An uncountable countable set
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From: Virgil on 25 Sep 2006 23:13 In article <45187d1c(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > David R Tribble wrote: > > [I'm a bit late in this subthread, but no matter...] > [ I just got back to it today too] > > > > Tony Orlow wrote: > >> If omega is the successor to the set of all finite naturals, ... > > > > It's not. It's a limit ordinal. > > > > Yes, it has no predecessor, but it is what comes immediately after the > complete set of finite naturals Omega IS the complete set of finite ordinals, so it cannot come AFTER itself. > But, my point is that it's "greater than any finite natural" in the > sense that, in the well-ordered set of ordinals, it comes after all > the finite successor ordinals. It contains each of the finite ordinals as both a member and as a subset, and is the first (smallest) ordinal to do so. > Why isn't it true that, if x comes before y in the ordinals, x<y is > not necessarily true? Because it is true that if x and y are ordinals with x a member of and a subset of y then necessarily x < y in both the membership ordering and the subset ordering. > > Nope. If w > x for all real x, then 1/w < 1/x for all real x as well. > > Which means that 1/w cannot be a real. > > Not a standard real. I'm talking here about nonstandard infinite real > values. Obviously those aren't standard reals. Any inverse would be > infinitesimal, and therefore also not a standard real. So? But there are no nonstandard reals in the standard reals. > >> To say that some count which is > >> greater than any finite count does not obey this general rule is a > >> kludge, like all the transfinite "arithmetic". To say that what is not a finite count does not necessarily obey all the rules of finite counts, is hardly a "kludge". > > > > Your "count" is not a well-defined term. Do you mean real, ordinal, > > cardinal, or something else? > > > > For the sake of this argument, we can talk about infinite reals, of > which infinite whole numbers are a subset. Only when we have an adequate description of all the necessary properties of these allegedly infinity reals. Note that there are axiom systems for the Archimedean ordered field of real numbers. Where is TO's corresponding axiom system for TO-reals? Are they like Abraham Robinson's hyperreals? If not, how do they differ?
From: Tony Orlow on 25 Sep 2006 23:15 cbrown(a)cbrownsystems.com wrote: > Tony Orlow wrote: >> mueckenh(a)rz.fh-augsburg.de wrote: >>> Mike Kelly schrieb: >>> >>>>>>> So lim [n-->oo] n/aleph_0 < 1 >>>>>> Division is not defined for infinite cardinal numbers. >>>>> Is that your only escape? If you dare to say that aleph_0 > n for any >>>>> n e N, then we can conclude the above inequality. >>>> No, because division is not defined on infinite cardinal numbers. The >>>> above inequality is meaningless. >>> For cardinals we have well defined n*aleph_0 = aleph_0 for any natural >>> n (see any book on set theory). Multiply with 1/n for any natural n, >>> then you get well defined >>> >>> A n e N: aleph_0 / n = aleph_0 > 1. Hence, my followin statement is >>> correct. >>>>> But remedy is easy. >>>>> Take lim [n-->oo] aleph_0 / n > 1 and reverse the following fractions >>>>> analogously. >>> [...] >>> >>>> Your position seems very inconsistent. You claim that numbers have no >>>> existence outside their representation. And now you are claiming there >>>> exists a "true" arithmetic. >>> It is obvious. You can verify it by experiment: II + III = IIIII. >>> >>> [...] >>>> It is not a proof. Division is not defined where either operand is an >>>> infinite cardinal number. >>> But you can conclude n / aleph_0 < 1 by inserting aleph_0 > n which is >>> definied *if aleph_0 is a number in trichotomy with natural numbers*. >>> >>> You cannot have both, assert that aleph_0 is a number larger than any n >>> but on the other hand prohibit that the inequality n < aleph_0 be >>> utilized. >>> >>> Regards, WM >>> >> I agree. If x is infinite, and that means greater than any finite, and >> trichotomy holds, then 1/x is in [0,1], and is real, though >> infinitesimal, of course. But, I am getting the feeling that set >> theorists now don't want to claim that infinite is greater than finite. > > I think it would be more correct to say that set theorists don't now > want, and have never wanted, to claim that some mathematical object is > greater than another mathematical object without an /explicit > definition/ of what is meant by "greater than" in the context of the > discussion. > >> Hmmm... > > Indeed. > > Cheers - Chas > How's about, "comes after in the well ordering"? Tony
