# Rational numbers, irrational numbers: each dense in real numbers

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### Ross A. Finlayson

Sep 18, 2007, 3:21:21 PM9/18/07
to
In ZFC, with standard definitions of the real, rational, and
irrational numbers, let p_i be an irrational number between zero and
one for i from a suitably large well-ordered index set X. With the
well-ordering of the index set, let the i'th element p_{i+1} be an
irrational number between zero and p_i, where i+1 is the least element
of the well-ordering X_i setminus i, that is defined to equal X_{i
+1}. There are uncountably many irrational numbers less than each p_i
and greater than zero, else the irrational numbers are countable (or
the real numbers are not standard). Define P to be comprised of the
p_i's. There exists a rational number q_i between p_i and p_{i+1},
else the rational numbers are not dense in the reals thus that between
any two irrational numbers there is a rational number. For each of
the irrational p_i's, there thus exists at least one unique rational
q_i between p_i and p_{i+1}, and infinitely many. Let the ordered
pair (p_i, q_i) be an element of a function, as a set, from P to Q.
If there is an uncountable set P of irrational numbers in (0,1), then
there is a 1-to-1 function defined by the set {(p_i, q_i), i E X} from
uncountable P to a subset of Q the rational numbers, and there is thus
an injection from an uncountable set of irrational numbers to a subset
of the rational numbers, a subset of a countable set is countable.

Contradiction ensues, from that an uncountable set injects into a
countable set. Which presupposition is false? Perhaps it is so that
the irrationals can not be well-ordered in their normal ordering, but,
via separation, for any given subset of the irrational numbers in (0,
p_i) there exists (0, p_{i+1}) for any p_i such that 0 < p_{i+1} <
p_i, or the irrationals are not dense in the reals. Perhaps it is so
that there are no uncountable subsets of the irrationals in (0,1), but
then the irrationals wouldn't be uncountable. The rationals are dense
in the reals so between any distinct p_i and p_{i+1} there exists a
q_i, or p_i = p_{i+1} and p_i =/= p_{i+1}. In ZFC there exists a
suitably large well-ordered index set.

Quantify over sets, ZF(C) and/or the standard definitions of the real,
rational, and irrational numbers are thus inconsistent.

Ross

--
Finlayson Consulting

### finite guy

Sep 18, 2007, 3:27:47 PM9/18/07
to

Uh huh...
Can you say the same thing without using 'infinity' since it is not
finite?
Fundamental point here...
It's simply why you can't have circles, spheres or cubes.
Regards

### Virgil

Sep 18, 2007, 4:00:16 PM9/18/07
to

"Ross A. Finlayson" <r...@tiki-lounge.com> wrote:

> In ZFC, with standard definitions of the real, rational, and
> irrational numbers, let p_i be an irrational number between zero and
> one for i from a suitably large well-ordered index set X.

I very much doubt that Ross can exhibit such an X explicitly.

> With the
> well-ordering of the index set, let the i'th element p_{i+1}

How is it that the ith element is not p_i??

[Remaining nonsense snipped]

### finite guy

Sep 18, 2007, 4:29:43 PM9/18/07
to
On Sep 19, 4:00 am, Virgil <vir...@comcast.net> wrote:

Don't take this as a criticism:
y'all believe in them circularity thingies, don ya? :-)

### tommy1729

Sep 18, 2007, 4:33:39 PM9/18/07
to
Ross wrote:

so ok.

between every pair of irrationals lies a rational.

and between every pair of rationals lies an irrational.

seems like the set 2n and 2n + 1 ( even and odd integers)

so implies equal density and both countable....

or both uncountable...

however

assume they are both uncountable.

then if the rationals are uncountable ; so are the integers...

that cant be right of course.

so assume irrationals are countable too.

but then there has to be a mapping from all rationals to all irrationals...

this means we need a function f(a,b) -> gives all irrationals for integer a and b.

f(a,b) does not exist.

even f(a,b,c) wont do.

for instance f(a,b,c) = a sqrt(2) + b sqrt(3) + c sqrt(5)

does not give all irrationals.

and if f was different then we would have left out the numbers of type a sqrt(2) + b sqrt(3) + c sqrt(5).

regards
tommy1729

### MoeBlee

Sep 18, 2007, 5:37:02 PM9/18/07
to
On Sep 18, 12:21 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
> In ZFC, with standard definitions of the real, rational, and
> irrational numbers, let p_i be an irrational number between zero and
> one for i from a suitably large well-ordered index set X. With the
> well-ordering of the index set, let the i'th element p_{i+1} be an
> irrational number between zero and p_i, where i+1 is the least element
> of the well-ordering X_i setminus i, that is defined to equal X_{i
> +1}. There are uncountably many irrational numbers less than each p_i
> and greater than zero, else the irrational numbers are countable (or
> the real numbers are not standard). Define P to be comprised of the
> p_i's. There exists a rational number q_i between p_i and p_{i+1},
> else the rational numbers are not dense in the reals thus that between
> any two irrational numbers there is a rational number. For each of
> the irrational p_i's, there thus exists at least one unique rational
> q_i between p_i and p_{i+1}, and infinitely many. Let the ordered
> pair (p_i, q_i) be an element of a function, as a set, from P to Q.
> If there is an uncountable set P of irrational numbers in (0,1), then
> there is a 1-to-1 function defined by the set {(p_i, q_i), i E X} from
> uncountable P to a subset of Q the rational numbers,

Wrong. Your homework assigment is to identify the fallacy in that
argument.

