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Topic: ran(EF) contains ~EF(n)
Replies: 9   Last Post: May 1, 2013 10:16 AM

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Registered: 2/15/09
Re: ran(EF) contains ~EF(n)
Posted: May 1, 2013 12:55 AM
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On Apr 30, 9:22 pm, William Elliot <> wrote:
> On Tue, 30 Apr 2013, Ross A. Finlayson wrote:
> > On Apr 30, 8:34 pm, William Elliot <> 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(d-oo) 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.


Ross Finlayson

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