Rational numbers, irrational numbers: each dense in real numbers
Ross A. Finlayson
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
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
Virgil
"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
> In article <1190143281.701445.175...@w3g2000hsg.googlegroups.com>,
Don't take this as a criticism:
y'all believe in them circularity thingies, don ya? :-)
tommy1729
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
> 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
> 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
> 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
se...@btinternet.com
> 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
> 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
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
MoeBlee
> 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
>
> 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
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
"infinitely many" : I presume that you presume 'infinitely many finite
numbers'? Wuh...?
finite guy
>
> - 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
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
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
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
> 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
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
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
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
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
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
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
> 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
Ross A. Finlayson
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
> ...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
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
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
Well, let's hear it. Don't be coy, if you see an error note it.
Ross
--
Finlayson Consulting
MoeBlee
> 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
MoeBlee
> 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
William Hughes
Yep iterations again. You will never get further than
countable using interations.
Call gibberish transfinite induction does not make it so.
Ross A. Finlayson
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
Ross A. Finlayson
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
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.
finite guy
MoeBlee
> 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
MoeBlee
> 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
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
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
finite guy
>
> - Show quoted text -...
>
> read more »
Will you take my Feramt bet?
neilist
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.
Ross A. Finlayson
It seems you made a reply that does not appear on the viewer here. If
you did please repost it.
Ross
--
Finlayson Consulting
Ross A. Finlayson
> On Sep 22, 5:53 am, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
>
>
>
> > On Sep 21, 1:12 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > > 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).
>
If the only difference in those definitions is that each subset is non-
empty, please see that as implied and accepted.
> > > > (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.
>
Yes, I'm referring to the notion that the universe is all-inclusive.
As well, it's defined by its elements.
Here the index i indicates an ordinal. In the well-ordering of the
elements of the set P, X is the function defined by ordered pairs from
P and ordinal X sufficiently large, (p_i, i), where p_m < p_n where
index m < n. The i+1'th element, p_{i+1} or p_i+, is the least
element of P setminus each p_m for each m < i+1 in the naturally well-
ordered ordinal X.
I still am missing the gist of your argument. You're not simply
referring to the notion that a well-ordering indicates a least member
for each non-empty subset, I would surmise, as that is implied.
For each of uncountably many interval in the unit interval, there is
defined a well-ordered sequence, separated at limit ordinals by any of
the uncountably many elements of the set, leading always to a non-
empty set left over, each with plenty of rational numbers among them.
> > > 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).
>
That's a key point of agreement.
> > > 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.
Are you simply presuming an error above? If not, on the face of it
that statement does not seem to agree with your position.
>
> > > 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!
In ZFC, if the reals are a set, each subset of the reals can be well-
ordered, including each (0, p) for irrational p E R.
Basically that is a restatement of the argument, then given
developments in the prequel, it is shown.
>
> > > 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
>
> > No. I don't mean "setminus (i.e. complement) {y+ | y R- less than
> > i}", which is obscure and immediately contextual,
>
> 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.
>
X is an ordinal, sufficiently large as to well-order P. It's
basically indistinguishable or interchangeable with the function.
Hopefully the above comment clarifies that adequately.
> 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's obvious.
> > 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.
>
Your point seems to be that an uncountable set can't be well-ordered.
In ZFC, that is not so, as a consequence of the axiomatization of the
existence of choice functions on any set, where subsets of sets are
sets, and the real numbers are a set.
> 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?
>
In the context of transfinite induction/transfinite recursion, it is
that if i is a successor ordinal then for each predecessor up to the
previous limit ordinal it's so, and for the prior limit ordinal it's
so.
> Let's have you concentrate just on that one question right now. Tell
> me what you come up with.
>
> MoeBlee
Any set can be well-ordered in ZFC. The rationals are dense in the
reals, as are the irrationals. For each of uncountably many disjoint
intervals of the reals, there exists an irrational number.
Ross
--
Finlayson Consulting
MoeBlee
> On Sep 24, 11:48 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Sep 22, 5:53 am, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
>
> > > On Sep 21, 1:12 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > > > 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).
>
> If the only difference in those definitions is that each subset is non-
> empty, please see that as implied and accepted.
The difference is that I specified that R is not just an ordering but
rather a strict total ordering.
But, granted, R could be a partial ordering and we'd get what some
people call a 'well ordering' but what some people call a 'weak well
ordering'. And, granted, if there is a weak well ordering on S, then
there a well ordering (in the sense of definition I gave above) too.
But, to be exact, if we're talking about the usual ordering on
ordinals, then that is a well ordering (in the sense of the definition
I gave above) and not MERELY a weak well ordering.
This is not crucial to your argument, but I did wanted to mention it
just for the sake of exactness.
In other words, this is not the important juncture now, let's move on
to the point I mentioned as the important one.
> > > > Hint here: What is missing at this point?
>
> Here the index i indicates an ordinal.
Good. That's even better. It simplifies things.
> In the well-ordering of the
> elements of the set P,
No, wait, you're jumping. P was the range of a certain function. Now
you're talking as if you have P, yet X came first in your setup.
This is not productive trying to sort out your confused formulations.
We can skip sorting through and even any controversy over whether your
formulations are confused if instead we go directly to the point where
we get down to a function p (on ordinals, we can add now) and P as the
range of that function.
In another post I asked you what we need to know about i to know that
p(i+) is defined. Well, we need to know that p(i) is defined. And to
know that p(i) is defined, we need to know what about i?
Since you're not even TRYING to solve this yourself, I I'll give it
away right here. You need that i itself is j+ for some j. So each of
the i's is either 0 (what I called 'c', but it's 0 now that we've
agreed we're working with ordinals directly) or a successor.
Each of the i's in the domain of p is either 0 or a successor of a
previous. Hmm, what system, what structure, of mathematical objects
does that remind you of? Yep, the natural numbers.
To put it more firmly, your definition of the function p makes sense
only for i=0 or for i+ where i is itself a successor. So you didn't
define p for i when i is a LIMIT ordinal. So we cannot infer that the
domain of p is uncountable (thus we can't conclude that P is
uncountable). All we know is that you haven't provided for a
definition of p(i) for i a LIMIT ordinal, so we at least cannot infer
that the domain of p is uncountable.
Again: To prove that you've defined the domain of p for some
UNCOUNTABLE ordinal, you must say what p(i) is for i when i is a LIMIT
ordinal, and you have not.
Then, you will find that when you try to specify what p(i) is for i a
LIMIT ordinal, then your argument falls all apart.
Please, if you respond, tell me that you now understand the mistake in
your argument as I just explained it to you (or, if you don't
understand this now, then I can't explain it for you any more
clearly). But then if you wish to try for a new argument in the same
vein, then please define your function so that it specifies not just
for 0 and successor ordinals but also for LIMIT ordinals.
MoeBlee
Ross A. Finlayson
You ask how there can be given a value for a limit ordinal, and its
simple.
Start with the first limit ordinal, zero, and select any irrational
p_0.
Then, for successor ordinals, select irrationals in a manner such that
they converge to another irrational. It doesn't matter which
irrational numbers they are, except they are in descending natural
order in well-ordering them. Then, for the next limit ordinal, say L,
let p_L be any value less than that to which the previous successor
ordinals converged. Repeat.
Continuing in that manner, then there exists for each limit ordinal L
a given value p_L, and for each successor of each limit ordinal a
given value, and so on, comprising a set of elements P of irrational
numbers, well-ordered in bijection to a suitably large ordinal X by
some implicit function X(p) mapping elements of P to elements of X.
There exists for each p E P at least one, and infinitely many,
rationals q.
Then, you might have that there could only be countably many of those,
yet, there are any of uncountably many irrationals always greater than
zero and less than any countable limit ordinal L's value p_L, thus a
well-ordering of an uncountable subset of the irrationals in this
manner exists up to and including some uncountable limit ordinal, and
that is sufficient for the proof's purposes that those thus describe
an injection from an uncountable set P to a subset of the rationals.
So, the irrationals are dense in the reals. The rationals are dense
in the reals. For each of uncountably many irrational numbers in the
real numbers, there exists a distinct rational number, due to the
denseness of the rationals in the reals, well-ordering the irrationals
in a simple manner as for example described above, and trichotomy of
the reals, in the unit interval of reals, a set.
In ZFC each of uncountably many sets of irrationals from subintervals
(0,1) can be well-ordered, and, between any ordinal's and successor's
p, between p_n and p_m for n < m E X, there exists any of many
distinct rationals that are not between p_j and p_k (j, k E X) for
either j < k <= n < m or n < m <= j < k.
I hope you would understand that the argument was originally framed as
it was in the context of a suitably large (limit, in ZFC uncountable)
ordinal.
So, well-order the reals.
Ross
--
Finlayson Consulting
MoeBlee
lounge.com>:
(1) Now that I've shown you the basic conceptual error that makes your
previous argument incorrect, it's good that you finally recognize that
error (I hope you do).
(2) You have a new argument now. I'll work with this one. Upon
identifying the error(s), I'll consider one more of your attempts, and
upon identifying the error(s) in it, I'll call you out on three
strikes swinging. You'll not have proven in ZFC that there exists an
uncountable set that bijects into a countable set.
(3) For your latest argument, let's be clear at each step. Let's
speficy exactly each formulation along the way, starting fresh. Please
tell me if you disagree with any of the following:
* You claim to prove, in ZFC, that there exists an uncountable ordinal
and a bijection from that ordinal into a subset of the rational
numbers in the real interval (0 1).
* Your method is to use some form of transfinite recursion on an
uncoutable ordinal such that you claim thereby to define a bijection
from that uncountable ordinal into a subset of the rational numbers in
the real interval (0 1).
* So, the first matter to settle is whether you are arguing that for
ANY arbitrary uncountable ordinal there is such a bijection, or
whether you have a particular uncountable ordinal in mind. It seems to
me, that, since you have not mentioned a particular uncountable
ordinal, then you're argument is about an arbitrary uncountable
ordinal.
So, let S be an uncountable ordinal. You are to show that there exists
a bijection from S into a subset of the rationals in the real interval
(0 1).
You begin this way, "Start with the first limit ordinal, zero, and
select any irrational p_0."
Just as terminological point, 0 is usually not taken to be a limit
ordinal.
Nevertheless, fair enough, we are definining a function p on S and
p(0) = z where z is some irrational in (0 1).
But notice that this function p is not into a subset of the rationals.
So whatever bijection that you claim from S into a subset of the
rationals will be some other function that you'll get to later.
Then you say, "Then, for successor ordinals, select irrationals in a
manner such that they converge to another irrational."
Already, it is not at all clear what kind of transfinite recursion
schema you have in mind. You need to say what p(i+) IS, not just say
that you've selected ALL the p(i+)'s to converge to some other
irrational.
Then you say, "It doesn't matter which irrational numbers they are,
except they are in descending natural order in well-ordering them."
Again, you need to specify which transfinite recursion schema you're
using, since I can't evaluate the legitimacy of your recursive
defintion if you don't tell me what recursion schema you're using.
Moreover, as to descending order of irrationals, the ordinary ordering
provides for that, but when you say "in a well ordering of them", you
haven't specified a well-ordering. So, we must suppose that from now
on, there is in play now a particular well-ordering B of (the whole
set of?) irrational numbers.
