Re: Dedekind Cuts, Fundamental Sequences: why?

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.

The fact that the reals are a complete ordered field can be derived
strictly from concepts related to rational numbers. What other
properties of the real numbers do you have in mind?

Define Pi strictly in terms of rational numbers.

Well, if you insist, pi can be expressed in many ways as the limit of a
Cauchy sequence of rational numbers.

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 + ...

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.

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.

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.

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. 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.

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

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.

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.

In cases where the Cauchy sequence (or Cantor sequence obeying the Cauchy
convergence criteria - I don't know who deserves the credit) does converge
to a rational number, we clearly have not added anything new to the domain
of numbers. In cases of irrational numbers, we have some expression, and
presumably some means of showing that the candidate is indeed irrational,
and also some means of showing that the candidate is admissible as a real
number. For example {x:x^2=-1} fails to meet the criteria for a real
number. The definition in terms of Cauchy sequences doesn't provide me a
means of determining the admissibility of that expression beyond what I
already had.

There is no rational number whose square is negative.

In the case of {x:2=x^2} a mathematician will tell me that now that we have
defined real numbers in terms of Cauchy sequences, we can say that sqrt(2)
is a supremum and infemum of the obvious sets of reals. I will also be
told that this enables the mathematicians to define addition and
multiplication using the rules that exist for rational numbers. I know I'm
not supposed to think this way, but I cannot help but perceive this as
saying 'what we really mean is that we can always find a rational
approximation of that thing we want to include, and treat that
approximation as if it were the real number itself'. That sends me back to
WTH was wrong with the definition in terms of infinite "decimal"
expressions?

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.

What I mean by loophole is that we are doing the following, if you will
forgive the informal language: After constructing the rational numbers
we conduct a few experiments and find out that there are expressions
which look as though they should represent (or at least imply) rational
numbers,
but they don't. The paradigm is {x:x^2=2}. So someone comes up with the
very impressive demonstration that the set of all rational numbers
bounded
from above by x has no supremum. That's all very good and well, except
that I don't have any formal way to talk about the place of x in the
ordering of \Q. Sure, there is an intuitively obvious meaning to the
notion, but it is formally undefined (in the developments I am aware of).

I have already addressed that point. You don't need to know what a real
number is in order to know that the set {x in Q: x^2 > 2} has no glb in
Q. Given any rational lower bound, we can find a larger one.

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.

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?

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.


--
Dave Seaman
Oral Arguments in Mumia Abu-Jamal Case heard May 17
U.S. Court of Appeals, Third Circuit
<http://www.abu-jamal-news.com/>
.

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