Re: Dedekind Cuts, Fundamental Sequences: why?

David C. Ullrich wrote:

On Mon, 04 Jun 2007 05:37:27 -0400, Hatto von Aquitanien
<abbot@xxxxxxxxxxxxxx> wrote:

David C. Ullrich wrote:

The fact that the reals are complete is used in many places - analysis
simply would not work without it. You're just at the start...

That begs the question.

Yes it does. When you know some analysis it will be very clear
to you why completeness is so important.

That still begs the question. The fact that something _is_ useful does not
explain _why_ it is useful.

Nor is it completely clear what it means to say that
the field of rational numbers does not exhibit the closure property of
boundedness.

Say S is the set of positive rational x such that x^2 < 2. Then
S does not have a rational least upper bound.

How does that add anything to the already established fact that x^2=2 does
not have a rational solution?

Make up your mind! A paragraph up you complain that I didn't answer
your question. Here I _did_ answer your question and you ask how
my answer adds to something else.

In previous posts in this thread the solution of x^2=2 was introduced as an
example of something which is not a member of the rational numbers. I am
told that requiring every set of rational numbers bounded from above to
have a least upper bound is necessary for the field of numbers to be
complete. I was then given the same example in terms of upper bounds, and
I asked how reformulating the example in terms of upper bounds adds
anything to the original example.

You _said_ it was not clear to you why the rationals were not
complete. I explained exactly why the rationals are not complete -
I didn't say anything about what I was or was not adding anything to.

Every definition I have consulted for supremum and infimum begins with
the
real numbers. So to tell me that the reason we need to extent the
rational numbers to the real numbers is so that the domain of numbers
has suprema and infima assumes the real numbers to be defined already.

Huh?

Definition: The ordered field F is complete if every nonempty
subset of F which is bounded above has a least upper bound.

That is not a definition of least upper bound.

I didn't say it was! It's a definition of completeness. You
know the definition of least upper bound, or I thought
you did.

"Every definition I have consulted for supremum and infimum begins with the
real numbers."


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