Virgil
Posts:
4,482
Registered:
1/6/11
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Re: The Distinguishability argument of the Reals.
Posted:
Jan 6, 2013 1:10 AM
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In article
<694aece8-90bf-400b-8663-dd86266adba9@ah9g2000pbd.googlegroups.com>,
"Ross A. Finlayson" <ross.finlayson@gmail.com> wrote:
> On Jan 5, 7:37 pm, fom <fomJ...@nyms.net> wrote:
> > On 1/5/2013 6:35 PM, Ross A. Finlayson wrote:
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > On Jan 4, 10:20 pm, Virgil <vir...@ligriv.com> wrote:
> > >> In article
> > >> <7850ae29-08d9-49ef-8c7b-e8979e037...@m4g2000pbd.googlegroups.com>,
> > >> "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
> >
> > >>> Consider the function that is the limit of functions f(n,d) = n/d, n =
> > >>> 0, ..., d; n, d E N.
> >
> > >> You mean the zero function?
> >
> > >> For every n, the limit of f(n,d) as d -> oo is 0, so your limit function
> > >> would have to be the zero function: f(n,oo) = 0 for all n.
> > >> --
> >
> > > No, none of those is the zero function, and each d->oo has it so that
> > > d/d = 1.
> >
> > That is true.
> >
> > The problem is that as d -> oo the value at any
> > given fixed n -> 0.
> >
> > 2/3, 2/4, 2/5, 2/6, 2/7, 2/8, 2/9, 2/10, ...
> >
> > So, the pointwise limit of the function is zero.
>
>
> lim_n->d n/d = 1
Since the set of values of n is finite for each value of d, no limit
process is required, or even defined.
One has f(d,d) = 1 for all d, but one does not have f(n,d) = 1 for any
n less than d, and one has the properly defined limit:
lim_(d -> oo) f(n,d) = 0 for every n in N
--
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