An uncountable countable set
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From: Tony Orlow on 2 Oct 2006 08:12 Virgil wrote: > In article <45205fa9(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <45203919(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: >>> >>> >>> >>>>> Since ordinals are, by definition, well ordered, they cannot contain any >>>>> endlessly decreasing sequences, which TO's models require. >>>> Neither can the reals. >>> How about the set of negative integers? >>> How is that not an endlessly decreasing sequence of reals? >> The origin is at a finite location. Order starts from the bottom, if >> "decreasing" has any meaning. > > The set of negative integers has no "bottom". > > TO seems to be changing his tune when it is used against him. > > According to TO every set of numbers has a natural order, and it is > within that natural order that we must view it, but now he wants to > reject the natural order because it runs counter to another of his > claims. > > TO blows hot and cold with the same reath. I'm saying the if you iterate the negative integers starting at 0, in that order, there is no infinite descending sequence. 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", and the axiom of choice aside, I don't see there being any well ordering of the reals. The closest one can come is the H-riffic numbers. :)
From: Tony Orlow on 2 Oct 2006 08:15 Virgil wrote: > In article <45205fc8$1(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <452039fb$1(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: >>> >>>> Virgil wrote: >>>>> In article <4520254e(a)news2.lightlink.com>, >>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>> >>>>> >>>>>> In the binary number circle, "100...000" is both positive and >>>>>> negative infinity. >>>>> What about "100...0001"? >>>> That is the actual greatest negative integer >>> Self-contradictory. >> not(what he said) > > Since it is clearly greater by 1 than "100...000" which is positive > according to TO, it must be more positive and thus not negative at all. 100...000 is both positive and negative. To find the negative, invert all bits and add 1. 011...111 + 1 = 100...000. That's +/- oo. :) > > And if it were an negative integer, subtracting 1 would give a more > negative integer, so it cannot be most negative. Subtracting 1 from the most negative number gives +/- oo. Counterintuitive, you say? Hmmm..... > > All in all, it is self-contradictory. Well, no, not self-contradictory, though counter your intuition, no doubt.
From: Tony Orlow on 2 Oct 2006 08:17 Virgil wrote: > In article <45206018(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >>> In article <45203f37(a)news2.lightlink.com>, >>> Tony Orlow <tony(a)lightlink.com> wrote: >>> >>>> Virgil wrote: >>>>> In article <452032b9(a)news2.lightlink.com>, >>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>> Virgil wrote: >>>>>>> In article <45201554(a)news2.lightlink.com>, >>>>>>> Tony Orlow <tony(a)lightlink.com> wrote: >>>>>>>> Virgil wrote: >>>>>>>>> A set is a container, and is not one of the objects that it contains. >>>>>>>> It is nothing more or less than its contents. >>>>>>> It is determined uniquely and entirely by its contents, as stated in >>>>>>> the >>>>>>> axiom of extentionality. >>>>>> So we agree. There is nothing besides the members. >>>>> To say that it is completely determined by its members is not to say >>>>> that it "is nothing besides its members". If it were nothing besides its >>>>> members we cold not give it a name as a thing of its own. >>>> So, it is completely determined by its members, but that's not all there >>>> is to its determination. Uh huh. >>> >>> TO tries so hard not to understand that the he almost always succeeds in >>> not understanding, as here. >> So, there's something else, besides the members of the set, which makes >> up the set? > > Belonging to a particular set distinguishes its members from all the > non-members, so the truth of "x is a member of set A" depends on which x > and which A one is considering. So, that would be a "no". Of course, a set can also be associated with a property which holds true for all members of the and not for non-members, but you wouldn't want to go there.
