Re: Dedekind Cuts, Fundamental Sequences: why?

In article <l4adnYr9ws-7PvrbnZ2dnUVZ_vyunZ2d@xxxxxxxxxxxxx>,
Hatto von Aquitanien <abbot@xxxxxxxxxxxxxx> wrote:

Dave Seaman wrote:

On Wed, 06 Jun 2007 18:33:17 -0400, Hatto von Aquitanien wrote:
Dave Seaman wrote:

On Wed, 06 Jun 2007 13:54:33 -0400, Hatto von Aquitanien wrote:
Dave Seaman wrote:

We don't prove definitions.

Read Weyl yourself.

I doubt that Weyl claims to be able to prove definitions. If he does,
then he is seriously mistaken.

I did not assert that he did so. That is your misinterpretation of what
I said.

Then my point stands: we don't prove definitions, your cryptic response
notwithstanding.

But you do prove theorems using definitions, do you not? As I understand
Weyl, he believed the definitions using either fundamental sequences or
Dedekind cuts were flawed because they assumed that the properties of the
members of the sets were given to the sets themselves without proof. As I
said, I don't necessarily agree with his view.

I don't know if there is a way of formally defining pi in terms of
rational numbers, but simply stating that it can be done as the limit of
a Cauchy sequence is not defining pi in terms of rational numbers.

pi/4 = 1 - 1/3 + 1/5 - 1/7 + ...

That is interesting. Thank you. After I made that statement I realized
that it was almost certainly possible using a Taylor series.

But you are misreading what I said. I specifically did not say that
every real number can be specified, either in terms of the rationals or
by any other means. After all, there are uncountably many reals and
only countably many specifications to go around.

I believe that observation was first made by Gödel. Nonetheless, it is
possible to specify infinite (or finite) categories of numbers. For
example, there are algebraic numbers, there are transcendental numbers,
etc. Since you were originally responding to my statement about being
able to specify some cases where the rational numbers are incomplete
using only the algebra of rational numbers, I believe it is you who has
failed to understand what the other person is saying.

Completeness is not an algebraic property. Completeness means that the
least upper bound axiom is satisfied.

Well there sure are a lot of example floating around showing that things
like sqrt(2) represent "holes" in the rational number that need to be
filled. That seems to be the source of the idea of the field of rational
numbers being incomplete.

I'm not sure what you mean by "axiom" in this context. It is unfortunate
that modern mathematics has corrupted the original definition in such a way
that "axioms" are not necessarily the most fundamental constituents of a
conceptual structure. This is a consequence of axiomatic approaches being
taken to develop such theories as the theory of groups. In that context it
is legitimate to talk about the properties of a group as axioms. But it is
wrong to talk about showing that something satisfies axioms. When
properties are used in a context where they can and must be demonstrated,
they are not, IMO, axioms.

If by "axiom" you mean some criteria which the structure can be tested
against, then I will contend that a supremum is an algebraic property. At
least the concepts which underlie it fall within the realm of algebra.
Specifically the concept of ordering.

What I said is that the set
of real numbers can be defined (and proved to be a complete ordered
field), using the rationals as a starting point. Defining a set does
not mean specifically defining every member of the set.

I am not convinced that this approach doesn't assume its conclusion as
one
of its unstated assumptions. That is distinct from saying that I am
convinced that it does.

1. The set C of Cauchy sequences of rationals, with
componentwise addition and multiplication, is a commutative
ring.

I believe this is where Weyl would object. It seems he accepted neither
that one could mix rational numbers with 'irrational Cauchy sequences', nor
replace the rational numbers with corresponding Cauchy sequences so that
every element in the set is a Cauchy sequence, and then assume the elements
follow the rules that rational number follow.

2. The ring C contains an ideal I consisting of all sequences
of rationals that converge to 0.

3. The ideal I is maximal in C, because for any sequence
{a_n} in C\I there is a {b_n} in I such that {a_n+b_n} has
an inverse in C.

4. Since I is maximal, it follows that the quotient ring
C/I is a field.

5. We say a sequence {a_n} in C is positive if there exists
epsilon > 0 and N > 0 such that |a_n| > epsilon for every
n > N.

