Re: infinity

In article <MPG.1d88ece773ad803c98a20c@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:

> Virgil said:
> > In article <MPG.1d879fe6c1c5af7398a1e5@xxxxxxxxxxxxxxxxxxxxxxxxx>,
> > Tony Orlow (aeo6) <aeo6@xxxxxxxxxxx> wrote:
> >
> > > Virgil said:
> > > > In article <1125629551.621618.60990@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> > > > "William Hughes" <wpihughes@xxxxxxxxxxx> wrote:
> > > >
> > > > > Tony Orlow (aeo6) wrote:
> > > >
> > > > > > The only contradiction arises from your obsession with a last
> > > > > > element, and conflation of it with finiteness for a set. I do
> > > > > > not accept that a last element necessarily indicates a finite
> > > > > > set,
> > > >
> > > > There are ordered sets with largest elements that are infinite,
> > > > such as the closed unit interval, but there are no non-empty
> > > > ordered sets without a largest element, at least according to the
> > > > Cantor definition of finiteness.
> > > >
> > > > I challenge TO to produce either a set which does not allow any
> > > > injections into proper subsets that is infinite by TO's definition
> > > > or to
> >
> >
> > > All whole numbers from 100...000 through 111...111.
> >
> > Now PROVE that there is no injection from the set of all those whatever
> > they are to any proper subset!
> >
> > > > produce any set which does allow an injection into a proper subset
> > > > that is finite by TO's definition.
> > > >
> > > > Absent TO's successful production of at least one of these two
> > > > types of examples, we are free to impose the Cantor criterion for
> > > > finiteness/infiniteness of sets on all his postings.
> > > >
> > > > The Cantor criteria are: A set is finite if and only if there do
> > > > not exist any injections from that set to any proper subset; a set
> > > > is infinite if and only if there exists at least one injection from
> > > > that set to some proper subset.
> > > >
> > > > Note that according to these criteria, a non-empty ordered set
> > > > without a maximum (or minimum) member, such as the set of (finite)
> > > > naturals, is necessarily infinite.
> >
> > > Yes, Cantor-infinite, as you say, but not infinite by other thinking.
> >
> > Cantor-infinite is the only infinite until we have a workable
> > alternative. So far none of TO's attempts have been workable
> > > >
> > > > > > therefore I see no contradiction between the set of finite
> > > > > > naturals being finite and not having a last element.
> > > >
> > > > Then TO must have in mind some some definition of finiteness versus
> > > > infiniteness of sets incompatible with Cantor's, and should
> > > > immediately provide that definition to us so that we can understand
> > > > what he is talking about.
> >
> > > The rules of finiteness that you seem to be ignoring are that a+b,
> > > a*b and a^b are all finite for finite a and b. Do you disagree with
> > > that statement?
> >
> > If a and b are any members of any ring, including the ring of integers
> > or the field of reals, a+b and a*b will be members of that same ring,
> > but a^b need not be.
> a^b=a*a*a*a..., so why do you say a*b will be a member of the same ring, but
> a*a*a... is not?


I did not say that, what I said was that a*b need not be a member of the
ring.

In the ring of integers, 2^(-2) is not a member of that ring.

In the ring of rationals, (2)^(1/2) is not a member of that ring.

In the ring of reals, (-1)^(1/2) is not a member of that ring.
> >
> > Since "standard" rings are of finite elements (their compactifications
> > are not, in general, rings), the sum or product of two members will be
> > as finite as their summands or factors.
> Are you saying a^b can be infinite with finite a and b?

I am saying that a^b need not be defined at all within the ring in
question.

Learn some mathematics before pontificating about it.
.

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