Re: Review of Mueckenheims book.

In article <1172131989.493718.251550@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
On 21 Feb., 02:46, "*** T. Winter" <***.Win...@xxxxxx> wrote:
>
> > And I think that the remark that Euclid used a
> > proof by contradiction is wrong
>
> ??? It is said (I didn't read the original text) that Euclid assumed
> that n primes exist and then he contradicted this assumption by
> proving the existence of prime number n+1. Many authors call this a
> proof by contradiction.

It would be a proof by contradiction, if there is an initial assumption
that is falsified by the reasoning. But see below.

Whether this is the case depends on the person. Euclid is said to have
assumed that.

Oh, as far as I know such an initial assumption was not made.

> > (actually the theorem and proof are quite
> > Cantoresque):
>
> Yes. Cantor assumed a complete list of reals and showed the existence
> of another real. Many authors call this a proof by contradiction.

Wrong. Cantor does not assume that the given list is complete. He shows
that given *any* list of reals there is a real not on the list. There is
*no* initial assumption of completeness.

So Cantor constructs one or the other list without attempting to
include all the reals? Then he finds another real and takes it as
evidence for what? That he personally is not able or was too lazy to
construct a complete list?

You apparently do not understand such proofs. Cantor assumes an arbitrary
given list and shows that it can not be complete. This proves (without
contradiction) that *no* list is complete.

> That however is not the wording used by Euclid.

What is the wording used by Euclid? (I may note that the Corollary
was indeed *not* by Euclid, but that is the modern corollary.)

Euclid is reported to have said (of course in Greek and slightly
differing in the thousands of copies of his Elements made by hand
before the 1500 printed editionsappeared): "There are more prime
numbers than any given set of primes contains" or "in any given set
of primes, there is prime number missing".

Ah. Where is the assumption that a collection of prime numbers is
*complete*? This is *indeed* extremely Cantoresque.

This is of course correct and can be generalized to natural numbers:
For any given set of natural numbers, there is a natural number not in
that set.

This is wrong. Note that in Euclids time any collection meant any finite
collection (and I wrote that in my review). In current times we also allow
infinite collections, so the original wording is no longer correct, but
should be adapted to "finite set".

> I think I told you already: 3/7 is 0.3 in base 7

But in my opinion 0.3 in base 7 is still a formula telling how the number
is calculated. It tells me that it is 3 * 1/7. And before Simon Stevin's
"De Thiende" from 1585, decimal numbers (and base 'n' notated rationals)
did not actually exist.

Once upon a time there were only unary numbers existing. Everything
else is abbreviation which has to be introduced by someone.

And in that time rational numbers did not exist. How would you express
3/7 in unary notation? But you refrain to answer to my original: "in
my opinion 0.3 in base 7 is still a formula telling how the number is
calculated".

But as Simon Stevin writes (parafrased):
"The importance of the proposal is not so much mathematical
profoundness, but more broad applicability."
And it was merely used as an aid to ease calculations.

Yes. Also decimal representation is only to ease calculation (and
recognition - if that is different).

But it is still a formula that tells me how to actually calculate the
number.


> I think I told you already: 3/7 is 0.3 in base 7 and this can be
> compared with any existing nunmber n/B (in natural base B) by 7n/7B <
> 3B/7B or 7n/7B = 3B/7B or 7n/7B > 3B/7B.

So for the comparison you do not use "0.3" but the original "3/7"?
Well, I can compare sqrt(2) with any rational number. Square the two.

How can a number be squared which does not exist? First it must exist.
Then it can be handled.

How comes 3/7 in existence?

> 3B/7B or 7n/7B = 3B/7B or 7n/7B > 3B/7B. So we need only find the
> natural numbers 7n and 3B which exist for every natural numbers n and
> B we can use. (In case all bits of our bit reservoir were required to
> establish n and B, then the reservoir of bits must be extended by 7 or
> 3 which should be possible in all practicable cases.)

For squaring not much more is needed.

Except the number to be squared.

How comes 3/7 in existence?

What quote of Robinson are you talking about? I do not find any such
quote. Pray state page number in your book.

Robinson says on page 110 (last lines) "Infinite totalities do not
exist in any sense of the word (i.e., either really or ideally). More
precisely, any mention, or purported mention, of infinite totalities
is, literally, meaningless" [ROB64].

That quote is followed by a bit more.