That last consideration leads me to wonder whether the ordinal S might
as well just be the cardinality of the continuum and that we consider
<R B> to be isomorphic to <S epsilon on S>, where 'R' stands for the
set of real numbers.
Then you say, "Then, for the next limit ordinal, say L, let p_L be any
value less than that to which the previous successor ordinals
converged. Repeat."
And again, I need to know what transfinite recursion schema you have
in mind for such a formulation, especially since I am not aware of a
formulation that takes "repeat" after defining for 0, successors, and
limits. After defining for 0, successors and limits, there should be
no need for nor no reliance upon a further instruction "repeat".
So, before going further, we need to be clear:
(1) Is S an arbitrary uncountable ordinal? Or is S a specific
uncountable ordinal, such as, perhaps, the cardinality of the
continuuum?
(2) What specific transfinite recursion schema are you using to define
the function p?
MoeBlee
Ross A. Finlayson
Hi,
In re-reading the original post I'm somewhat content that it largely
represents the argument as it is now.
I'm under the impression that zero is a limit ordinal. It's not an
infinite limit ordinal, as are all the other limit ordinals, in a
theory with regular infinite ordinals such as ZFC, where each
transfinite cardinal is equivalent to a particular limit ordinal.
I'll accept that it's not a limit ordinal by definition, it just has
no predecessor, sharing that property with all the other limit
ordinals.
S, which I was calling X, is any sufficiently large ordinal thus that
P is uncountable.
When you say "transfinite recursion schema", that was addressed to
some extent in an earlier reply. How would you see it structured?
Consider again http://en.wikipedia.org/wiki/Transfinite_induction .
The element p_0 is readily selected (specified), p_{alpha+1} from the
sequence up to p_alpha readily selected, and p_lambda readily
selected, for the zero, successor, and limit cases.
Then, there are uncountably many irrationals are not, and where there
are, P is uncountable, then as described above for each element p_i E
P, i E X, there is a distinct rational number, q_i.
In the class function representation again the domain of the class
functions is simply any uncountable ordinal.
Ross
--
Finlayson Consulting
MoeBlee
> I'm under the impression that zero is a limit ordinal. It's not an
> infinite limit ordinal, as are all the other limit ordinals,
You may find authors who take 0 to be a limit ordinal, but I think
you'll find that great majority of authors do not to take 0 as a limit
ordinal. So there are three kinds of ordinals: 0, successor ordinals,
and limit ordinals; and every ordinal is one and not more than one of
those three.
But this is merely terminological. It shouldn't affect the substance
of this discussion.
> S, which I was calling X, is any sufficiently large ordinal thus that
> P is uncountable.
"sufficiently large" has no apparent definition in ZFC. I said that
you need to stipulate whether X (I'll call it 'X' now; but then please
don't refer to X as a function) is an arbitrary uncountable ordinal or
whether X is a some specific uncountable ordinal, such as I suggest,
taking X to be the cardinality of the set of real numbers.
Please answer that question: arbitrary or specific?
And it seems to me that there is nothing gained by NOT using a
specific uncountable ordinal, since doing so only gives you additional
ammo in your premises. I suggest you just say that X is the
cardinality of the set of real numbers, since that would line things
up quite neatly.
> When you say "transfinite recursion schema", that was addressed to
> some extent in an earlier reply. How would you see it structured?
> Consider againhttp://en.wikipedia.org/wiki/Transfinite_induction.
That link URL says 'transfinite induction'. Maybe the article also
discusses transfinite recursion too, but I at least have to point out
that I hope you know the difference. It's transfinite recursion, not
transfinite induction, that is at issue here.
Anyway, there's nothing for me to do by following such links. Please
just tell me what transfinite recursion schema you are using to define
p. There's nothing to evaluate in your argument until you do.
> The element p_0 is readily selected (specified), p_{alpha+1} from the
> sequence up to p_alpha readily selected, and p_lambda readily
> selected, for the zero, successor, and limit cases.
>
> Then, there are uncountably many irrationals are not, and where there
> are, P is uncountable, then as described above for each element p_i E
> P, i E X, there is a distinct rational number, q_i.
>
> In the class function representation again the domain of the class
> functions is simply any uncountable ordinal.
MoeBlee
Ross A. Finlayson
In terms of the transfinite recursion schema, I guess that means three
"class functions" labelled G_1, G_2, G_3, such that G_1 is for the
zero, G_3 for infinite limit ordinals, and G_2 is for the successor
ordinals.
In sketch, the idea is to have a way to store, for a given limit
ordinal and its successors, a value to which the sequence of
successors is convergent. So, then that involves a coordinate pair,
where for a given limit ordinal the value for the immediately previous
successor ordinals (those successors of the previous limit ordinal) is
used to derive a value for the current limit ordinal. That is where
the specification (selection, the definition) of an element p_i (for i
being 0, alpha, or lambda) is based upon being less than all the
elements p_m for m < i, and also, greater than any irrational
arbitrarily greater than zero.
So, for example for successor ordinals, there is an ordered pair,
(p_i, t_i), for p the element, and t the target which is a constant
for all given successor ordinals of a given limit ordinal. Then,
p_alpha+1 is defined in terms of p_alpha, and t_alpha, so it is not
less than or equal to t_alpha, nor greater than or equal to p_alpha.
Then, for limite ordinals, p_lambda is less each element (arbitrarily
equal to t_alpha the target constant) of all p_alpha's for alpha <
lambda, and greater than zero, and it has a target t_lambda less than
p_lambda and greater than zero.
As far as specific vis-a-vis arbitrary ordinal X, I don't really care
(it's immaterial) as long as for any countable limit ordinal lambda
that p_lambda > 0, thus there exists uncountably many in the interval
(0, p_lambda).
So, to directly specify p_{alpha+1} given p_alpha, or p_lambda, there
is any choice function on the irrationals in the interval (t_alpha,
p_alpha), and p_{alpha+1} is the value of that function.
This type of machinery is implicit in the original posting.
Does that answer your questions?
Ross
--
Finlayson Consulting
MoeBlee
> In terms of the transfinite recursion schema, I guess that means three
> "class functions" labelled G_1, G_2, G_3, such that G_1 is for the
> zero, G_3 for infinite limit ordinals, and G_2 is for the successor
> ordinals.
> In sketch, the idea is to have a way to store, for a given limit
> ordinal and its successors, a value to which the sequence of
> successors is convergent. So, then that involves a coordinate pair,
> where for a given limit ordinal the value for the immediately previous
> successor ordinals (those successors of the previous limit ordinal) is
> used to derive a value for the current limit ordinal. That is where
> the specification (selection, the definition) of an element p_i (for i
> being 0, alpha, or lambda) is based upon being less than all the
> elements p_m for m < i, and also, greater than any irrational
> arbitrarily greater than zero.
I get the general idea of what you want to do. It's just that you
don't know how to do it (and you won't anyway, unless set theory is
inconsistent).
But please just state a transfinite recursion schema and your
transfinite recursive definition in that schema. That's how proof by
transfinite recursion works. If you don't know any specific
transfinite recursion schemata, then look in such textbooks on set
theory such as Suppes's 'Axiomatic Set Theory' or Enderton's 'Elements
Of Set Theory'.
> So, for example for successor ordinals, there is an ordered pair,
> (p_i, t_i), for p the element, and t the target which is a constant
> for all given successor ordinals of a given limit ordinal.
Now you've introduced t as some kind of function when you haven't even
properly defined p.
If you have a transfinite simultaneous recursion schema in mind, then
please state it.
> Then,
> p_alpha+1 is defined in terms of p_alpha, and t_alpha, so it is not
> less than or equal to t_alpha, nor greater than or equal to p_alpha.
> Then, for limite ordinals, p_lambda is less each element (arbitrarily
> equal to t_alpha the target constant) of all p_alpha's for alpha <
> lambda, and greater than zero, and it has a target t_lambda less than
> p_lambda and greater than zero.
All of that is meaningless without specification of an actual
recursive definition using an actual recursion schema.
> As far as specific vis-a-vis arbitrary ordinal X, I don't really care
> (it's immaterial) as long as for any countable limit ordinal lambda
> that p_lambda > 0, thus there exists uncountably many in the interval
> (0, p_lambda).
You can't specify that X has such properties until you properly define
p, but to properly define p, you need to have specified as to your use
of the variable 'X', which is circular.
So, since you won't coherently answer the question, I'll just take X
to be an arbitrary uncountable ordinal.
> So, to directly specify p_{alpha+1} given p_alpha, or p_lambda, there
> is any choice function on the irrationals in the interval (t_alpha,
> p_alpha), and p_{alpha+1} is the value of that function.
>
> This type of machinery is implicit in the original posting.
>
> Does that answer your questions?
No. But if you studied a textbook or two in set theory then I wouldn't
have to ask questions about your incoherent formulations.
MoeBlee
Ross A. Finlayson
> All of that is meaningless without specification of an actual
> recursive definition using an actual recursion schema.
>
I'm under the impression that a transfinite recursion schema is the
provision of class functions for the ordinal zero, (other) limit
ordinals, and successors, as ordinals, as is done above.
Let there be a family of choice functions F_(x,y) for each x,y E (0,1)
such that each C_(x,y) E F_(x,y) returns an irrational number p such
that y < p < x and p is irrational.
Then, let p_0 = C_(1, 0), and t_0 = C_(p_0,0). (Hopefully that
notation is clear: y < C_(x,y) < x, C_(x,y) is irrational.) Then,
the class function for zero is defined. For each successor ordinal:
t_alpha+ = t_alpha, where alpha+ is shorthand for alpha+1 in ordinal
arithmetic, and p_alpha+ = C_(p_alpha, t_alpha). Then, for limit
ordinals lambda, where the alpha's are from those successors of the
previous limit ordinal, p_lambda is a function of the previous limit
ordinal's successors, p_lambda = C_(t_alpha, 0), and t_lambda is a
C_(p_lambda, 0).
The domain of those class functions has definition for ordinals less
than the initial ordinal of the cardinality greater than the
cardinality of the irrationals.
Then, if the irrationals are uncountable, there is a well-ordering of
the uncountable set {(p_i, i)} by i's natural well-ordering as an
ordinal.
So, there is a transfinite recursion schema.
Furthermore, for each p_i there exists a distinct q_i from the
rationals Q that are in (0,1). The value of q_0 is a rational between
p_0 and 1, then q_alpha+ is a rational between p_alpha+ and p_alpha,
and q_lambda is a rational between t_alpha and p_lambda. Given
trichotomy of the reals the q_i's are distinct.
(Here, there is a slight difference from the intermediate development,
with the limit ordinals' p_lambda's being strictly less than t_alpha,
although that was indicated in allusion, and q_i > p_i to simplify
definition.)
Then, construct a function f:P->Q as the set of ordered pairs {(p_i,
q_i), i E X}, defined for an arbitrarily large uncountable ordinal X.
That's an injection, a 1-1 mapping, from an uncountable set of
irrationals to a subset of the rationals. In your terminology, for i
E S, {(i,q_i)} is a bijection from S to a subset of the rationals.
