The Definition of Points
in [Math]
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From: Tony Orlow on 31 Mar 2007 20:47 Lester Zick wrote: > On Fri, 30 Mar 2007 12:04:33 -0500, Tony Orlow <tony(a)lightlink.com> > wrote: > >> Lester Zick wrote: >>> On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <tony(a)lightlink.com> >>> wrote: >>> >>>>>>> Okay, Tony. You've made it clear you don't care what anyone thinks as >>>>>>> long as it suits your druthers and philosophical perspective on math. >>>>>>> >>>>>> Which is so completely different from you, of course... >>>>> Difference is that I demonstrate the truth of what I'm talking about >>>>> in mechanically reduced exhaustive terms whereas what you talk about >>>>> is just speculative. >>>> You speculate that it's agreed that not is the universal truth. It's not. >>> No I don't, Tony. It really is irritating that despite having read >>> E201 and E401 you call what I've done in those root threads >>> "speculation". What makes you think it's speculation? I mean if you >>> didn't understand what I wrote or how it demonstrates what I say then >>> I'd be happy to revisit the issue. However not questioning the >>> demonstration and still insisting it's speculation and no different >>> from what you say is just not okay. >> I've questioned that assumption all along. We've spoken about it plenty. > > What assumption, Tony?You talk as if there is some kind of assumption. > That "not not" is self-contradictory, as if "not" is a statement.... >>> I don't speculate "it's agreed" or not. I don't really care whether >>> it's agreed or not and as a practical matter at this juncture I'd have >>> to say it's much more likely not agreed than agreed. What matters is >>> whether it's demonstrated and if not why not and not whether it's >>> agreed or not since agreements and demonstrations of truth are not the >>> same at all. Agreements require comprehension and comprehension >>> requires study and time whereas demonstrations of truth only require >>> logic whether or not there is comprehension. >>> >>> ~v~~ >> Demonstrate what the rules are for producing a valid one of your logical >> statements from one or more other valid ones of your logical statements, >> because "not not" and "not a not b" are not standard valid logic >> statements with known rules of manipulation. What are the mechanics? As >> far as I can tell, the first is not(not(true))=true and the second is >> or(not(a),not(b)), or, not(and(a,b)). > > Or you could demonstrate why the standard valid logic you cite is > standard and valid. > > ~v~~ Okay, I'll take that as a disinclination and failure to comply. You have the right to remain silent... ;) 01oo
From: Tony Orlow on 31 Mar 2007 20:50 Lester Zick wrote: > On Fri, 30 Mar 2007 12:22:30 -0500, Tony Orlow <tony(a)lightlink.com> > wrote: > >> 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? > > I have no idea what you think science is, Tony. Declare what and then > measure what and figure out the rules of what, right, when you've got > nothing better to do of an afternoon? > > ~v~~ I've been dropping feathers and bowling balls out my window all morning.... What do YOU think science is? 01oo
From: Tony Orlow on 31 Mar 2007 20:55 Mike Kelly wrote: > On 1 Apr, 00:36, Tony Orlow <t...(a)lightlink.com> wrote: >> Lester Zick wrote: >>> On Fri, 30 Mar 2007 12:10:12 -0500, Tony Orlow <t...(a)lightlink.com> >>> wrote: >>>> Lester Zick wrote: >>>>> On Thu, 29 Mar 2007 09:37:21 -0500, Tony Orlow <t...(a)lightlink.com> >>>>> wrote: >>>>>>>> Their size is finite for any finite number of subdivisions. >>>>>>> And it continues to be finite for any infinite number of subdivisions >>>>>>> as well.The finitude of subdivisions isn't related to their number but >>>>>>> to the mechanical nature of bisective subdivision. >>>>>> Only to a Zenoite. Once you have unmeasurable subintervals, you have >>>>>> bisected a finite segment an unmeasurable number of times. >>>>> Unmeasurable subintervals? Unmeasured subintervals perhaps. But not >>>>> unmeasurable subintervals. >>>>> ~v~~ >>>> Unmeasurable in the sense that they are nonzero but less than finite. >>> Then you'll have to explain how the trick is done unless what you're >>> really trying to say is dr instead of points resulting from bisection. >>> I still don't see any explanation for something "nonzero but less than >>> finite". What is it you imagine lies between bisection and zero and >>> how is it supposed to happen? So far you've only said 1/00 but that's >>> just another way of making the same assertion in circular terms since >>> you don't explain what 00 is except through reference to 00*0=1. >>> ~v~~ >> But, I do. >> >> I provide proof that there exists a count, a number, which is greater >> than any finite "countable" number, for between any x and y, such that >> x<y, exists a z such that x<z and z<y. No finite number of intermediate >> points exhausts the points within [x,z], no finite number of >> subdivisions. So, in that interval lie a number of points greater than >> any finite number. Call |R in (0,1]| "Big'Un" or oo., and move on to the >> next conclusion....each occupies how m,uch of that interval? >> >> 01oo > > So.. you (correctly) note that there are not a finite "number" of > reals in [0,1]. You think this "proves" that there exists an infinite > "number". Why? (And, what is your definition of "number")? > > -- > mike. > There are not zero, nor any finite number of reals in (0,1]. There are more reals than either of those, an infinite number, farther from 0 than can be counted. If there were a finite number, then some finite number of intermediate points would suffice, but that leaves intermediate points unincluded. What is a "number"? Good question. It's really the symbolic representation of a quantity. That's why folk like Han and WM discount unrepresentable numbers. I don't. I allow infinite strings, like the T-riffics and adics, and the uncountable sequence of the real H-riffics. tony.
From: Tony Orlow on 31 Mar 2007 20:56 Lester Zick wrote: > On Fri, 30 Mar 2007 12:25:24 -0500, Tony Orlow <tony(a)lightlink.com> > wrote: > >> 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. > > Which tells us what exactly, Tony, infinite sequences are infinite? > > ~v~~ It tells us "actual" means "uncountable" in the context of "infinite". 01oo
From: Tony Orlow on 31 Mar 2007 20:57
Lester Zick wrote: > On 30 Mar 2007 21:17:38 -0700, "Brian Chandler" > <imaginatorium(a)despammed.com> wrote: > >>> Under what definition of sequence? >> Oh come on... definition schmefinition. This is Tony's touchy-feely >> statement of what he feels it would be for a sequence to be "actually >> infinite". Actually. > > The same could be said for your touchy feely definitions, Brian. Six > of one half dozen of the other. > >> You're just being disruptive, trying to inject some mathematics into >> this stream of poetry... > > Mathematics? What mathematics did you have in mind exactly, Brian? > SOAP operas? Zen? What pray tell? > > ~v~~ Brian feels better. That's what really matters, to me at least... 01oo |