The Definition of Points
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From: Tony Orlow on 13 Apr 2007 15:47 cbrown(a)cbrownsystems.com wrote: > On Apr 13, 10:13 am, Tony Orlow <t...(a)lightlink.com> wrote: >> Virgil wrote: >>> In article <461e8...(a)news2.lightlink.com>, >>> Tony Orlow <t...(a)lightlink.com> wrote: >>>> MoeBlee wrote: >>>>> On Mar 31, 5:18 am, Tony Orlow <t...(a)lightlink.com> wrote: >>>>>> Virgil wrote: >>>>>>> In article <1175275431.897052.225...(a)y80g2000hsf.googlegroups.com>, >>>>>>> "MoeBlee" <jazzm...(a)hotmail.com> wrote: >>>>>>>> On Mar 30, 9:39 am, Tony Orlow <t...(a)lightlink.com> wrote: >>>>>>>>> They >>>>>>>>> introduce the von Neumann ordinals defined solely by set inclusion, >>>>>>>> By membership, not inclusion. >>>>>>> By both. Every vN natural is simultaneously a member of and subset of >>>>>>> all succeeding naturals. >>>>>> Yes, you're both right. Each of the vN ordinals includes as a subset >>>>>> each previous ordinal, and is a member of the set of all ordinals. >>>>> In the more usual theories, there is no set of all ordinals. >>>> Right. Ordinals are...ordered. Sets aren't. >>> Ordinals have a unique ordering by reason of their being ordinals. >>> Sets in general have all sorts of orderings, but none which is as >>> inherent in their being sets as the ordinal order is in sets being >>> ordinals. >> Once they are ordered in whatever manner, they become sequences, trees, >> or other structures, and it is only with such a recursive definition >> that such an infinite structure can be created. > > Why is a recursive definition required? Given any set S, the set of > all subsets of S can be (partially) ordered as follows: for subsets A, > B of S, define A <= B iff every member of A is a member of B. What is > recursive about that definition? That structure doesn't look like a tree to you, which each node having as children the elements of its power set? That's the picture that appears to me. > > Alternatively, S be the set of all subsets of the naturals (note that > S is not countable). If A, B are in S, define A <= B if there is a > natural number m in B such that m is not in A, and for all n < m, n in > A and n in B. What is recursive about that definition? > The Archimedean principle, as far as I can tell. >> In that sense, there is >> no pure infinite set without some defining structure, so whatever >> conclusions one thinks they have come to regarding infinite sets without >> structure have no basis for comparison. Powerset(S) is 2^|S| sets, no >> matter the size of S. That is a specific case of N=S^L, which applies to >> symbolic strings and alphabets, as well as power sets where elements can >> have S different levels of truth, not just 2. There are 3^log2(n) as >> many ternary strings of length n as there are binary strings of length >> n, be n finite or infinite. But, that involves a discussion of structure. >> >>>>>> Anyway, my point is that the recursive nature of the definition of the >>>>>> "set" >>>>> What recursive definition of what set? >>>> Oh c'mon! N. ala Peano? (sigh) What kind of question is that? >>> Does TO seem to thing that N is the only set defineable recursively or >>> that "successor" is the only recursively defineable operations on sets? >> Does Virgil forget what he cuts from the post? What do you think we were >> discussing? I thought it was N specifically. >> > > I thought it had something to do with the real line, and orderings. > Points and lines, anyway. And points on R in N. Or, whatever. >>>>>> Order is defined by x<y ^ y<z -> x<z. >>>>> Transitivity is one of the properties of most of the orderings we're >>>>> talking about. But transitivity is not the only property that defines >>>>> such things as 'partial order', 'linear order', 'well order'. >>>> It defines order, in general. >>> Only to TO. For everyone else, other properties are required. >>> For example, in addition to transitivity, >>> ((x>y) and (y>x)) -> x = y >>> is a necessary property /every/ ordering. >> Um, that one is blatantly self-contradictory. x>y -> not y>x, always. > > I don't see how this follows only from your assertion "x < y and y < z > -> x < z". You stated: > >>>>>> Order is defined by x<y ^ y<z -> x<z. > > Or do you mean that there is /more/ to the definition of an order "<" > than "x < y and y < z -> x < z"? If so, that was exactly Virgil's > point. > Actually I corrected this response to Virgil. I misspoke a little, but he's still wrong. :) >> suppose you meant: >> ((x>=y) and (y>=x)) -> x = y >> or: >> (~(x>y) and ~(y>x)) -> x = y >> > > These two statements are not equivalent. In some situations, the first > can hold, while the second does not. > Please do elaborate. >> Yes, if neither x<y or y<x is true, that is, if no order can be >> determined, then x=y for the purposes of that order. > > Let's order the subsets of {a,b,c} by inclusion. Then: {a,b} < > {a,b,c}. {a,c} < {a,b,c}. {a} < {a,b}. {a} < {a,c}. But not ({a,b} < > {a,c}); and not ({a,c} < {a,b}). Does that mean that {a,b} = {a,c} > "for the purposes of that order"? > Where b<c, {a,b}<{a,c}. Where