From: Tony Orlow on 2 Oct 2006 09:05 cbrown(a)cbrownsystems.com wrote: > Tony Orlow wrote: >> cbrown(a)cbrownsystems.com wrote: >>> Tony Orlow wrote: >>>> cbrown(a)cbrownsystems.com wrote: >>>>> Tony Orlow wrote: >>> <snip> >>> >>>>>> Like I said, there were >>>>>> terms in my infinitesimal sections of moving staircase which differed by >>>>>> a sub-infinitesimal from those in the original staircase. So, they could >>>>>> be considered to be two infinitesimally different objects in the limit. >>>>> Here's a thing that confuses me about your use of the term "limit". >>>>> >>>>> In the usual sense of the term, every subsequence of a sequence that >>>>> has as its limit say, X, /also/ has a limit of X. >>>>> >>>>> For example, the sequence (1, 1/2, 1/2, 1/3, ..., 1/n, ...) usually is >>>>> considered to have a limit of 0. And the subsequence (1/2, 1/4, 1/6, >>>>> ..., 1/(2*n), ...) which is a subsequence of the former sequence has >>>>> the same limit, 0. >>>>> >>>>> But the way you seem to evaluate a limit, the sequence of staircases >>>>> with step lengths (1, 1/2, 1/3, ..., 1/n, ...) is a staircase with >>>>> steps size 1/B, where B is unit infinity; but the sequence of >>>>> staircases with step lengths (1/2, 1/4, 1/6, ..., 1/(2*n), ...), which >>>>> is a subsequence of the first sequence, would seem to have as its limit >>>>> a staircase with steps of size 1/(2*B). >>>>> >>>>> Unless steps of size 1/B are the same as steps of size 1/(2*B), I don't >>>>> see how that can be possible. >>>>> >>>>> Cheers - Chas >>>>> >>>> It's possible because no distinction is currently made between countable >>>> infinities, even to the point where a set dense in the reals like the >>>> rationals is considered equal to a set sparse in the reals like the >>>> naturals. Where there is no parametric understanding of infinity, >>>> infinity is just infinity, and 0 is just 0. >>> Uh, OK. I assume that you somehow resolve this lack of "parametric >>> understanding" in /your/ interpretation of T-numbers. >> Well, yes. That's the whole point, but I'm not sure which particular >> numbers you mean. Maybe the T-riffics, which are center-indefinite >> digital numbers? >> > > I mean, whatever type of number the /particular/ number "x" is; when > you state "/the/ limit of the staircases has steplength 1/x". > If you say you have some particular infinite number of steps x, then you can consider 1/x to be the specific infinitesimal size of each. As long as 1/x is considered nonzero, there is a ratio between the x and y offsets of any given segment which gives a direction at that point. It is clear that when defined this way the directions of the segments in the staircase do not at all approach the directions of the corresponding segments in the diagonal, even the locations of the endpoints of the segments do become arbitrarily close. >>>> Where there is a formulaic >>>> comparison of infinite sets as n->oo, the distinction can be made. The >>>> fact that you have steps of size 1/n as opposed to steps of size 1/(2*n) >>>> is a reflection of the fact that the first set has twice the density on >>>> the real line as the first. As a proper superset, it SHOULD be larger. >>>> So, it's quite possible to make sense of my position, with a modicum of >>>> effort. >>> Well, let me ask you this: >>> >>> Suppose we have the original sequence of staircases, with step lengths >>> (1, 1/2, 1/3, ..., 1/n, ...). Let S be the T-limit staircase; you claim >>> that it has step sizes 1/B, where B is unit infinity. >> Well, where B is some infinite number of iterations for both staircase >> and diagonal. > > "Some" number? Do you mean that there is no "the limit" of the > staircases, but instead many possible "a limit" to the staircases? No, the limit is the limit. Once you have applied an infinite number of steps you have become infinitesimally close to the limit, and have essentially reached it. Any infinite value will do. In the limit, as a segment sequence, the infinite staircase continues to have a length of 2/x for each step, and x steps, for a length of 2. > >> We can call it the unit infinity if you want, especially >> since it covers a space of 1 unit. :) >> > > One might as well start somewhere. > > Starting from 0 and 1, we can construct the naturals. Add the concept > of "subtraction", and (arguably) we get the intgeres. Add the concept > of multiplication and division, and we get