Re: Dedekind Cuts, Fundamental Sequences: why?

Hatto von Aquitanien wrote:

Dave Seaman wrote:

On Fri, 08 Jun 2007 17:09:24 -0400, Hatto von Aquitanien wrote:
Dave Seaman wrote:

Convergence means convergence to a limit.

I do not agree. At least in the sense of a_n->L. If by limit you also
include abs(a_n-a_m)->0 as N->infinity and n,m>N.

I don't care whether you agree or not. "Convergence" implies the
existence of a limit.

<http://en.wikipedia.org/wiki/Convergence>
<http://mathworld.wolfram.com/ConvergentSequence.html>

and particularly note:

"Because Cauchy sequences require the notion of distance, they can only
be defined in a metric space. Their utility lies in the fact that in a
complete metric space ( one where all such sequences are known to
converge to a limit), they give a criterion for convergence which
depends only on the terms of the sequence itself."

<http://en.wikipedia.org/wiki/Cauchy_sequence>

Note the qualification "in a complete metric space". That's what your
many references leave out, which makes them inaccurate and misleading.

The page you referenced supports my original assertions.

You did not mention completeness in your assertions. The quote above
mentions that completeness is required for the two conditions to be
equivalent.

Completeness is not stated as a requirement for convergence, and quite
frankly the statement you quoted is misleading. You do not need
completeness in order that all Cauchy sequences converge. A valid synonym
for Cauchy sequence is "convergent sequence". A Cauchy sequence is a
sequence that meets the Cauchy criterion for convergence. A sequence
converges iff it is a Cauchy sequence. There is no mention of the limit
to which the sequence converges in that criterion.

<several bad references snipped>

Convergence makes sense in an arbitrary topological space, but to speak
of a Cauchy sequence, one needs a metric.

Which exists in \Q. abs(r_1-r_2).

Precisely. After all, my point is that there is a difference between the
Cauchy criterion and the concept of convergence (they apply to different
kinds of spaces), and in fact there are Cauchy sequences in the rationals
that do not converge.

There are no Cauchy sequences which do not converge.

That implies a metric on the rationals, of course.

"The rationals" define a metric.

I do not agree. One can talk about convergence to a rational limit.
It is possible to use the above quoted definitions of convergence in
conjunction with a test for abs(a_n-L)->0.

I specified that L is not rational in the paragraph above.

But that is not required in order to talk about limits and convergence.

I didn't say it was required. Here is my original statement, which you
snipped:


Actually, what you said is this:

"The number L that appears in the definition is called the limit of the
sequence.  I have been assuming that you were familiar with the
definitions, since you claimed to have taken a senior-level real analysis
course.  (Senior level in what, I might ask.  High School?)"

The point I was making is that if we don't know what the reals are yet,
then the definition of convergence makes no sense, since we don't know
what L is supposed to be if it is not rational.

You responded to this as if I were claiming that convergence never makes
sense in the rationals. That was not my claim. I was pointing out that
there are examples of Cauchy sequences in the rationals that do not
converge,

There are no Cauchy sequences which do not converge. There are no
sequences which are convergent and do not converge.

and that it makes no sense to speak of a limit of such a
sequence in the rationals. The fact that some other sequences happen to
converge is irrelevant.

"The Cauchy Convergence Criterion gives a _definition_ of _convergence_
for sequences of rational numbers _without_ _reference_ _to_ _real_
_numbers_ . Hence the Cauchy Criterion for convergence tells which
sequences of rational numbers 'represent real numbers': A sequence q_1,
q_2,... of rational numbers is said to be convergent if it is true that
for every rational number epsilon>0 there is a natural number N such that
for any two natural numbers n,m>N the condition abs(q_n-q_m)<epsilon
holds." - Advanced Calculus: A Differential Forms Approach By Harold M.
Edwards



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