The Definition of Points
in [Math]
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From: Lester Zick on 29 Mar 2007 17:55 On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >>> It's the same as Peano. >> >> Not it isn't, Tony. Cumulative addition doesn't produce straight lines >> or even colinear straight line segments. Some forty odd years ago at >> the Academy one of my engineering professors pointed out that just >> because there is a stasis across a boundary doesn't necessarily mean >> that there is no flow across the boundary only that the net flow back >> and forth is zero.I've always been impressed by the line of reasoning. > >The question is whether adding an infinite number of finite segments >yields an infinite distance. I have no idea what you mean by "infinite" Tony. An unlimited number of line segments added together could just as easily produce a limited distance. ~v~~
From: Lester Zick on 29 Mar 2007 17:56 On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >> In other words modern mathematikers just assume that because the Peano >> and suc( ) axioms produce successive straight line segments between >> numbers there is some kind of guarantee that the successive straight >> line segments will themselves line up colinearly on straight line >> segments and that we can thus just assume or infer the existence of >> straight line segments and straight lines from those axioms.Doesn't >> happen that way because even if we assume the existence of straight >> line segments between numbers that doesn't demand successive segments >> align in any particular direction colinearly along any common straight >> line segment. Same principle as above, different application. >> > >"Straight" doesn't even seem to mean anything in the context of Peano... "Straight" certainly seems to mean something to mathematikers who talk about straight lines and geometry. ~v~~
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~~ |