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 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 }