By the way: There is a comprehensive index at the end of my book,
including the data of about 170 scholars and the page numbers where
they are mentioned.

Yes, but if you quote somebody, referring only to a book, it is difficult
to find the place that you actually did quote.

Well, when I did learn geometry the reality was that you could actually
not state that a longer distance had more points.

No, you cannot, unless you believe that all real numbers do actually
exist. Then you must think that the points are all there.

Yes, but a longer distance has still not more points. There is a bijection,
as I did show by the geometric construction to divide a line in five equal
parts. In that construction the bijection is used.

> Better: He fails to fail to see that it gives trichotomy of sets.

Pray elaborate, I do not understand this.

Bolzano was correct when he said that set inclusion gives trichotomy.

Ok, set inclusion gives trichotomy. Now use set inclusion to find how
to compare the following two sets: {a, b} and {c, d}. If there is
trichotomy the outcome should be <, = or >.

A long line has more points than a short line, notwithstanding
bijection. (Of course, all this is only correct, if reals actually
exist.)

By set inclusion? Strange. If the lines are at angles to each other, there
is no set inclusion at all.

And you are stating here that a well-ordering can not be defined in ZFC.

Yes, not without additional axioms. Easiest by the axiom that a well
order has been defined.

I do not know of such an axiom. Can you state it? AC merely states that
a well-ordering does exist, not that it can be defined. So, pray, elaborate.

So, what is it. Can you show the proof that a well-ordering can not be
defined? Or at least a source for such a proof? I state that ZFC is
*not* inconsistent with a definable well-ordering of R.

It has been proven by forcing, AFAIK, but being in vacations, I have
no access to literarture presently.

In that case you should seriously review your sources. It has been shown
that a definable well-ordering of the reals is not inconsistent in ZF (see,
even C not needed). AC only states that a well-ordering exists, not that
a definable well-ordering exists. But there are models of ZFC *and* of
ZF that provide a definable well-ordering of the reals.

> Further, all your arguing is in vain unless you can present a well
> ordering working in reality. And you cannot.

What reality? Did you read the reference I presented?

I would like to read the well-ordering.

Work in the model. And look at the reference I gave.

> > But that is wrong, because when a map 'f' is viewed as a set,
> > it is defined as:
> > { (n, s) | n in N, s in P(N), f(n) = s }
> > So when 'f' can not be defined in a model, neither can the set
> > representing 'f'.
>
> When P(N) exists in the model,

Of course it exists if it is a model of ZF: the axiom of the powerset.

> then at least a function f from P(N)
> to ... exists there.

To what?

The identity mapping of P(N) exists when P(N) exist. This is including
the mapping from the singletons of N to N to P(N).

Yes. Such mappings do indeed exist. They are trivial because the properties
of 'f' defining such mappings are almost certainly present in the model.
But what I was stating is that the set of subsets of N is not the *same*
in each model. Each subset is (in ZF) *defined* by a property "phi".
Whether a particular property "phi" is valid depends on the model used.
A subset of N may be a subset in one model, but that is not necessarily
the case in another model.

> It is the similar to the nodes and edges of the
> tree. Every existing node is the end of an edge. There are not less
> edges than nodes. If all elements of P(N) exist, then all elements of
> f exist. And the existence of P(N) is guaranteed by the power set
> axiom.

No argument here, it does not contradict what I stated.

So you think that a set may exist but no identity mapping?

No. Pray try to understand what a model of ZF does do. It may give
severe restrictions of what can be seen as a subset of a set.

I agree.
There is no mapping from N or R. But most set theorists deny that (In
fact, I never met a set theorist who missed the identiy mapping R -->
R in real mathematics. How could it get lost in some model which obeys
the same set of axioms?)

Strawman.

What now follows is a clear example of bad cut and paste. In general a
repeat of what was stated above. One new thing though (I think):

What are the subsets of N in any model containing N? Every combination
of natural numbers is a subset of N. Your "axiom of specification"
seems to be introduced to have a tool to exclude some subsets?

The subsets of N in any model are the subsets that the model allows to
specify. You clearly have no idea how models work, so all your ramblings
about Skolem's countable model of ZF have exactly no validity. And, no,
the "axiom of specification" was not introduced for that purpose. That
axiom was introduced to allow the definition of subsets. Without such
a definition, subsets could not be properly defined. Your statement
"Every combination of natural numbers is a subset of N"
is *not* a definition.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.

Quantcast