```Date: May 1, 2013 2:58 AM
Author: William Elliot
Subject: Re: ran(EF) contains ~EF(n)

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(d-oo) 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 reitteratingyour 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 Nor f(n,d) = min{ n/d, 1 }
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