An uncountable countable set
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From: Virgil on 28 Aug 2006 15:34 In article <1156765872.764714.246070(a)i42g2000cwa.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <44eef04d(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > > If the naturals are not a subset of the reals in ZFC and NBG, then those > > > theories are even more screwed up than they already seemed. > > > > While there are sets of objects within the rationals and reals which > > look very similar to the urset of naturals, they are not quite the same. > > They are merely copies of the naturals within the larger systems. > > That is an opinion. It is an opinion held by a great many mathematicians, those who understand the foundations on which that opinion is based. I have the opposite opinion. > my opinion is the geometric interpretation. Another one is the > mathematically correct equation 1 = 1.000... (in physics it would not > hold). What will or will not hold in physics is irrelevant in mathematics, and validity the equation cited will depend on context. If "1" refers to a natural number in some set theory axiom system like ZF, then the equation is false, since 1.000... does not refer to such a natural number, but a real, or at least rational number.
From: Virgil on 28 Aug 2006 15:40 In article <1156765996.907742.198440(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Tony Orlow schrieb: > > > > But since no natural requires an infinite bit string, that is irrelevant. > > > > If no natural requires an infinite bit string, even the very largest, > > and all bit positions are included in it, then no infinite set of bit > > positions is required. > > Tony, please drop the "very largest". The rest is ok. If no natural > requires an infinite set then all naturals together do not require an > infinite set. If the contrary is asserted (and if infinity is a number > larger than any finite number), then we may ask which natural is the > first such that all of its predecessors require an infinite number of > bit positions. > > Regards, WM That is the wrong question, as it can be shown that some predecessors of every natural which has any predecessors can be represented with only finitely many bits. A better question is to ask for the first natural requiring more that any finite number of bits. Since the naturals are well ordered, every non-empty set has a first member, so if any natural requires "infinitely many" bits, there must be a first one. So that now TO must find THAT one.
From: David R Tribble on 28 Aug 2006 18:25 Tony Orlow wrote: >> Yes, true. Absolute 0, the origin, is a single point, and >> infinitesimals, neighboring points, are something ever so slightly >> different. But then again, oo as a concept and a limit is something >> absolute and different from, say, the number of points per unit of >> space, or some other infinite unit. To have infinitesimal numbers, you >> need to have non-absolute infinities also. > David R Tribble wrote: >> That's not correct. To have infinitesimals, where >> for infinitesimal e, 0 < e < x for all real x, >> it is sufficient to define them as reciprocals of "illimited" numbers, >> where >> for infinitesimal e, e = 1/h, where h is illimited; >> for illimited h, x < h for all real x. > Tony Orlow wrote: > Okay, but how is that different from what I said. Specific infinities, > or illimited numbers, go hand in hand with the notion of infinitesimals. My point is that illimited numbers are not infinite numbers. It is more correct to think of them as another set of real-like numbers outside the reals. David R Tribble wrote: >> These illimited numbers may have properties like the reals (e.g., >> order, addition, multiplication, etc.), but are not reals because they >> exist outside R. But these illimited (or "unreal") numbers do not >> have be "infinite" numbers - they just have to be numbers with >> magnitudes larger than any real. (Obviously, these numbers do >> not exist in standard arithmetic.) > Tony Orlow wrote: > No, but they are more compatible with standard finite mathematics than > the transfinite variety of infinity. No, they are simply a different kind of "number" that the standard reals. As such, they are not part of standard arithmetic, but can be included by adding a few axioms to create non-standard arithmetic (or internal set theory). Tony Orlow wrote: > Do you consider them compatible > with limit ordinals and infinite cardinalities? Of course. Why wouldn't they be? Adding non-standard illimited numbers in a consistent fashion (axiomatically) to a consistent system does not produce an inconsistent system. Non-standard sets have cardinalities just like standard sets. David R Tribble wrote: >> Indeed, the non-real numbers in non-standard analysis are often >> called "illimited" rather than "infinite". One reason is that they do >> not act like infinite numbers (infinite ordinals or cardinals), so it's >> misleading to call them "infinite". > Tony Orlow wrote: > Hmmm. On the other hand, since they behave like other numbers, it's > fitting to call them numbers in the first place, which is not clearly > the case with limit ordinals and transfinite cardinalities. Obviously you're having trouble with the concept of "number". Perhaps you should use different terms that you are more comfortable with, such as "quantity", "length", "measure", and "cardinality", to distinguish between the various kinds of "numbers". Trying to shoehorn all of these kinds into the same mold won't work and only confuses things. > I would imagine they were called "illimited" so as not to step on Cantorian > toes, since transfinitology already claimed the first definition of > "infinite". No, "infinite" was around a long time before Cantor and Dedekind. The invention of infinite ordinals and cardinals did not change the existing meaning of infinity in, say, calculus and analytic geometry. I'm guessing, but it would seem that it's a safer presumption that they are called "illimited" simply because it's more correct than calling them "infinite", since they don't act like the infinities in other areas of math, including standard arithmetic and, yes, set theory.
