An uncountable countable set
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From: Lester Zick on 4 Oct 2006 17:48 On Wed, 04 Oct 2006 13:11:12 -0600, Virgil <virgil(a)comcast.net> wrote: >In article <4523c954$1(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> David R Tribble wrote: >> > Tony Orlow wrote: >> >>> On the other hand >> >>> I don't know why I said "neither can the reals". In any case, the only >> >>> way the ordinals manage to be "well ordered" is because they're defined >> >>> with predecessor discontinuities at the limit ordinals, including 0. >> >>> That doesn't seem "real" >> > >> > Virgil wrote: >> >>> In what sense of "real". There are subsets of the reals which are order >> >>> isomorphic to every countable ordinal, including those with limit >> >>> ordinals, so until one posits uncountable ordinals there are no problems. >> > >> > Tony Orlow wrote: >> >> The real line is a line, with >> >> each point touching two others. >> > >> > That's a neat trick, considering that between any two points there is >> > always another point. An infinite number of points between any two, >> > in fact. So how do you choose two points in the real number line >> > that "touch"? >> > >> >> They have to be infinitely close, so actually, they have an >> infinitesimal segment between them. :) > >But any "infinitesimal segment" within the reals is bisectable. So, in the words of a famous mathematician, what? ~v~~
From: Lester Zick on 4 Oct 2006 17:52 On Wed, 04 Oct 2006 12:43:42 -0600, Virgil <virgil(a)comcast.net> wrote: >In article <4523c1ac(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: [. . .] >> No, "is finite" is not an equality. > >Being finite is a property. Then being infinite is not a property. ~v~~
From: Lester Zick on 4 Oct 2006 18:22 On 4 Oct 2006 12:09:43 -0700, imaginatorium(a)despammed.com wrote: >Lester Zick wrote: >> On 4 Oct 2006 11:36:27 -0700, imaginatorium(a)despammed.com wrote: >> >Lester Zick wrote: >> >> On 3 Oct 2006 23:15:04 -0700, imaginatorium(a)despammed.com wrote: > >> ><snip> >> ><snop> > >> >Very disappointed with your answer, Lester. You didn't use the word >> >"technically" even once. >> >> "Technically" is reserved for answers, Brian, not questions. And >> needless to say you've gotten so snippy with questions lately it's a >> little difficult to tell if any answers qualify for a "technically". > >You can be irksome sometimes Herr-Professor Zick. OK, here's what you >said: You'll find unanswered questions considerably more irksome than me, sensei. >> Well perhaps not quite so easy as you might imagine, Brian. Perhaps >> you'd care to spell out exactly how you choose a first point that's so >> easy? Now please don't try to beg off from the problem under the >> pretext that actually choosing points is easy since it's an excercise >> in physics where you don't excel but not in the zen of mathematics >> where you don't seem to excel either. Just tell us how the trick is >> done that's so easy that's anything other than you're saying it's so. > >Is the first sentence of this not a statement? An answer? Preamble to a question establishing context. > Is it not an >answer that might be raised to something more resembling your usual >intellectual level by saying - for example - Not an answer to a question certainly but a response to a simplistic assertion on your part. >"Well technically, perhaps not quite so easy as you might imagine, >Brian." You claimed it was easy. I'm not sure it is easy. Certainly it seems to be more than a little difficult for you. If memory serves I first used the term "technically" in response to Bob Kolker's original answer to a question as to how lines were composed of points. His reply as I recollect was that one integrates points into lines which is of course technically incorrect since one integrates infinitesimals and not points. And as soon as you vouchsafe incorrect answers to questions I shall be happy to vouchsafe "technically"s. At the moment you seem content to vouchsafe non answers to unasked questions. >Incidentally, you think "choosing a point" is something in physics? I can't tell what you think it is. Points are chosen in physics and math. You seem to choose your points in math. I can't tell why you think selecting the first is any more easy than others. >Like "choosing a refrigerator" is something in domestic science? Hmm, >well, I confess: I really hadn't thought of that. If I merely chose >pi+2 as a point on the real line what would that be then? I can tell you that it wouldn't be, a point on an imaginary real number line. > (Don't be too >zen about it if possible. Wouldn't intuit it. > Of course, since in the end we all learn by >observing a master at work, if there's something unsatisfactory about >my choice, I'm sure you'll demonstrate your own techniques for doing >it.) I'm only trying to intuit why you think choosing a point is easy. ~v~~
From: David R Tribble on 4 Oct 2006 19:08 Virgil wrote: >> If TO does not like to be able to tell one ball from another, he does >> not have to play the game, but he should not ever try to pull that in >> games of pool or billiards. > Tony Orlow wrote: > If distinguishing balls gives a less exact answer, and a nonsensical one > to boot, then that attention can be judged to be ill spent, and not > contributing to a solution at all. It is clear that sum(x=1->oo: 9) > diverges, is infinite, not 0. It's ridiculous to think otherwise. What about: sum{n=0 to oo} (10n+1 + ... + 10n+10) - sum{n=1 to oo} (n) The left half specifies the number of balls added to the vase, and the left half specifies those that are removed.
