The Definition of Points
in [Math]
|
Prev: Guide to presenting Lemma, Theorems and Definitions
Next: Density of the set of all zeroes of a function with givenproperties
From: Tony Orlow on 30 Mar 2007 13:22 Lester Zick wrote: > 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. > If you define the infinite as any number greater than any finite number, and you derive an inductive result that, say, f(x)=g(x) for all x greater than some finite k, well, any infinite x is greater than k, and so the proof should hold in that infinite case. Where the proof is that f(x)>g(x), there needs to be further stipulation that lim(x->oo: f(x)-g(x))>0, otherwise the proof is only valid for the finite case. That's my rules for infinite-case inductive proof. It's post-Cantorian, the foundation for IFR and N=S^L. :) >> 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~~ Well sure, that's science. Declare a unit, then measure with it and figure out the rules or measurement, right? 01oo
From: MoeBlee on 30 Mar 2007 13:23 On Mar 30, 9:39 am, Tony Orlow <t...(a)lightlink.com> wrote: > They > introduce the von Neumann ordinals defined solely by set inclusion, By membership, not inclusion. > and > yet, surreptitiously introduce the notion of order by means of this set. "Surreptitiously". You don't know an effing thing you're talking about. Look at a set theory textbook (such as Suppes's 'Axiomatic Set Theory') to see the explicit definitions. MoeBlee
From: Tony Orlow on 30 Mar 2007 13:24 Lester Zick wrote: > 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. In my own behalf, yes. 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~~ He has some valid points about the condition of the patient, but of course he and I have different remedies. 01oo
From: Tony Orlow on 30 Mar 2007 13:25 Lester Zick wrote: > 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~~ An actually infinite sequence is one where there exist two elements, one of which is an infinite number of elements beyond the other. 01oo
From: Tony Orlow on 30 Mar 2007 13:27
Lester Zick wrote: > 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~~ Cantorians try with their lame "aleph_0". Better you get used to the fact that there is no more a smallest infinity than a smallest finite, largest finite, or smallest or largest infinitesimal. Those things simply don't exist, except as phantoms. 01oo |