The Definition of Points
in [Math]
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From: cbrown on 1 Apr 2007 03:07 On Mar 31, 10:22 pm, step...(a)nomail.com wrote: > In sci.math Tony Orlow <t...(a)lightlink.com> wrote: > > > step...(a)nomail.com wrote: > > >> None of the options mention "size" Tony. What does "size" have > >> to do with a, b or c? > > > Ugh. Me already tell you, nth one is n, then there are n of them. So > > easy, even a caveman can do it. Size is difference between. > > Brilliant Tony. Act like an idiot when backed into a corner. So easy, even a caveman can do it. Cheers - Chas
From: Mike Kelly on 1 Apr 2007 06:40 On 1 Apr, 01:33, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > 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? > > Linear means x<y ^ y<z ->x<z Linear means transitive? Very, very strange use of language. Not all transitive orderings are linear (by the normal definitions of those words). So why use that word? This is just going to confuse people. > Continuous means x<z -> Ey: x<y ^ y<z So the ordering on {a, b, c} of a < b, a < c, b < a, b < c, c < a, c < b is "countinuous"? Or were there some more conditions you didn't mention.. > Discrete means not continuous, that is, given x and z, y might not exist. So the set { [0,1] U [2,3] } with the usual ordering on the reals is a discrete ordering? > For something to "occur", it must happen "at some time". In a sequence, > this is defined as after some set of events and before some other > mutually exclusive set, in whatever order is under consideration. This paragraph is completely incoherent. You've used many undefined terms in an attempt to define something. I get the feeling you are *never* going to be able to explain your ideas. There has been a *lot* of work done on desribing various types of orderings. Maybe if you understood the common language used to describe orderings you'd have better luck getting people to understand your ideas. You're violently opposed to the idea of actually learning any math though.. > >>>> 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. > > Explain away. What? I don't know what your ideas are. Nobody but you does. It's not even clear that *you* know exactly what you're trying to say, given that you're incapable of explaining your ideas to anyone else. -- mike.
From: Mike Kelly on 1 Apr 2007 06:46 On 1 Apr, 01:45, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > On 31 Mar, 16:47, Tony Orlow <t...(a)lightlink.com> wrote: > >> Mike Kelly wrote: > >>> On 31 Mar, 13:48, Tony Orlow <t...(a)lightlink.com> wrote: > >>>> step...(a)nomail.com wrote: > >>>>> In sci.math Virgil <vir...(a)comcast.net> wrote: > >>>>>> In article <460d4...(a)news2.lightlink.com>, > >>>>>> Tony Orlow <t...(a)lightlink.com> wrote: > >>>>>>> An actually infinite sequence is one where there exist two elements, one > >>>>>>> of which is an infinite number of elements beyond the other. > >>>>>> Not in any standard mathematics. > >>>>> It is not even true in Tony's mathematics, at least it was not true > >>>>> the last time he brought it up. According to this > >>>>> definition {1, 2, 3, ... } is not actually infinite, but > >>>>> {1, 2, 3, ..., w} is actually infinite. However, the last time this > >>>>> was pointed out, Tony claimed that {1, 2, 3, ..., w} was not > >>>>> actually infinite. > >>>>> Stephen > >>>> No, adding one extra element to a countable set doesn't make it > >>>> uncountable. If all other elements in the sequence are a finite number > >>>> of steps from the start, and w occurs directly after those, then it is > >>>> one step beyond some step which is finite, and so is at a finite step. > >>> So (countable) sequences have a last element? What's the last finite > >>> natural number? > >>> -- > >>> mike. > >> As I said to Brian, it's provably the size of the set of finite natural > >> numbers greater than or equal to 1. > > > Provable how? > > Look back. The nth is equal to n. The nth what is equal to n? Do you really think you're explaining anything here? > Inductive proof holds for equality in the infinite case Oh, another meaningless mantra. > >> No, there is no last finite natural, > > > You keep changing your position on this. > > Nope, I don't. > > >> and no, there is no "size" for N. Aleph_0 is a phantom. > > > When we say that a set has cardinality Aleph_0 we are saying it is > > bijectible with N. Are you saying it's impossible for a set to be > > bijectible with N? Or are you saying N does not exist as a set? > > Something else? > > I have been saying that bijection alone is not sufficient for measuring > infinite sets relative to each other. Who said that bijection was the only way to compare infinite sets? Bijectibiltiy is one way to compare infinite sets. Cardinality is used to denote bijectibility. This is sometimes useful. There are also, of course, many other ways of comparing infinite sets. Why do you have such a huge problem understanding this? > > I find it very hard to understand what you are even trying to say when > > you say "Aleph_0 is a phantom". It seems a bit like Ross' meaningless > > mantras he likes to sprinkle his posts with. > > Yes, NeN, as Ross says. I understand what he means, but you don't. Where > taking away makes something less, aleph_0-1<aleph_0, and there is no > smallest infinity, except in the nonlogical imagination. Chase that tail! But subtraction is not defined on infinite cardinals. Are you *really* incapable of understanding this? -- mike.
