The Definition of Points
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From: Lester Zick on 29 Mar 2007 18:05 On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >> Finite addition never produces infinites in magnitude any more than >> bisection produces infinitesimals in magnitude. It's the process which >> is infinite or infinitesimal and not the magnitude of results. Results >> of infinite addition or infinite bisection are always finite. >> >>> Wrong. >> >> Sure I'm wrong, Tony. Because you say so? >> > >Because the results you toe up to only hold in the finite case. So what's the non finite case? And don't tell me that the non finite case is infinite because that's redundant and just tells us you claim there is a non finite case, Tony, and not what it is. > You can >start with 0, or anything in the "finite" arena, the countable >neighborhood around 0, and if you add some infinite value a finite >number of times, or a finite value some infinite number of times, you're >going to get an infinite product. If your set is one of cumulative sets >of increments, like the naturals, then any infinite set is going to >count its way up to infinite values. Sure. If you have infinites to begin with you'll have infinites to talk about without having to talk about how the infinites you have to talk about got to be that way in the first place. ~v~~
From: Lester Zick on 29 Mar 2007 18:09 On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >>> Add 1 n >>> times to 0 and you get n. If n is infinite, then n is infinite. >> >> This is reasoning per say instead of per se. >> > >Pro se, even. If the first natural is 1, then the nth is n, and if there >are n of them, there's an nth, and it's a member of the set. Just ask >Mueckenheim. Pro se means for yourself and not for itself. I don't have much to do with Mueckenheim because he seems more interested in special pleading than universal truth. At least his assumptions of truth don't seem especially better or worse than any other assumptions of truth. ~v~~
From: Lester Zick on 29 Mar 2007 18:10 On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >>> If n is >>> infinite, so is 2^n. If you actually perform an infinite number of >>> subdivisions, then you get actually infinitesimal subintervals. >> >> And if the process is infinitesimal subdivision every interval you get >> is infinitesimal per se because it's the result of a process of >> infinitesimal subdivision and not because its magnitude is >> infinitesimal as distinct from the process itself. > >It's because it's the result of an actually infinite sequence of finite >subdivisions. And what pray tell is an "actually infinite sequence"? > One can also perform some infinite subdivision in some >finite step or so, but that's a little too hocus-pocus to prove. In the >meantime, we have at least potentially infinite sequences of >subdivisions, increments, hyperdimensionalities, or whatever... Sounds like you're guessing again, Tony. ~v~~
From: Lester Zick on 29 Mar 2007 18:12 On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >> Just ask yourself, Tony, at what magic point do intervals become >> infinitesimal instead of finite? Your answer should be magnitudes >> become infintesimal when subdivision becomes infinite. > >Yes. Yes but that doesn't happen until intervals actually become zero. >But the term >> "infinite" just means undefined and in point of fact doesn't become >> infinite until intervals become zero in magnitude. But that never >> happens. > >But, but, but. No, "infinite" means "greater than any finite number" and >infinitesimal means "less than any finite number", where "less" means >"closer to 0" and "more" means "farther from 0". Problem is you can't say when that is in terms of infinite bisection. ~v~~
From: Lester Zick on 29 Mar 2007 18:18
On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >>And until it does the magnitude of subdivisions remains >> finite. The fact that there is a limit to the process doesn't mean the >> process itself ever reaches that limit or ever can reach that limit. >> > >There is a definable limit, but that limit is not reached. Because you >cannot convey a point in every language with a finite alphabet in a >finite string, this means that point doesn't exist? Then, no points >exist, since there is always a number system which requires an infinite >number of bits to specify that point. What bits? Do limits exist only because there are bits? Why don't you stop beating around the bush and just come right out and say "points are bits and as long as we restrict ourselves to binary logic bits everything'll be just honky dorry and we can play mathematics on a Nintendo GameCube instead of having to demonstrate truth logically"? ~v~~ |