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Topic: Simple analytical properties of n/d
Replies: 20   Last Post: Mar 11, 2013 11:01 PM

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ross.finlayson@gmail.com

Posts: 1,179
Registered: 2/15/09
Re: Simple analytical properties of n/d
Posted: Mar 10, 2013 1:43 PM
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On Mar 10, 7:29 am, Shmuel (Seymour J.) Metz
<spamt...@library.lspace.org.invalid> wrote:
> In <Pine.NEB.4.64.1303070027340.14...@panix3.panix.com>, on 03/07/2013
>    at 12:38 AM, William Elliot <ma...@panix.com> said:
>

> >What's a pointed disk?
>
> (D,d), where d \in interior D[1].
>

> >R^2\(0,0) is neither a disk nor pointed; it's a punctured plain.
>
> ITYM punctured plane; it's homeomorphic to a punctured disk.
>

> >No. The image of f is R.
>
> What's f^{-1}(0)?
>
> [1] The definition works for both open and closed disks.
>
> --
> Shmuel (Seymour J.) Metz, SysProg and JOAT  <http://patriot.net/~shmuel>
>
> Unsolicited bulk E-mail subject to legal action.  I reserve the
> right to publicly post or ridicule any abusive E-mail.  Reply to
> domain Patriot dot net user shmuel+news to contact me.  Do not
> reply to spamt...@library.lspace.org



f_d(0) = 0. Then, for f_\infty being f, if f is invertible (it's one-
to-one) then what is f^-1 (0)? Here there exists the inverse: for
elements of the range, f^-1 ( f(n) ) = n. Clearly: standardly lim d-
>oo n/d = 0 for n < d.

This is about the characteristics of the range given the domain, that
for f_d, it is constant monotone increasing, and ranges from zero to
one. Then, a convention might have that f^-1(0) = 0 where f(n>0) =/=
0, that
f(n>0) > 0
f(n<m) < f(m)
f(m>n) - f(n) = f(m-n)
and that
f^-1(f(n)) = n.

Then, proofs (in their structure, soi-disant proofs) that use that f
is one-to-one, would use the properties of the function that the
inverse exists for elements of ran(f), and that f^-1(0) is undefined,
without the stipulation: that it is f^-1(a) where a = f(0).

Simply, the inverse exists for elements of ran(f), where ran(f) has
the properties as above of that f(n=/=m) =/= f(m).

Each f_d(n) is in ran(f_d), for each region r of length 1/d [x, x+1/d]
in [0,1] there is at least one element of ran(f_d) in r. Then, for
each x in [0,1] there exists f^-1(x) e N, is considered for that for
each x in {0, (0<n<=d)/d} there exists f_d^-1(x) = n.

Basically for that m>n => f(m>n) > f(n), f is constant monotone
strictly increasing, which follows from constant monotone increasing
and f -> 1: f^-1(0) = 0.

Regards,

Ross Finlayson



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