An uncountable countable set
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From: imaginatorium on 31 Oct 2006 00:23 Tony Orlow wrote: > stephen(a)nomail.com wrote: > > Tony Orlow <tony(a)lightlink.com> wrote: > >> stephen(a)nomail.com wrote: > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>> stephen(a)nomail.com wrote: > >>>>> Tony Orlow <tony(a)lightlink.com> wrote: > >>>>>> stephen(a)nomail.com wrote: > >>>>> <snip> > > If somebody presents another definition of set size, I will > > consider it. You have not presented such a definition. > > I have presented an approach that works for the majority of infinite > bijections, and explained some of the exceptions. IFR works for all > numeric sets mapped from a common set. N=S^L works for all languages, > including those that express the first set. Both work on a parameteric > basis, using infinite case induction to finely order the values of > formulas for a specific infinite n. Rare exceptions include the set 1/n > for neN, whose inverse is itself, which IFR ends up saying has size 1, > but that's because the natural indexes and fractional mapped reals only > share one point in their range, 1. So, I think Bigulosity is worth > considering. Assuming (charitably) that you intend to claim this is a "definition of set size" (and that the mention of "bijections" in the first sentence is some sort of brainfart), in what sense do you claim that this is applicable to the _majority_ of anything? I think the most reasonable mathematical interpretation would be that your ideas apply to "almost nothing" (in a precise sense, that's beyond you). I've given this - extremely elementary - example before, but which of your ideas would you apply to considering the "size" of the set of topologically distinct polyhedra? Would IFR "work"? Would N=S^L "work"? Or would you need to have another idea? > It subsumes Ross's EF, and asymptotic set density, and > works perfectly for finite sets, so there should be no issue trying to > generalize it, as a sound foundation. Most set definitions which aren't > handled by Bigulosity are poorly formed, or can be restated in ways that > make them easily handled. Gosh. Grandiose claims. Well, you'd better do the one above pretty quick then. Brian Chandler http://imaginatorium.org
From: David Marcus on 31 Oct 2006 00:51 Randy Poe wrote: > Lester Zick wrote: > > It doesn't? My mistake. So there's "no least real" > > There's no least real. > > > but a "least integer real"? > > There's no least integer. > > There's a least POSITIVE integer. Is there some reason you > keep ignoring the critical word POSITIVE? Conjecture: People not trained in mathematics don't realize that each and every word is important. -- David Marcus
From: Virgil on 31 Oct 2006 00:52 In article <1162268163.368326.64650(a)m73g2000cwd.googlegroups.com>, imaginatorium(a)despammed.com wrote: > Virgil wrote: > > In article <45462ba0(a)news2.lightlink.com>, > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > stephen(a)nomail.com wrote: > > > > Tony Orlow <tony(a)lightlink.com> wrote: > > > >> stephen(a)nomail.com wrote: > > > >>> Tony Orlow <tony(a)lightlink.com> wrote: > > > >>>> stephen(a)nomail.com wrote: > > <snipola> > > > > >> OH. So, sets don't have sizes which are numbers, at least at particular > > > >> moments. I see.... > > > > > > > > If that is what you meant, then you should have said that. > > > > And technically speaking, sets do not have sizes which are numbers, > > > > unless by "size" you mean cardinality, and by "number" you include > > > > transfinite cardinals. > > > > > > So, cardinality is the only definition of set size which you will > > > consider.....your loss. > > > > It is the only definition of set size that is known to produce a valid > > partial ordering on sets. > > Huh? I thought cardinality produced a valid *total* ordering on sets. The cardinalities are totally ordered, but the sets are not. A total order on sets would require that when neither of two sets was "larger than" the other that they must be the same set, not merely the same size. > As I have pointed out when all this started, of course everyone knows > that the subset relation produces a perfectly valid *partial* ordering > on sets. > > Brian Chandler > http://imaginatorium.org
From: Ross A. Finlayson on 31 Oct 2006 01:04 Virgil wrote: > ... > To suppose one to be talking over TO's head is hardly self flattery, > except possibly for Ross or Zick or their ilk. Say what? I'm trying to figure out dividing R into (-oo, 0] and [0, oo), to actually have the mereological boundary at zero, not to the left or right, for symmetry about the origin. This is "sci.math", not "Tony and Virgil rip on each other." If you can't sustain a reasonable argument then you're not. With Tony's semi-glossolalia on the one side and Virgil's vicious character stabs bait-switched with feigned academic innocence, so he feels OK, on the other, I prefer Tony, because I'm not here for the mainstream except for the end, of the Hilbert program. Tony, you make some sense, please endeavor to do so in general. Have a point, remain honest. Mathematics actually demands conformance. Orthodoxy I can get out of a book, and meanness I don't really stand in person, toad. Stuff it, jive turkey. That witch's brew talk is a weak Halloween reference. Here's something that's frightening, to some: Gorin and Kukushkin divide by zero and evaluate Gamma of negative integers to integrate the Devil's "Scare"-case. P(P(V)): set? Infinity: less than nothing. Non-archimedean? Ross
From: imaginatorium on 31 Oct 2006 01:30
Virgil wrote: > In article <1162268163.368326.64650(a)m73g2000cwd.googlegroups.com>, > imaginatorium(a)despammed.com wrote: > > > Virgil wrote: > > > In article <45462ba0(a)news2.lightlink.com>, > > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > > > > stephen(a)nomail.com wrote: > > > > > Tony Orlow <tony(a)lightlink.com> wrote: > > > > >> stephen(a)nomail.com wrote: > > > > >>> Tony Orlow <tony(a)lightlink.com> wrote: > > > > >>>> stephen(a)nomail.com wrote: > > > > <snipola> > > > > > > >> OH. So, sets don't have sizes which are numbers, at least at particular > > > > >> moments. I see.... > > > > > > > > > > If that is what you meant, then you should have said that. > > > > > And technically speaking, sets do not have sizes which are numbers, > > > > > unless by "size" you mean cardinality, and by "number" you include > > > > > transfinite cardinals. > > > > > > > > So, cardinality is the only definition of set size which you will > > > > consider.....your loss. > > > > > > It is the only definition of set size that is known to produce a valid > > > partial ordering on sets. > > > > Huh? I thought cardinality produced a valid *total* ordering on sets. > > The cardinalities are totally ordered, but the sets are not. > A total order on sets would require that when neither of two sets was > "larger than" the other that they must be the same set, not merely the > same size. Oh, right. But - and I'm not quite sure how to say this, but the cardinalities _are_ totally ordered; for any two sets A and B, c(A) < c(B), or c(A) = c(B), or c(A) > c(B). If you "reduce" the sets by the cardinality equivalence relation, they are totally ordered. The subset relation doesn't lead to an equivalence relation, only a partial ordering: so there is no s(A) = s(B) unless A=B; but for most pairs of sets, the subset relation simply says nothing at all. (Until His Master's Voice is heard, telling us something totally arbitrary.) Anyway, your claim was clearly wrong, since the subset relation provides a valid partial ordering on sets. Brian Chandler http://imaginatorium.org > > > As I have pointed out when all this started, of course everyone knows > > that the subset relation produces a perfectly valid *partial* ordering > > on sets. > > > > Brian Chandler > > http://imaginatorium.org |