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

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 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
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(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 }

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 point-wise limits: there are different features of the function
than of the point-wise 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(d-n), and n and d-n are in N.
This establishes that for each value r less than 1/2 in ran(f), there
is 1-r 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