If you think that's funny I also described a method to select natural
integers at uniform random by selecting real numbers from the unit
interval at uniform random by fair coin tosses, and EF is the CDF of a
uniform probability distribution over the naturals. (The real numbers
are more than standard.)
Ross
--
Finlayson Consulting
Ross A. Finlayson
is the (not fundamentally) heuristic notion that if one were to try to
sample at uniform random from the real numbers in [0,1] by flipping
fair coins (independent Bernoulli trials) to form the binary expansion
of a real number, that it is extremely unlikely to have the sequence
terminate in ending with all zeros or ones, or some repeating
sequence.
Consider this then, besides that each particular sequence has the same
probability of selection as any other which is among reasons whyit is
said that there exists a uniform probability distribution of the reals
of the unit interval, that a random sample of the second binary digit
(bit) of a real number's expansion is as well independently a sample
of another real number's first digit. Then, for the i'th binary
sample, there is a bag (multiset) of i random real numbers, to the
precision i, i-1, ..., 1. In sampling one real number, infinitely
many _not-necessarily distinct_ values are sampled.
Then, if for example zero was a sample from the unit interval of
reals, it's sampled infinitely many times. If a rational is sampled,
then its trailing repeating sequence is sampled infinitely many
times. If the number terminates (in zeros) then zero is sampled
infinitely many times. If an irrational is sampled, then infinitely
many distinct irrationals are sampled. So, where it might seem that
the probability of a rational being sampled is infinitely smaller than
that of an irrational being sampled, once sampled that rational number
represents infinitely much more about the population than a particular
irrational number being sampled, as an artifact of the process it
recurred infinitely many times.
Then, particular rationals of smaller numerators, after one and zero,
and denominators, as well in these infinite expansions have various
considerations of why they are sampled more strongly upon occurrence,
with more weight, in partly casual terms.
Consider further, with regards to this notion of an existence of an
injection from an uncountable subset of the irrationals to the
rationals, an alternate proof:
http://groups.google.com/group/sci.math/msg/f218848b9bbe4830
Borel vs. Combinatorics, anyone?
Ross
--
Finlayson Consulting
MoeBlee
> I'm under the impression that a transfinite recursion schema is the
> provision of class functions for the ordinal zero, (other) limit
> ordinals, and successors, as ordinals, as is done above.
Each transfinite recusrion schema is proven and use of transfinite
recursion requires using a schema. If the schema in use is not
obvious, then you need to mention which schema you intend. And nothing
you've mentioned so far is an obvious schema (not to me at least). I
can't verify the correctness of a purported proof that purports to use
transfinite recursion unless I know which specific transfinite
recursion schema is purported to be deployed. Any work I would do to
figure out how to fit your argument into a particular transfinite
recursion schema is work that you might as well perform yourself so
that I, as the reader, could easily follow your argument. So, for
about the fourth time, it ain't my job to try to figure out how your
argument might be put into a correct transfinite recursion schema.
After this post, if you continue to argue as if by transfinite
recursion but don't provide a schema, then I'm not going to waste any
more of my time on this.
> Let there be a family of choice functions F_(x,y) for each x,y E (0,1)
> such that each C_(x,y) E F_(x,y) returns an irrational number p such
> that y < p < x and p is irrational.
Now you're taking yet another formulation.
I started trying to figure out what you might mean, but later down in
your post I found that I need to ask you whether by '(0,1)' you mean
the real inveral (0 1) or whether you mean the set {0 1}.
Let me know. Then I'll see as to the rest.
MoeBlee
Ross A. Finlayson
That's not really a new formulation, it's basically the same thing as
selecting any irrational between two real numbers.
In the above (0,1) indicates the open unit interval of the (standard)
real numbers. If I meant the set containing only zero and one I would
write {0,1}, with the braces instead of parentheses. In the previous
post (Mon, 01 Oct 2007 18:48:46 -0700), all the other occurrences of
pairs in parentheses indicate ordered pairs, not open interval
endpoints.
"More informally, the transfinite recursion theorem tells us that to
define a function on a well-ordered set < W,< >, it suffices to give a
definition of each function value F(t) in terms of all the previous
function values F(x) where x< t." --
http://www.mscs.dal.ca/~selinger/courses/582W99/handouts/ordinals.pdf
Those notes indicate that Herb Enderton's text would indicate
requirements of a proof by transfinite recursion schema, besides
indicating that class functions as illustrated above are sufficient.
To some extent I wonder why you seem set against the use of
transfinite induction, vis-a-vis "recursion schema." I used
transfinite induction first, then for your satisfaction I would also
like to use "transfinite recursion schema", for induction's sake. I'm
content using transfinite induction, and, transfinite induction is a
generally accepted method of proof for each case of an index set that
has a well-ordering by that ordering.
http://eom.springer.de/t/t093700.htm
In the provision of class functions G_1, G_2, G_3, for zero,
successor, and limit ordinals, that seems an adequate outlay of
machinery for transfinite induction and transfinite recursion. So,
where the transfinite induction schema is to define, here each p_i, it
is done as p_i = F(i) = G( {F(j), j < i} ).
Aatu defines a transfinite recursion scheme: "If G is a definable
class function, there is a unique definable class function F such
that, provably in ZFC, F(alpha) = G(F restricted to alpha) for all
ordinals alpha", in http://groups.google.com/group/sci.logic/msg/9b5c2beeeaba1d92
. Then, a transfinite recursion scheme is illustrated above.
If you have something else in mind, please indicate an appropriate
template. MoeBlee, I think definition of the recursive class
functions suffices.
Ross
--
Finlayson Consulting
MoeBlee
> On Oct 2, 1:23 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > > Let there be a family of choice functions F_(x,y) for each x,y E (0,1)
> > > such that each C_(x,y) E F_(x,y) returns an irrational number p such
> > > that y < p < x and p is irrational.
>
> > Now you're taking yet another formulation.
>
> > I started trying to figure out what you might mean, but later down in
> > your post I found that I need to ask you whether by '(0,1)' you mean
> > the real inveral (0 1) or whether you mean the set {0 1}.
> In the above (0,1) indicates the open unit interval of the (standard)
> real numbers.
Okay. So when you write, "for each x,y E (0,1) such that each C_(x,y)
[...]", I take it that you mean 'for each x and y such 0 =< x < y =<1
such that C((x y)) [...]'. In other words, C is a function whose
domain is the set of open intervals that are subsets of (0 1). Is that
correct?
> "More informally, the transfinite recursion theorem tells us that to
> define a function on a well-ordered set < W,< >, it suffices to give a
> definition of each function value F(t) in terms of all the previous
> function values F(x) where x< t." --http://www.mscs.dal.ca/~selinger/courses/582W99/handouts/ordinals.pdf
Indeed that is quite informal. The exact manner in which F(t) is
defined in terms of function values F(x) where x<t is given by various
of the schemata.
> Those notes indicate that Herb Enderton's text would indicate
> requirements of a proof by transfinite recursion schema, besides
> indicating that class functions as illustrated above are sufficient.
Do you have Enderton's set theory book? If you do, then look in it and
you will see exactly what I'm talking about.
> To some extent I wonder why you seem set against the use of
> transfinite induction, vis-a-vis "recursion schema."
Transfinite induction and transfinite recursion are very closely
related, but they are different things. And I haven't said anything
about being against any correct application of either of them.
> I used
> transfinite induction first, then for your satisfaction I would also
> like to use "transfinite recursion schema", for induction's sake. I'm
> content using transfinite induction, and, transfinite induction is a
> generally accepted method of proof for each case of an index set that
> has a well-ordering by that ordering.
You're very confused about the entire business. Though, below you do
make some progress:
> http://eom.springer.de/t/t093700.htm
>
> In the provision of class functions G_1, G_2, G_3, for zero,
> successor, and limit ordinals, that seems an adequate outlay of
> machinery for transfinite induction and transfinite recursion. So,
> where the transfinite induction schema is to define, here each p_i, it
> is done as p_i = F(i) = G( {F(j), j < i} ).
Ah, now we're getting closer.
What you probably mean is:
For each ieX, F(i) = G({F(j) | j<i}).
If I'm not mistaken, that's a perfectly legitimate definition by
transfinite recursion (as long as we also have a correct definition of
G).
Now tell me precisely what G is.
MoeBlee
Ross A. Finlayson
> On Oct 2, 8:49 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
> > In the provision of class functions G_1, G_2, G_3, for zero,
> > successor, and limit ordinals, that seems an adequate outlay of
> > machinery for transfinite induction and transfinite recursion. So,
> > where the transfinite induction schema is to define, here each p_i, it
> > is done as p_i = F(i) = G( {F(j), j < i} ).
>
> Ah, now we're getting closer.
>
> What you probably mean is:
>
> For each ieX, F(i) = G({F(j) | j<i}).
>
> If I'm not mistaken, that's a perfectly legitimate definition by
> transfinite recursion (as long as we also have a correct definition of
> G).
>
> Now tell me precisely what G is.
>
> MoeBlee
Hi,
I did already, G_1, G_2, and G_3 as above, for the zero ordinal,
successor ordinals, and limit ordinals. So, then there is a well-
ordered and more than countable subset of the irrationals P = {p_i, i
E X, |X| > |N|}. Then, for each p_i it is shown, via another
transfinite induction, that there is a distinct rational q_i, which
readily leads to a contradiction if the rationals are countable, yet,
if there's not a distinct q_i for each p_i, that as well readily leads
to a contradiction if the rationals are dense in the reals.
Now, it is well-known that the existence of uncountably many non-empty
and non-degenerate (single point) partitions of the unit interval into
subintervals would be contradictory to ZFC, inconsistent with ZFC and
standard definitions of the real numbers. While that may be so, via
the transfinite recursion schema as described above, the class
functions are defined up to ordinals equivalent to the irrationals.
Were they not defined, up to uncountably many ordinals, then there
wouldn't be uncountably many irrationals in some interval of the
lwa...@lausd.net
> don't know how to do it (and you won't anyway, unless set theory is
> inconsistent).
>
> But please just state a transfinite recursion schema and your
> transfinite recursive definition in that schema. That's how proof by
> transfinite recursion works.
I know that, like RF, Tony Orlow is also concerned with a new
type of induction schema he calls "infinite case induction." I have
attempted to find a schema that would satisfy RF's and TO's
intuition, but I still wasn't able to do so without a contradiction.
For our first attempt, we begin with ZF (or even ZF-Infinity) and
introduce a new primitive, N, in addition to the sole primitive
already present in the theory of ZF, namely "in," i.e., the
membership relation. We choose the letter N because it's
supposed to remind us of the set of natural numbers.
Now let us add some axioms to describe N. Since N is supposed
to be similar to the natural numbers, why don't we try some
Peano-like axioms to describe N:
(PA1) 0 in N.
(PA2) forall n(n in N -> successor(n) in N)
Note that zero and successor are not primitives in our new theory,
but actually have their von Neumann definitions. So zero is really
the empty set (expanded so that only the primitive "in" is used)
and successor(n) is defined as n union {n}.
(PA3) forall m(forall n(successor(m)=successor(n) -> m=n))
(PA4) forall n(not(successor(n) = 0))
This one is actually redundant, since given the von Neumann
definitions, zero is empty, yet successor(n) is nonempty, since
it is n union {n}, containing n as an element.
