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

From: Shmuel (Seymour J.) Metz (spamtrap_at_library.lspace.org.invalid)
Date: 01/06/05 

Date: Thu, 06 Jan 2005 18:13:32 -0500

In <craehu$ap$1@gander.coarse.univie.ac.at>, on 01/06/2005
   at 05:04 AM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:

>Yes, I already figured out that there would exist a bijection between
>an infinite set and its powerset if the notion of "function" would
>not be limited to countable steps with finite amount of operations,
>or something like that.

Not in ZFC there isn't.

>and of course extending natural numbers into infinity

What do you mean by that? Completing the Rationals is not extending
them to infinity.

>such that some numbers simply have an infinite length

Numbers don't have lengths, infinite or otherwise.

>with the additional attribute that all such infinite numbers would be taboo in usage

What do you mean by taboo? If you can define them then you can use
them. Arm waving, of course, doesn't count, but there is no taboo
against coining new nomenclature.

>Real numbers do circumvent this taboo

No. There's nothing to circumvent.

>However, I do have a problem in understanding why an open set does not have a minimum,

Because of the way that you define the standard topology for an
ordered set. If S contained in Q has a minimum, then any open interval
containing that minimum would also contain a rational number less than
the minimum, showing that the set is not open.

>simply because I imagine each Real number as lying between 2 unique
>rational numbers

No, the bracketing rational numbers are not unique.

>but then this would defy the idea that some rational numbers are taboo

Where do you get the idea that any well defined construct is taboo?

>while their irrational equivalent is not,

What do you mean by equivalent?

>and again we would get Real numbers which are taboo (and which
>require a powerset of R to make them accessible)

No.

>and the notion of completeness would lose its absoluteness since it
>would only be relative to the level of taboos I wish to get rid of...

The notion of completeness has nothing to do with taboos.

>I just would like to learn methods of how to express above paragraph
>in a much more logical and comprehensible way.

You can't, because it is not logical.

In <crampv$b2$1@gander.coarse.univie.ac.at>, on 01/06/2005
   at 05:04 AM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:

>I clearly said "above paragraph" since I was referring to the
>paragraph about Q being dense in R.

But the denseness of Q in R does not prevent the intersection of the
intervals in Q from being empty.

>The point is that according to the definition of R all countable
>sequences of (open) intervals do converge towards a single number

No. The definition of R doesn't include any definition of convergence
of a sequence of open intervals. Further, if you defined such a notion
you would still have to prove that it had any relevance to the
intersection of the intervals in the sequence. Finally, what you have
are sequences of intervals in Q, which are not the same as sequences
of intervals in R.

>If there would exist an uncountable "sequence" of intervals,

What do you mean by uncountable sequence? How do you prove that the
properties you need carry over from sequences?

>However, this prove for either R being countable or the intersection
>being empty, does most probably again contain many errors, and
>therefore I did not even mention it, but merely assumed that some of
>the readers might have come up with a similar proof. For those
>readers who do believe in this proof for f(x) being empty I did add
>the possibility of a countable intersection of uncountable sets
>which according to the definition of R should contain an element of
>R

No. Again, the sequences that you were talking about were sequences of
intervals in Q; the intersections are empty but the intersections of
the closures in R are non-empy.

>Doh! I completely forgot! First I wanted to write x instead of y,
>but then I noticed that G(x,z) isn't defined on the irrational x
>already given as an argument to g, since no guarantee does exist for
>the existance of any irrational number r lying between another
>irrational and a rational number,

The guaranty is that Q is dense in R.

>Therefore I chose

Chose a constant or chose a function. If the latter then you need to
write out what you intend it to be a function of, e.g., G(y(z),z).

>Or do you think that such a y would not exist?

Clearly.

>why?

Because Q is dense in R. Consider any rational number in (y,(y+z)/2).

>(while only using "<" on Q plus some well-chosen irrational numbers
>which are far enough away from the element to get compared)?

That would be *ALL* irrational numbers.

>Anyway, since such a common element is too good to be true I should
>rather define g(x):=intersection of F(x,C(s))

That's undefined because

>(where s is the intersection of all G(y,z) over all rational numbers
>y) over all rational numbers z.

That's empty.

>No, the issue is the rational number q>x where all irrational
>numbers returned by C(G(y,z)) are larger than q.

Write it all out in Mathematical notation, with all of the pieces in
place, and not forgetting any functional dependencies. It may be
clearer ro you then. 

-- 
     Shmuel (Seymour J.) Metz, SysProg and JOAT
     ISO position; see <http://patriot.net/~shmuel/resume/brief.html> 
We don't care. We don't have to care, we're Congress.
(S877: The Shut up and Eat Your spam act of 2003)

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