MoeBlee

### MoeBlee

Sep 18, 2007, 5:42:24 PM9/18/07
to
On Sep 18, 1:33 pm, tommy1729 <tommy1...@gmail.com> wrote:

> between every pair of irrationals lies a rational.
>
> and between every pair of rationals lies an irrational.
>
> seems like the set 2n and 2n + 1 ( even and odd integers)
>
> so implies equal density and both countable....
>
> or both uncountable...

Whatever your definition of 'equal density' (or even if you just mean
that between any two distinct rationals there is an irrational and
between any two distinct irrationals there is a rational), you have
not shown any proof from axioms and a specified logic (especially here
where the context is ZFC which is the set of sentences derivable by
first order logic from the ZFC axioms) that equal density entails both
countable or both uncountable.

MoeBlee

### Ross A. Finlayson

Sep 18, 2007, 5:43:39 PM9/18/07
to
On Sep 18, 1:00 pm, Virgil <vir...@comcast.net> wrote:

> "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
>
> > In ZFC, with standard definitions of the real, rational, and
> > irrational numbers, let p_i be an irrational number between zero and
> > one for i from a suitably large well-ordered index set X.
>
> I very much doubt that Ross can exhibit such an X explicitly.
>

Any ordinal equivalent to the set of irrationals would do. (Ordinals
are sets of lesser ordinals.)

> > With the
> > well-ordering of the index set, let the i'th element p_{i+1}
>
> How is it that the ith element is not p_i??
>
> [Remaining nonsense snipped]

Ah, mea culpa, "{i+1}'th".

Ross

### se...@btinternet.com

Sep 18, 2007, 5:51:01 PM9/18/07
to
On 18 Sep, 21:33, tommy1729 <tommy1...@gmail.com> wrote:
> so ok.
>
> between every pair of irrationals lies a rational.

More than one.

>
> and between every pair of rationals lies an irrational.

Again, more than one. In fact even more than in the previous case.

>
> seems like the set 2n and 2n + 1 ( even and odd integers)

No it does not.

>
> so implies equal density

Again no.

### William Hughes

Sep 18, 2007, 6:33:57 PM9/18/07
to
On Sep 18, 3:21 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
> In ZFC, with standard definitions of the real, rational, and
> irrational numbers, let p_i be an irrational number between zero and
> one for i from a suitably large well-ordered index set X. With the
> well-ordering of the index set, let the i'th element p_{i+1} be an
> irrational number between zero and p_i, where i+1 is the least element
> of the well-ordering X_i setminus i, that is defined to equal X_{i
> +1}. There are uncountably many irrational numbers less than each p_i
> and greater than zero, else the irrational numbers are countable (or
> the real numbers are not standard). Define P to be comprised of the
> p_i's. There exists a rational number q_i between p_i and p_{i+1},
> else the rational numbers are not dense in the reals thus that between
> any two irrational numbers there is a rational number. For each of
> the irrational p_i's, there thus exists at least one unique rational
> q_i between p_i and p_{i+1},

This is the fallacy. q_i is not unique. There are an uncountable
number of pairs
p_i and p_{i+1} and only a countable number of q_i available (q_i
must be a rational in [0,1]). Thus, although we can choose a q_i for
each index i, we cannot choose the q_i in such a way that
q_i =/= q_j for i =/= j.

>and infinitely many. Let the ordered
> pair (p_i, q_i) be an element of a function, as a set, from P to Q.
> If there is an uncountable set P of irrational numbers in (0,1), then
> there is a 1-to-1 function defined by the set {(p_i, q_i), i E X} from
> uncountable P to a subset of Q the rational numbers,

since q_i is not different for each index i, this function is not 1-
to-1.

- William Hughes

### Ross A. Finlayson

Sep 18, 2007, 7:23:55 PM9/18/07
to

the rationals, only the density (denseness) of the rationals in the
reals. That there is a contradiction is so, I agree, but, to avoid
that contradiction by denying the denseness of the rationals in the
in the reals.

The "uncountably many pairs" of distinct irrational numbers are
generated in a particular way thus that the open intervals (each
containing infinitely many rationals) connecting them are disjoint.
Then, where there are at least that many rational numbers, at least
one for each of those intervals, the collection of ordered pairs of
the irrational upper limit (least upper bound) of those intervals and
any rational number in the interval, where at least one exists,
describes a 1-to-1 function, from each of an uncountable collection of
irrationals to a subset of the rationals.

I'm reminded of Friedman's quote that well-ordering the rationals
requires an arbitrarily large countable ordinal. The physical
universe, as an infinite collection of items, mathematically is its
own powerset, presenting a realizable counterexample to the powerset
result.

So, why else could there not be constructed uncountably many
subintervals of the unit interval, of the form (0, p_i), for
irrational p_i? That's a different form from the problem here posed
but represents an alternative form of proof of that an uncountable
subset of the irrationals injects into the rationals, although again
observing the denseness property of the rationals in the reals.

Ross

--
Finlayson Consulting

### MoeBlee

Sep 18, 2007, 7:47:34 PM9/18/07
to
On Sep 18, 4:23 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:

> No presumptions were made about the countability or lack thereof of
> the rationals, only the density (denseness) of the rationals in the
> reals. That there is a contradiction is so,

You've shown no contradiction, since you've not shown a bijection from
an uncountable set of irrationals to a set of rationals. You think
you've shown such a bijection only because you don't know how to make
a mathematical argument. You continue to fail Real Analysis 101 until
you recognize the mistake in your argument.

> I'm reminded of Friedman's quote that well-ordering the rationals
> requires an arbitrarily large countable ordinal.

> So, why else could there not be constructed uncountably many

> subintervals of the unit interval, of the form (0, p_i), for
> irrational p_i?