6. If {a_n} is positive according to definition 5, and if
{b_n} is such that {a_n-b_n} is in I, then {b_n} is likewise
positive.

7. The equivalence classes in R = C/I that consist of positive
sequences are closed with respect to addition and multiplication,
and satisfy the trichotomy law.

I don't know what you mean here. Do you mean each equivalence class is
closed individually, or do you mean that the set of equivalence classes is
closed collectively?

Each equivalence class determines a distinct real number, and both the
'sum" and then "Product" of such equivalence classes are themselves
equivalence classes. Dave Seaman's analysis above shows that one can
find a lub rational Cauchy sequence for any set of rational Cauchy
sequences that is bounded above.

8. Thus R = C/I is an ordered field.

9. It remains to prove that R is complete. Let A = {A_i:i
in I} be a nonempty indexed subset of R = C/I that is bounded
above by a real number b.

It is not clear if you intend by "real number" an equivalence class in C, or
something else defined prior to this development.

Each real is a distinct equivalence class.

Let {b_n} be a representative
of the equivalence class for b. For each a_i in A, we can
choose a representative Cauchy sequence {a_{i,n}} such that
a_{i,n} <= b_n for each i in I and each n > 0. We now need
to produce a sequence {d_n} that will represent a least
upper bound for A. For each n, we can choose d_n such that
d_n <= b_n and also such that 0 <= inf({|d_n-a_{i,n}|: i in I})
< 1/n. Then the sequence {d_n} is the required representative
of a lub for A.

I have to think about this more. There are many assumptions in this last
step which I have not given much thought to in the past. If I am not
mistaken you have what might be called second order Cauchy sequences. That
is Cauchy sequences of Cauchy sequences.

Each equivalence class contains an infinity of Cauchy sequences of
rationals. To show that a particular real exists, one only need show
that an appropriate rational Cauchy sequence exists.


Is there a step here that "assumes its conclusion"?

If there is, it would be something along the lines of assuming that a
sequence representing an irrational number and constructed using only
rational numbers will converge

The completeness proof (step 9) is tricky when dealing with Cauchy
sequences, but is much simpler for Dedekind cuts. If A = {(L_i,R_i) : i
in I} is a nonempty indexed collection of Dedekind cuts that is bounded
above by B = (L,R), then the lub is given by D = (L',R'), where L' =
U{L_i : i in I} and R' = Q\L'. Some of the other steps, however, are
much simpler for Cauchy sequences, thanks to ring theory.

There is no problem in software design which cannot be solved by introducing
another level of indirection. In this case there appears to be
self-referential recursion going on.


There is a layered construction going on.

The base layer is the arithmetic of (positive) naturals.
The next layer builds, say, the positive rationals from the naturals.
The next layer adds on the negative rationals.
The top layer, so far, builds the reals from the rationals.

An common alternate sequence, though a bit less elegant, goes from
naturals to integers to rationals to reals,

That is, the structure refers to
itself from a level of construction above the level being referred to.

Such references are always to the structure "below" the one being
defined.

There is something called the Xerox effect which says that copies of copies
of copies, ..., are never as good as the original.

Each of the "copies" here contains an exact image of its source within
it, but contains more.

The set of Cauchy sequences of rationals, with elementwise addition and
multiplication, forms a ring. There is a maximal ideal in this ring,
consisting of all those sequences that converge to 0. The quotient ring
is therefore a field, and it is provably a complete ordered field. This
is what I meant by the statement above about proving the properties of
the reals strictly from the rationals.

The place where my doubts arise is when we start testing the model by
plugging in such expressions as sqrt(2). You say that there is a Cauchy
sequence that will converge to sqrt(2). I sincerely believe that it is
possible to write an algorithm which will produce a rational number with
every iteration which when squared will be closer to 2 than the square of
the result of the previous iteration. I just have never been able to
convince myself that this is adding anything to what I had already
assumed.

Then perhaps I should switch to the Socratic method and simply get you to
"remember" the answers to all your questions.

Do you know what the Newton-Raphson method is? Do you know about
continued fractions? Those are two ways of producing such a sequence.

