Re: Dedekind Cuts, Fundamental Sequences: why?

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

David C. Ullrich wrote:

On Sun, 03 Jun 2007 22:57:37 -0400, Hatto von Aquitanien
<abbot@xxxxxxxxxxxxxx> wrote:

Bob Kolker wrote:

Hatto von Aquitanien wrote:

This is the motivation stated by Pickert and Görke, but it is not clear to
me why it matters.

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.

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.

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.

For every rational number I can very easily divide the rational numbers
into two disjoint sets by asserting that every rational number greater
than the selected number is a member of the set whose lower bound is the
selected
number. Likewise for the symmetrically opposite case. I then arbitrarily
chose one side of the bifurcation to include the chosen rational number.

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.

That definition does not begin with the real numbers. If you've
only seen the definition in the context of the real numbers
that's because the reals _are_ complete, while, say, the
rationals are not. It does not follow that "to say that the reason
we need to extend 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."

http://www.cuyamaca.edu/bruce.thompson/Fallacies/circ_justification.asp


************************

David C. Ullrich


************************

David C. Ullrich
.

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