b=c, {a,b}={a,c}. If not b<c and not c<b, then b=c. >> That defines '=' in >> terms of '<'. It defines a point on the line, more or less, to get back >> to the original question. >> > > Could you state what the definition of "<= totally orders the set S" > is again? There are three simply stated properties, IIRC. > Lookitup. Transitivity does not define "total order", but is that start of order. >> >>> Also there are lots of transitive relations which are not orderings, at >>> least as usually understood. E.g., universal relations, which hold true >>> for all x and y in the relevant set. >>> So that TO's notion of an ordering does not necessarily order anything. >> You're missing the point. All I said was that one starts with inequality >> defining the line itself. > > Is every set a line? Is the set of all triangles in the Euclidean > plane a line? > That's a bunch of lines. >> Then one defines equality. Defining equality >> where there is no relative order doesn't make sense. > > So, it makes no sense to say that the set of all finite subsets of the > naturals having a prime number of elements is equal to itself? > > Cheers - Chas > Again, there is order to the elements themselves. Tally Ho! Tony
From: Tony Orlow on 13 Apr 2007 15:51 MoeBlee wrote: > On Apr 13, 11:25 am, Tony Orlow <t...(a)lightlink.com> wrote: > >> I've discussed that with you and others. It doesn't cover the cases I am >> talking about. The naturals have a "measure" of 0, no? So, measure >> theory doesn't address the relationship between, say, the naturals and >> the evens or primes. It's not as general as it should be. So, what do >> you want me to say? > > Nothing, really, until you learn the mathematics you're pretending to > know about. > I didn't bring up "measure theory". >>> It's very easily provable that if "size" means "cardinality" that N >>> has "size" aleph_0 but no largest element. You aren't actually >>> questioning this, are you? >> No, have your system of cardinality, but don't pretend it can tell >> things it can't. Cardinality is size for finite sets. For infinite sets >> it's only some broad classification. > > Nothing to which you responded "pretends" that cardinality "can tell > things it can't". What SPECIFIC theorem of set theory do you feel is a > pretense of "telling things that it can't"? > AC >>> It has CARDINALITY aleph_0. If you take "size" to mean cardinality >>> then aleph_0 is the "size" of the set of naturals. But it simply isn't >>> true that "a set of naturals with 'size' y has maximum element y" if >>> "size" means cardinality. >> I don't believe cardinality equates to "size" in the infinite case. > > Wow, that is about as BLATANTLY missing the point of what you are in > immediate response to as I can imagine even you pulling off. > > MoeBlee > What point did I miss? I don't take transfinite cardinality to mean "size". You say I missed the point. You didn't intersect the line. ToeKnee
From: Tony Orlow on 13 Apr 2007 16:07 cbrown(a)cbrownsystems.com wrote: > On Apr 13, 10:38 am, Tony Orlow <t...(a)lightlink.com> wrote: >> MoeBlee wrote: >>> On Apr 12, 2:36 pm, "MoeBlee" <jazzm...(a)hotmail.com> wrote: >>>> Zermelo's motivation was to prove that every set is well ordered. >>> Since that phrasing might be misunderstood, I should say that I mean: >>> Zermelo's motivation was to prove that for every set, there exists a >>> well ordering on it. >>> MoeBlee >> I am not sure how the Axiom of Choice demonstrates that. >> > > AoC says (roughly) that we have a way to unambiguously choose an > element from any set: i.e., for any set S, there exists a function f > such that for any subset A of S, f(A) is a member of A (and of course > therefore, a member of S). > > "S is well ordered by <=" states that "<=" is a total order on S, and > for any subset A of S, there is a specific least element of A. > > So if "<=" well-orders S, then "the least element of A" is a function > like the one that AoC tells us exists: it chooses a specific element > from any subset A of S. So "every set can be well-ordered" implies > "for every set S, there exists a choice function for S". > > The converse ("for every set S, there exists a choice function for S" > implies "every set can be well-ordered") is a bit more complicated. A > proof online is at > > http://planetmath.org/?op=getobj&from=objects&id=3359 > > but you'll have to accept certain facts about ordinals (e.g., that > they exist, that any set of them is well-ordered by inclusion, that > transfinite induction over a well-ordered set is possible, etc.) which > you have previously balked at. > > The basic idea is to let f(S) be the smallest element of S, then f(S\ > {f(S)}) be the next smallest element, and so on. Of course, > transfinite induction is required for sets which are not countable. > > Cheers - Chas > > Thanks for the reply, Chas. You should see mine to MoeBlee. I don't have a lot of time now, because I have to tool on outta here, but I took a quick look at your link, and found it rests on transfinite induction. I don't recall exactly how that works right now, but I definitely recall feeling it was very kludgy. So, I'm not sure I have to accept that proof. I'll try to take a look at it in more dtail when I get a chance. Just to recap what I said to MoeBlee, I can't help feeling that, since a countable union of countable sets is countable, either there will be an uncountable number of partitions of an uncountable set, in which case there will exist infinite descending chains before the first "limit ordinal" in the well order (besides the first element of the first partition), or there will exist at least one uncountable partition which will produce infinite descending chains within itself, after all countable partitions have been exhausted. DC or ACC seem acceptable, but not AC, to me. Tony
