Re: (sketch of a) Proof that the set of Real Numbers doesn't exist

From: Piotr Sawuk (piotr5_at_unet.univie.ac.at)
Date: 02/25/05 

Date: 25 Feb 2005 03:10:01 GMT

In article <4217d965$9$fuzhry+tra$mr2ice@news.patriot.net>,
        "Shmuel (Seymour J.) Metz" <spamtrap@library.lspace.org.invalid> writes:
> In <cv1c7q$cb$1@gander.coarse.univie.ac.at>, on 02/17/2005
> at 05:45 PM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:
>
>>exactly! I have seen several definitions of "open set" in R, and
>>since the representation of R does differ, so does this definition.
>
> No. You have seen several models of R, but they are canonically
> isomorphic and homeomorphic. You have not seen multiple topologies on
> R; the definitions of open sets on each are equivalent.

well, I do have a problem with the word "equivalent", especially
since I do not properly know the meaning of the word "canonically".
fact is that in maths there is the concept of "equivalence-relation"
and there is some notion of equivalence beyond any definitions.
several models of R might be isomorphic and homeomorphic, but
this does merely guarantee a certain set of properties to be
in common (most probably them all being complete archimedean fields
and Q being dense), this does merely guarantee some kind of
equivalence-relation for each pair of models. That's why one
needs to ask if your set is an open set in all of them, no matter
how you choose a and b!
>

> complete Archimedean field. Once you have done so, then the fact that
> each non-empty open set contains both rational and irrational numbers
> is a simple theorem and the proof does not need to go back to your
> construction.

thanks for stating the obvious. But as I said, my question is if
the fact that this model is defined to have some Q being dense in
it, if this property is already capable of proofing that each open
set (defined as F(x) for each cauchy-filter F and each element x)
does contain rational and irrational numbers, and that it actually
is an archimedean field. as of yet I haven't seen any such proof.
>
>>What I wish is an algorithm for creating a bunch of models for R.
>
> I don't see the utility.

but I do: it's another tool useful for porting results from one
area of mathematics and its language, to another.
>
>>what Cauchy-filters do offer me is a description of the properties a
>>model of R should fulfill,
>
> No. The properties that a model of R should[1] fulfil are those of a
> complete Archimedean field, nothing more.

nice theorem, unfortunately the set-theoretical approach doesn't follow
this path, neither the model with cauchy-filters nor the model with
dedekind-cuts are proven to fulfill this axiom, as far as I could see.
the person with the tinyurl did attempt to prove something in that
direction and used another theorem you attempted to prove by relying
on the property that (a + (b-a)/3, b - (b-a)/3) is an open set, which
I still can not believe.

since the beginning of my studies I had a problem with these things,
then I discovered something noone ever told me: that R is defined to
fulfill the archimedean axiom. then I found other models, and again
the proof of them fulfilling the archimedean axiom was somehow missing.
>

> There is no such thing. You need to make explicit what quantifiers are
> involved and what their scopes are. You can do that in English, but it
> takes more care than you have been willing to invest in your articles.

well, I'm trying to.
>
>>I3_x=intersection(I3_x_n),I3_x_n=(x,q1(n)), q1_x(n)>x and {q1_x(n)}
>>is {q>x in Q}.
>
> Did you mean
>
> I3_x=intersection(I3_x_n),I3_x_n=(x,q1_x(n)), q1_x(n)>x and {q1_x(n)}
                                        ^^
> in {q>x in Q}
>
> ? Why specify ">x" twice?

apart from my little omission marked above these two differ by the word
"in" (I assume you meant "subset" just like I did below) replacing the
equivalence I used. "for each interval with rational endpoint" was what
tinyurl used for defining his intersection I3, and thereby the range
of q1_x needs to *be* the set {q>x in Q} and not merely be contained.
that {q>x in Q} is a subset of the range of q1_x would already follow
from "q1_x:N->Q and q1_x(n)>x", while I need q1_x to be surjective!
writing only "q1_x(n)>x" does not guarantee that {q>x in Q} is a subset
of the range of q1_x!

so, again:

I asked
>>>>>>>> can you prove that the intersection of all open intervals with a
>>>>>>>> fixed left point x and all rational numbers as the right-most points
>>>>>>>> wouldn't differ from my f(x)?

where f(x) was defined as intersection of F(x,y) with y over all irrational
numbers, and F(x,y) being defined by
>>>>>>>>>> we have a function F(x,y):={q element of Q|x<q<y} (and empty
>>>>>>>>>> otherwise) defined on RxR.

to which tinyurl replied with something in the spirit of:

Below n ranges over natural numbers and x is a fixed irrational number:

I3_x=intersection(I3_x_n),I3_x_n=(x,q1_x(n)),q1_x(n)>x & {q1_x(n)} == {q>x in Q}.
I4_x=intersection(I4_x_n),I4_x_n=(x,r_x(n)),r_x(n)>q1_x(n)>x & r_x(n) in R\Q
I3_x_n included in I4_x_n => I3 subset of I4.
I1_x=I5_x=intersection(F(x,y)),y>x in R\Q

to which I had the
>> question: does r_x(n) exist such that for-all F(x,y) with y>x in R\Q
>> there exists I4_x_n intersected with Q subset of F(x,y)? Anyway,
>> even if it exists, then only I3_x intersected with Q subset of f(x)
>> has been proven, and not I3_x subset of f(x)!
>>

tinyurl's proof continued:

I2_x=intersection(I2_x_r),I2_x_r=(x,q2_x(r)),q2_x(r)>r>x,q2_x(r) in Q,r in R\Q.
...=> I1_x subset of I2_x.
>>
>> question: does there exist above q2_x:R\Q -> Q such that for all I3_x_n
>> there exists some I2_x_r subset of I3_x_n? I think it does: it's *the*
>> surjective function from the uncountable subset of R\Q to the countable
>> subset of Q which does exist according to some proof in set-theory. 

-- 
Better send the eMails to netscape.net, as to
evade useless burthening of my provider's /dev/null...
P