Virgil
Posts:
4,482
Registered:
1/6/11
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Re: The Distinguishability argument of the Reals.
Posted:
Jan 6, 2013 8:19 PM
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In article
<90d0b714-c19e-406c-ba9b-28ba2650ca27@r10g2000pbd.googlegroups.com>,
"Ross A. Finlayson" <ross.finlayson@gmail.com> wrote:
> On Jan 6, 2:59 pm, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <7a163160-c36a-46d0-ab7a-97cf0fa11...@q16g2000pbt.googlegroups.com>,
> > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > On Jan 6, 11:55 am, Virgil <vir...@ligriv.com> wrote:
> > > > In article
> > > > <1038fe29-f169-4511-bd13-c7ade7fd1...@pd8g2000pbc.googlegroups.com>,
> > > > "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
> >
> > > > > > lim_(d -> oo) f(n,d) = 0 for every n in N
> > > > > > --
> >
> > > > > No
> >
> > > > If, as Ross defined it, f(n,d) = n/d for all d in N and all n in
> > > > {0,1,2,...,d}, then for any n, lim_(d -> oo) f(n,d) = 0
> >
> > > > And no amount of denial by Ross will alter that fact.
> > > > --
> >
> > > Wait, aren't you going to misquote Einstein? Because, you have quite
> > > the practice of misquoting me. Now, I'm no Einstein, but, I generally
> > > heartily agree with him, of the rather conscientious sort.
> >
> > > d/d = 1
> > > lim_(n->d) n/d = 1
> > > lim_(n->d, d->oo) n/d = 1
> >
> > That last one is false because
> > lim_(n -> d) [ lim_(d -> oo) n/d ] =lim_(n -> oo) 0 = 0
> > lim_(d -> oo) [ lim_(n -> d) n/d ] =lim_(n -> oo) 1 = 1
> > SO
> > lim_(d->oo)(lim_(n->oo) n/d =/= lim_(n->oo)(lim_(d->oo) n/d
> >
> > SO Ross's lim_(n->d, d->oo) n/d does not exist!
> >
> > At least in stndard mathematics.
> >
> > But he might try in in WMytheology or RAFeology.
> > --
>
>
> http://en.wikipedia.org/wiki/Order_of_integration_(calculus)
>
> It's not quantifier dyslexia so much as that there's only one free
> parameter modeling the function: d.
>
> There d, for denominator, for n, numerator: is free and unbounded, as
> is n, as it ranges through elements simply enough in d. For all
> values of d, the range is [0,1].
For each value of d, the range is a finite subset of d+1 values in [0,1].
You may properly may say that the codomain for each d is [0,1], but the
range is the set of values values actual taken by the function, and for
each value of d the range is a set of d+1 values, not an interval.
Wrong again, Ross!
--
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