From: Tony Orlow on 13 Apr 2007 16:11 Lester Zick wrote: > On Fri, 13 Apr 2007 13:40:24 -0400, Tony Orlow <tony(a)lightlink.com> > wrote: > >> Lester Zick wrote: >>> On Thu, 12 Apr 2007 14:12:56 -0400, Tony Orlow <tony(a)lightlink.com> >>> wrote: >>> >>>> Lester Zick wrote: >>>>> On Sat, 31 Mar 2007 18:05:25 -0500, Tony Orlow <tony(a)lightlink.com> >>>>> wrote: >>>>> >>>>>> Lester Zick wrote: >>>>>>> On Fri, 30 Mar 2007 12:06:42 -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: >>>>>>>>> >>>>>>>>>>>> You might be surprised at how it relates to science. Where does mass >>>>>>>>>>>> come from, anyway? >>>>>>>>>>> Not from number rings and real number lines that's for sure. >>>>>>>>>>> >>>>>>>>>> Are you sure? >>>>>>>>> Yes. >>>>>>>>> >>>>>>>>>> What thoughts have you given to cyclical processes? >>>>>>>>> Plenty. Everything in physical nature represents cyclical processes. >>>>>>>>> So what? What difference does that make? We can describe cyclical >>>>>>>>> processes quite adequately without assuming there is a real number >>>>>>>>> line or number rings. In fact we can describe cyclical processes even >>>>>>>>> if there is no real number line and number ring. They're irrelevant. >>>>>>>>> >>>>>>>>> ~v~~ >>>>>>>> Oh. What causes them? >>>>>>> Constant linear velocity in combination with transverse acceleration. >>>>>>> >>>>>>> ~v~~ >>>>>> Constant transverse acceleration? >>>>> What did I say, Tony? Constant linear velocity in combination with >>>>> transverse acceleration? Or constant transverse acceleration? I mean >>>>> my reply is right there above yours. >>>>> >>>>> ~v~~ >>>> If the transverse acceleration varies, then you do not have a circle. >>> Of course not. You do however have a curve. >>> >>> ~v~~ >> I thought you considered the transverse acceleration to vary >> infinitesimally, but that was a while back... > > Still do, Tony. How does that affect whether you have a curve or not? > Transverse a produces finite transverse v which produces infinitesimal > dr which "curves" the constant linear v infinitesimally. > > ~v~~ Varying is the opposite of being constant. Checkiddout! http://dictionary.reference.com/browse/constant 01oo
From: Tony Orlow on 13 Apr 2007 16:19
Lester Zick wrote: > On 13 Apr 2007 11:24:48 -0700, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: > >> On Apr 13, 10:56 am, Tony Orlow <t...(a)lightlink.com> wrote: >>> Well, of course, Moe's technically right, though I originally asked >>> Lester to define his meaning in relation to his grammar. Technically, >>> grammar just defines which statements are valid, to which specific >>> meanings are like parameters plugged in for the interpretation. >> That is completely wrong. You have it completely backwards. What you >> just mentioned is part of semantics not grammar. Grammar is syntax - >> the rules for formation of certain kinds of strings of symbols, >> formulas, sentences, and other matters related purely to the >> "manipulation" of sequences of symbols and sequences of formulas, and >> of such objects. On the other hand, semantics is about the >> interpretations, the denotations, the meanings of the symbols, strings >> of symbols, formulas, sentences, and sets of sentences. Mathematical >> logic includes the study of these two things - syntax and semantics - >> both separately and in relation to each other. >> >>> I asked >>> the question originally using truth tables to avoid all that, so that we >>> can directly equate Lester's grammar with the common grammar, on that >>> level, and derive whether "not a not b" and "not a or not b" were the >>> same thing. They seem to be. >> Truth tables are basically a semantical matter. Inspection of a truth >> table reveals the truth or falsehood of a sentential formula per each >> of the assigments of denotations of 'true' or 'false' to the sentence >> letters in the formula. > > If any and all these things are not demonstrably true and merely > represent so many assumptions of truth why would anyone care what you > think about what they are or aren't? I mean it really isn't as if > truth is on your side to the exclusion of what others claim, Moe(x). > > ~v~~ Define "assumption". Do you "believe" that truth exists? Is there a set of statements S such that forall seS s=true? Is there such a thing as truth, or falsity? Does logic "exist". There exists a set of assumptions, A, which are true. True? 01oo |