Aside from whatever you might personally mean by 'constructed', there
do exist uncountably many invervals of the form (0 p) for p in {p | p
irrational and 0<p<1}.

MoeBlee

### lwa...@lausd.net

Sep 18, 2007, 8:46:34 PM9/18/07
to
On Sep 18, 1:33 pm, tommy1729 <tommy1...@gmail.com> wrote:
> between every pair of irrationals lies a rational.
>
> and between every pair of rationals lies an irrational.

OK, so here we go again with the common argument,
which I admit is rather intuitive, that given two
sets X and Y and an order < on X union Y, if
between every two elements of X there exists an
element of Y and vice versa, then X and Y must
have the same cardinality.

It is indeed true in ZFC that if < is a wellorder
then the cardinalities of X and Y must differ by
at most one.

(Proof sketch: since < is a wellorder, every
element of X union Y has a successor except the
maximum element of the union. The successor of
an element of X must be an element of Y, since
if it were an element of X there would be two
elements of X without an intervening element of
Y, and similarly the successor of an element of
Y must be in X. The successor function itself
would be a bijection between X and Y, perhaps
with a single element -- the maximum.)

But if < is not a wellorder, as is
the case of R in ZFC, then the proof fails.

But RF, if I recall correctly, believes that < is
indeed a wellorder on R (and he gives the name
"iota" to the smallest positive real). If he is
working in a set theory in which < is actually a
wellorder on R, then the proof works. But in a
set theory in which < doesn't wellorder R, such
as ZFC, then the proof fails.

### William Hughes

Sep 18, 2007, 9:05:40 PM9/18/07
to

This is not possible. Let's check the construction

let p_i be an irrational number between zero and
one for i from a suitably large well-ordered index set X. With the
well-ordering of the index set, let the i'th element p_{i+1} be an
irrational number between zero and p_i, where i+1 is the least
element
of the well-ordering X_i setminus i, that is defined to equal X_{i
+1}.

You are trying to iterate through an uncountable set. You will not
get very far. In particular, p_j will not be defined if j is not
equal to i+1 for some i in X. Nor will p_k, where k>j. So p_i will
not
be defined for i greater than or equal to the first limit ordinal.

- William Hughes

### finite guy

Sep 18, 2007, 9:05:47 PM9/18/07
to

"infinitely many" : I presume that you presume 'infinitely many finite
numbers'? Wuh...?

### finite guy

Sep 18, 2007, 9:13:59 PM9/18/07
to
> - William Hughes- Hide quoted text -
>
> - Show quoted text -

'suitably large' : that's inane really... because you end up calling
it infinite when it is finite. Uncountable does not imply infinite
anywhere, does it?... It just means you 'aint got a clue of the size
of the number.'
How many people are alive in the world... NOW...? Must be infinite
because we can't count them. Yuh, ok, sure thing... Oh, is yours
bigger than mine? Even if so, that only makes 'two'.
Are you a zealot of some sort that will not reason? :-)

### Ross A. Finlayson

Sep 19, 2007, 12:26:10 AM9/19/07
to

That's why there is "transfinite induction."

http://en.wikipedia.org/wiki/Transfinite_induction

With any interval of the reals containing uncountably many
irrationals, and each subinterval bounded by irrationals as well
having that many (using standard definitions of the numbers and ZFC),
as there are that many, well-ordered via choice, then the first
infinite limit ordinal and following infinite limit ordinals have for
each a distinct p_i.

Ross

--
Finlayson Consulting

### Ross A. Finlayson

Sep 19, 2007, 12:31:13 AM9/19/07
to

There's a trivial injection from the rationals to a subset of the
irrationals. Thus, with an injection either way, in basically trivial
augmentations of the set, the Cantor-Bernstein theorem gives a
bijection, where Cantor-Schroeder-Bernstein gives a bijection for
surjections either way. That is to say: contradiction ensues, rather
obviously, clearly, as an exercise, etcetera.

(Infinite sets are equivalent.)

Ross

--
Finlayson Consulting

### Ross A. Finlayson

Sep 19, 2007, 12:43:23 AM9/19/07
to

In this context it seems X would be P and Y would be the set {q_i:
(p_i, q_i) E F}, where F is the putative injection from P to Q. Well-
ordered by the index set X as previously labelled, then as you write
above "X and Y must have the same cardinality."

About non-standard notions of the real numbers, for example as I see
them the dually partially ordered ring with a rather restricted
transfer principle and complete ordered field, those real numbers
which must fulfill all qualities of the continuum of real numbers,
that's in a post-Cantorian, post-Goedelian, and post-Euclidean sense,
i.e. A theory the null axiom theory. Then, in consideration of the
structure of the continuum of real numbers as a contiguous sequence of
points, in the ultimate space wherein they reside, a particular
infinity (which I was calling the unit scalar infinity but now would
call simply the unit infinity as it's not necessarily a scalar) for
which iota as a displacement multiplied by that particular unit
infinity yields a unit vector.

Here, I'm more concerned with the careful use of the standard
definitions of the numbers and illustrating these properties to hold
within ZFC, even though I say ZF is inconsistent because
quantification over sets yields its universe which is non-well-
founded.

Ross

--
Finlayson Consulting

### lwa...@lausd.net

Sep 19, 2007, 1:51:41 AM9/19/07
to
On Sep 18, 12:27 pm, finite guy <adamle...@amnet.net.au> wrote:
> It's simply why you can't have circles, spheres or cubes.

