An uncountable countable set
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From: imaginatorium on 7 Sep 2006 15:48 Mike Kelly wrote: > Tony Orlow wrote: > > Mike Kelly wrote: > > > Tony Orlow wrote: > > >> Mike Kelly wrote: > > >>> Tony Orlow wrote: > > >>>> Virgil wrote: > > >>>>> In article <44fd9eba(a)news2.lightlink.com>, > > >>>>> Tony Orlow <tony(a)lightlink.com> wrote: > > >>>>> > > >>>>>> Dik T. Winter wrote: > > >>>>>>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow <tony(a)lightlink.com> > > >>>>>>> writes: <Hubble bubble double trouble> I wish I were a little grub, With whiskers round my tummy. I'd climb into a honey pot, And make my tummy gummy. > > >> So, while ...111 may not be considered a standard natural, I see no > > >> reason why it should not be considered, say, an extended natural. > > > > > > And you think this is a better model of "natural numbers" than standard > > > ones? You think it accords better with the intuitive picture people > > > have of what "counting numbers" are? > > > > > > > Yes, in the infinite realm, it does. It's precisely the binary signed > > integer in the limit as the number of bits approaches oo. > > I laughed out loud here. You think anybody but you on the PLANET would > think that this is relevant to the intuitive, naive idea of what > "numbers" are? Dunno why exactly. He seems at last to have adopted what I pointed out he was talking about months ago - a binary register representation of numbers, with the length indefinitely increased so there are no edge effects. I think I pointed out various problems with this too. Brian Chandler http://imaginatorium.org
From: imaginatorium on 7 Sep 2006 15:52 Mike Kelly wrote: > Tony Orlow wrote: > > David R Tribble wrote: > > > Virgil wrote: <bam!> > > > Tony Orlow wrote: <biff!> > > > Virgil wrote: <bom!> > > > Tony Orlow wrote: <splat!> > > That's my point. You can biject the naturals with the bit posiitons, and > > also with the strings, but the set of strings is the power set of the > > bit positions. Therefore you have a bijection between a set and its > > power set, and no cardinality which is appropriate for the set of bit > > positions. > > Bullshit. There is no bijection between the naturals and the set of all > binary strings. There might be if the universe is expanding so fast that all the infinite integers just happen to get included between the times of the two bijections. Brian Chandler http://imaginatorium.org
From: Virgil on 7 Sep 2006 16:10 In article <45001a98(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <44fe1725(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> MoeBlee wrote: > >>> Tony Orlow wrote: > >>>> And yet, you are saying it's not technically correct. So, 1 is a natural > >>>> but not a rational or real? If rigorous formulations come to that > >>>> conclusion, then rigor does not ensure correctness. > >>> You're hopeless. You didn't understand a thing I said, which may be my > >>> fault for not providing adequate explanation in the context of your > >>> ignorance of the subject; but it is no my fault that you won't even > >>> look at a book on mathematics, such as even a introductory text in real > >>> analysis. > >>> > >>> MoeBlee > >>> > >> Your whole point here is ludicrous. You are arguing that the naturals > >> are not a subset of the rationals, which are not a subset of the reals. > > > > In which argument he is perfectly correct. That there is an isomorphic > > image of the system of naturals in some other systems does not mean that > > the original naturals are in those systems. > > In terms of points on the line, there is no difference. There are differences. That these differences do not affect the arithmetical nor ordinal properties of the different sets is immaterial and irrelevant, as their members are not the same objects. Analogously, different chess sets are all the same game with the same rules, but are still not the same sets. > > > The empty set is, at least in the von Neumann model, a natural number > > but it is not a rational number nor a real number, as the corresponding > > rational and real numbers are both much more complicated objects. > > The empty set is nothing. The von Neumann ordinals are a misleading > model of the naturals. They did not mislead von Neumann and they do not mislead me. if they mislead TO, that is his problem. > That's all about specifying points on the line. If you specify the same > point, it's the same number, no matter how you got there. > > > > > For someone of TO's talents, learning the details of this would involve > > at lest a semester's serious and guided study, but more likely a year > > or two. After which he just might understand why the set of naturals is > > not directly a subset of either the set of rationals or the set of reals. > > It's really a pedantic point. Constructions differ for the three, but a > point is a point is a point. The rules of chess are the same regardless of which set of chessmen one plays with, but that does not mean there cannot be different chess sets. The rules of the "game" of naturals are the same regardless of which set of pieces one plays with, the von Neumann set, the rationals set, the reals set, or any other set which follows those rules.
