The Definition of Points
in [Logic]
|
Prev: On Ultrafinitism
Next: Modal logic example
From: Lester Zick on 1 Apr 2007 19:21 On Sat, 31 Mar 2007 22:42:19 -0600, Virgil <virgil(a)comcast.net> wrote: >In article <460f1ef1(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Virgil wrote: >> > In article <460f0317(a)news2.lightlink.com>, >> > Tony Orlow <tony(a)lightlink.com> wrote: >> > >> > >> >> There are not zero, nor any finite number of reals in (0,1]. >> > >> > There are every finite and more of reals in (0,1]. >> >> You mean any sequential ordering of the reals in (0,1] will contain >> elements in finite positions, plus more. Same thang. Tru dat, yo. > >What I mean, and what TO misrepresents me to mean have nothing in common. Yeah well I'd like to say that sentence reads just like you have more in common with great apes than anyone else except I have no interest in further demeaning great apes. ~v~~
From: Lester Zick on 1 Apr 2007 19:23 On 1 Apr 2007 03:49:59 -0700, "Mike Kelly" <mikekellyuk(a)googlemail.com> wrote: >On 1 Apr, 01:55, Tony Orlow <t...(a)lightlink.com> wrote: >> 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]. > >OK so far. > >> 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. > >OK, so there are *not* a finite number of reals. > >Now, apparently, this *proves* that this thing called "BigUn" exists >to denote how many reals there are. And, apparently, this BigUn >behaves just like those good old finite numbers we're used to. We can >perform all the usual arithmetical operations on it. However, it's not >defined in any way other than "it's larger than finite". All we know >about it is that it's a "symbolic representation of quantity" and that >it's "larger than any finite". And yet we can do arithmetic with it >like it was a natural number. And yet modern mathematikers like Bob make straight lines out of points without blushing. Go figure. >Do you not see *any* problem with this picture? ~v~~
From: Lester Zick on 1 Apr 2007 19:24 On Sat, 31 Mar 2007 18:40:56 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >Lester Zick wrote: >> On Fri, 30 Mar 2007 12:11:23 -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: >>>> >>>>>> Equal subdivisions. That's what gets us cardinal numbers. >>>>>> >>>>> Sure, n iterations of subdivision yield 2^n equal and generally mutually >>>>> exclusive subintervals. >>>> I don't know what you mean by mutually exclusive subintervals. They're >>>> equal in size. Only their position differs in relation to one another. >>>> >>>> ~v~~ >>> Mutually exclusive intervals : intervals which do not share any points. >> >> What points? We don't have any points not defined through bisection >> and those intervals do share the endpoints with consecutive segments. >> >> ~v~~ > >Okay, lay off the coffee. > >Sure. Now subdivide the line so that the left endpoint is always >included and the right never. [x,y). Then each is mutually exclusive of >all others. Except the ones in the middle. ~v~~
From: Lester Zick on 1 Apr 2007 19:40 On Sat, 31 Mar 2007 18:47:45 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >Lester Zick wrote: >> On Fri, 30 Mar 2007 12:13:57 -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: >>>> >>>>>>> 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~~ >>> Not unless the vast majority are infinitesimal. >> >> No that isn't what I'm talking about. You seem to assume consecutive >> segments would have to be colinear and lie along a straight line. I've >> already tried to explain why this isn't so. They could all connect in >> completely different directions even though mathematikers commonly >> assume they somehow for some reason would very plolitely line up in >> one direction alone. Line segments are only connected by points, Tony. >> And their direction is not determined by those points because there is >> no definable slope at point intersections. >> > > >I'm sorry Lester. Perhaps I misunderstood. When you used the word >"added", I assumed you meant addition. That assumes a linear >construction. But, perhaps, you meant some other form of addition. >Addition is linear, as commonly understood... As is commonly pretended, Tony, and not however as is mechanically required. Modern mathematikers merely wish to assume the guise of geometry without the necessity of geometry. They just assume whatever they want because they think if they dress it up with enough mathspeak no one will notice the implicit assumptions of truth they're too lazy or stupid to conceal much less avoid. >>> If there is a nonzero >>> lower bound on the interval lengths, an unlimited number concatenated >>> produces unlimited distance. >> >> And if segments were all of equal finite size we could make a finite >> plane hexagon out them which would be quite limited in distance. >> >> ~v~~ > >Not exactly, but that's a complicated topic.... Sure exactly, Tony. And where exactly do all these exhalted self righteous modern mathematikers get off assuming whatever they feel like without so much as being able to demonstrate the truth of what they assume or even to define or identify what they assume as true? It looks to me like modern mathematikers collectively suffer from HBSE or Holy Bovine Spongiform Encephalopathy or Mad Sacred Cow Disease. ~v~~
From: Lester Zick on 1 Apr 2007 19:41
On Sat, 31 Mar 2007 19:50:10 -0500, Tony Orlow <tony(a)lightlink.com> wrote: >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? Exactly what I said it was in E201: the demonstration of truth. That and nothing more or less. ~v~~ |