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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.

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

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

Adam

Sep 18, 2007, 4:00:16 PM9/18/07

to

In article <1190143281.7...@w3g2000hsg.googlegroups.com>,

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

> 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]

Sep 18, 2007, 4:29:43 PM9/18/07

to

On Sep 19, 4:00 am, Virgil <vir...@comcast.net> wrote:

> In article <1190143281.701445.175...@w3g2000hsg.googlegroups.com>,

> In article <1190143281.701445.175...@w3g2000hsg.googlegroups.com>,

Don't take this as a criticism:

y'all believe in them circularity thingies, don ya? :-)

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

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,

> 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

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

Sep 18, 2007, 5:43:39 PM9/18/07

to

On Sep 18, 1:00 pm, Virgil <vir...@comcast.net> wrote:

> In article <1190143281.701445.175...@w3g2000hsg.googlegroups.com>,

> "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.

>

> In article <1190143281.701445.175...@w3g2000hsg.googlegroups.com>,

> "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

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.

> 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.

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},

> 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

Sep 18, 2007, 7:23:55 PM9/18/07

to

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, I agree, but, to avoid

that contradiction by denying the denseness of the rationals in the

reals leads to another contradiction, in that the rationals are dense

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

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

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.

>

> 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.

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

Sep 18, 2007, 9:05:47 PM9/18/07

to

"infinitely many" : I presume that you presume 'infinitely many finite

numbers'? Wuh...?

Sep 18, 2007, 9:13:59 PM9/18/07

to

>

> - 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? :-)

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

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

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

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.

> 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.

Sep 19, 2007, 7:33:46 AM9/19/07

to

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

wrote:

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

Sep 19, 2007, 7:41:02 AM9/19/07

to

David C. Ullrich a écrit :

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

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

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

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.

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

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

your argument.

MoeBlee

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

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

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

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

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

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

in your argument.

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

> > your argument.

> 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

> > your argument.

> 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

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.

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.)

What's your point?

Ross

--

Finlayson Consulting

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

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.

Please don't waffle on about integers etc. because the spin-off is

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,

Adam Lewis, Perth.

Sep 21, 2007, 5:15:45 AM9/21/07

to

Message has been deleted

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.

> What's your point?

Your 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}."

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

in your construction so far.

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

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

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)".

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.

> What's your point?

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

Sep 24, 2007, 4:09:44 PM9/24/07

to

>

> - Show quoted text -...

>

> read more »

Will you take my Feramt bet?

Sep 24, 2007, 4:16:49 PM9/24/07

to

You're off topic, you crank.

There! I said it. Get back to your thread, and post your

"solution",

or you keep the "crank" appellation.

Sep 25, 2007, 4:24:07 PM9/25/07

to

Hi William,

It seems you made a reply that does not appear on the viewer here. If

you did please repost it.

Ross

--

Finlayson Consulting

Sep 25, 2007, 4:40:50 PM9/25/07

to

On Sep 24, 11:48 am, MoeBlee <jazzm...@hotmail.com> wrote:

> On Sep 22

> On Sep 22