From: Virgil on 7 Sep 2006 16:26 In article <45001e94(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <44fe1d38(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > So that if TO wants justifications, they are available, but they are > > often too technical for someone of TO's level of understanding to > > comprehend, and even if not, tend to be buried in papers that are > > otherwise highly technical in precisely the ways TO object to all the > > time. > > > > In other words, you justify the viability of the axioms by the fact that > they have gone through a long evolutionary process and survived in their > current form, rather than having died out. The same can be said for the > platypus. Quack! ;) The platypus, despite having what looks much like a duck's bill, does not quack. If one chooses to view it as a sort of survival of the fittest of axiom sets, mathematicians wouldn't have it any other way. Those axiom sets which do not deserve to survive don't for any length of time. Note that we still have, in a somewhat evolved form, Euclidean geometry, clearly a survivor. I predict that TO's, if it ever gets finished at all, will not long survive. > > > > > > >> As I said below, IFR is justified geometrically given > >> the graph of a function > > > > On the contrary, there are lots of "graphs" for which it is nonsense. > > Not of invertible functions, which are required for bijections between > sets of reals in the first place. Even for them. In order for the IFR to be of any value at all, it must be restricted to order preserving (or reversing) continuous functions with bounded domain and codomain. > > > >> Somewhere in the statement of an > >> axiom should be included the intent > > > > The intent is discussed in detail in all those books that TO declines > > to read. > > Uh huh. > > > > > > >>> Oooh boy. You don't even have a logicistic system, logical axioms, nor > >>> rules of inference, and yet you're telling me that you can prove your > >>> mathematical axioms, let alone your notion of inductive proof as > >>> provable by logic by forming an infinite loop shows you really have no > >>> idea what induction is. > >> If you say so. What is it, then, in a nutshell? I mean, you can't define > >> "mathematics", but maybe something a little more restrictive could be a > >> good place to start. So, why don't you explain induction? > > > > It has been defined quite clearly any number of times for TO, but he has > > such a weak memory that nothing sticks. > > > > Induction says: Given a subset S of the set of all naturals,N, > > if the first natural is in S and the successor of every element of S is > > also an element of S, then S = N. > > > > And that is all it says. Is that so difficult to remember, TO? > > That's not all it says, or it wouldn't be applicable to proofs. > P(1) ^ (P(n)->P(n+1)) -> A neN P(n), for any property or statement P > regarding n. Sure it would. Let S = {n \in N: P(n)} and it follows immediately. > It's used to prove statements regarding the naturals. The > mistake in the standard treatment is that it's never taken to be valid > for infinite n. There is a form of induction valid for infinite ordinals, but even it does not allow one to prove what TO wants to prove.
From: Virgil on 7 Sep 2006 16:36
In article <45001fa0(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <44fe2642(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > >>> In article <44fd9eba(a)news2.lightlink.com>, > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> > >>>> Dik T. Winter wrote: > >>>>> In article <44ef3da9(a)news2.lightlink.com> Tony Orlow > >>>>> <tony(a)lightlink.com> > >>>>> writes: > >>>>> Your axiom uses things that are not defined. What is the *meaning* of > >>>>> "x<z"? > >>>> Geometrically it means that x is left of z on the number line. > >>> > >>> And for someone standing on the other side of the number line would x be > >>> on the right of z? > >>> > >>> And does the line stay horizontal as one moves around earth? Which way > >>> is larger if the line ever goes vertical. And how does the "larger" work > >>> at antipodes? > >>> > >> Silly questions. > > > > In response to a silly definition. > >>> > >>>> It means > >>>> A y (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y). That says about all it > >>>> needs to, wouldn't you say? > >>> Not hardly. > >>> A y (y = x) v (y = z) v (y<x ^ y<z) v (x<y ^ y<z) v (x<y ^ z<y) > >>> is a bit better but still insufficient. > >> True, I should have specified y<>x and y<>z. I guess it's usually done > >> using <= for this reason, eh? > >> > >>>>> > > That is not a definition, because it makes no sense. "The set of > >>>>> > > naturals > >>>>> > > is as large as every natural"? > >>>>> > > >>>>> > It is not larger than all naturals > >>>>> > >>>>> That is something completely different again. > >>>> It's not LARGER than every finite. > >>> Which natural(s) is it "not larger" than", in the sense of not being a > >>> proper superset of that natural or having that natural as a member? > >> ....11111 binary (all bit positions finite) > > > > Unless that string has only finitely many bit positions as well as only > > finite bit positions, it is not a natural at all, as it is then neither > > the first natural nor the successor of any natural, and every natural > > has to be one or the other. > > It is the successor to ....11110. Duh. But ....11110 is not a natural either, and no member of any sequence of its predecessors is either. Let S be the set of naturals representable with only finitely many digit positions. The first natural is in S, and the successor of every member of S, requiring no more that one more digit, is also a member of S, so that, by induction, every natural is a member of S. I.e., by induction, every member of N is represented by a FINITE sequence of digits, and anything else is NaN (not a natural). This proof holds equally well in binary, trinary, or any larger natural number base. |