An uncountable countable set
in [Math]
|
Prev: integral problem
Next: Prime numbers
From: Virgil on 1 Oct 2006 23:13 In article <45205e7d$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > mueckenh(a)rz.fh-augsburg.de wrote: > >> This is an extremely good example that shows that set theory is at > >> least for physics and, more generally, for any science, completely > >> meaningless. > > > > Except that set theory axiomatizes the portions of standard mathematics > > used by physics and the sciences. > > > > You say that set theory is meaningless as you work on a computer that > > depends on the advent of technology hastened from the theoretical > > contexts of mathematical logic and set theory. > > > > MoeBlee > > > > Transfinitology is an unsound extension of what "works" in the finite case. As TO is the only purveyor of "transfinitology" hereabouts, he should then stop purveying it,
From: Virgil on 1 Oct 2006 23:19 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.
From: Virgil on 1 Oct 2006 23:23 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. And if it were an negative integer, subtracting 1 would give a more negative integer, so it cannot be most negative. All in all, it is self-contradictory.
From: Virgil on 1 Oct 2006 23:26 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.
From: cbrown on 2 Oct 2006 01:04
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". > > > >> 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? > 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. 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. > > > > 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? > Smaller and larger infinities can be expressed with B > formulaically. That is my point. If B is something you consider a "number", then it appears that B+1 is also a "number", B-1 is also a "number", 1/(B-1) is also a "number", etc. To be more specific, you seem to recognize any real number r in R as a "number"; and that you additionally accept that there is (at least one) "number" B, which is not a member of R, and for which the relatioin B>r for all r in R. So if x is a "number", and r is some real number, then the following are also "numbers": (i) x + r (ii) r*x (iii) x+B (iv) B*x It follows that for every B-polynomial p of the form: p = a_0 + a_1*B + a_2*B^2 + ... + a_n*B^n where n is some natural number, is a "number". Since you have previously opined that "every infinite has an inverse", we can then assume that (at the least), every rational function f of the form p/q, where p and q are B-polynomials, is a "number". Given that B > r for all r in R, we can even give this system a total ordering: i.e., it can be made into an ordered field, with results that are "intuitive". E.g., B^2 > r*B for all r in R, 1 > B/(B+r) for all r In R > 0, etc. > I am simply saying that, for infinite n, as for finite n, > the ratio between 1/n and 1/n remains 1 (slope of the diagonal). In other words, you presume that your "numbers", which includes at least some /particular/ infinite number B, also have the property that they form a field: i.e. for all "numbers" x, exists a number 1/x such that x*(1/x) = x/x = 1 for all "numbers". > The > ratio between 0 and 1/n is 0 or infinite, if 1/n>0, and these are the > slopes of the segments comprising the staircase. You might conclude > that, in the limit, 1/n=0... It would be far more accurate to say: the limit as n -> oo of 1/n = 0, by which we mean that for all /strictly positive real numbers e/, it is the case that there is some real number m such that for all m'>m, |1/m' - 0| = |1/m'| < e. < , and the staircase breaks down to segments of > (0,0). In that case, your measure is lost. > You have never justfied this statement. The best I can int |