the rationlas. Throw in exponentiation, logs and roots, and we get the reals. I suppose I should get back to my H-riffic numbers. > > Starting from the reals and B, one can talk about B+1, B-1, 1/(B-1), > etc. These may not perhaps cover /every/ infinite "number", but from > your previous statements, certainly for /some/ B these "numbers" exist. Well, I think by declaring such a unit infinity as the number of reals in the unit interval has its applications. Add to that the application of induction to such numbers, as if they really exist on the same continuum with the finite reals, and you have a rich means of distinguishing infinite sets. I guess my approach seems too "normal" for infinite numbers, but it seems to work. > >>> Now, suppose we just forget about the very first staircase, but >>> otherwise continue normally. Now we have step lengths (1/2, 1/3, 1/4, >>> ..., 1/(n+1), ...). Do you claim that the T-limit staircase has step >>> size 1/(B+1)? Do you propose that 1/B is or is not equal to 1/(B+1)? >> Oh, I see where this is going. I'm not saying that every problem breaks >> down into B steps. > > Well, what about those problems which /do/ break down into B steps? > Shouldn't we expect some sort of consistency between those problems? Indeed. Let's start with the base set of naturals {1,2,3...}, and say there are B naturals in our base set. We want to map this set to each of our sets above, and compare the mapping formulas which are order-isomorphic. We map the first set using f1(x)=1/x and the second using f2(x)=1/(x+1). The inverse of f1 is g1(x)= 1/x, and the inverse of f2 is g2(x)=1/x - 1. Clearly, the first set has one more element, the difference between 1/x and 1/x -1. Now, this is almost an application of general IFR, except that the mapping function is the same in both directions. This would indicate a direct equivalence between these naturals and the first set. General IFR doesn't work here becau
From: Tony Orlow on 2 Oct 2006 09:22
Han de Bruijn wrote: > Tony Orlow wrote: > >> Han.deBruijn(a)DTO.TUDelft.NL wrote: >> >>> Randy Poe wrote: >>> >>>> Noon exists. >>> >>> Sure. By dogma. Randy Pope is infallible. >> >> See? Your position now is that noon doesn't exist. > > Yes. That has been my position all the time. > > Consider the ideal gas law: p.V = constant ; p = pressure, V = volume. > Imagine a finite container with gas and start compressing. Then V is the > independent and p is the dependent variable. The more you compress (V <) > the higher the pressure (p >). Can the volume ever become zero (V = 0) ? > No, it can't. So a zero volume doesn't exist in this problem. Mind that > V is the _independent_ variable, though! Yes, well that's because the dependent variable becomes infinite, in theory, and it would require infinite energy, like reaching c for a particle rest mass. So, sure, infinite energy is not realizable, and even infinite subdivision of space or time is not physically achievable. > >>>> But in order for the vase to transition from not-empty >>>> to empty, there would have to be a last non-empty >>>> moment. That would be the last time before noon. >>> >>> Aha. As clear as a klontje. >> >> What's that, a type of mud? > > It's English-Dutch for "het is zo klaar als een klontje" => "it is as > plain as the nose on your face". See? (I can be more espressive in my > mother's tongue) So, that was sarcastic... :) > >>> Nature does not jump, Leibnitz said. >> >> Leibniz lived in reality. Is that a necessary requirement? > > Yes. Typo. > >>>> Noon is the first moment at which the vase is empty. >>> >>>> But noon is not the transitional moment. There's no >>>> time just before noon where the transition happened. >>> >>> Wow ! And _that_ calls himself a physicist ... >> >> The time was not before that moment, but it was not after it either. >> Therefore we cannot make any judgments as to when it happened. That's >> transfinitology for you. > > Uhm? Dunno what you mean. > > Han de Bruijn > They are saying that the vase empties because every ball inserted is removed. They agree that this does not occur before noon, when there are always balls in the vase, but by noon the vase is empty. But they cannot say that, even though there are balls before noon, and none at noon, that the vase "became empty" at noon, because they are claiming "the limit doesn't exist". So, don't ask me what they mean. I can't figure it out. TOny |