OK, it seems as if there have been many so-called
"cranks" who have appealed to geometry in order
to justify their claims regarding set theory. And
there has been a wide variety of views on this, from
RF's belief that a line segment contains only
countably many points -- and that points may have
successors on the line -- to WM's "set geometry,"
and now finite guy's belief that circles and spheres
cannot exist.

Let us consult both Euclid's original postulates, and
the modern postulates given by Hilbert, and see how
the so-called "cranks" and their geometries measure
up to Euclid and Hilbert:

http://www.friesian.com/space.htm

We begin with RF and his claim that points on a line
segment may have successors -- infinitesimals, to
be sure. Several posters have made such a claim,
and the usual response is that it would contradict
Euclid's First Postulate:

First Postulate: To draw a line from any point to any point.

To me, this is a bit vague. Was Euclid merely saying
for every pair of points, there exists a line that
contains both points? Or was Euclid implying that
_between_ every two points (on the line that
contains them, of course) there must exist yet
another point?

If the former, then not only may points have
adjacent successors, but even _finite_ geometries
are not excluded:

http://en.wikipedia.org/wiki/Finite_geometry

Even in the latter case, clearly if between every two
points there exists another point, then a line must
be dense and consist of infinitely many points. But
still, that doesn't imply that a line must have
_uncountably_ many points. (Of course not, since
uncountability didn't exist in Euclid's day.) Indeed,
recall that Euclid dealt mainly with constructions. I
fail to see how any of Euclid's axioms imply the
existence of any segment whose length is not a
constructible number, such as cbrt(2) (Doubling the
Cube) or pi (Squaring the Circle). So Euclid's
Axioms imply the existence of only countably
many points.

(Before we leave Euclid, one can't help but wonder
about Euclid's Fifth Axiom -- that's _Axiom_, not his
Fifth _Postulate_:

Fifth Axiom: The whole is greater than the part.

which seems to imply the nonexistence of
Dedekind infinite sets, or even the Third Axiom:

Third Axiom: If equals be subtracted from equals, the remainders are
equal.

applied to Dedekind infinite sets. It's interesting
how no one seems to bring that one up!)

Of course, the defects in Euclid are supposed to
be remedied in Hilbert. Returning to the link:

http://www.friesian.com/space.htm

we see that if we only admitted the axioms
labeled "Axioms of Incidence," it is obvious that
a finite model exists. The "Axioms of Order" do
prove that infinitely many points exist. And
finally, Archimedes' Axiom implies that no
non-Archimedean segments (infinitesimals)
exist, so RF geometry is no longer a model.

On the surface, the Axiom of Line Completeness
appears to imply that there exist uncountably
many points (just like the complete metric
space R):

(Axiom of Line Completeness) An extension of a
set of points on a line with its order and congruence
relations that would preserve the relations existing
among the original elements as well as the
fundamental properties of line order and congruence
that follow from Axioms I-III, and from V,1 is impossible.

In other words, if L is a line that satifies the
previous axioms, then one can't add points
to L and still satisfy the axioms. It's not
obvious to me how this must imply that there
are uncountably many points. Indeed, since the
other axioms imply only the existence of
segments whose lengths are constructable, one
could argue that the Axiom of Line Completeness
prevents us from adding points to a line whose
distance from each other is not constructable
(like cbrt(2) and pi).

So where does one get the idea that there exist
uncountably many points anyway? When I took
high school geometry, the very first postulate in
the book is the Ruler Postulate:

http://www.mnstate.edu/peil/geometry/C2EuclidNonEuclid/3Ruler.htm

The Ruler Postulate clearly states that there
exists a bijection between the set of real numbers
and the set of points on a line, and since the
former is uncountable, so is the latter. But the
Ruler Postulate is not listed among the axioms of
either Euclid or Hilbert, unless the latter's Axiom
of Line Completeness is intended to be
equivalent to the Ruler Postulate.

Perhaps I'm wrong, and I'd like to see how exactly
one implies the existence of uncountably many
points (not real numbers, but _points_) using only
Euclid or Hilbert (not the Ruler Postulate, unless,
once again, Line Completeness is intended to be
its equivalent).

What about finite guy? FG's claim that no circles
exist clearly contradicts Euclid's Third Postulate,
yet Hilbert says nothing about circles. In the
model of geometry in which only constructable
points exist, circles may exist, but a segment of
measure pi doesn't (Squaring the Circle). FG also
tells us that cubes don't exist, and he presents
Fermat's Last Theorem as his "justification," but
FLT has nothing to do with geometry.

### David C. Ullrich

Sep 19, 2007, 7:33:46 AM9/19/07
to
On Tue, 18 Sep 2007 16:33:39 EDT, tommy1729 <tomm...@gmail.com>
wrote:

No, it doesn't imply that. "Seems like" is not a proof.

>however
>
>assume they are both uncountable.
>
>then if the rationals are uncountable ; so are the integers...
>
>that cant be right of course.
>
>so assume irrationals are countable too.
>
>but then there has to be a mapping from all rationals to all irrationals...
>
>this means we need a function f(a,b) -> gives all irrationals for integer a and b.
>
>f(a,b) does not exist.
>
>even f(a,b,c) wont do.
>
>for instance f(a,b,c) = a sqrt(2) + b sqrt(3) + c sqrt(5)
>
>does not give all irrationals.
>
>and if f was different then we would have left out the numbers of type a sqrt(2) + b sqrt(3) + c sqrt(5).
>
>regards
>tommy1729

************************

David C. Ullrich

### Denis Feldmann

Sep 19, 2007, 7:41:02 AM9/19/07
to
David C. Ullrich a écrit :

> On Tue, 18 Sep 2007 16:33:39 EDT, tommy1729 <tomm...@gmail.com>
> wrote:
>
>> Ross wrote:

The first mistake in tommy1729 post is answering to Ross. That mistake
is HUGE

>>
>>> Ross
>>>
>>> --
>>> Finlayson Consulting
>>>
>> so ok.
>>
>> between every pair of irrationals lies a rational.
>>
>> and between every pair of rationals lies an irrational.
>>
>> seems like the set 2n and 2n + 1 ( even and odd integers)
>>
>> so implies equal density and both countable....
>>
>> or both uncountable...
>
> No, it doesn't imply that. "Seems like" is not a proof.

