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Tony Orlow wrote:
> imaginatorium@xxxxxxxxxxxxx wrote:
> > Virgil wrote:
> >> In article <1162268163.368326.64650@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> >> imaginatorium@xxxxxxxxxxxxx wrote:
> >>
> >>> Virgil wrote:
> >>>> In article <45462ba0@xxxxxxxxxxxxxxxxxxx>,
> >>>> Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
> >>>>
> >>>>> stephen@xxxxxxxxxx wrote:
> >>>>>> Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
> >>>>>>> stephen@xxxxxxxxxx wrote:
> >>>>>>>> Tony Orlow <tony@xxxxxxxxxxxxx> wrote:
> >>>>>>>>> stephen@xxxxxxxxxx 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.
>
> If that's what a partial ordering vs. a total ordering is, Bigulosity is
> a partial ordering on sets, not total ordering. Different sets can have
> the same Bigulosity.
Virgil was muddling everything up, as usual, but the difference between
a partial and total ordering is basically whether there are pairs of
elements for which the order is undetermined. The subset relation is a
very obvious example, where (for example) the set of reals in [0, 1]
and the set of prime integers cannot be compared, because neither is a
subset of the other.
"Bigulosity" has never been sufficiently clearly defined to tell, but
since you get very steamed up about subsets, and since the only known
coherent claim is that A proper subset of B -> b(A) < b(B), it's
extremely unlikely Bigulosity could be extended to become a total
ordering.
Please compare the Bigulosities of the set of polygons with vertices on
integral x-y coordinates and the set of topologically distinct
polyhedra. Show your working. (Of course you don't need to come up with
an "answer" like "ratio of 5 pi^2", but you need to show how such a
task would be approached. One of the things you still don't seem to
have realised is that before anything can be "maths" it has to be
teachable to other people. I don't think anyone but you has the
faintest idea what Bigulosity is really supposed to be, except in a
ragbag of specific cases.)
Brian Chandler
http://imaginatorium.org
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