Now instead of the usual induction schema of PA, we attempt to
formulate TO's and RF's version of induction. TO and RF tend to
believe that if every finite natural number has a property, then
there exists an infinite set with the property. So we have the
following schema:
(TO/RF) (phi(0) and forall n(phi(n) -> phi(successor(n)))) -> phi(N)
so if every element of N has a property phi, then the set N itself
would have property phi.
Now here's the contradiction. Let phi be the formula "n in N." Now
clearly phi(0) (from PA1) and phi(n) implies phi(successor(n))
(from PA2), so we conclude from (TO/RF) that phi(N), i.e., that
N in N.
Now that blatantly contradicts Foundation/Regularity. Then again,
some so-called "cranks" are prepared to abandon Foundation, so
suppose we considered Z (without Foundation) rather than ZF. But
now let psi be the formula "not(n in n)." Clearly psi(0) holds, and
psi(n) implies psi(successor(n)) since the successor of a
wellfounded set is still wellfounded. So we conclude psi(N), and
suddenly both phi(N) (N in N) and psi(N) (not(N in N)) holds, a
contradiction reminiscent of Russell's Paradox.
I attempted to remedy this problem by assuming that phi is a
formula that doesn't use the symbol N (just as in the Transfer
Principle, which uses formulae without the word "standard").
But now we let phi be the formula:
"n=0 or exists m(n=successor m)."
Once again phi(0) is trivial, and phi(successor(n)) is also
trivial for any n. So we conclude phi(N), so that either N=0
or exists m(N=successor m). Obviously N is not 0, so N
must be a successor to some set m. Let M be the
instantiation of this existential quantifier (we know that such
m must be unique by PA3).
So N=successor(M), so by the von Neumann definition of
successor, N=M union {M}, so M in N. Then by (PA2), we
have successor(M) in N, so N in N once again. And by
repeating the same argument with psi above, we find that
not(N in N), thus leading to the same contradiction again.
I have not found a way to remove the contradiction. And
so as for now at least, "infinite case induction" is not
yet consistent. I still wonder whether there is a way to
make something similar to the above be consistent.
MoeBlee
> On Oct 1, 11:28 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > I get the general idea of what you want to do. It's just that you
> > don't know how to do it (and you won't anyway, unless set theory is
> > inconsistent).
>
> > But please just state a transfinite recursion schema and your
> > transfinite recursive definition in that schema. That's how proof by
> > transfinite recursion works.
>
> I know that, like RF, Tony Orlow is also concerned with a new
> type of induction schema he calls "infinite case induction." I have
> attempted to find a schema that would satisfy RF's and TO's
> intuition, but I still wasn't able to do so without a contradiction.
Finlayson claims to have a proof in ZFC. And his claim is that he puts
an uncountable ordinal 1-1 with a subset of the rationals. So, if he's
correct, then he's proven ZFC inconsistent, thus ZF inconsistent. I
don't think any amount of tyring to demonstrate some "infintite-case
induction" is going to prevail here. And after indulging his several
attempts, I'm ready to quit with him if he doesn't give a correct
formulation in his next attempt.
> For our first attempt, we begin with ZF (or even ZF-Infinity) and
> introduce a new primitive, N, in addition to the sole primitive
> already present in the theory of ZF, namely "in," i.e., the
> membership relation. We choose the letter N because it's
> supposed to remind us of the set of natural numbers.
Then this is non-conservative extension of ZF? Then game's over as far
as what Finlayson claimed. Moreover, if this extension is inconsistent
(which it would be if it supports the claims of the likes of Finlayson
and Orwell), then what's the point? So, if it's inconsistent with ZF,
then indeed it had better be an extension of ZF-I rather.
> Now let us add some axioms to describe N. Since N is supposed
> to be similar to the natural numbers, why don't we try some
> Peano-like axioms to describe N:
>
> (PA1) 0 in N.
> (PA2) forall n(n in N -> successor(n) in N)
Then this is already looking like it's pretty much taking us from ZF-I
right back to ZF.
> Note that zero and successor are not primitives in our new theory,
> but actually have their von Neumann definitions. So zero is really
> the empty set (expanded so that only the primitive "in" is used)
> and successor(n) is defined as n union {n}.
>
> (PA3) forall m(forall n(successor(m)=successor(n) -> m=n))
> (PA4) forall n(not(successor(n) = 0))
>
> This one is actually redundant, since given the von Neumann
> definitions, zero is empty, yet successor(n) is nonempty, since
> it is n union {n}, containing n as an element.
>
> Now instead of the usual induction schema of PA we attempt to
> formulate TO's and RF's version of induction. TO and RF tend to
> believe that if every finite natural number has a property, then
> there exists an infinite set with the property. So we have the
> following schema:
>
> (TO/RF) (phi(0) and forall n(phi(n) -> phi(successor(n)))) -> phi(N)
Maybe it could be: (P(0) & AneN(P(n) -> P(n+))) -> P(N).
> so if every element of N has a property phi, then the set N itself
> would have property phi.
>
> Now here's the contradiction. Let phi be the formula "n in N." Now
> clearly phi(0) (from PA1) and phi(n) implies phi(successor(n))
> (from PA2), so we conclude from (TO/RF) that phi(N), i.e., that
> N in N.
>
> Now that blatantly contradicts Foundation/Regularity. Then again,
> some so-called "cranks" are prepared to abandon Foundation, so
> suppose we considered Z (without Foundation) rather than ZF. But
> now let psi be the formula "not(n in n)." Clearly psi(0) holds, and
> psi(n) implies psi(successor(n)) since the successor of a
> wellfounded set is still wellfounded.
Wait. You didn't assume n is well founded. You assumed ~nen. You're
quite sure that ZF-I+new_axioms entails (~n=0 & ~nen) -> Exen n/\x=0 ?
Or, you could argue ~nen -> ~ n+ e n+. But have you established that
in your theory?
> So we conclude psi(N), and
> suddenly both phi(N) (N in N) and psi(N) (not(N in N)) holds, a
> contradiction reminiscent of Russell's Paradox.
Simpler, let P(n) <-> n is finite. Then N is finite. So let N be 1-1
with some natural number k. But k+ is a natural number. So k+ injects
into N. And N injects into k+. So N bijects with k+. So k bijects with
k+. Violates pigenohole, so contradiction.
And all definitions, such as 'finite', 'natural number' 'injects',
'bijects' and theorems used, such as pigeonhole and Schroder-
Bernstein, are all doable in Z-I-regularity.
And this gets to the heart of why "infinite-case induction" is so
basically unintuitive and incompatible not just with the axiom of
infinity, but even ZF-I-regularity.
> I attempted to remedy this problem by assuming that phi is a
> formula that doesn't use the symbol N (just as in the Transfer
> Principle, which uses formulae without the word "standard").
Okay, but you think that just precluding the symbol 'N' might be
enough? What about terms t such that ZF-I+new_axioms |- t = N ?
But note that you can't have a rule that precludes such terms t from
being in P in the schema, since it's not decidable whether ZF-I
+new_axioms |- t = N.
> But now we let phi be the formula:
>
> "n=0 or exists m(n=successor m)."
>
> Once again phi(0) is trivial, and phi(successor(n)) is also
> trivial for any n. So we conclude phi(N), so that either N=0
> or exists m(N=successor m). Obviously N is not 0, so N
> must be a successor to some set m. Let M be the
> instantiation of this existential quantifier (we know that such
> m must be unique by PA3).
>
> So N=successor(M), so by the von Neumann definition of
> successor, N=M union {M}, so M in N. Then by (PA2), we
> have successor(M) in N, so N in N once again. And by
> repeating the same argument with psi above, we find that
> not(N in N), thus leading to the same contradiction again.
If you meet my questions about well foundedness earlier such that this
argument does go through, then it looks pretty much like the same
underlying problem is exposed as in my own P(n) <-> n is finite
argument.
> I have not found a way to remove the contradiction. And
> so as for now at least, "infinite case induction" is not
> yet consistent. I still wonder whether there is a way to
> make something similar to the above be consistent.
Maybe you'll roll the right dice eventually, but, were I a betting
man, I'd put my money on Don't Pass here.
Perhaps better to start completely from scratch. Forget about Z
anything and just come up with all new primitives and axioms. Though,
still, I wouldn't be optimistic that you'd derive an alternative real
analysis.
MoeBlee
MoeBlee
> On Oct 3, 11:19 am, MoeBlee <jazzm...@hotmail.com> wrote:
> > What you probably mean is:
>
> > For each ieX, F(i) = G({F(j) | j<i}).
>
> > If I'm not mistaken, that's a perfectly legitimate definition by
> > transfinite recursion (as long as we also have a correct definition of
> > G).
>
> > Now tell me precisely what G is.
> I did already, G_1, G_2, and G_3 as above
Okay, we're done. There is only G in the schema you mentioned, not G1,
G2 and G3 (which I'm not going to search back in your posts for what
they are even supposed to be). There are schemas such that we define
separately for 0, successors, and limits, but you've not shown any
recognizably proven schema that you use for that. My time is not
unlimited. You're out. But keep peddling your wares; there are always
people who know nothing about this subject who might think you talk
just like a real live mathematician.
MoeBlee
lwa...@lausd.net
> Then this is non-conservative extension of ZF? Then game's over as far
> (which it would be if it supports the claims of the likes ofFinlayson
> and Orwell), then what's the point? So, if it's inconsistent with ZF,
> then indeed it had better be an extension of ZF-I rather.
Obviously, ZF-Infinity is what I meant. Forget about ZF.
> > (TO/RF) (phi(0) and forall n(phi(n) -> phi(successor(n)))) -> phi(N)
>
> Maybe it could be: (P(0) & AneN(P(n) -> P(n+))) -> P(N).
Which is once again what I meant.
> > Now that blatantly contradicts Foundation/Regularity. Then again,
> > some so-called "cranks" are prepared to abandon Foundation, so
> > suppose we considered Z (without Foundation) rather than ZF. But
> > now let psi be the formula "not(n in n)." Clearly psi(0) holds, and
> > psi(n) implies psi(successor(n)) since the successor of a
> > wellfounded set is still wellfounded.
>
> Wait. You didn't assume n is well founded. You assumed ~nen. You're
> quite sure that ZF-I+new_axioms entails (~n=0 & ~nen) -> Exen n/\x=0 ?
> Or, you could argue ~nen -> ~ n+ e n+. But have you established that
> in your theory?
OK, then, let psi(n) be the formula "n is wellfounded" (expanded out
to primitives, of course). Of course psi(0) holds, since the empty set
is obviously wellfounded. Now we may validly conclude that psi(n)
implies psi(successor(n)), since the successor of a wellfound set is
still wellfounded.
Surely one can prove that every natural number (finite ordinal) is
wellfounded in Z-Foundation/Regularity, right? This is, of course,
the topic of another active thread, Adam Burley's "Principles of
Induction in non well-founded set theories?"
> > So we conclude psi(N), and
> > suddenly both phi(N) (N in N) and psi(N) (not(N in N)) holds, a
> > contradiction reminiscent of Russell's Paradox.
