Uncountability of the Rationals?
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From: Virgil on 11 Jun 2010 16:19 In article <1f9d45ad-1a81-4b97-97a3-d5b2e8735fdb(a)r20g2000vba.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 10, 10:51�pm, David R Tribble <da...(a)tribble.com> wrote: > > Tony Orlow wrote: > > > Consider f(x) and g(x) to be any two given arithmetic formulae on x. If: > > > > > (1) It can be classically inductively proven that f(x)>g(x) for every > > > real value greater than some finite x (the base case), and > > > > > (2) The difference between the two terms which establishes the > > > inequality does not have a limit of zero as x approaches infinity, > > > then > > > > > (3) If omega is greater than any finite counting number, or c or > > > aleph_1 (whether they are the same or not), then the same rule also > > > applies to such large numbers. > > > > > Please let me know which step seems vague. > > > > How do you get from any finite x to any non-finite x? > > Through an uncountable number of finite additions, multiplications, > exponentiations or whatever, or through an uncountable addition, > multiplication, exponentiation, etc. I don't believe in any smallest > infinity, so please don't shove that in my mouth. Standard math has some difficulties even with a countably infinity of additions or multiplications, not to mention whatevers, so without TO providing considerably more groundwork, his assertion of uncountable numbers of such operations is mere nonsense. > > > > > Your assumption is, by fiat, that this is the case. However, > > you have given no arithmetic justification for this. > > I have dozens of times. You just don't like my perspective on the > matter. > > > > > Specifically, given an arithmetic formula f(x) on all real x, > > on what basis can you assert that it automatically applies > > to any non-real x? > > On the basis that a difference without a limit of zero in the infinite > case remains in the infinite case, and stands as the basis for > quantitatively ordering two infinite values expressed in terms of such > functions on the variable. There are all sorts of ways a limit may fail to be zero. And until you can say exactly what to do in each of them, your claims are nonsense. > > > > > Explaining how this is the case for a simple example would > > help immensely. For example, take a simple real function: > > �f(x) = x + 1 > > Okay. > > > Now explain how this function retains its arithmetic meaning > > when extended to non-real (infinite) x. Or even how it can > > be extended in this way at all. > > Okay. You would use this function, say, to map the naturals starting > at 0 with the naturals starting at 1. For every element x of the first > set exists an element y in the second such that y=x+1, right? Now, the > inverse function is x=y-1, correct? In the infinite case (barring > limit ordinals, which do not figure into my theory) this difference > does not decrease to 0, but stays at a constant value of 1. Using IFR, > then, we can say that set y has 1 fewer elements than set x, which is > what we would intuitively desire, since it is missing element '0'. For > me, x+1>x for all x, not just for the finite. > > I'm not sure if that answered your question exactly, but hopefully it > clarified it a bit. It would appear that you methods might possibly work with a limited number of functions, but needed is the set of functions for which it is claimed to work and proof that it works for all such functions, but no such proof is in evidence. The progress of mathematics is not based on mere claims , but on proofs of such claims.
From: Virgil on 11 Jun 2010 16:20 In article <9a25d7d8-e5a2-4531-9140-aabec8dbacfd(a)5g2000vbf.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > On Jun 10, 11:20�pm, FredJeffries <fredjeffr...(a)gmail.com> wrote: > > On Jun 9, 6:51�am, Tony Orlow <t...(a)lightlink.com> wrote: > > > > > > > > > > �http://en.wikipedia.org/wiki/Schnirelmann_density > > > > > This looks a little heady to wade through right now but, "the > > > Schnirelmann densities of the even numbers and the odd numbers, which > > > one might expect to agree, are 0 and 1/2 respectively". That's nothing > > > like Bigulosity. > > > > If I may please ask, according to Bigulosity, are there the same > > number of even natural numbers as odd, which would seem to be > > indicated by the figure: > > > > 0 <--> 1 > > 2 <--> 3 > > 4 <--> 5 > > � �. > > � �. > > � �. > > > > OR is there one more even number than odd, which the following seems > > to indicate: > > > > 0 > > 2 <--> 1 > > 4 <--> 3 > > 6 <--> 5 > > � �. > > � �. > > � �. > > > > ? > > It ultimately doesn't matter. Then TO's pseudomath doesn't either.
From: Virgil on 11 Jun 2010 16:23 In article <737eb4f2-cc72-4d17-a68c-6c35631e42f9(a)b29g2000vbl.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > > It's Tony who keeps claiming that his bigulosity, T-riffics, IFC, and > > all the rest fit well within (or as simple extensions of) standard > > theory, > > in spite of all continuing evidence to the contrary. > > I never claimed any such thing. My ideas are clearly at odds with > transfinite set theory. Then TO should either retract his 'ideas' or provide some justification for preferring them to standard ideas. So far TO has done neither satisfactorily.