From: cbrown on 4 Oct 2006 19:11
imaginatorium(a)despammed.com wrote: > Tony Orlow wrote: > > Hmm, it seems to me, Tony, that this post illustrates rather well just > how close to total is your ignorance of what mathematics is, and your > inability to grasp the notion of a formal argument. (So why do I > bother?...) > > The theme running through almost all of your comments here is one we > see a lot in JSH arguments. When little children learn arithmetic at > school, it's common to start with positive integers (which form a > semigroup under both addition and multiplication), so lots of things > "can't be done": 3-5, or 11 / 6. Later they learn further concepts, > such as positive rationals, which form a group under multiplication, so > things which previously "couldn't be done" now can. Similarly they > learn about negative numbers, so that now 5-23 can be calculated. In > the context of school arithmetic as a sort of "calculation > engineering", it's reasonable to see this as just learning more and > more powerful ways of "working out the answer". This may well include > what I call "Javascript arithmetic", in which "Infinity" is one > additional value, allowing us to calculate things to do with lenses in > a very useful way. I think this is exacly the problem that occurs with Tony, HdB, and many others. (Ross, WM and Zick have other, uh, issues). What's interesting to me is to consider exactly where this approach breaks down. In some sense, Tony is asking "why can't we simply /define/ some number B which is infinite (i.e., B > r for all r in R), and then simply evaluate formulas such as f(x) = (x+1)/(x^2+1) at x = B?" And of course we /can/ construct such a system (which I will call here the T-numbers); as I outlined in a previous post, for which his ideas hold as long as f is a rational function of polynomials over R. Even his "infinite induction" principle holds in its most straightforward mode: if f is a rational function of polynomials over R, and for all real numbers r > 0, f(r)>0, then f(B)>0. The system starts to fall apart when we say: what if f is /not/ a rational function of polynomials over R? For TO and others, "a function is a formula"; which loosely means: a rational function of polynomials + "other well-known functions, such as sin, cos, e^x, log, etc.". This is tied into his confusion over what the meaning of a "limit" is, and why R itself is very different from Q or the algebraic numbers. By TO's lights, to find the limit of a function f(x) as x->oo, one simply "plugs-in" B (or "any infinite") into the "formula", and voila! the resulting f(B) is the limit of f(x) as x->oo. This is a sort of "plug-and-chug" approach to limits that we are taught as a sort of rule of thumb: if a limit has a form of (n+1)/(n^2 - n), then we apply: oo+1 = oo, oo^2 - oo = oo^2, oo/oo^2 = 1/oo = 0; so "therefore" the limit is 0. In the more, um, extended domain of the T-numbers, this would read as "B+1 is negligibly different from B, B^2 - B is negligibly different from B^2, so (B+1)/(B^2 - B) is negligibly different from 1/B, which is different from 0 by an infinitesimal amount". But how does one evaluate the expression "sin(x)" when x = B? One presumes that it would be the same way that one would evaluate the function at any real number x as the limit of as n->oo of the sum:: x - x^3/3! + x^5/5! - x^7/7! + ... + (-1)^n*x^(2*n+1)/(2*n+1)! This is not the sort of "formula" where one can simply "plug in" B for n and "get the answer". TO cannot even answer the question "what is (-1)^B?" without getting into logical contortions regarding whether B is "prime or composite" (whatever that may mean), let alone deciding what is meant by "B!". All of his confusions seem to boil down to this problem of a "limit" as some kind of formulaic substitution of a symbol representing "the infinite case" into some kind of algebraic formula, which then exists as a "number". This isn't typically true for limits which are not taken as n->oo. For example, lim x->0 (sin(x)/x) isn't "sin(0)/0 = 0/0", where "0/0" is some new "number" that we then append to the reals (N.B.: by L'hopital's rule, it's 1). Can the T-numbers be saved? I.e., can we define "limits" on the T-numbers in such a way that we can still calculate, e.g., lim 1/x to be some specific T-number, while also allowing quantities such as B and 1/B, and still remaining a field with the total order so beloved of Tony? Looking at y = lim x->oo 1/x, where y is some T-number, I think we have three choices: (i) The limit is understood to mean that for all positive real number e, there is a real number x such that for all real numbers z > x, |1/z - y| < e. First, y = 1/B clearly satisfies the conditions. If y is any infinitesimal satisfying the above limit condition, then 2y > y, and 1/z - y > 1/z - 2y > 0 follows (since 1/z is real), and so it is also the case that |1/z - 2y| < e; and in fact 2y is actually "closer" to the limit than y is! (i.e., for any e > 1/z - y, there is a smaller e' such that 1/z - y > e' > 1/z - 2y). We cannot even say "let y be the largest infinitesimal satisfying the limit condition", becuase there is no largest infinitesimal in the T-numbers (just as there is no smallest infinite in the T-numbers). So in definition (i), the lim x->oo 1/x is not defined as a particular T-number. (ii) The limit is understood to mean that for all positive T-numbers e, there is a real number x such that for all such that for all real numbers z > x, |1/z - y| < e. Again in this case, the limit is undefined, by similar logic to the above. (iii) The limit is understood to mean that for all positive T-numbers e, there is a T-number x such that for all such that for all T-numbers z > x, |1/z - y| < e. In this case, the limit is y = 0; because if y > 0, then let e = y/2, Then for all z > 2/y, |1/z - y| = y - 1/z > e. On the other hand, if y = 0, then for all e, set x = 1/e; then for all z > x, 1/z < e. I see no "fourth way" of preserving the idea of a limit, that yields lim x->oo 1/x = 1/B for some infinite B. Cheers - Chas |