Yes, this is the second mistake

>
>> however
>>
>> assume they are both uncountable.
>>
>> then if the rationals are uncountable ; so are the integers...
>>
>> that cant be right of course.
>>
>> so assume irrationals are countable too.
>>
>> but then there has to be a mapping from all rationals to all irrationals...
>>
>> this means we need a function f(a,b) -> gives all irrationals for integer a and b.
>>
>> f(a,b) does not exist.
>>
>> even f(a,b,c) wont do.
>>
>> for instance f(a,b,c) = a sqrt(2) + b sqrt(3) + c sqrt(5)
>>
>> does not give all irrationals.
>>
>> and if f was different then we would have left out the numbers of type a sqrt(2) + b sqrt(3) + c sqrt(5).
>>
>> regards
>> tommy1729
>

But you (David) missed a third one : tommy1729's proof shows that
(irrational) algebraic numbers are uncountable either : see : f is not
onto, and every other function g would miss the range of f. In fact,
this technique gives an easy proof that no (infinite) set is ever in
bijection with any other set, not even with itself...
>
> ************************
>
> David C. Ullrich

### William Hughes

Sep 19, 2007, 8:42:18 AM9/19/07
to

For transfinite induction you need more than a definition of p_j
when j is equal to i+1 and p_i is defined. You need a definition of
p_j when
you have a definition for p_i for all i<j, even when there is no i
such that j=i+1. You do not have this.

- William Hughes

### tommy1729

Sep 19, 2007, 8:51:07 AM9/19/07
to

and i did not claim i did ...

in fact implie means by the logic of the OP.

but the post continues to show that he is incorrect.

of course i cant prove anything true , that is false :-)

regards
tommy1729

### tommy1729

Sep 19, 2007, 9:06:01 AM9/19/07
to
david wrote :

darn it david , thats exactly my point !!!

seems like is the view of the OP

and i disagree.

of course there's no proof.

you know this very well.

and so do I.

you just enjoy saying im wrong.

### tommy1729

Sep 19, 2007, 9:10:07 AM9/19/07
to

not true for countable sets.

well at least your smart and honest enough to know that

"seems like " does not imply proof ; not even that i agree with the OP

David just loves to say im wrong ; even if im right.

> >
> > ************************
> >
> > David C. Ullrich

tommy1729

### MoeBlee

Sep 19, 2007, 12:09:00 PM9/19/07
to
On Sep 18, 2:43 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:

> Any ordinal equivalent to the set of irrationals would do. (Ordinals
> are sets of lesser ordinals.)

Yes, any ordinal equinumerous with the set of irrationals would do.
That's fine. But let us know when you identify the mistake later in

MoeBlee

### Ross A. Finlayson

Sep 20, 2007, 3:15:43 PM9/20/07
to

About the application of transfinite induction, consider as above
where even MoeBlee will accept that there can be uncountably many
intervals (0, p_i) in the unit interval. Those exist, then there is
the consideration of whether the mapping of the irrationals to
elements of an ordinal can be so imposed in manner that happens to
correspond to the reverse of the normal ordering.

Due to the denseness of the irrationals in the reals, a well-ordered
sequence of irrationals can be constructed, even if not explicitly
listed, for the successor ordinals of zero, the first limit ordinal.
In this context each subset of P will have a particular greatest
element, in the normal ordering, that is a least element in the
reverse ordering which is an artifact of the direction. (For clarity
it would be simple to use the normal ordering and build the intervals
upwards from a small irrational instead of downwards from an
irrational near one.) Then, that sequence can be selected to converge
to but not reach a p_lambda-1 > p_lambda, irrational, which as well
given the uncountability of the irrationals and denseness of the
irrationals in the reals, establishes the first infinite limit
ordinal's case as a consequence of the finite successor ordinals'
cases, because there is a difference between them thus that a rational
exists between them.

Then, following limit ordinals' cases are established in a similar
manner, where again due their denseness and presumed uncountability,
the irrationals are inexhaustible. Is that not sufficient? If not,
please explain why you would think not.

Ross

--
Finlayson Consulting

### Ross A. Finlayson

Sep 20, 2007, 3:25:52 PM9/20/07
to
On Sep 18, 10:51 pm, lwal...@lausd.net wrote:

> ...we see that if we only admitted the axioms

> labeled "Axioms of Incidence," it is obvious that
> a finite model exists. The "Axioms of Order" do
> prove that infinitely many points exist. And
> finally, Archimedes' Axiom implies that no
> non-Archimedean segments (infinitesimals)
> exist, so RF geometry is no longer a model.
>

What if "real" infinity isn't Archimedean, or there are non-
Archimedean infinities? Consider for example the analog from
physics: absolute zero is generally recognized as the lower bound on
temperature, but there is a condition where there can be negative
temperature, yet, the negative temperature is hotter than any finite
temperature. Then, it is similar to a finite modulus, clock
arithmetic, that nothing is greater than infinity so oo < 0.