>
> Simpler, let P(n) <-> n is finite. Then N is finite. So let N be 1-1
> with some natural number k. But k+ is a natural number. So k+ injects
> into N. And N injects into k+. So N bijects with k+. So k bijects with
> k+. Violates pigenohole, so contradiction.
Notice that WM does conclude that N is "finite." TO/RF, on the
other hand, accept that N is infinite, and so if one uses infinite
case induction to prove N to be finite, then this is an obvious
blow to the concept of infinite case induction. They do not
want to allow phi to be the formula n is finite.
It's interesting that you mentioned the Pigeonhole Principle here,
since in the end, what TO (not sure about RF) ultimately wanted
to prove using "infinite case induction" was that the Pigeonhole
Principle holds for infinite sets as well as finite sets. These
so-called "cranks" want to prove the nonexistence of a bijection
between an infinite set and a proper subset of itself, just as the
PHP states that there exists no bijection between a finite set and
a proper subset of itself.
> And all definitions, such as 'finite', 'natural number' 'injects',
> 'bijects' and theorems used, such as pigeonhole and Schroder-
> Bernstein, are all doable in Z-I-regularity.
I'm curious as to what definition of finite is being used here, since
there is no AC to equate the definitions.
If by finite you mean "Dedekind finite," then once again, the
so-called cranks want infinite sets to be "Dedekind finite" in
that they wouldn't admit bijections between themselves and
their own proper subsets.
If on the other hand, you mean a set to be finite if there exists
a bijection between the set and a natural, then one must define
the predicate "is a natural number." In our theory, we intend N
to be the set of natural numbers (so that n is a natural number
<-> n in N), so N in N is not a contradiction. (N would be indeed
bijectable with an element of N, namely N itself.)
But you say that you define naturals in ZF-Infinity-Regularity, so
that has nothing to do with our primitive N.
> And this gets to the heart of why "infinite-case induction" is so
> basically unintuitive and incompatible not just with the axiom of
> infinity, but even ZF-I-regularity.
Of course, I already knew that this attempt is doomed.
> > I attempted to remedy this problem by assuming that phi is a
> > formula that doesn't use the symbol N (just as in the Transfer
> > Principle, which uses formulae without the word "standard").
>
> Okay, but you think that just precluding the symbol 'N' might be
> enough? What about terms t such that ZF-I+new_axioms |- t = N ?
>
> But note that you can't have a rule that precludes such terms t from
> being in P in the schema, since it's not decidable whether ZF-I
> +new_axioms |- t = N.
Now I see why my analogy with Transfer Principle/"standard" fails --
we know that no term t, as you put it, can equal the class of all
standard sets because this is a proper class, but we want N to be
a set, so we can't preclude a term t such that t = N.
So in other words, this attempt fails again. And since you already
told RF that you're done with this induction schema, I have to
abandon this attempt as well.
Ross A. Finlayson
I detect quite a negative turn in your attitude.
Anyways G is G_1 for the zero ordinal, G_2 for successor ordinals, and
G_3 for limit ordinals.
{ G_0 for i is the zero ordinal
G(i) = { G_1 for i is a successor ordinal
{ G_2 for i is a limit ordinal
According to each of the sources above, except you, the composite
class function G or its separate class functions G_1, G_2, and G_3 are
adequate machinery for a scheme of transfinite recursion, in
transfinite induction. So, I aready did.
If there are more than countably many irrationals in the interval (0,
r), then in ZFC one can always select any of more than countably many
irrationals in that interval, say p_i, and that is as well so for the
resulting interval (0, p_i), ad infinitum. Then, in considering
whether that is considerable "more than infinitely" many times, in a
sense of the use of transfinite induction, as above in what seems
quite an standard manner the class function is defined for more than
countably many ordinals (in ZFC), else there aren't more than
countably many irrationals. That is justification that the class
function is defined for more than countably many irrationals. The
class function G in this context just describes an uncountable subset
of the irrationals.
The rationals are dense in the reals.
As far as ZF(C) being inconsistent, in naive set theory the universe
of ZF(C) is the Russell set. As the SOASTDNCT, set of all sets that
don't contain themselves, it contains itself, casually ZFC E ZFC, in
describing a theory by all of its objects. Sets are defined by their
elements, etcetera. (Aren't proper classes defined by their
elements?) Furthermore in terms of ZF being inconsistent, consider
any set of physical objects as mathematical objects, and functions
between them as well physical objects, and so on, then there are
infinitely many and the physical universe is its own powerset,
providing via mere recognition of existence a rational justification
that ZF is inconsistent, in terms of the powerset result.
Pick an irrational between zero and one, then select an irrational
between that and zero, then between that and zero, and so on. In your
ZFC, for any you select there are uncountably many left and each can
be selected. Zero is not irrational.
Ross
--
Finlayson Consulting
MoeBlee
> > Now let us add some axioms to describe N. Since N is supposed
> > to be similar to the natural numbers, why don't we try some
> > Peano-like axioms to describe N:
>
> > (PA1) 0 in N.
> > (PA2) forall n(n in N -> successor(n) in N)
>
> Then this is already looking like it's pretty much taking us from ZF-I
> right back to ZF.
I just realized that I didn't put that strongly enough and that even
my previous proof of inconsistency is superflous for the following
reason: Indeed, the new system you described subsumes ZF with the
axiom of infinity. The axiom of infinity is trivially derivable in the
new system (I mean, all you've really done is state the axiom of
infinity for a particular primitive constant instead of for an
existentially bound variable). So then adding "infinite-case
induction" is clearly inconsistent.
MoeBlee
MoeBlee
> On Oct 4, 1:15 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > Then this is non-conservative extension of ZF? Then game's over as far
> > as whatFinlaysonclaimed. Moreover, if this extension is inconsistent
> > (which it would be if it supports the claims of the likes ofFinlayson
> > and Orwell), then what's the point? So, if it's inconsistent with ZF,
> > then indeed it had better be an extension of ZF-I rather.
>
> Obviously, ZF-Infinity is what I meant. Forget about ZF.
>
> > > (TO/RF) (phi(0) and forall n(phi(n) -> phi(successor(n)))) -> phi(N)
>
> > Maybe it could be: (P(0) & AneN(P(n) -> P(n+))) -> P(N).
>
> Which is once again what I meant.
>
> > > Now that blatantly contradicts Foundation/Regularity. Then again,
> > > some so-called "cranks" are prepared to abandon Foundation, so
> > > suppose we considered Z (without Foundation) rather than ZF. But
> > > now let psi be the formula "not(n in n)." Clearly psi(0) holds, and
> > > psi(n) implies psi(successor(n)) since the successor of a
> > > wellfounded set is still wellfounded.
>
> > Wait. You didn't assume n is well founded. You assumed ~nen. You're
> > quite sure that ZF-I+new_axioms entails (~n=0 & ~nen) -> Exen n/\x=0 ?
> > Or, you could argue ~nen -> ~ n+ e n+. But have you established that
> > in your theory?
>
> OK, then, let psi(n) be the formula "n is wellfounded" (expanded out
> to primitives, of course).
Perhaps my lapse, but I'm sometimes unclear what people mean by
'wellfounded' if not explicitly defined (it seems some people use the
word differently from other people?). So what's the definition we're
using here?
> Of course psi(0) holds, since the empty set
> is obviously wellfounded. Now we may validly conclude that psi(n)
> implies psi(successor(n)), since the successor of a wellfound set is
> still wellfounded.
Just for my own sense of certainty, let's see what proof we can devise
(with your definition of 'wellfounded').
> Surely one can prove that every natural number (finite ordinal) is
> wellfounded in Z-Foundation/Regularity, right?
Right, finite ordinals have all of the properties that would fall
under virtually any common definition of 'wellfounded'.
But, if this is part of a proof of "n wellfounded then n+ well
founded", you don't know that n is a finite ordinal.
Anyway, this is all nugatory, since the system is inconsistent on even
simpler argument. The system proves the axiom of infinity and is thus
not consistent with "infinte-case induction".
> > Simpler, let P(n) <-> n is finite. Then N is finite. So let N be 1-1
> > with some natural number k. But k+ is a natural number. So k+ injects
> > into N. And N injects into k+. So N bijects with k+. So k bijects with
> > k+. Violates pigenohole, so contradiction.
>
> Notice that WM does conclude that N is "finite." TO/RF, on the
> other hand, accept that N is infinite, and so if one uses infinite
> case induction to prove N to be finite, then this is an obvious
> blow to the concept of infinite case induction. They do not
> want to allow phi to be the formula n is finite.
I'm sorry to say this, but I really don't see the point of what you're
doing here. We already know refutations of "infinite-case induction".
So far all you've done is go through different arguments that are
superfluous to what we already know. I mean you might as well say,
"Okay, you people have been claiming that the cranks are wrong, so now
here I am to prove for you that indeed the cranks are wrong (but this
is only a corrective step toward rehabiliting the crank notions
somehow)." And so I'm saying, "Yeah, well, duh, we already know such
proofs that the cranks are wrong. We hardly need to see them rehashed
yet again for our OWN benefit. So, if you have a rehabiliation, then
let's see it; meanwhile, we don't need proofs that the concept even
NEEDS rehabilitation."
> It's interesting that you mentioned the Pigeonhole Principle here,
> since in the end, what TO (not sure about RF) ultimately wanted
> to prove using "infinite case induction" was that the Pigeonhole
> Principle holds for infinite sets as well as finite sets. These
> so-called "cranks"
Why "so-called"? They're cranks.
> want to prove the nonexistence of a bijection
> between an infinite set and a proper subset of itself, just as the
> PHP states that there exists no bijection between a finite set and
> a proper subset of itself.
No, they want it to be the case that there is no bijection. They have
no SERIOUS agenda of actually PROVING anything either from ordinary
axioms or axioms of their own.
> > And all definitions, such as 'finite', 'natural number' 'injects',
> > 'bijects' and theorems used, such as pigeonhole and Schroder-
> > Bernstein, are all doable in Z-I-regularity.
>
> I'm curious as to what definition of finite is being used here, since
> there is no AC to equate the definitions.
The ordinary Tarski definition ['P' stands for 'power set of']:
n is finite
<->
AS((S subset of Pn & ~S=0) -> EmeS AxeS ~ x proper subset of m)),
which is equivalent with the other ordinary definitions. It's not
equivalent with 'Dedekind finite' without using a choice principle,
but Dedekind finite is not at issue here anyway.
> If by finite you mean "Dedekind finite," then once again, the
> so-called cranks want infinite sets to be "Dedekind finite" in
> that they wouldn't admit bijections between themselves and
> their own proper subsets.
I don't mean Dedekind finite.
> If on the other hand, you mean a set to be finite if there exists
> a bijection between the set and a natural, then one must define
> the predicate "is a natural number."
I define 'is a natural number' as 'is finite and is an ordinal'. Then
it's a theorem that every finite set is equinumerous with a unique
natural number.
> In our theory, we intend N
> to be the set of natural numbers (so that n is a natural number
> <-> n in N), so N in N is not a contradiction. (N would be indeed
> bijectable with an element of N, namely N itself.)
The theory is inconsistent anyway.
> But you say that you define naturals in ZF-Infinity-Regularity, so
> that has nothing to do with our primitive N.