From: Jesse F. Hughes on 11 Jun 2010 16:17 Tony Orlow <tony(a)lightlink.com> writes: > On Jun 11, 8:47 am, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> Tony Orlow <t...(a)lightlink.com> writes: >> >> > (1) It can be classically inductively proven that f(x)>g(x) for every >> >> > real value greater than some finite x (the base case), and >> >> >> Does that imply that these are real functions? >> >> > Yes, they can be any invertible arithmetic functions. Does that pose a >> > problem for you? Probably, since I am applying them to infinite >> > values. However, that's what infinite case induction is for. >> >> No, your description of infinite case induction only gave certain >> sufficient conditions for concluding that f(omega) > g(omega). It does >> *not* give any reason to believe that a function f defined on natural >> (or real) numbers can be extended to infinite sets in some unique way. >> You are presuming that without stating it explicitly. > > Please reread what I wrote above. I am saying that for IFR the two > mapping functions are mutual inverses, which means they form a > bijection. I was saying that your problem with my idea occurs when I > apply these functions to infinite values, because you don't consider > them to be defined for nonstandard values. I am telling you that this > application to the infinite for the purposes of IFR is acheived > through the application of ICI. Do you understand what I wrote now? No. Your description of ICI only allowed one to conclude that f(w) > g(w), given certain conditions. It did not allow one to conclude that, if f is defined for finite values (and some other property?), then there is a unique extension of f to infinite values (such that so-and-so is true?). > Okay, do you disagree with this statement? > > AneN n<omega Yes. > Do you disagree with this one? > > AneN 1<n -> n<n^2 Okay. > Not so far, right? > > How's about we talk about the set of all set sizes, or "counts", and > call it N+? Now, you undoubtedly disagree with this statement: > > AneN+ 1<n -> n<n^2 Of course I disagree if the operations and < relation are the obvious ones, but this isn't really the meat of my question. I know what the function n |-> n^2 means for infinite values. I'm asking you know about the function sqrt. What is sqrt(omega)? What properties does it satisfy? > It runs completely counter to cardinality, but is a much cleaner > extension of finite induction than transfinite induction IMHO. Just > pointing out exactly where our differences lie. There's a concrete > example. > >> >> What you really want, but are not stating, is something like this: >> >> For every (invertible?) function f:R -> R, there is a unique extension >> f':? -> ? (Ord -> Ord?) such that ... (what?). > > Exactly!!!! :D You hit the nail on the head. Or, with your head. > > For every countably infinite set S whose membership is defined by > monotonically increasing function f:N -> R, which of necessity has > real inverse function g:seS -> N, the count of elements in the range > [a,b] is given by floor(g(b))-ceiling(g(a))+1. Do you care to dispute > whether this works, in the finite case? Please do comment. Let's see if I understand. Let f:N -> R be monotonically increasing and S = range(f). Let g:S -> N be the inverse of f. Then S n [a,b] = floor(g(b))-ceiling(g(a))+1. As written, this doesn't make sense, because you didn't restrict a and b to be elements of S, but dom(g) = S. You have assumed (without argument) that g has a unique extension to R (satisfying something?). But let's look at a case where g is indeed definable on R in a natural way. Let f:N -> R be defined by f(n) = n^2 and g:R -> N be the sqrt function. Then, for instance, S is the set of squares of natural numbers and S n [a,b] = floor(sqrt(b)) - ceiling(sqrt(a)) + 1. Yes, I think that looks correct, but for the assumption that g is somehow naturally extended to R. I think that what you have in mind really is a function f:R -> R with inverse g:R -> R. To see why your statement doesn't make much sense, let f:N -> R be defined by f(n) = nth prime. This is an increasing function, and hence there is an inverse g:Prime -> N, but it makes no sense to ask what g(b) is unless b is a prime. There are various ways to extend g to R. Here are three: g_1(b) = number of primes less than or equal to b. g_2(b) = 1 + number of primes less than b g_3(b) = g_1(b) if b <= 4 g_2(b) if b > 4 So, as we can see, you have not quite correctly stated the context of your claim. It will be another thing entirely to tell me what the claim means when we move to infinite arguments. -- "So I speak before a crowd of the damned, cursed to be unloved throughout time, with only their hatred and bile to comfort them now, having betrayed what should have been their one true lover: Mathematics." -- James Harris reaches a bit
From: Virgil on 11 Jun 2010 16:27
In article <34f47542-ff10-4880-9057-620fd12c49eb(a)30g2000vbi.googlegroups.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Um, excuse me, but is the list of all finite digital strings not > ultimately countably infinite in width? As it is not a geometrical object, it does not have a "width". At least until one is defined. If you mean that there is no finite upper bound on the number of digits in a digital string, say so. |