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 03:10:31 GMT

In article <4217da77$11$fuzhry+tra$mr2ice@news.patriot.net>,
        "Shmuel (Seymour J.) Metz" <spamtrap@library.lspace.org.invalid> writes:
> In <cv1jfb$cl$1@gander.coarse.univie.ac.at>, on 02/17/2005
> at 05:45 PM, piotr5@unet.univie.ac.at (Piotr Sawuk) said:
>
>>again I forgot to explicitely write down the dependencies:
>
>>1) for-all r1 and r2 non-equal in R there exists q in Q such that
>>either r1 <_R q <_R r2 or r2 <_R q <_R r1 with <_R being transitive
>>and anti-reflexive.
>
>>2) for-all r1 and r2 in R there exists q in Q such that from r1 <_R
>>r2 follows that r1 <_R q <_R r2, again with <_R being transitive and
>>anti-reflexive.
>
> Why not the simpler for all r1 and r2 in R with r1 <_R r2, there
> exists q in Q with r1 <_R q <_R r2?

you're right, since <_R is transitive it actually doesn't make a
difference to 2). however, I wished to emphasize that r1 <_R q <_R r2
doesn't neccessarily hold forall r1,r2 in R, that r1 and r2 need
to be compareable!
>
>>the difference is that 1) does enforce anti-symetry, linearity and
>>extensionality from <_Q (since pred_(<_R)(x) non-equal pred_(<_R)(y)
>>for x<y is a result of some q between x and y) onto <_R restricted
>>to QxQ. the point is that if <_R is not a linear order, then 2)
>>could still hold!
>
> Please clarify your nomenclature.

you mean "anti-symetry", "linearity", "extensionality", "pred_<"
and "linear order"? Well, I guess you could read them up in any
resource which attempts to explain the basic properties of ordinal
numbers in set-theory. So, I'll assume you meant "<_R" and "<_Q":

"<_Q" is the set {(f(x),f(y))|x<y both in Q} for a given injective
f:Q->R and a given "<" on Q. Q is the set of rational numbers and
is supposed to fulfill the archimedean axiom with "<".

"<_R" is any transitive and anti-reflexive relation on R with the
property that field(<_R)==R iff field(<_Q)==f(Q). (again f is the
injection from Q to the model of R in question, and f(Q) does denote
its range.)

A model of R is a set of the cardinality of the powerset of N
together with some "<_R" defining the topology on R
.
the question is what other properties a model of R is supposed
to fulfill (other than the existance of an injective f:Q->R
under which Q is dense in R, and under which all converging
sequences do converge in R). the question is if "<_R" does
need to be defined as a linear order.
>
>>a strict quasi-order is defined to be only transitive and
>>anti-reflexive, this does imply that some pair of elements could
>>exist such that neither "<" nor ">" nor "=" is true. the members of
>>such a pair are defined as "incompareble", written as x _|_ y, and
>>otherwise they are compareable.
>
> R is an ordered field, so all elements are comparable in that sense.

that's the question! maybe it is true for dedekind-cuts, but other
models merely define some injective f:Q->R with "<" only defined
on Q to be an ordered field, nothing whatsoever is said about
the existance of a linear order on R! maybe you could create a
well-ordering on R according to AC, but how do you prove that
some linear order on R does exist for all models of R?

> With only a partial order it wouldn't be R. But even if you were
> talking about some R1 that has only a partial order, a map of Q->R'
> could only preserve the orders of elements of Q, not of R'.

since "<_R'" is transitive and Q is dense in R' under "<_R'", so naturally
a map of Q->R' could preserve the orders of elements of R'\Q in some sense.

however, I was not talking about any mapping preserving order of
elements of R'\Q, I was talking about a mapping preserving order
of *compareable* elements of f(Q)!
>

>>I don't know how you define homomorphism,
>
> Then maybe you should do some reading. A homomorphism of fields is a
> function that preserves zero, unity, addition and multiplication.

so you actually meant "homomorphism of fields"? then how do you
prove that those properties you mentioned already are sufficient
for showing that they preserve the order of Q and are injective?
>
>>but an isomorphism is a bijection, and it is called
>>structure-preserving when both directions (i.e. also the inverse
>>function) are structure-preserving,
>
> No. A bijection that is not structure preserving is, by definition,
> not an isomorphism.

again, I think you are talking about some kind of "isomorphism of fields",
while in set-theory "structure-preserving isomorphism" was explicitely
defined as a structure-preserving bijective function where its inverse
is structure-preserving too.
>
>>I am using the language from Graph-theory since I do not know enough
>>about the vocabulary on trees.
>
> Then why did you write "tree" and "root"?

