The Definition of Points
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From: Tony Orlow on 31 Mar 2007 19:36 Lester Zick wrote: > On Fri, 30 Mar 2007 12:10:12 -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: >>> >>>>>> 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
From: Tony Orlow on 31 Mar 2007 19:40 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. 01oo
From: Mike Kelly on 31 Mar 2007 19:41 On 31 Mar, 16:46, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > On 31 Mar, 13:41, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > >>> On 30 Mar, 18:25, 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: > >>>>>>>> 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 > >>> Under what definition of sequence? > >>> -- > >>> mike. > >> A set where each element has a well defined unique successor within the > >> set. > > > So any set is a sequence? For any set, take the successor of each > > element as itself. > > There is no successor in a pure set. That only occurs in a discrete > linear order. What does it mean for an ordering to be "discrete" or "linear"? What does it mean for something to "occur in" an ordering? > >> Good enough? > > > You tell me. Did you mean to say "a sequence is a set"? If so, good > > enough. > > > -- > > mike. > > Not exactly, and no, what I said is not good enough. > > A set with an order where each element has a unique successor is a > forward-infinite sequence. Each can have a unique predecessor, and then > it's backward-infinite. And if every element has both a unique successor > and predecessor, then it's bi-infinite, like the integers, or within the > H-riffics, the reals. One can further impose that x<y ->~y<x, to > eliminate circularity. > > Good enough? Probably not yet. So when you say "sequence" you're refering to a set and an ordering on that set? There are some conditions on the properties of the ordering. You're not, as yet, able to coherently explain what those conditions are. So when you say "sequence" you're using an undefined term. As such, it's rather hard to your evaluate claims such as "There are actually infinite sequences". I have literally no idea what you are even trying to say. -- mike.
From: Tony Orlow on 31 Mar 2007 19:44 Virgil wrote: > In article <460e8251(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Mike Kelly wrote: >>> On 31 Mar, 13:41, Tony Orlow <t...(a)lightlink.com> wrote: >>>> Mike Kelly wrote: >>>>> On 30 Mar, 18:25, 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: >>>>>>>>>> >>>>>>>>>> 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 >>>>> Under what definition of sequence? >>>>> -- >>>>> mike. >>>> A set where each element has a well defined unique successor within the >>>> set. >>> So any set is a sequence? For any set, take the successor of each >>> element as itself. >> There is no successor in a pure set. That only occurs in a discrete >> linear order. >> >>>> Good enough? >>> You tell me. Did you mean to say "a sequence is a set"? If so, good >>> enough. >>> >>> -- >>> mike. >>> >> Not exactly, and no, what I said is not good enough. >> >> A set with an order where each element has a unique successor is a >> forward-infinite sequence. Each can have a unique predecessor, and then >> it's backward-infinite. And if every element has both a unique successor >> and predecessor, then it's bi-infinite, like the integers, or within the >> H-riffics, the reals. One can further impose that x<y ->~y<x, to >> eliminate circularity. >> >> Good enough? Probably not yet. > > > You are right, not yet. > I'm right!!! :D heh. > Every "sequence" must be a totally ordered set which is order isomorphic > either to the ordered set of naturals, if it has a first element, or to > the ordered set of integers, if it does not have a first element. Okay. How would you boil down that statement, and in which cases would you say it applies? > > Note that since the obvious mapping between the natural numbers and the > negative integers is an order isomorphism with order reversal, one need > not include that third case separately. Sure. One not even really distinguish between '<' and '>', but it helps. :) Tony
From: Tony Orlow on 31 Mar 2007 19:47
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... >> 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.... 01oo |