The Definition of Points
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From: Virgil on 31 Mar 2007 22:46 In article <460f1b3e(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > In article <460ef839(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> Virgil wrote: > >>> In article <460ee056(a)news2.lightlink.com>, > >>> Tony Orlow <tony(a)lightlink.com> wrote: > > > >>>> Please do expliculate what the contradiction is in an uncountable > >>>> sequence. What is true and false as a result of that concept? > >>> A mathematical sequence is a function with the naturals as domain. > >>> If TO wishes to refer to something which is not such a function, he > >>> should not refer to it as a sequence if he wishes to be understood in > >>> sci.math. > >>> > >>> > >> Pray tell, what term shall I use???? > > > > TO is so inventive in so many useless ways that I cannot believe that > > his imagination will fail him in such a trivially useful way. > >>>>> I know you are incapable of actually thinking about all the elements of > >>>>> N, > >>>>> but that is your problem. In any case, N is not an element of N. > >>>>> Citing Ross as support is practically an admission that you are wrong. > >>>>> > >>>>> Stephen > >>>>> > >>>> Sure, of course, agreeing with someone who disagrees with you makes me > >>>> wrong. I'll keep that in mind. Thanks.. > >>> > >>> It is not so much that Ross disagrees with one person, it is that he > >>> disagrees with everyone, frequently including himself. > >> Ross has a vision, even if not axiomatically expressed. In fact, he's > >> entirely honest about that, expounding an axiom free system. I like > >> Ross. So do you. Admit it. :) > >> > > > > > > Like Russell? > > > > What is there about him to like? > > You don't like Russell? I don't know him well enough to like or dislike. I dislike his anti-mathematical idiocies.
From: Tony Orlow on 31 Mar 2007 22:52 Virgil wrote: > In article <460f00a0(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> wrote: > >> Look back. The nth is equal to n. Inductive proof holds for equality in >> the infinite case > > Not in vN. I know that statement is not generally acceptable. I don't care. It's true. Infinite-case induction has not been disproved, despite Chas' very excellent effort, and some other that was lame. In other words, it's not contradictory. And inductive proofs do not work that way. One can prove by > induction that something is true for each natural, but that does not > create any infinite naturals for which it is true. You can agree with your buddies these are the rules of your club, and you can hang around the pickup drinking cans of beer and talking about trucks and ladies, but somes of us gots interstates to travel, and, you know, big cities to deliver to. Where an equality is proven true for all x>k, then any positive infinite value is greater than any finite k, and the proof holds. Simple clear logic. Refute, please.
From: Tony Orlow on 31 Mar 2007 22:54 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. Tony
From: cbrown on 31 Mar 2007 23:11 On Mar 31, 5:33 pm, 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 Funny; everyone else calls that "a transitive relation". Let S = {a,b,c,f,g, h} Impose the following ordering: b < a c < b c < a f < a g < f g < a if x <> h, then h < x. The equivalent Hasse diagram: a / \ b f / \ c g \ / \ / \ / h This ordering satisfies, for all x,y,z in S: if x<y and y<z, then x<z. This is not what most people mean when they say "a linear ordering". Instead, it's an example of what people usually call a partial order. See: http://en.wikipedia.org/wiki/Partial_order > Continuous means x<z -> Ey: x<y ^ y<z Is "<" a partial order? a pre-order? a total order? Unless you specify, I might say that in a triangle, the third vertex is "between" any two given distinct vertices. > Discrete means not continuous, that is, given x and z, y might not exist. So [0,1) u (1,2], with the usual ordering of the reals, is a "discrete" ordering? > For something to "occur", it must happen "at some time". Does "1 + 5" "occur", i.e., happen, "at some time" different than when "2 + 4" "occurs"? > 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. > Is "1+1" an "event" which "occurs" or "happens" at some "time"? When is that time? Has it already "occured"? > > 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. > > Oh gee, there has to be some word for it... > There almost certainly is; but as usual, it depends on what the /heck/ you're talking about. Perhaps the words "well-order" or "total order" actually already satisfy your requirements; or some particular proper subset of all non-isomorphic total orders satisfy your requirements. Or not. But how will you ever know if you refuse to /learn/ what these words refer to? Cheers - Chas
