and with it, the end of days. The world will be > destroyed, and then remade, as foretold. You will be lost, with your > children, and then there will be others, and one day they will be tested, and > will pass, but that is another story." --James S Harris gets a bit excited.- Hide quoted text - > PS - If you wouldn't mind laying off, the JSH references get a tad tiring.
From: Tony Orlow on 11 Jun 2010 18:11 On Jun 11, 12:45 pm, David R Tribble <da...(a)tribble.com> wrote: > Tony Orlow wrote: > > I would like you consider a slightly modified goal: To exactly order > > sets according to size, and to distinguish between those that have an > > absolute size and those that do not. For me, no countably infinite set > > has an absolute size, but only one relative to other countably > > infinite sets, especially the standard N, starting at 0. > > Wouldn't it be easier to say that some chosen countably infinite > set has a standard "absolute" size, and then that all other such > sets have a size that is some fraction of that size? Thus every > set then has an "absolute" size, based on the standard set size. That might be easier, and would lead, as suggested, to natural density. However, that doesn't address the relations between infinite sets that have other ratios or offsets besides the finite. It is much more fruitful to look for a unit infinity based on uncountable infinity, which can be much more easily combined with measure and topology. > > For convenience, you'd want to choose the "largest" countably > infinite set for the standard "absolute" set size. This would > probably be N, since it contains all of the countable naturals. > (Another logical choice would be Z.) Given this, you could then > say that every countably infinite set does indeed have an > "absolute" set size.\ But, for me, they do not. If N starts at the first location with 1, and then the second with 2, etc, then the nth natural is n, and unless a set has n elements, there is no nth one. If there are omega naturals, then there is an omegath, which by the order=value definition of that set must equal omega. However omega cannot exist within the set. Consider this. Are the following two statements logically equivalent? AneN n<omega ~EneN n>=omega The first says there is no ntaural that can be the size of the entire set of naturals. That's true. The second says that there is no n in N that is greater than or equal to this "size". Does either imply, actually, that such a size exists? No. Both assert that no natural number will suffice. However, this is not to say that there exists ANY number which does. Logically, you must concede, neither statement proves that omega exists as a number. I am not obligated to imagine such a number, or give it any credibility. When it comes to the countably infinite, we can only look at this as some kind of limit, and for most discriminating results, consider them as Big-O measures or something similar. > > Of course I've glossed over all the important details, but that's > really your job to provide. Of course. Love, TOny
From: Tony Orlow on 11 Jun 2010 18:15
On Jun 11, 12:44 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: > Tony Orlow <t...(a)lightlink.com> writes: > > On Jun 11, 12:21 am, David R Tribble <da...(a)tribble.com> wrote: > >> Tony Orlow wrote: > >> > For, "none > >> > shall drive us from the Garden which Cantor has created for us". If it > >> > doesn't produce fruit, it's time to plant a new bed, or at least > >> > fertilize. > > >> Well, it's done pretty well for the last 100+ years. Have any > >> of your ideas produced any results of consequence yet? > > >> I'm still waiting to hear what bigulosity/IFR says about the size > >> of the set of natural squares. Any results yet? > > > I already did that. According to IFR it's sqrt(omega), which according > > to ICI is less than omega. > > So, it is a(n ordinal) number x such that x * x = omega, is that right? Ordinals have nothing to do with my theory. I've called the von Neumann ordinals "schlock" for years. You should know that, at the very least. > > I'm not clever like you, but I'd wager that one can prove no such > ordinal exists, when * stands for ordinal multiplication. Does that > bother you? Not much, nope, not at all. > > As well, one can prove that omega is the least infinite ordinal (where > "least" is defined in terms of cardinality, of course). That is, if > alpha is an infinite ordinal, then there is an injection from omega to > alpha. Is this contrary to your claims? You sometimes say that you > don't think omega is the smallest infinite ordinal, but I'm not sure > what you mean when you say that. I don't use the word "ordinal" in my arguments (sure, go find one mention from 1996 or whatever). I say there are a wide spectrum of countable and uncountable infiniites, given the right techniques. |