The Definition of Points
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From: Lester Zick on 13 Apr 2007 18:39 On Fri, 13 Apr 2007 16:52:21 +0000 (UTC), stephen(a)nomail.com wrote: >In sci.math Tony Orlow <tony(a)lightlink.com> wrote: >> Lester Zick wrote: >>> On Mon, 2 Apr 2007 16:12:46 +0000 (UTC), stephen(a)nomail.com wrote: >>> >>>>> It is not true that the set of consecutive naturals starting at 1 with >>>>> cardinality x has largest element x. A set of consecutive naturals >>>>> starting at 1 need not have a largest element at all. >>>> To be fair to Tony, he said "size", not "cardinality". If Tony wishes to define >>>> "size" such that set of consecutive naturals starting at 1 with size x has a >>>> largest element x, he can, but an immediate consequence of that definition >>>> is that N does not have a size. >>> >>> Is that true? >>> >>> ~v~~ > >> Yes, Lester, Stephen is exactly right. I am very happy to see this >> response. It follows from the assumptions. Axioms have merit, but >> deserve periodic review. > >> 01oo > >Everything follows from the assumptions and definitions. And since definitions are considered neither true nor false everything follows from raw assumptions which are considered neither true nor false. ~v~~
From: Lester Zick on 13 Apr 2007 18:42 On Fri, 13 Apr 2007 10:06:51 -0400, Bob Kolker <nowhere(a)nowhere.com> wrote: >Mike Kelly wrote: >> >> a) A consecutive set of naturals starting with 1 with size X can not >> have any maximum other than X. >> b) A consecutive set of naturals starting with 1 with size X has >> maximum X. > >Whoa! If a -consectutive-(!!!) set of naturals with 1 as its least >element has cardinality X (an integer), then its last element must be X. > >A simple induction argument will show this to be the case. > >Can you show a counter example? You mean a true counter example, Bob, or just a counter example whose truth is assumed true because you're too lazy or stupid to consider the truth of what you say but not too lazy or stupid to say it anyway. ~v~~
From: Lester Zick on 13 Apr 2007 18:43 On Fri, 13 Apr 2007 10:08:14 -0400, Bob Kolker <nowhere(a)nowhere.com> wrote: >Bob Kolker wrote: > >> Mike Kelly wrote: >> >>> >>> a) A consecutive set of naturals starting with 1 with size X can not >>> have any maximum other than X. >>> b) A consecutive set of naturals starting with 1 with size X has >>> maximum X. >> >> >> Whoa! If a -consectutive-(!!!) set of naturals with 1 as its least >> element has cardinality X (an integer), then its last element must be X. >> >> A simple induction argument will show this to be the case. >> >> Can you show a counter example? > >I should have said, for I assumed it, that X is finte. Sorry about that. Or you regret your assumptions of truth in this particular instance but not in general? ~v~~
From: Lester Zick on 13 Apr 2007 18:45 On 13 Apr 2007 07:16:42 -0700, "Mike Kelly" <mikekellyuk(a)googlemail.com> wrote: >On 13 Apr, 15:08, Bob Kolker <nowh...(a)nowhere.com> wrote: >> Bob Kolker wrote: >> > Mike Kelly wrote: >> >> >> a) A consecutive set of naturals starting with 1 with size X can not >> >> have any maximum other than X. >> >> b) A consecutive set of naturals starting with 1 with size X has >> >> maximum X. >> >> > Whoa! If a -consectutive-(!!!) set of naturals with 1 as its least >> > element has cardinality X (an integer), then its last element must be X. >> >> > A simple induction argument will show this to be the case. >> >> > Can you show a counter example? >> >> I should have said, for I assumed it, that X is finte. Sorry about that. >> >> Bob Kolker > >Ah. Fair enough. Thanks, sport, that's swell. From now on we'll just consider where assumptions of truth are concerned Bob is sorry, very sorry. ~v~~
From: cbrown on 13 Apr 2007 18:45
On Apr 13, 12:47 pm, Tony Orlow <t...(a)lightlink.com> wrote: > cbr...(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. Perhaps that works for you; although your "tree" has a maximal member (every subset A of S satisfies A < S) and and a minimal member (the empty sets satisfies {} < A for all subsets of A). Most times, when people talk about trees, they mean a set having either a maximal element or a minimal element; but not both. To me, that "looks" more like a fish net or a giant diamond than a tree. At any rate, my question was: what is recursive about that definition? > > > > > 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 what way does the assertion: For all positive real numbers r, there exists a natural number n such that n*r > 1? apply to definition above, which is about sets of naturals, and never mentions real numbers at all? > > > >> 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. On R, and yet in N? I don't understand what you mean. > Or, whatever. Indeed. > > > > >>>>>> 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. > See the example I posted below regarding subsets of {a,b,c}. > >> 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}. Since "<" indicates subset inclusion, it is the case that not b < c. > Where b=c, {a,b}={a,c}. Since b and c are distinct elements of the set {a,b,c}, not b=c. > If not b<c and not c<b, > then b=c. > Since "<" indicates subset inclusion, it is the case that not c < b. As noted earlier, also not b < c. And yet not b=c. I should note that "<" is therefore a partial order, and not a total order. > >> 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. > Of course. But when you stated as you did earlier: >>>>>> Order is defined by x<y ^ y<z -> x<z. then you were stating a falsehood; and you knew it was a falsehood. Liar! Pants on fire! > > > >>> 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. > Come, Mr. Tony mon, tally me triangles. Daylight come, and me wan go home. Isoceles, acute, obtuse, bunch! Daylight come, and me wan go home. > >> 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. > I'm confused. Do you mean that the elements of the set of all (sets of subsets of N) are ordered with respect to each other? If so, how? Or do you mean that any given set of (subsets of N) is always, itself, an ordered set? If so, how? Or do you mean that any subset of the naturals is an ordered set? If so, do you mean the usual order, or some other ordering? In each of these cases, there of course are /many/ possible such orders, not all of them isomorphic. Which one are you speaking of? Cheers - Chas |