From: Virgil on 25 Sep 2006 23:16 In article <45187f82(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > Mike Kelly schrieb: > > > >>>>> So lim [n-->oo] n/aleph_0 < 1 > >>>> Division is not defined for infinite cardinal numbers. > >>> Is that your only escape? If you dare to say that aleph_0 > n for any > >>> n e N, then we can conclude the above inequality. > >> No, because division is not defined on infinite cardinal numbers. The > >> above inequality is meaningless. > > > > For cardinals we have well defined n*aleph_0 = aleph_0 for any natural > > n (see any book on set theory). Multiply with 1/n for any natural n, > > then you get well defined > > > > A n e N: aleph_0 / n = aleph_0 > 1. Hence, my followin statement is > > correct. > >>> But remedy is easy. > >>> Take lim [n-->oo] aleph_0 / n > 1 and reverse the following fractions > >>> analogously. > > [...] > > > >> Your position seems very inconsistent. You claim that numbers have no > >> existence outside their representation. And now you are claiming there > >> exists a "true" arithmetic. > > > > It is obvious. You can verify it by experiment: II + III = IIIII. > > > > [...] > >> It is not a proof. Division is not defined where either operand is an > >> infinite cardinal number. > > > > But you can conclude n / aleph_0 < 1 by inserting aleph_0 > n which is > > definied *if aleph_0 is a number in trichotomy with natural numbers*. > > > > You cannot have both, assert that aleph_0 is a number larger than any n > > but on the other hand prohibit that the inequality n < aleph_0 be > > utilized. > > > > Regards, WM > > > > I agree. If x is infinite, and that means greater than any finite, and > trichotomy holds, then 1/x is in [0,1], and is real, though > infinitesimal, of course. But, I am getting the feeling that set > theorists now don't want to claim that infinite is greater than finite. > Hmmm... Set theorists have no problems with Abraham Robinson's hyperreals, which include both infinite and infinitesimal members. Set theorists, and most others, have difficulty with TO's ambiguities for which he has no system.
From: Tony Orlow on 25 Sep 2006 23:23 Virgil wrote: > In article <45187409(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <451149ef(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: >>> >>> >>>> Consider the equally spaced staircase from (0,0) to (1,1), as the number >>>> of steps increases from 1 without bound. Is it the same as the diagonal >>>> line? Inductively we can prove that the length of the staircase is 2 at >>>> every step. Does it really suddenly become sqrt(2) in the infinite case? >>> There is no "infinite case", there is only a limit case. >>> >> Then noon never comes and the vase is never empty, > > > Both are limit cases, which are what actually occurs, given the models. Is this state actually achieved? If so, then why do you say "There is no infinite case"? If not, then why do you say the limit case "actually occurs"? >>> If the cases for a finite number of steps are sets of points, so is the >>> limit case. >> Correct. There is no measure. >> >>> If the finite cases are sets of segments with specific directions >>> determined by their endpoints, the limit case will only contain pairs of >>> identical points which do not determine any direction at all, and so is >>> ill defined. >> Incorrect. First of all, each segment was defined by a xy-offset pair, >> starting at point (0,0). Each such pair denotes, through these relative >> coordinates, a length and a direction. > > So what length and direction are indicated by the pair {(0,0),(0,0)} ? That's a pair of pairs, first of all. {0,0} would represent a null segment, really a point, without size or direction. For all x, {1/x,1/x} always has a slope of 1 and a length of sqrt(2)/x. The corresponding section of the staircase, though, is two segments, {0,1/x} and {1/x,0}, one with infinite slope, and one with zero slope. They are clearly differently defined segments, direction-wise, and length-wise. Tony
From: Tony Orlow on 25 Sep 2006 23:28