From: David R Tribble on 28 Aug 2006 18:41 Virgil wrote: >> Then show that the set of all natural numbers does not have cardinal number >> aleph-0 and ordinal number w. A proof please. > Tony Orlow wrote: >> I have already shown how the set of bit positions in the binary naturals >> has no cardinal or ordinal that can be assigned to it. > Virgil wrote: >> TO has claimed it, but not proved it. >> TO claims a lot but never proves any of it. >> Since the set of digit positions in any positional notation for the >> members of N is indexed by N, the cardinality of the set of bit >> positions equals the cardinality of its index set. > Tony Orlow wrote: > Since the bit strings are the power set of the set of bit positions, > each natural being a unique subset of 1 positions, the set of naturals > is power set to the set of bit positions. That sounds reasonable at first glance. But further study shows that there is a direct one-to-one correspondence between the bitstrings and the naturals. So the logical conclusion is that they are the same size. If you associate all the (finite) bitstrings with naturals, they are in fact the same set. So there can't be an in-between cardinality because the sets have the same cardinality. > In your concoction, the naturals are power set to nothing. Exactly. To be so, there would have to be an infinite set smaller than N that could not be bijected with N. But there is no such set. Specifically, like I said above, the set of bit positions bijects with the set of bitstrings. For every natural, there is a bit index, and vice versa. For every natural, there is a bitstring, and vice versa. So every bit index is a natural, and every bitstring is a natural. Hence they are exactly the same set.
From: David R Tribble on 28 Aug 2006 18:53
Tony Orlow wrote: >> In fact, mapping the naturals in [1,Big'un] to the reals in [Lil'un,1] >> using the mapping function f(x)=x/Big'un yields Ross' Finlayson numbers, >> and is perfectly consistent with IFR. Not only do we obviously have >> Big'un^2 reals on the line because we have the sum of Big'un unit >> intervals each containing Big'un reals, ... > David R Tribble wrote: >> That implies that there are BigUn naturals in the real number line, >> so that |N| = |R|. You state this as though it's a fact, but what is >> your proof? > Tony Orlow wrote: > What? No. I mapped the reals in [Lil'un,1] (Lil'un is successor to 0, in > the Finlayson system. He calls it iota, which is finite) to the naturals > in [1,Big'un] (the real line). The reals in the unit interval are the > image of the naturals on the entire line. Which implies that there are the same number of naturals in the real number line (or in N) as there are reals in (0,1]. That's provably false. Either that or you're omitting an awful lot of reals in (0,1] to get your mapping to the naturals to work. Or you're using some alternate version of the "real number line" that is not dense in the reals. (Based on your previous alleged well-ordering of the reals, and your acceptance of Ross's iota ordering, that could very well be the case.) > You are mistakenly applying > the standard cardinalistic fact that the number or reals in [0,1] is the > same as the number of all reals. That is false in my system. Since it's rather easily proved true in standard theory, you need to provide that missing proof of yours showing that it's false in your system. For instance, you have to demonstrate that there are more reals in (0,2] than in (0,1] while showing that both intervals are dense. Good luck with that. (Hint: assume that there are only a finite number of reals in any interval.) Of course, if it's false in your system, your system cannot be compatible with standard arithmetic. You see why, don't you? |