From: Mike Kelly on 1 Apr 2007 06:49 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. Do you not see *any* problem with this picture? -- mike.
From: Mike Kelly on 1 Apr 2007 06:58
On 1 Apr, 02:29, Tony Orlow <t...(a)lightlink.com> wrote: > Mike Kelly wrote: > > On 1 Apr, 00:58, Tony Orlow <t...(a)lightlink.com> wrote: > >> step...(a)nomail.com wrote: > >>> In sci.math Brian Chandler <imaginator...(a)despammed.com> wrote: > >>>> step...(a)nomail.com wrote: > >>>>> In sci.math Tony Orlow <t...(a)lightlink.com> wrote: > >>>>>> If all other elements in the sequence are a finite number > >>>>>> of steps from the start, and w occurs directly after those, then it is > >>>>>> one step beyond some step which is finite, and so is at a finite step. > >>>>> So you think there are only a finite number of elements between 1 and > >>>>> w? What is that finite number? 100? 100000? 100000000000000000? > >>>>> 98042934810235712394872394712349123749123471923479? Which one? > >>>> None of the ones you've mentioned. Although it is, of course, a > >>>> perfectly ordinary natural number, in that one can add 1 to it, or > >>>> divide it by 2, its value is Elusive. Only Tony could actually write > >>>> it down. > >>> These Elusive numbers have amazing properties. According to > >>> Tony, there are only a finite number of finite naturals. > >>> There exists some finite natural Q such that the set > >>> { 1,2,3,4,.... Q} > >>> is the set of all finite natural numbers. But what of Q+1? > >>> Well we have a couple of options: > >>> a) Q+1 does not exist > >>> b) Q+1 is not a finite natural number > >>> c) { 1,2,3,4, ... Q} is not the set of all finite natural numbers > >>> Tony rejects all these options, and apparently has some fourth > >>> Elusive option. > >>> Stephen > >> Oy. The "elusive" option is that there is no acceptable "size" for N. > >> That was really hard to figure out after all this time... > > > Lucky, then, that set theory don't refer to the "size" of sets but > > rather to their "cardinality". > > > You still haven't figured that out after all this time. It's a very > > strange mental block to have. > > > -- > > mike. > > hi. mike. good thing, that. lucky, in fact. > > block. mental. to have, yes. I, nope don't. > > The question is to what extent set theoretical conclusions can be > trusted, when the notion of order is introduced through such a suspect > mechanism as the von Neumann ordinals. Just give '<' equal status with > 'e' as a fundamental operator, and get real already! Sheesh! Heh! ;) This appears to have nothing to do with my post. I guess this kind of babble means that you're literally incapable of recognising that your arguments are completely bunk. Your mind will not allow you to realise that you've been wasting a lot of time spewing meaningless garbage. If it did, you might have to admit you were wrong. Maybe you'd even learn something from your errors. And Tony Orlow doesn't want that. Tony Orlow doesn't want to learn how to communicate mathematically. Tony Orlow doesn't want to be able to explain his ideas to other people. Tony Orlow just wants to pontificate endlessly on usenet and play at being a mathematician. You complain about set theory giving a wrong size to N when there is no "acceptable" size to give it. It's pointed out that set theory *doesn't* give a size to N. You utterly refuse to acknowledge your error and start babbling incoherently about only tangentially related stuff. Stop being such an intellectual coward. -- mike. |