Yes. I am familiar with both of these.

I distinctly recall having a conversation with a man many years ago who
insisted that he was not convinced that a sequence of rational numbers
could converge to an irrational number. At the time, it seemed obvious to
me that such a convergence was possible. I have much more sympathy for
that man now than I did then.

No, we do not treat the rational number as if it were the real. We
choose a representative Cauchy sequence and perform the required
operations on that. Then we show that the result (as an equivalence
class) does not depend on which representative was chosen, i.e. that the
operation is well defined.

And that is based on an argument by mathematical induction.

So when you are done defining real numbers in terms of Cauchy sequences
you tell me to put sqrt(2) where I was going to put it all along.

There is an ordering relation on the set of equivalence classes in R =
C/I, which I described in point (7) above.

The entity I am calling sqrt(2) was specifically rejected for membership in
the set used to define the Cauchy sequences, the equivalence classes of
which represent the real numbers.

Given any positive rational, a_0, the recursion formula
a_(n+1) = (a_n^2 + 2)/(2*a_n) generates a Cauchy sequence of rationals
which "converges" to sqrt(2).



It arose from the algebra of the field
of rational numbers. It may seem ridiculous to raise this objection, but
it seems to me worth considering. I can now talk about {x:2=x^2, x \in
\R}. Is that the same {x:2=x^2, x \in \Q}?

No. The former set contains two members, the latter none.

This seems to be along the lines of what Weyl was object to. That is, there
is no proof that these are actually the same. I'm not saying that is a
correct objection, but it certainly comes to mind.

Once we have finished constructing the reals, we have a perfectly good
order relation on R, which is what we need to talk about the place of x
= sqrt(2) on
the real line. If x and y are reals, represented as Dedekind cuts, then
by definition we take x < y to mean that the left set of x is strictly
contained in the left set of y.

You ever see a guy reach out an grab at a fly and claim he caught it.
You
saw the fly before he grabbed at it. You don't see the fly while he
claims
to be holding it. When he opens his hand you don't see the fly fly away,
but a moment later, there's the fly. You ever been the guy who tried the
grab the fly and weren't sure if you had it or not? That's about how I
feel about all of this.

Do points (1)-(9) help?

They help clarify what is going on. Unfortunately, I believe this is one of
those cases where we either accept the identity of entities at different
levels of abstraction, or fall into infinite regress. There has to be a
Cauchy sequence of representative Cauchy sequences of Cauchy sequences
hiding around the corner.

Not when you finally sort things out properly.

But the problem we were trying to solve was the case where the set of
rational numbers is _not_ closed. "Shshshshs... Don't tell anybody.
Just stick that sqrt(2) thing in the hole and pretend like it belongs
there...."

No, that's not what was done. Given either the Dedekind cuts or the
Cauchy sequence construction, we explicitly construct the real number x

0 such that x^2 = 2. There is nothing up anyone's sleeve. It is no
more mysterious than constructing a rational x such that x + x = 1.

I still have trouble convincing myself that a sequence of rational
numbers will converge to something irrational.

Think about Newton-Raphson.

In the sense of solving pragmatic problems, it doesn't bother me. In the
case of absolute rigor, I may simply have to accept it on faith.
.

Relevant Pages

  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... What is the step of logic which leads one to seek an extention of the ... But when they go from the rationals to the reals, ... class of Cauchy sequences that converge to that same point. ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... What is the step of logic which leads one to seek an extention of the ... But when they go from the rationals to the reals, ... class of Cauchy sequences that converge to that same point. ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... using the rationals as a starting point. ... that one could mix rational numbers with 'irrational Cauchy sequences', ... or do you mean that the set of equivalence classes is ... the reals strictly from the rationals. ...
    (sci.math)
  • Re: Dedekind Cuts, Fundamental Sequences: why?
    ... What is the step of logic which leads one to seek an extention of the ...  But when they go from the rationals to the reals, ... class of Cauchy sequences that converge to that same point. ... built up, step by step, from rationals, to sequences ...
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  • Re: Heine-Borel Lemma-
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