ran(EF) contains ~EF(n)

[Ross A. Finlayson]

Basically this EF is the limit of a function. Not-a-real-function,
it's modeled by real functions and in fact very simply n/d. Then, the
properties of its range have that: it is symmetric about some
midpoint in its range 1/2, for f^-1(1/2), defined for even d. Just as
ran(f) starts from .000... and increases in constant monotone, in
reverse it starts with .111... (or for example in decimal, .999...)
and decreases. Then, as there is a less than finite difference
between elements of the range for consecutive elements of the domain
by their natural ordering, there is in the range an element that
starts with .1, .11, .111, ..., and each finite initial segment of
integers as expansion. This is from seeing that the elements of
ran(f) are the same elements of ran(REF) for REF the "reverse"
equivalency function.

EF: n/d
REF: (d-n)/d

As there exists d-n for each n and d, there exists here ~EF(n) =
REF(n) constructively, and each element in ran(EF) is in ran(REF).

Then, here, constructively: ~EF(n) e ran(EF), quod erat
demonstrandum.

There are replete properties of this function and its range that its
range is R_[0,1].

Regards,

Ross Finlayson

[William Elliot]

On Tue, 30 Apr 2013, Ross A. Finlayson wrote:

> Basically this EF is the limit of a function.

Basically this all is nonsense without a clear and precise definition
of EF (electromagnetic frequency?) instead of hand waving woo woo voodoo.

[Ross A. Finlayson]

It has an established definition.

http://groups.google.com/groups/search?safe=off&q=EF+definition+author%3AFinlayson&btnG=Search&sitesearch=

Pretty much the same definition is used throughout.

Regards,

Ross Finlayson

[William Elliot]

I'm not going to go chasing around a lengthy web page for some imagined
definition, that may or may not be there and that may or may not be a
definition instead of some vagueness. Post your definition here, clearly
and accurately or failing that, post your pseudo definition in philosophy.

[Ross A. Finlayson]

On Apr 30, 8:34 pm, William Elliot <ma...@panix.com> wrote:
> On Tue, 30 Apr 2013, Ross A. Finlayson wrote:
> > It has an established definition.
>

> >http://groups.google.com/groups/search?safe=off&q=EF+definition+autho...

>
> > Pretty much the same definition is used throughout.
>
> I'm not going to go chasing around a lengthy web page for some imagined
> definition, that may or may not be there and that may or may not be a
> definition instead of some vagueness.  Post your definition here, clearly
> and accurately or failing that, post your pseudo definition in philosophy.

"EF_d(n) is n/d for 0 <= n/d <= 1 for natural integers n, d, as d goes
to infinity."

[William Elliot]

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.

[Ross A. Finlayson]

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.

Regards,

Ross Finlayson

[William Elliot]

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 }

[Virgil]

On Tue, 30 Apr 2013, Ross A. Finlayson wrote:



>
> lim_d->oo lim_n->d n/d = 1

But lim_n->d lim_d->oo n/d = 0
--

[Ross A. Finlayson]

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