
Re: ran(EF) contains ~EF(n)
Posted:
May 1, 2013 12:55 AM


On Apr 30, 9:22 pm, William Elliot <ma...@panix.com> wrote: > On Tue, 30 Apr 2013, Ross A. Finlayson wrote: > > On Apr 30, 8:34 pm, William Elliot <ma...@panix.com> wrote: > > "EF_d(n) is n/d for 0 <= n/d <= 1 for natural integers n, d, as d goes > > to infinity." > > EF, or f for short, isn't a function, it's a group of functions f_d > with the property, f_d(n) = n/d provided n <= d. Thus to make some > sense of you hand waving: > > For d in N, f_d is a function from { 1,2,.. d } into [0,1] > that maps n to n/d. > > How are you defining f = lim(d>oo) f_d. Pointwise? > Then f(n) = lim(d>oo) n/d = 0, except that f_d isn't defined > for all n in N. > > So you'e defining f_d to be over N mapping n to min{ n/d, 1 }. > Thus f(n) = lim(doo) min{ n/d, 1} = 0 for all n in N.
lim_d>oo lim_n>d n/d = 1 (f is increasing) m > n > f(m) > f(n), for all d (f is monotone increasing) lim_d>oo f(n+1)f(n) = lim_d>oo f(m+1)  f(m) (f is constant monotone increasing)
These are sufficient that ran(f) <= [0,1], then that ran(f) is dense in [0,1].
Then the kicker is that convergent elements of ran(f) have convergent elements of dom(f) (with 1 the special case), where it is so because as r_a_n+1  r_a_n diminishes or is bounded, so is the difference f^1(r_a_n+1)  f^1(r_a_n), that goes to zero, and f^1(r) e N because N is closed to addition, and, f is defined for all n e N.
So: ran(f) = R_[0,1].
Of course, yours is a perfectly sensible argument, which is why the properties of the function, constant monotone increasing for each value of d, in the limit as a function, instead of in the limit for each value, are used to see the forest: for the trees.
Regards,
Ross Finlayson