In the context of the non-Euclidean geometries, for example one where
the ends of a line in the geometry eventually meet, thus that the line
becomes a large circle of sorts. That some physicists think that
space is that way where geometers using real numbers might consider
those to be the real number lines at least give an example of a
consideration of a kind of non-Archimedean infinity.

It is also to be considered that the Archimedean infinity is the
potential infinity vis-a-vis the non-Archimedean actual/completed
infinity, here that is to reflect that there are a variety of notions
of "Archimedean" with regards to "infinity." When you refer to an
Archimedean property of an infinity or an infinite set, what do you
have in mind?

With regards to the notion of infinitesimals in the real numbers,
consider the notion that as the real numbers are complete, that if
there exist infinitesimals, then they exist in the real numbers, and
if there don't exist infinitesimals, then there is no true
differential. (The closer fundamental physical particles' masses are
measured the smaller they appear to be, yet another physical effect
having to do with infinity/infinitesimals, where as well the more the
universe is inspected the larger it appears to be.)

You bring an interesting perspective. Consider instead of defining
geometry in terms of points, then lines, then planes, etcetera, each
eventually defined terms or primitives in the geometry, in the
Euclidean manner, where set-theoretically they're generally
represented in the predicate defining the set as point sets instead of
in the analytic geometrical way, that they are point sets, thus
presuming some space in which they all reside, i.e., a geometrical
universe.

Thanks,

Ross

--
Finlayson Consulting

### William Hughes

Sep 20, 2007, 3:57:21 PM9/20/07
to

Yes, there are uncountable many (0,p_i) in the unit interval.
They are not disjoint. This has zilch to do with transfinite
induction.

> Those exist, then there is
> the consideration of whether the mapping of the irrationals to
> elements of an ordinal can be so imposed in manner that happens to
> correspond to the reverse of the normal ordering.

Nope. the consideration is whether you can find an uncountable
set of intervals that are disjoint. For this you need
a lot more than just a reversal of the normal ordering.

>
> Due to the denseness of the irrationals in the reals, a well-ordered
> sequence of irrationals can be constructed, even if not explicitly
> listed, for the successor ordinals of zero, the first limit ordinal.
> In this context each subset of P will have a particular greatest
> element, in the normal ordering, that is a least element in the
> reverse ordering which is an artifact of the direction. (For clarity
> it would be simple to use the normal ordering and build the intervals
> upwards from a small irrational instead of downwards from an
> irrational near one.) Then, that sequence

A sequence is countable, so far we are talking about a countable
subset of P.

> can be selected to converge
> to but not reach a p_lambda-1 > p_lambda, irrational, which as well
> given the uncountability of the irrationals and denseness of the
> irrationals in the reals, establishes the first infinite limit
> ordinal's case as a consequence of the finite successor ordinals'
> cases, because there is a difference between them thus that a rational
> exists between them.

So we have now definied a countable number of irrationals,
and defined a value for the first countable limit ordinal.
(note in your original construction, you did not have any limits
at all).

>
> Then, following limit ordinals' cases are established in a similar
> manner,

No, you can handle a countable number of limit ordinals in this
fashion, but you do not get to an uncountable limit ordinal.

> where again due their denseness and presumed uncountability,
> the irrationals are inexhaustible. Is that not sufficient? If not,
> please explain why you would think not.

Because you have defined values for a countable number of countable
sequences. This means you have defined values for a countable number
of ordinals. That is all you have done. You still have not defined
values for an uncountable set.

Note this is not trasfinite induction.

- William Hughes

### Ross A. Finlayson

Sep 20, 2007, 5:54:10 PM9/20/07
to

For any of countably many iterations, given that the irrationals are
not countable, there yet exists a p_lambda_1 greater than zero. Then,
for all those successor ordinals of all those (regular, ZFC, von
Neumann) limit ordinals, the case is shown that there is a difference
containing rational numbers. Then, if there are uncountably many
irrationals in the unit interval, and rationals and irrationals are
each dense, there are the uncountably many distinct well-ordered
p_i's, and that many ordered pairs with distinct q_i's. It's shown
that for any successor ordinal of a countable limit ordinal, the case
holds, and that thus the case for the next limit ordinal does as
well. Then, for the same reason that there are enough irrationals for
there to be countably infinitely many, they can always be selected in
a way leaving an interval containing a dense subset of them, for the
countable limit ordinals, that they can be selected in a way not
exhausting the irrationals (leaving an interval of the reals
containing only countably many irrationals), via transfinite induction
the result is shown to hold.

So, in ZFC,with standard definitions of the numbers: the irrationals
are uncountable, irrationals are dense in the reals, or rationals are
dense in the reals. Pick two.

Ross

--
Finlayson Consulting

### Ross A. Finlayson

Sep 20, 2007, 5:55:10 PM9/20/07
to

Well, let's hear it. Don't be coy, if you see an error note it.

Ross

--
Finlayson Consulting

### MoeBlee

Sep 20, 2007, 6:31:47 PM9/20/07
to
On Sep 20, 12:15 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:

> About the application of transfinite induction, consider as above
> where even MoeBlee will accept that there can be uncountably many
> intervals (0, p_i) in the unit interval.

That there are uncountably many intervals (0 x) that are subsets of [0
1] doesn't need mentioning transfinite recursion. It doesn't even need
the axiom of replacement.

Meanwhile, it seems you still have not figured out what is the mistake

MoeBlee

### MoeBlee

Sep 20, 2007, 6:34:58 PM9/20/07
to
On Sep 20, 2:55 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
> On Sep 19, 9:09 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Sep 18, 2:43 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
>
> > > Any ordinal equivalent to the set of irrationals would do. (Ordinals
> > > are sets of lesser ordinals.)
>
> > Yes, any ordinal equinumerous with the set of irrationals would do.
> > That's fine. But let us know when you identify the mistake later in

> Well, let's hear it. Don't be coy, if you see an error note it.