It doesn't matter. Call mine a definition of 'schmaturals' if you
like. Or revert everything in my argument to primitives. My argument
proves the theory inconsistent. But then I realized that even my
argument is superfluous since the theory is clearly inconsistent as
soon as you added the two axioms.
> > And this gets to the heart of why "infinite-case induction" is so
> > basically unintuitive and incompatible not just with the axiom of
> > infinity, but even ZF-I-regularity.
>
> Of course, I already knew that this attempt is doomed.
Then why post all the rigmarole to prove it? We already knew it and
have proven it too.
> > > I attempted to remedy this problem by assuming that phi is a
> > > formula that doesn't use the symbol N (just as in the Transfer
> > > Principle, which uses formulae without the word "standard").
>
> > Okay, but you think that just precluding the symbol 'N' might be
> > enough? What about terms t such that ZF-I+new_axioms |- t = N ?
>
> > But note that you can't have a rule that precludes such terms t from
> > being in P in the schema, since it's not decidable whether ZF-I
> > +new_axioms |- t = N.
>
> Now I see why my analogy with Transfer Principle/"standard" fails --
> we know that no term t, as you put it, can equal the class of all
> standard sets because this is a proper class, but we want N to be
> a set, so we can't preclude a term t such that t = N.
>
> So in other words, this attempt fails again. And since you already
> told RF that you're done with this induction schema, I have to
> abandon this attempt as well.
Good call. Though, it would be interesting to me any progress you make
toward a systemm that upholds "infinite-case pigeonhole" and is
adequate for proving an alternative real analysis.
MoeBlee
MoeBlee
> I detect quite a negative turn in your attitude.
No, I told you several posts ago that I'm not going to indulge your
mathematical illiteracy indefinitely.
> Anyways G is G_1 for the zero ordinal, G_2 for successor ordinals, and
> G_3 for limit ordinals.
>
> { G_0 for i is the zero ordinal
> G(i) = { G_1 for i is a successor ordinal
> { G_2 for i is a limit ordinal
Okay, I'll give it just one more try to see whether you can put this
into the recursion schema you mentioned earlier.
But we still have previous matters.
We have that X is an arbitrary uncountable ordinal. Okay.
Then before we get to G, we still need to be clear about your C and F,
which you tried to define previously in a definition that tangled the
two. Please define one or the other by itself, then you can define the
second in terms of the first.
Now, are you going to define C first and F second, or vice versa?
I asked previously:
C is a function whose domain is the set of open intervals that are
subsets of (0 1). And for any open interval (x y) that is a subset of
(0 1), we have C((x y)) is an irrational in (x y). Is that correct?
PLEASE keep your response uncluttered with anything about G (we'll get
to that later, IF I continue with this) or about anything else other
than whether you're defining C first and F second or vice versa and
then PRECISE defintions of the first and then the second.
MoeBlee
Ross A. Finlayson
I appreciate that.
Please don't call me a crank.
F_(x,y) is a collection of choice functions C_(x,y) such that C_(x,y)
returns an irrational number p (in the standard meaning, a real
irrational number) such that y < p < x, with an element C of F for
each p, y < p < x. F is defined by its elements. (There are various
ways to consider/denote the well-orderings of the irrationals in (y,x)
and for a particular p in (y,x) all those well-orderings with minimal
element p, choice functions returning p.)
Then, C( (x,y) ), or C_(x,y)((void)) is an irrational in (y,x).
(Again, for clarity the ordering is immaterial and could be arranged
C( (x,y) ) is an irrational in (x,y) by relabelling, the function is
evaluated for (p_i, t_i) as described where t_i < p_i, the ordered
pair could instead be (t_i, p_i), as described, for x< y. )
That's basically interchangeable, with the correct semantic shims,
with the quantified notion "for any x, y E R, x=/=y, there exists
irrational p s.t. y < p < x." It's also basically an observation of
the denseness of the irrationals in the reals.
Ross
--
Finlayson Consulting
MoeBlee
> I appreciate that.
>
> Please don't call me a crank.
Why shouldn't I?
> F_(x,y) is a collection of choice functions C_(x,y) such that C_(x,y)
> returns an irrational number p (in the standard meaning, a real
> irrational number) such that y < p < x, with an element C of F for
> each p, y < p < x. F is defined by its elements. (There are various
> ways to consider/denote the well-orderings of the irrationals in (y,x)
> and for a particular p in (y,x) all those well-orderings with minimal
> element p, choice functions returning p.)
You are EXASPERATING! I asked you please to define C first,
standalone, then define F or vice versa, then you say "I appreciate
that", then you just come back to repeat the same jumble you'd posted
previously.
That's it. I'm done.
> Then, C( (x,y) ), or C_(x,y)((void)) is an irrational in (y,x).
> (Again, for clarity the ordering is immaterial and could be arranged
> C( (x,y) ) is an irrational in (x,y) by relabelling, the function is
> evaluated for (p_i, t_i) as described where t_i < p_i, the ordered
> pair could instead be (t_i, p_i), as described, for x< y. )
>
> That's basically interchangeable, with the correct semantic shims,
> with the quantified notion "for any x, y E R, x=/=y, there exists
> irrational p s.t. y < p < x." It's also basically an observation of
> the denseness of the irrationals in the reals.
Enough is enough. I'm done.
MoeBlee
Ross A. Finlayson
I plainly described C, then F, above. I'd hoped you would see that as
a sincere, straightforward, besides correct, answer. I'm not a crank,
nor a troll.
G, the composite class function defined by G_1, G_2, G_3, (not
G_0, ... as above, a slip) is a transfinite recursion scheme
describing the existence of an uncountable set of irrationals P. (P
in blackboard bold refers to all the irrational reals, here P is a
proper subset of the irrationals.) For each p_i E P, for i E X an
uncountable ordinal, there exists a distinct q_i E Q, thus there
exists an injective (1-to-1) function f:P->Q from an uncountable set
to a subset of the rationals defined by the ordered pairs (p_i, q_i),
such that f(p_i) = q_i, in ZFC using standard constructions of the
real, irrational, and rational numbers.
Then, ZFC is inconsistent.
Bye.
Ross
--
Finlayson Consulting
lwa...@lausd.net
> > It's interesting that you mentioned the Pigeonhole Principle here,
> > since in the end, what TO (not sure about RF) ultimately wanted
> > to prove using "infinite case induction" was that the Pigeonhole
> > Principle holds for infinite sets as well as finite sets. These
> > so-called "cranks"
>
> Why "so-called"? They're cranks.
I dislike the use of the word "crank" to describe the opponents of
ZFC, but since there's no simple word to describe collectively
RF, TO, WM, and the others, I use the phrase "so-called."
Yesterday, I posted for the first time in the thread of yet another
opponent of mainstream mathematics, Archimedes Plutonium, and
he was very upset at my use of the word "crank" to describe him,
even with the tag "so-called." Therefore nothing is accomplished by
repeated name-calling (both sides are guilty of this).
> > Of course, I already knew that this attempt is doomed.
>
> Then why post all the rigmarole to prove it? We already knew it and
> have proven it too.
You ask what I'm trying to accomplish by posting all of this.
As I've said before, much of what the opponents of ZFC wish to show
can be made quite rigorous using Nonstandard Analysis. But then
the mainstream mathematicians point out that Nonstandard Analysis
is derived from Standard Analysis and is rooted in ZFC.
What the opponents of ZFC wish to do is come up with a new set
theory that somehow isn't based in ZFC at all. They do not wish to
derive infinitesimals by first coming up with the standard set of real
numbers and using the Transfer Principle. They'd rather show that
infinitesimals must a priori exist without going through the standard
real numbers at all. (As one such person -- not a sci.math poster --
once put it, they want infinitesimals to be "naturally derived," not
"forcibly derived" from the real numbers.)
Indeed, I posted in the Archimedes Plutonium thread about how his
p-adics resembled hyperreals, AP argued that he didn't want them to
be derived from hyperreals at all. He considered the hyperreals to be
part of an algebra-based theory (and he considers all mainstream
mathematics to be algebra-based), while he wants his p-adics to be
more geometry-based. Thus AP completely rejects standard theory
and its reliance on algebra, just as many others reject mathematics
that's based on set theory (especially ZFC).
So obviously a new set of axioms is needed. These attempts to
come up with "infinite-case induction" fail. Yet still, I can't help
but
wonder whether there is a rigorous set theory in which:
-- no set, not even an infinite set, can be bijected with a proper
subset of itself
-- given infinity, one can find infinity-1, infinity-2, infinity/2,
sqrt(infinity), log(infinity) (mentioned by TO and others)
-- infinitesimals exist (TO, RF, etc.)
-- "cardinality" and "measure" are more intimately related (perhaps
by simply multiplying the former by an infinitesimal to derive the
latter, hinted at by several people)
but every attempt to come up with such a theory has failed.
Not to exclude WM, but some of his theories appear to be
even easier to axiomatize. WM doesn't believe in the existence
of uncountable sets. Obviously there are some theories, such
as ZF-Infinity and ZFC-Powerset, in which it is impossible to
prove the existence of uncountable sets. And in others, such
as Pocket Set Theory (which I mentioned in previous threads),
in which every set is countable -- uncountable classes turn out
to be proper classes.
> Good call. Though, it would be interesting to me any progress you make
> toward a systemm that upholds "infinite-case pigeonhole" and is
> adequate for proving an alternative real analysis
That's still my goal, but of course it won't be easy (otherwise TO,
RF,
etc., would already have accomplished it).
Ross A. Finlayson
That's an interesting goal, one held by many, who aren't necessarily
cranks, and I'm not a crank.
I think that in the infinite the universal quantifier can be
specialized into forms that reflect whether the transfer principle
applies, and, that might have some meaning.
Consider for example, if there was a set of only all the finite
ordinals, that for each n-set of finite ordinals ({0, 1, ..., n}),
it's a finite ordinal, but for all n-sets of finite ordinals, it's not
a finite ordinal.
As another example, for each finite integer n there exists a value
between 1/n and zero, but for all finite integers n, there doesn't
exist a value between 1/n and 0.
Obviously, sometimes the transfer principle applies.
I think an adequate theory is the null axiom theory because it can be
both complete and consistent. Things exist no matter what we say, so
why demand to proscribe them? I think that the null axiom theory,
which corresponds to a variety of long-held creation stories, has
objects in its universe, and that they're not inscrutable.
Then, in terms of discovering meaningful interactions among these soi-
disant mathematical objects, largely in the sense of what could be
done with them, instead of generation of primary/primitive elements
along the lines of 0, 1, ..., instead it is 0, universe, ....
(Interactions among mathematical objects are discovered moreso than
invented, yet of course methods of expression might be moreso invented
than discovered, in terms of novel mathematics.)
So, then there's the empty set (as a set, null, void, nothing) and the
universal set (everything, being). As sets, they contain other
things, or not, with variously the empty set containing nothing and
universal set containing everything, in a set theory.
The empty set is the universe, then there's everything else,
everything other, everything, in the universe. Or, the universe is
empty, then ....
With some principles along the lines of sufficiency and necessity, the
infinitely many elements uniquify themselves, differentiate, only to
the point yielding a continuum of them as they wouldn't more. The
continuum of elements is recognized as having the properties of a
number-theoretic continuum, and much analysis derives from it.
