Re: Possible proof of Gabriel's Theorem?

From: William Hughes (wpihughes_at_hotmail.com)
Date: 03/11/05 

Date: 10 Mar 2005 16:51:34 -0800

Jason wrote:
> William Hughes wrote:
> > Jason wrote:
> > > No. That's where the definition is a load of non-sense because
the
> > > function is integrated
> > > over the entire interval. This means that delta must eventually
be
> > > zero.
> >
> > Poppy***. There is nothing in the definition that says
> > that delta must eventually become zero, however, according to the
> > definition constant functions are integrable.
> >
> > "the function is integrated over the entire interval"
> > is nonsense. What is the difference between integrating
> > over an interval and integrating over an "entire interval" ?
> >
> >
> > -William Hughes
>
> When you integrate over an interval, there are *no holes* in the
> interval, therefor delta is eventually zero regardless of what
> the definition says. In fact we are only interested in the result
> once delta is zero, else we do not obtain the correct result.

More drivel. What are "holes"? Why does making delta 0 get
rid of them? To repeat: What is the difference between integrating
over an interval and integrating over an "entire interval" ?

>
> The problem with *epsilon-delta* logic is that its not supposed
> to use *infinitesimals* but every epsilon-delta argument does
> exactly this: "Give me any delta/epsilon and I can always find a
> smaller delta/epsilon greater than zero..."

No. Every epsilon-delta argument has the form:

    Given an epsilon > 0 I can find a delta > 0
    such that if foo < delta, bar < epsilon.

Finished. Two real numbers greater than 0. No infinitesimals.
Nothing indeed about epsilon being small.

We only talk about epsilon being small when discussing the
meaning/motivation of the argument. We say: because the argument
works with any epsilon, it must work with any
arbitrarily small epsilon. But by arbitrarily small we mean

  -given any real number t>0 we can find an epsilon (possibly
   different for each t) such that epsilon < t

We do not mean

   -we can find an epsilon > 0 (not dependent on t) such that
    given any real number t>0, epsilon < t

(Indeed, in the second case epsilon must be an infinitesimal.
However, we don't go there.)

Note again: There is nothing about epsilon being small in the
original argument. The argument states that epsilon is arbitrary
as long as it is a real number greater than 0. There
is no mention of infinitesimal.

> But this is what an
> infinitesimal was initially defined to be: a number greater than zero
> but less than every positive number.

Correct (almost. replace "every positive number" with something like
"1/N for every integer N", infinitesimals can be positive and
they are numbers) but irrelevant. Epsilon-delta arguments do not talk
about
"a number greater than zero but less than every positive number"

        <more drivel snipped>

                            -William Hughes