No, I do that so often with other cranks. I think it would be more
interesting to lead you to it. So a hint: look more closely at the
definition of 'well ordering'.

MoeBlee

### William Hughes

Sep 20, 2007, 6:37:15 PM9/20/07
to

Yep iterations again. You will never get further than
countable using interations.

Call gibberish transfinite induction does not make it so.

### Ross A. Finlayson

Sep 21, 2007, 1:17:16 AM9/21/07
to

A well ordering is an ordering relation on elements of a set such that
each subset of the set has a least element by the ordering.

(At least I consider each element to be in the universe, and when
collections are defined by their elements and have the element-of and
subset defined that they're sets.)

Ross

--
Finlayson Consulting

### Ross A. Finlayson

Sep 21, 2007, 1:22:55 AM9/21/07
to

There are more than countably many irrationals or not. If only
countably many irrationals can be selected then there are only
countably many. Where there are not countably many, then, there are
uncountably many intervals, and thus uncountably many rationals. In
the context of transfinite induction, if there are uncountably many
irrationals then uncountably many members of the set of irrationals
are mapped to an ordinal, and that is sufficient, where the condition
doesn't have to hold for higher limit ordinals, i.e., ordinals larger
than the irrationals. In the context of transfinite recursion,
consider this notion from the reference: "frequently a construction
by transfinite recursion will not specify a _unique_ value for A_{alpha
+1}, given the sequence up to alpha, but will specify only a
_condition_ that A_{alpha+1} must satisfy, and argue it is possible to
meet this condition." Due to the denseness of the irrationals they
can be selected from an arbitrarily small interval, for example any of
the "uncountably many" in any given interval of the reals.

In the original argument the induction being as necessary trans-finite
was considered and is implied by the index being a suitably large
ordinal.

What form or structure would you expect to see in an argument of
transfinite induction or transfinite recursion?

Ross

--
Finlayson Consulting

### finite guy

Sep 21, 2007, 5:12:59 AM9/21/07
to

Wow. I really appreciate your reply. It is thorough but open to
issues.
As FG I believe in quanta. And I understand it's non-smooth
outworking.

We are discussing 'mathematically and geometrically perfect' circles
and spheres.

In simple terms, my true and proven statements are your own
mathematics.

Consider a circle, ANY size:
If you know the mathematically EXACT circumference, what is the
equally exact radius? You cannot work it out, it involves Pi.
If the universe were a sphere and you could measure it exactly in
diameter, you could determine nothing equally exact about ANY area or
volume... Pi...
That makes any circle or sphere a strange indeed quantum machine...
can't never measure no center, sir.. look...>> o (circle -
approximate representation) :-)
I hold not attachments to Pi (or 23) BUT Pi is perfectly represented
in mathematics (consider it's properties c a r e f u l l y). Yes, and
it is calculated at magnitude. Must be something wrong, somewhere,
true?

Anyway a universal sized 'sphere' has a pretty long decimal tail when
we regard it from a Planck (not 'quantum') level.

BUT what if we measure it as: r = half universal diameter.
Doesn't that mean that Pi is 3 exactly?
Heh, Pi is in the Universal Constant... heh, heh, heh...

Boys, Infinite means 'finite not' not 'finite more and more and one
more no one more more'. Finite can't approach infinite mathematically
but it can approach N. After all, N*N^N^N is still finite, no matter
how big it's size. Even a super-schlong has a finite limit. And YOUR
big big finite number (or trans/finite number) is flacid compared to
my super-shlong number. Always will be. :-)

Now, cubes and squares. This is more surprising, but true.

Cubes have been disproved by Fermat/Wiles... mathematically of course.
that there CAN BE NO quantised cube in our reality: mathematical or
physical.
The simple a^2 + b^2 = c^2, n>2 (bad) destroys the usual way that XYZ
representations are used. But...
We can't go from (1,1,1) to even (2,2,2) in the accepted way...!!

Squares are 'real' though.
Very straight forward:
they are achievable 'through' magnitudes but not 'in' magnitudes. A 2n+
(-)1 thing. No problem at all to show in Google SketchUp if someone
wants a copy.

Regards,

### finite guy

Sep 21, 2007, 5:15:45 AM9/21/07
to
On Sep 19, 1:51 pm, lwal...@lausd.net wrote:

FLT is geometrical if Pythagorus is... keep it simple.

Message has been deleted

### MoeBlee

Sep 21, 2007, 4:12:37 PM9/21/07
to
On Sep 20, 10:17 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:

> A well ordering is an ordering relation on elements of a set such that
> each subset of the set has a least element by the ordering.

Close.

R is a well ordering on S <-> (R is a strict total ordering on S &
every nonempty subset S has an R-least member).

> (At least I consider each element to be in the universe,

Obviously any object mentioned in a theory is a member of any universe
of a model of the theory.

> and when
> collections are defined by their elements

The axiom of extensionality stipulates that a set is determined only
by its elements, if that's what you mean.

> and have the element-of and
> subset defined that they're sets.)

I don't know what that is supposed to mean.

"In ZFC, with standard definitions of the real, rational, and
irrational numbers, let p_i be an irrational number between zero and
one for i from a suitably large well-ordered index set X. With the
well-ordering of the index set, let the i'th element p_{i+1} be an
irrational number between zero and p_i, where i+1 is the least
element
of the well-ordering X_i setminus i, that is defined to equal X_{i
+1}."