As mathematicians there are many perfectly reasonable theories like
geometry and number theory. They're great. Since Goedel and the 50's
many would believe that there is no set of all true statements about,
say, the natural integers, or, points and lines, yet via a simple
counterargument, true statements about those objects can only be
expressions of the theory, those that aren't inconsistent, otherwise
they would be provable. That's not necessarily to say that according
to Goedel, or rather others as many refer to Goedel, there is no set
of only true statements about, say, the natural integers or points and
lines. Then there wouldn't be a collection of false statements
either. Given all the statements, true and false, Goedel can't
separate them.
Anyways, in terms of infinite sets being irregular, consider the
casual notion that infinity + 1 = infinity. It is presumed in
transfinite ordinal arithmetic that omega + 1 = omega, while in
transfinite cardinal arithmetic aleph_0 + 1 = aleph_0. So, consider
infinity as an ordinal, and adding another means to include the set
itself, when infinity + 1 = infinity, the infinity already included
itself: infinity is irregular. Consider some countable universe, if
the universe contains itself because it contains everything, then
according to Russell's paradox some least infinite set is irregular,
so all of them are.
In terms of an alternative analysis, it has much to do with geometry,
and geometrical mutations of a sort in the large and small, and
polydimensional points and the space containing them.
So, do you think the above argument holds, about the "transfinite
recursion schema" describing an uncountable set and for each element
of it a distinct rational? If not, why is not the "transfinite
recursion scheme" defined for more than countably many irrationals
unles there are not more than countably many?
Half of the integers are even.
I think I already have indicated the theory, because it's the only one
that could possibly be, the question then is how to make theorems of
it.
Ross
--
Finlayson Consulting
MoeBlee
My response to this post doesn't seem to have appeared. I'll wait to
see if it does.
MoeBlee
MoeBlee
>
> > > It's interesting that you mentioned the Pigeonhole Principle here,
> > > since in the end, what TO (not sure about RF) ultimately wanted
> > > to prove using "infinite case induction" was that the Pigeonhole
> > > Principle holds for infinite sets as well as finite sets. These
> > > so-called "cranks"
>
> > Why "so-called"? They're cranks.
>
> I dislike the use of the word "crank" to describe the opponents of
> ZFC,
That's reasonable. And I don't use the word 'crank' to mean 'opponent
of ZFC'.
> but since there's no simple word to describe collectively
> RF, TO, WM, and the others, I use the phrase "so-called."
Okay. Meanwhile, they are cranks.
> Yesterday, I posted for the first time in the thread of yet another
> opponent of mainstream mathematics, Archimedes Plutonium, and
> he was very upset at my use of the word "crank" to describe him,
> even with the tag "so-called." Therefore nothing is accomplished by
> repeated name-calling (both sides are guilty of this).
For me, it's not a matter of name-calling. I don't begrudge anyone
eschewing, rejecting, opposing, critiquing, or criticizing any
particular mathematics, including ZFC. But if someone does that in a
crank manner, then they're being a crank.
> > > Of course, I already knew that this attempt is doomed.
>
> > Then why post all the rigmarole to prove it? We already knew it and
> > have proven it too.
>
> You ask what I'm trying to accomplish by posting all of this.
>
> As I've said before, much of what the opponents of ZFC wish to show
> can be made quite rigorous using Nonstandard Analysis. But then
> the mainstream mathematicians point out that Nonstandard Analysis
> is derived from Standard Analysis and is rooted in ZFC.
>
> What the opponents of ZFC wish to do is come up with a new set
> theory that somehow isn't based in ZFC at all.
Perhaps better to say "some" or "many" of the opponents, since not all
of them do wish to devise a theory at all.
> They do not wish to
> derive infinitesimals by first coming up with the standard set of real
> numbers and using the Transfer Principle. They'd rather show that
> infinitesimals must a priori exist without going through the standard
> real numbers at all. (As one such person -- not a sci.math poster --
> once put it, they want infinitesimals to be "naturally derived," not
> "forcibly derived" from the real numbers.)
I'm familiar all of this. My point was not to ask what is your long
range objective. I know your long range objective already, as you've
posted it. What I was asking in this particular instance is what is
your point of posting such failed attempts as your recent one, when we
already know the basics of such failures?
> Indeed, I posted in the Archimedes Plutonium thread about how his
> p-adics resembled hyperreals, AP argued that he didn't want them to
> be derived from hyperreals at all. He considered the hyperreals to be
> part of an algebra-based theory (and he considers all mainstream
> mathematics to be algebra-based), while he wants his p-adics to be
> more geometry-based. Thus AP completely rejects standard theory
> and its reliance on algebra, just as many others reject mathematics
> that's based on set theory (especially ZFC).
>
> So obviously a new set of axioms is needed. These attempts to
> come up with "infinite-case induction" fail. Yet still, I can't help
> but
> wonder whether there is a rigorous set theory in which:
>
> -- no set, not even an infinite set, can be bijected with a proper
> subset of itself
> -- given infinity, one can find infinity-1, infinity-2, infinity/2,
> sqrt(infinity), log(infinity) (mentioned by TO and others)
> -- infinitesimals exist (TO, RF, etc.)
> -- "cardinality" and "measure" are more intimately related (perhaps
> by simply multiplying the former by an infinitesimal to derive the
> latter, hinted at by several people)
>
> but every attempt to come up with such a theory has failed.
Yes, I know all of this. It's not to the question I asked.
> Not to exclude WM, but some of his theories appear to be
> even easier to axiomatize. WM doesn't believe in the existence
> of uncountable sets.
If you devise formal axioms to capture his notions, then you're not in
his greater framework, since he rejects the formal method. Though, of
course, there's no law that says you can't capture certain of his
notions while not keeping to his larger views. And, he doesn't just
reject UNCOUNTABLE sets.
> Obviously there are some theories, such
> as ZF-Infinity and ZFC-Powerset, in which it is impossible to
> prove the existence of uncountable sets. And in others, such
> as Pocket Set Theory (which I mentioned in previous threads),
> in which every set is countable -- uncountable classes turn out
> to be proper classes.
So, again, why are you recounting all of this business we already
know. Of course ZF-I doesn't prove the existence of uncountable sets.
> > Good call. Though, it would be interesting to me any progress you make
> > toward a systemm that upholds "infinite-case pigeonhole" and is
> > adequate for proving an alternative real analysis
>
> That's still my goal, but of course it won't be easy (otherwise TO,
> RF,
> etc., would already have accomplished it).
I'm interested in any progress you make. But your last few posts
seemed pointless to me, for the reasons I've said.
MoeBlee
MoeBlee
> On Oct 6, 9:14 pm, lwal...@lausd.net wrote:
>
> > On Oct 5, 10:41 am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > Good call. Though, it would be interesting to me any progress you make
> > > toward a systemm that upholds "infinite-case pigeonhole" and is
> > > adequate for proving an alternative real analysis
>
> > That's still my goal, but of course it won't be easy (otherwise TO,
> > RF,
> > etc., would already have accomplished it).
>
> That's an interesting goal, one held by many, who aren't necessarily
> cranks,
Right. To work toward a particular alternative mathematical goal is
not itself to be a crank. What is crank is in the manner of operation
and communication regarding such manners.
> and I'm not a crank.
You're quintessentially a crank.
Here's quintessentially crank writing:
Let me interject your nonsense at this particular point to say: WRONG.
Godel said no such thing.
> or, points and lines, yet via a simple
> counterargument, true statements about those objects can only be
> expressions of the theory, those that aren't inconsistent, otherwise
> they would be provable. That's not necessarily to say that according
> to Goedel, or rather others as many refer to Goedel, there is no set
> of only true statements about, say, the natural integers or points and
> lines. Then there wouldn't be a collection of false statements
> either. Given all the statements, true and false, Goedel can't
> separate them.
>
> Anyways, in terms of infinite sets being irregular, consider the
> casual notion that infinity + 1 = infinity. It is presumed in
> transfinite ordinal arithmetic that omega + 1 = omega,
I do hate to interrupt again when you've got such a great rhythm
going, but WRONG. In ordinal arithmetic it is NOT the case that w+1 =
w.
> while in
> transfinite cardinal arithmetic aleph_0 + 1 = aleph_0. So, consider
> infinity as an ordinal, and adding another means to include the set
> itself, when infinity + 1 = infinity, the infinity already included
> itself: infinity is irregular. Consider some countable universe, if
> the universe contains itself because it contains everything, then
> according to Russell's paradox some least infinite set is irregular,
> so all of them are.
>
> In terms of an alternative analysis, it has much to do with geometry,
> and geometrical mutations of a sort in the large and small, and
> polydimensional points and the space containing them.
>
> So, do you think the above argument holds, about the "transfinite
> recursion schema" describing an uncountable set and for each element
> of it a distinct rational? If not, why is not the "transfinite
> recursion scheme" defined for more than countably many irrationals
> unles there are not more than countably many?
>
> Half of the integers are even.
>
> I think I already have indicated the theory, because it's the only one
> that could possibly be, the question then is how to make theorems of
> it.
The question is how you got so mixed up. Or rather, why in the world
you think that rambling inchorently about mathematics should move
anyone to think anything other than that you're a huge nutcase or a
huge phoney.
MoeBlee
a_plutonium
lwal...@lausd.net wrote:
>
> Indeed, I posted in the Archimedes Plutonium thread about how his
> p-adics resembled hyperreals, AP argued that he didn't want them to
Okay, you say "resembled"
> be derived from hyperreals at all. He considered the hyperreals to be
Now you say "derived". How do you jump from something as weak as
resembles or hints of something else to that of "deriving".
It is fine for mathematicians to dive into proofs where every step is
a logic follow through.
But then the bigger picture of "working in mathematical thoughts"
where
in one sentence we have resembles and then the next sentence derives,
is a huge gap of logic.
> part of an algebra-based theory (and he considers all mainstream
> mathematics to be algebra-based), while he wants his p-adics to be
Now you are "putting words into my mouth" for I never said that. I
said there is
too much hype over algebra in mathematics which has lead to alot of
erroneous
mathematics. Algebra hype has entered areas of geometry where the
geometry
should be leading the way, not algebra.
We see this in common ordinary life, where in the decade of the 1990s
many went
overboard in stockmarket and technology companies which crashed in the
early 2000s
then in the 2000s real estate became the next hype and ballooned far
too large and
now is deflating. We saw this phenomenon in physics of the 20th
century where Einstein
and his General Relativity became the hype that led the vast majority
of physicists astray.
What I am saying is that 20th century mathematics hyped Algebra so far
out that it
led most astray and wasted the careers of alot of mathematicians.
Finite Groups, what
a bunch of poppycaca. It is human nature to get swept up in
sensationalism bandwagonish
malarkey and this bandwagon phenomenon is not limited to finances and
social waves
but gets into the natural sciences, and even mathematics is infested
with hype that leds
to false mathematics.
I did not say that above, and the author intrepreted it as such, and I
do not like it when people misinterpret
what I mean or say. So the above author should be more careful.
