
Re: Simple analytical properties of n/d
Posted:
Mar 10, 2013 1:43 PM


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 Email subject to legal action. I reserve the > right to publicly post or ridicule any abusive Email. 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 toone) 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(mn) and that f^1(f(n)) = n.
Then, proofs (in their structure, soidisant proofs) that use that f is onetoone, 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