Take it step by step:

Let R be a well ordering of X non-empty.

Let p be a function from X into {r | r is irrational & r e (0 1)}.

After that, your formulation is incoherent (but it's still not the
fatal flaw, since we could still fix your incorehent formulation and
find a deeper mistake): You say "i+1 is the least element of the well-
ordering X_i setminus i". In other words, you're defining a 1-place
operation on X. But the notation 'X_i' indicates that X itself is also
a function. But you've not said what function X is. Moreover, you say
"setminus i" when you must mean "setminus (i.e. complement) {y+ | y R-
less than i}.

So we might address that mess this way (and from now on instead, of 'R-
less' and things like that, I'll just say '<' etc. instead of 'R-less'
as R is understood by context:

What you want is to say what "i+" is for any element i of X.

So, since R is a well ordering of X, there is a successor relation in
X (whether or not it's the ordinary successor relation of the ordinals
- i u {i} - is not crucial; what's important is that we can define 'R-
successor' and thus have our successor relation on X on that basis.
(By context, we can just say 'successor' instead 'R-successor)

So i+ = the successor of i.

Then, let c be the least member of X. So p(c) is some irrational in (0
1).

Then, for any i in X, p(i+) is some irrational in (0 1)\{p(y) | y <
i}.

And we'll add a stipulation about p that you mentioned previously: For
every i+, we have p(i+) < p(i).

Hint here: What is missing at this point?

Then you say, "There are uncountably many irrational numbers less than
each p_i and greater than zero".

Yes, for any r>0, there are uncountably many irrational numbers in any
interval (0 r). But, back to the hint, something is not accounted for

Then you define P as the range of p.

Okay, but think about the hint again now.

Then you say, "There exists a rational number q_i between p_i and p_{i
+1}"

Yes, for any i in X, there is a rational number between p(i) and p(i
+1).

You continue, "For each of the irrational p_i's, there thus exists at
least one unique rational q_i between p_i and p_{i+1}, and infinitely
many."

Uncountably many indeed.

Then you say, "Let the ordered pair (p_i, q_i) be an element of a

function, as a set, from P to Q."

I'd make that:

Let q be a choice function on P such that q(p(i)) = an irrational
between p(i) and p(i+).

So p(i+) < q < p(i+).

If you want this to be on X, then:

Let q' be a choice function on X such that q(p(i)) = an irrational
between p(i) and p(i+).

Then you say, "If there is an uncountable set P of irrational numbers
in (0,1)"

IF. Is P uncountable? P is the range of p. Go back to the hint now!

You finish, "then there is a 1-to-1 function defined by the set {(p_i,

q_i), i E X} from uncountable P to a subset of Q the rational numbers"

I don't even have to check whether the function is 1-1. You just need
to go back to the hint, which will lead you to reexamine your claim
that P is uncountable.

MoeBlee

### MoeBlee

Sep 21, 2007, 5:14:43 PM9/21/07
to
On Sep 21, 1:12 pm, MoeBlee <jazzm...@hotmail.com> wrote:

> R is a well ordering on S <-> (R is a strict total ordering on S &
> every nonempty subset S has an R-least member).

I didn't mean to leave out the word 'of' between 'subset' and 'S'. So,
should be:

R is a well ordering on S <-> (R is a strict total ordering on S &

every nonempty subset of S has an R-least member).

MoeBlee

### Ross A. Finlayson

Sep 22, 2007, 8:53:47 AM9/22/07
to
On Sep 21, 1:12 pm, MoeBlee <jazzm...@hotmail.com> wrote:

No. I don't mean "setminus (i.e. complement) {y+ | y R- less than
i}", which is obscure and immediately contextual, I mean p_i.
It is not "Then, for any i in X, p(i+) is some irrational in (0 1)\
{p(y) | y < i}", instead "Then, for any i in X, p(i+) is some
irrational in (0 1)\ (p_i, 1)".

### MoeBlee

Sep 24, 2007, 2:48:05 PM9/24/07
to

You may not mean it, but it's not obscure; it's perfectly well
defined. And I don't know what you mean by "immediately contextual"
other than that, of course, it depends on R.

Anyway you said, ""i+1 is the least element of the well-ordering X_i
setminus i"

I explained why that is mixed up. First of all, you haven't said what
way X is a function that takes arguments i.

> I mean p_i.

So it was i now it's p_i. Okay. But it's still incoherent as you
haven't said in what way X is a function that takes arguments i.

> It is not "Then, for any i in X, p(i+) is some irrational in (0 1)\
> {p(y) | y < i}", instead "Then, for any i in X, p(i+) is some
> irrational in (0 1)\ (p_i, 1)".

Later I made another stipulation that, on my merely cursory glance,
reduces to just what you said (actually going in downwards instead of
upwards, but that's not of any consequence). This point does not
matter much, since the error in your argument is even more basic.

I gave you the hint at the EXACT point where your fundamental error
(not just some problem in notation or formulation, but rather a
fundamental misconception) begins.

I'll state it even more specifically this time (but I'll still leave
it for you to reason out the rest of it):

When you define p(i+) in terms of p(i), you're assuming that p(i) is
defined. But for p(i) to be defined, what must be established about i?

Let's have you concentrate just on that one question right now. Tell
me what you come up with.

MoeBlee

### finite guy

Sep 24, 2007, 4:09:44 PM9/24/07
to
> are mapped to an ordinal, and that is sufficient, where the- Hide quoted text -
>
> - Show quoted text -...
>

Will you take my Feramt bet?