From: Tony Orlow on 31 Mar 2007 23:11
cbrown(a)cbrownsystems.com wrote: > On Mar 31, 3:30 pm, Tony Orlow <t...(a)lightlink.com> wrote: >> cbr...(a)cbrownsystems.com wrote: >>> On Mar 31, 5:30 am, Tony Orlow <t...(a)lightlink.com> wrote: >>>> Virgil wrote: >>>>> In standard mathematics, an infinite sequence is o more than a function >>>>> whose domain is the set of naturals, no two of which are more that >>>>> finitely different. The codmain of such a function need not have any >>>>> particular structure at all. >>>> That's a countably infinite sequence. Standard mathematics doesn't allow >>>> for uncountable sequences like the adics or T-riffics, because it's been >>>> politically agreed upon that we skirt that issue and leave it to the >>>> clerics. >>> That's false; >> Please elucidate on the untruth of the statement. It should be easy to >> disprove an untrue statement. >> > > I did in the continuation of that sentence; but I'll repeat myself. > > You claimed that mathematics doesn't "allow for uncountable > sequences" (for which you agree you've given no real useful > definition). But on the contrary, "people" (aka, mathematicicians) > have studied all sorts of ordered sets, finite, countable, and > uncountable; and functions from them (whether they use the term > "uncountable sequence" or not). > > So your claim that ordered sets which are not countable have not been > studied is false; and therefore your comments that the reasons /why/ > they have not been studied (political or religious) are non-sequiturs. > > The obvious question is why haven't /you/ studied them; instead of > making vague and uninformed statements about them (regardless of what > you choose to call these ordered sets). > The question is, "is there an acceptable term with which to refer to such uncountable linearly ordered sets?" >> people have examined all sorts of orderings, partial, >> >>> total, and other. The fact that you prefer to remain ignorant of this >>> does not mean the issue has been skirted by anyone other than >>> yourself. >> There have always been religious and political pressures on this area of >> exploration. >> > > How would you know what has "always" been the case in this area? Is > there an example please of a /religious/ or /political/ pressure that > you can cite? Besides the quite reasonable recommendation to educate > yourself regarding the subject matter, using the freely available > material relating to the subject, e.g., > > http://en.wikipedia.org/wiki/Order_theory > > ? > Google it up. >>>> However, where every element of a set has a well defined >>>> successor and predecessor, it's a sequence of some sort. >>> Let S = {0, a, 1, b, 2, c}. >>> Let succ() be defined on S as: >>> succ(0) = 1 >>> succ(1) = 2 >>> succ(2) = 0 >>> succ(a) = b >>> succ(b) = c >>> succ(c) = a >> Okay you have two sequences. >> > > Why two? Why not one, or three, or six? Your definition fails to say. > > Is S = {0,1}, succ(0)=1, succ(1)=0 a "sort of sequence"? It has a well- > defined successor and predeccessor for each element. How about S = > {0}, and succ(0) = 0? > >> >>> Every element of S has a well-defined successor and predecessor. What >>> "sort of sequence" have I defined? Or have you left out some parts of >>> the /explicit/ definition of whatever you were trying to say? >>> Cheers - Chas >> Yes, I left out some details. > > Given that you are claiming that your definition is somehow being > surpressed by religious or political forces, why not take the > opportunity to provide these details (in which we all know the devil > resides)? > > As it stands, I have no real idea what you're talking about; and quite > frankly, I doubt you yourself have a clear idea of what you are > talking about. > > Other people have thought through similar ideas and presented a > comprehensive structure for understanding. See, e.g., > > http://en.wikipedia.org/wiki/Order_theory > > in case you'd actually like to try to /learn/ something about the > subject. > > Cheers - Chas > Gee, thanx. |