"tree" is a notion of graph-theory, and "root" I know from some
combinatorical lecture where we talked about counting how many
trees one would obtain under certain circumstances if one would
declare one node of the tree to be the root.
>
>>a tree is a graph without circles,
>
> ITYM acyclic DIRECTED graph.

no, "DIRECTED" is a consequence from the fact that removing a
node does result in 2 seperate trees iff that node had more than
one neighbours, and that a single node which I chose to call root
is only a node in one of those two trees.
>
>>additionally I put a "direction" on the set of neighbours which are
>>seperated from the root by the given node (what you called
>>children).
>
> If they're separated then they're not neighbors.

since there seems to be a misunderstanding, let me re-phrase that.
>

> Recipe for tiger stew: first catch the tiger. You've sort of kind of
> defined a structure that does not include all elements of R, or even
> all elements of Q. Whether you could construct other structures that
> included one of the missing elements is irrelevant.

ok, again:

>>>> for example: consider a tree of infinite depth and width, each node
>>>> is a real number in [0,1] written in base 10. the root is 0, and
>>>> neighbours of 0 are 0.1, 0.2, ..., 0.9. apart from those 0 does
>>>> also have 0.01, ..., 0.09 and 0.001, ..., 0.009 ad infinitum as
>>>> neighbours.

apart from all finite paths between the root and some rational
numbers I also allow for paths of infinite length between the
root 0 and elements of R with infinitely many digits after "0."
and define that those nodes do not have any children.

additionally, for each node x I put a "direction" on the set of
neighbours of x which are seperated from the root by the given
node x (those neighbours are what you called
>>children).

>>>> the original ordering "<" is defined such that all
>>>> brances of 0 are larger than 0, 0.01 and its subtree is larger
>>>> than 0.001, ..., 0.009 and their subtrees, and so on. each node
>>>> does again have a similar structure as 0, except that the first
>>>> digits are fixed to be the same as the digits of the node. then

with such a "<" which does define sort of a direction on some of the
children of each node,
>>>> one could define a quasi-order-preserving f2:Q->R such that Q
>>>> is quasi-dense in R. what is still missing is a relation between
>>>> 0.1, ..., 0.9 and similar elements of subtrees. f2 might map
>>>> 1/10 to any such incompareable number, and one could re-arrange
>>>> that mapping such that 0.1 is selected and define a new "<" for
>>>> that number. because of denseness and whole R being the image
>>>> of the original "<", the transitivity is fulfilled at each step.
>>>> the result is f1 together with an ordering defined on RxQ and QxR
>>>> and QxQ when seen through this function f1. is the ordering on
>>>> RxR the unique extension of this constructed ordering? does it
>>>> really fulfill the same axioms as the ordering on Q? when one
>>>> does assume that the same axioms are fulfilled, does this then
>>>> make the ordering unique? has this topic been reserched in the
>>>> past? where could I read up on such things?
>

>>if you do have equivalence-classes then you must walk the long way
>>through proofing that properties are independant of the
>>representants used,
>
> So? That involves less work than dealing with ultrafilters.
>
>>and you need to say something about the size and number of
>>equivalence-classes.
>
> No I don't; I need only show that the construction yields a complete
> Archimedean field.

OK, then please show me that dedekind-cuts can be proven to fulfill
these axioms without first proving that only equivalence-classes
of size 2 are required once for each rational number. you can't!

actually you still did not prove that dedekind-cuts *do* have no
other ambiguity-issues than "whether to use (-oo,q] and (q,oo) or
to use (-oo,q) and [q,oo)"! as I said, I suspect that there might
be some ambiguity with some nodes of the tree I constructed above
who's path to the root 0 is of infinite length. for example 1/3
is such a node, it is a child-node of another node x, and that
node x does have other children which have a distance of 0... 

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

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