Virgil wrote: > In article <45187864(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Han de Bruijn wrote: >>> Tony Orlow wrote: >>> >>>> Mike Kelly wrote: >>>> >>>>> Han de Bruijn wrote: >>>>> >>>>>> Mike Kelly wrote: >>>>>> >>>>>>> Han de Bruijn wrote: >>>>>>> >>>>>>>> Mike Kelly wrote: >>>>>>>> >>>>>>>>> What the hell are you talking about? Arguing with someone who can't >>>>>>>>> speak English is getting aggravating. >>>>>>>> My English is much better than your Dutch. >>>>>>> So what? Your English is still too poor for this discussion to be >>>>>>> fruitful. >>>>>> Still don't get the point, huh? >>>>>> >>>>>> You are lacking even the most elementary form of politeness. It's very >>>>>> impolite to cut of a discussion with somebody from a foureign country - >>>>>> somebody who is doing his best to communicate with you - only because >>>>>> you are obviously superior in expressing your thoughts within your own >>>>>> mother's tongue. >>>>> You're a very rude person yourself, Han. I generally don't feel the >>>>> need to be civil to those who won't reciprocate. >>>> I don't think I have ever found Han to be rude, except when he >>>> referred to my "babbling" recently. Ahem. But anyway, while we >>>> disagree on the actuality of any infinity, we have the open mind of >>>> spirited debate, and feel no need to get nasty. >>> I may be rude sometimes, but I never get _personal_ by calling somebody >>> an "idiot" or a "crank". Tony's "babbling" translates with Euroglot as >>> "babbelen" in Dutch, which is a word I can use here in the conversation >>> with my collegues without making them very angry (if I say "volgens mij >>> babbel je maar wat"). But, of course, I cannot judge the precise impact >>> of the word in English. Apologies if it is heavier than I thought. >> Do you have the saying, "Shallow brooks babble, and still waters run >> deep"? I figured you picked up the usage from this forum, actually. It's >> meant, in English, to mean you aren't making any sense. :) >> >>>> Furthermore, I have never had any trouble understanding what Han is >>>> saying, except where he is using some mathematical construct with >>>> which I am not familiar. His English is not bad, and blaming your >>>> disagreement on his inability to communicate is kind of low. >>> Thank you very much, Tony, for this sort of defense. >> My pleasure. It seemed like a vacuous excuse. I get pretty sick of those >> diversionary tactics. >> >>>> So, let's engage in lively debate, and maintain our civility, while >>>> chopping each other's arguments to pieces. Of course, this can only >>>> happen if we don't consider our arguments to be part of our anatomy. >>>> Otherwise, it gets personal. >>>>>>> You are misinterpreting virtually all my posts. You claim that you're >>>>>>> not dishonest so I have to conclude you're simply incapable of >>>>>>> comprehending written English. This makes this whole subthread >>>>>>> pointless. >>>>>> I have only this kind of trouble with _you_ and nobody else on the web. >>>>> Really? You've never had anybody else other than me complain that you >>>>> misinterpret their posts? I suppose I must have hallucinated dozens of >>>>> posts I've seen of just that, then. >>>>> >>>>> You've never had anyone other than me struggling to understand what the >>>>> devil you mean by your broken English? I must have hallucinated, for >>>>> example, "A little physics would be no idleness in mathematics", then >>>>> :)? >>>> Well, that's a difficult type of quote. Han - I wouldn't mind working >>>> on exactly how you want to say that in English, if you like. :) >>> Uhm, since litteraly everybody is complaining ... Let it be an encrypted >>> message then :-) >> Well, it seems to me that perhaps you're saying something like, "Those >> with their heads in the abstract should keep their feet in the >> concrete", though that sounds a little funny. >> >> Math=Science? > > > Scientists, particularly those in the sciences most dependent on > mathematics, tend to think that all mathematics is, or should be, a > subservient to their particular fragment of science. > > Mathematicians know better. Define "better". Those that work in various areas of science share a notion which defines science. Theories which have no means of verification are not science, but philosophy. In mathematics, verification really consists of corroboration by other means, agreement between different approaches. In science, where you find a contradiction with your theory, it needs revision. So, the scientific approach to mathematics requires some criterion for universal consistency, as measured by the predictions of the various theories that comprise it. Where two theories collide, one or both is in error. I think that's better. Tony |