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Topic: Matheology � 258
Replies: 53   Last Post: May 11, 2013 10:07 PM

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 ross.finlayson@gmail.com Posts: 1,974 Registered: 2/15/09
Re: Matheology 258
Posted: May 11, 2013 8:18 PM
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On May 11, 4:43 pm, Virgil <vir...@ligriv.com> wrote:
> In article
> <105e4062-9b00-46e8-acf2-142600826...@k8g2000pbf.googlegroups.com>,
>  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>

> > To see a corresponding example, integrate: S_0^1 1 dx.  The
> > differential vanishes as b-a -> 0  but the sum over them = 1.

>
> The integral from a to b certainly vanishes (does to 0) as b=a -> 0, but
> the "differential" does not change all, it remains dx.
> --

You see how ubiquitous the notions of summation and the differential
in x are, as are a and b for the bounds of integration, or here
casually. Then, as delta x goes to zero: is not the area still equal
to the sum of the differential areas: only for all of them? And, is
it not so only for no finite nor zero value, generally, for all of
them together?

Seems hypocritical to have real analysis for no finite differential,
and not have non-zero infinitesimals in the reals. Of course, that is
where standard real analysis doesn't depend on non-zero
infinitesimals, instead as to the synthesis of the area: under the
limit.

Leibniz' notation in the integral calculus, the infinitesimal
analysis, with the integral bar for summation of the less-than-any-
finite and non-zero differential quantities, and only all of them,
well remains quite standard.

Then, to ignore the application of these tools as to simply defining
the uniformly divided unit interval as to the constant differences of
successive integers seems rather closed-minded.

Of course you'd be thanked to haul out all the replete applications
for all of higher mathematics, as applied, solely due transfinite
cardinals, why do you refuse?

The tools of higher mathematics that are of the integral calculus
remain largely unaffected by notions of transfinite cardinals. Then
it is a theorem of having ZF and real analysis: additivity is but
countable. Is that not a theorem of objects of ZF, not consequent its
axioms and simply to remain, if not relevant, not incongruous?

Seems it's let that LUB outweighs ZFC.

Regards,

Ross Finlayson

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