
Re: ran(EF) contains ~EF(n)
Posted:
May 1, 2013 10:16 AM


On Apr 30, 11:58 pm, William Elliot <ma...@panix.com> wrote: > On Tue, 30 Apr 2013, Ross A. Finlayson wrote: > > On Apr 30, 9:22 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're 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) > > That limit is true irrespective of f because lim(n>d) n/d = 1 > > > m > n > f(m) > f(n), for all d (f is monotone increasing) > > For all n, f(n) = 0 as was shown above. > > > lim_d>oo f(n+1)f(n) = lim_d>oo f(m+1)  f(m) (f is constant > > monotone increasing) > > f is not afferect by the value of d. f_d however is. > > > These are sufficient that ran(f) <= [0,1], then that ran(f) is dense > > in [0,1]. > > No, not at all; it's not even sufficient mathematically. > > > 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. > > The kicker is that you don't know what your talking about. > > > So: ran(f) = R_[0,1]. > > No, range f = {0}. > > > 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. > > Your arguement doesn't make sense. > > Clearly you don't agree with my definition for f, ignore the roll of f_d > nor given a cogent definition for f by which you can make your unfounded > claims. Until you address these lapses, there's no point in reitterating > your claims, which seems to be all that you're capable of. > > Perhaps you want a two place function f defined > over NxN, such as f(n,d) = n/d for all n,d in N > or f(n,d) = min{ n/d, 1 }
Here again it's the properties of the function that are used to illustrate the range, for _all_ the values of the function, instead of as to pointwise limits: there are different features of the function than of the pointwise limits. It is the properties of the function as d is unbounded that are used to note what may well be generally interesting.
For example, for all d, f(n) = 1  f(dn), and n and dn are in N. This establishes that for each value r less than 1/2 in ran(f), there is 1r in ran(f). That is so for all values of d, in N. Then, another property is that there is no interval of length 1/d in [0,1] without an element of ran(f). The elements of ran(f) are dense in [0,1], as f is: constant monotone strictly increasing, for all d in N.
As d is unbounded, then as above for this note: ran(EF) contains ~EF(n) for each n in dom(EF).
Then yes I maintain that.
Regards,
Ross Finlayson