The greatest fault of mathematics of the 20th century and the wee few
7 years of the 21st century
is that most mathematicians have not yet come to the realization or
the fact, or even the idea, that
their subject-- mathematics-- is a tiny morsel of the subject that is
Physics. I have come to that realization
in 1990. So the problem of Algebra is one tiny symptom. Another
symptom is what the above author
is working on --Set theory and ZFC.
> more geometry-based. Thus AP completely rejects standard theory
> and its reliance on algebra, just as many others reject mathematics
> that's based on set theory (especially ZFC).
I do not do that, you are beginning to really annoy me with what you
think I believe.
There is a proof (according to my intuition) that Galois theory is
restricted to only the Reals and
Euclidean Geometry. So that the system of P-adics by Hensel and modern
day mathematicians
and the system of HyperReals and the system of Surreals, all these are
merely costumed Reals.
Reals that are deformed and pretending to be something new and
independent, when in fact they
are nothing but deformed Reals.
The P-adics should have been constructed not from Algebra, but from
symmetry and Geometry.
That you have the Reals as infinite rightwards, so the world must have
something that is altogether
new, different and independent that is infinite leftwards. And to
guide us into what P-adics are, we
rely on Geometry, not Algebra.
>
> So obviously a new set of axioms is needed. These attempts to
> come up with "infinite-case induction" fail. Yet still, I can't help
> but
> wonder whether there is a rigorous set theory in which:
>
> -- no set, not even an infinite set, can be bijected with a proper
> subset of itself
I am busy on other things and have no time to discuss Set theory. But
Set theory is another misguidance
by modern day mathematicians.
I said that Mathematics is heading for the true and correct path which
is Physics. So we ask
ourselves, does Physics in its long history have what we can consider
"its set theory"? Is there
a theory in Physics that acts central to Physics such as what Set
theory is to modern mathematics.
And the answer is that yes, Physics has a key theory that acts as Set
theory for Physics and it
is the Atomic theory begun some 2,500 years ago and its modern day
version is the Quantum
Mechanics.
Now, look at the history of mathematics and does set-theory of
mathematics have any shape or form
of that history that physics has had for Atomic theory? In other
words, is Set theory really important
and central for mathematics? What is set theory by the way, other than
a pitiful and measly concept
of membership? And if Set theory is so important, would it not have
been discovered 2,500 years ago
when Atomic theory was discovered and that in modern times, set theory
would have advanced
comparable to the advances that Democritus to Bohr-Heisenberg-Dirac
would advance.
But no, Set theory has been one measly little concept-- membership.
So, does this suggest to a thinking person that Set theory is some
great edifice of mathematics
since it was never needed for 2,500 years and only shows up on the
radar about 1890? Is set theory
really just modern day math-hype, like Internet stocks of the late
1990s.
It is my opinion that 99 percent of mathematicians today are caught up
in hype, not in real mathematics,
and they are mostly wasting their time and career and spinning their
wheels, simply because they
cannot stand back and look at all of math.
I would appreciate the above author, if he mentions my name to at
least by honest as to quoting me
and "not putting words in my mouth".
As for his worry about ZFC set theory, it is my opinion that since all
of mathematics is a tiny subset of
physics, that to worry about ZFC which is a minor detail of
mathematics, but that mathematics
is only a minor subset of physics, that there is no worry about Set
theory at all. It would be
my advice that if you want to fix Set theory and the problems it has,
you simply have to go out and
learn what Physics is, because physics is going to tell you what Set
theory is. And from my view,
there is nothing in Set theory that is important, for it is like a
gang of philosophers who like
Wittgensteins fly in the bottle were banging their head for a solution
when all they had to do is look
up and fly out of the bottle. ZFC is create your own problem and spend
and waste your life in
trying to work out a pseudo problem.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
Ross A. Finlayson
No, in denial, when Goedel claims that no consistent theory can be
complete, and that thus there are "true" statements about the objects
of the theory that are not theorems of the theory, I disagree with
that, because I think "true" means "provable."
Then, consider a collection of all theorems, consistent and
inconsistent, about the objects of a theory strong enough to represent
the, say, natural integers. That's all the theorems about those
objects. If Goedel could separate that collection into consistent and
inconsistent theorems, then the collection of consistent theorems
would represent all facts about the objects. There couldn't be more
facts about those objects because they are completely and only defined
in terms of the theory already. Any new fact would represent new
objects, or objects augmented with new properties, and a different
theory, not some truism about the objects of the initial theory. So,
Goedel can't generally separate consistent from inconsistent theorems
of a theory.
I was in the process of explaining to Larry here some of my positions
on various statements having to do with philosophical foundations of
mathematics and so on. So, Moe, while you might be an expert in what
it means to be a crank, the above is not an inaccurate representation
of consequences of inexistence of complete and consistent theories
accorded to Goedel's rresults. Larry, I hope you don't mind if I call
you Larry, you don't have a very accurate understanding of what I say
about these things, although it is agreeably in the right vein.
Generally I don't disagree, in various specific technicalities there
is very much material in terms of foundations and the infinite but in
your initial analyses not so much to directly contradict. For the
most part, I think the majority of people here have little idea, or
care, of the contents of discussion. I appreciate your comments,
because they illustrate a refreshing open-mindedness as opposed to
knee-jerk jingoism. I think that you recognize that there are
perceived and addressable deficiencies in standard theories is a sign
of philosophical maturity, Larry. Larry, I'd appreciate having a name
to call you.
About omega + 1 > omega, that was a typographical error, excuse me.
As is hopefully obvious from the context, in illustrating a difference
between regular ordinal arithmetic and an ordinal arithmetic with
irregular ordinals, of a sort, where infinity = infinity + 1 for a
mechanistic reason, the standardly correct statement (fact) was meant.
I don't ramble incoherently about mathematics, I ramble in a very
concise and specific manner. I'm direct, dammit. That I saw reason
to inspect the status quo's foundation of mathematics and find them
lacking, is, I would agree, not a conventional perspective.
Moe, you would deny in a set theory where everything is a set that
everything is a set. Here's another one, consider for example G(zero)
= G(alpha) = G(lambda), G(i) defined up to On, the "proper class" of
ordinals, to be "the union restricted to i and i is an ordinal",
there's Burali-Forti again, the paradox. You can't have the simple
predicate x=x to define a set in ZFC. That is to say, one of the most
direct truisms: tautology, identity, can't be used to apply to sets in
ZFC, because there is no universe in ZFC. Yet, in theories containing
the universe of ZFC, ZFC contains itself. When you read the above,
does it not make sense?
The above defined transfinite recursion schema about sets dense in the
reals shows ZFC inconsistent.
Ross
--
Finlayson Consulting
MoeBlee
> On Oct 8, 6:38 pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Oct 8, 5:58 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
> > > As mathematicians there are many perfectly reasonable theories like
> > > geometry and number theory. They're great. Since Goedel and the 50's
> > > many would believe that there is no set of all true statements about,
> > > say, the natural integers,
>
> > Let me interject your nonsense at this particular point to say: WRONG.
> > Godel said no such thing.
> No, in denial, when Goedel claims that no consistent theory can be
> complete,
But Godel did not say that.
> and that thus there are "true" statements about the objects
> of the theory that are not theorems of the theory, I disagree with
> that, because I think "true" means "provable."
Aside from the fact that you're incorrectly summarizing Godel, now you
go on to say that your point is based on taking a different
definition. That's no interesting. Anyone can get different results
just by changing definitions.
> Then, consider a collection of all theorems, consistent and
> inconsistent, about the objects of a theory strong enough to represent
> the, say, natural integers. That's all the theorems about those
> objects. If Goedel could separate that collection into consistent and
> inconsistent theorems, then the collection of consistent theorems
> would represent all facts about the objects.
Except (1) a union of consistent theories is not necessarily a
consistent theory and (2) there is no effective procedure for
determining whether a theory is consistent, so there'd be no effective
procedure for determining the axioms of such a union theory.
> There couldn't be more
> facts about those objects because they are completely and only defined
> in terms of the theory already. Any new fact would represent new
> objects, or objects augmented with new properties, and a different
> theory, not some truism about the objects of the initial theory. So,
> Goedel can't generally separate consistent from inconsistent theorems
> of a theory.
>
> I was in the process of explaining to Larry here some of my positions
> on various statements having to do with philosophical foundations of
> mathematics and so on. So, Moe, while you might be an expert in what
> it means to be a crank,
I'm not an expert on the subject of cranks, but I do know enough about
it that I can see that certain people are indeed cranks through and
through.
> [...]
> About omega + 1 > omega, that was a typographical error, excuse me.
Thanks. That's a bit of progress. Though, relatively, not much more
than closing the curtains in face of a hurricaine.
> As is hopefully obvious from the context, in illustrating a difference
> between regular ordinal arithmetic and an ordinal arithmetic with
> irregular ordinals, of a sort, where infinity = infinity + 1 for a
> mechanistic reason, the standardly correct statement (fact) was meant.
>
> I don't ramble incoherently about mathematics, I ramble in a very
> concise and specific manner. I'm direct, dammit. That I saw reason
> to inspect the status quo's foundation of mathematics and find them
> lacking, is, I would agree, not a conventional perspective.
>
> Moe, you would deny in a set theory where everything is a set that
> everything is a set.
Now you're not just rambling, but also lying about me.
> Here's another one, consider for example G(zero)
> = G(alpha) = G(lambda), G(i) defined up to On, the "proper class" of
> ordinals, to be "the union restricted to i and i is an ordinal",
> there's Burali-Forti again, the paradox. You can't have the simple
> predicate x=x to define a set in ZFC. That is to say, one of the most
> direct truisms: tautology, identity, can't be used to apply to sets in
> ZFC, because there is no universe in ZFC. Yet, in theories containing
> the universe of ZFC, ZFC contains itself. When you read the above,
> does it not make sense?
No.
> The above defined transfinite recursion schema about sets dense in the
> reals shows ZFC inconsistent.
Sure it does, Ross, sure it does. Now time for the rec room, Ross, to
meet all your friends there.
MoeBlee
Ross A. Finlayson
Goedel has a theory about theories that no consistent theory strong
enough to represent, say, the natural integers would be complete. So,
is Goedel's theory incomplete? Then, where that's the only statement
of the theorem, for there to be any different property of his theory,
it is that there is a complete theory. Otherwise, according to
himself, it's inconsistent.
How else would you describe truth, in terms of the objects of a
theory, except in terms of facts proven by theorems of the theory? Is
it not true that 1+1 = 2? Is it not a fact? What do you see as the
difference between a truth and a fact?
Perhaps it's similar to the difference between a proper class, and a
set, where each are defined by their elements.
If everything (each thing) in a set theory is a set, then why not
everything (all things)? That the universe in ZFC is called a proper
class in ZFC with classes instead of a set in ZFC, a set theory,
doesn't help quantifying over sets in the universe (domain of
discourse, all sets) of ZFC. I certainly am not misattributing you,
I'm not a liar, instead simply illustrating via simple plain language
statements the inconsistency of a variety of your assumptions about
(all) the objects of a theory to which you claim adherence (ZFC).
Where Goedel claims that no consistent theory strong enough to
represent the natural integers could be complete, where a variety of
trivial theories are obviously consistent and complete (eg x = x), the
above is not an incorrect summarization of Goedel, just partial and in
context.
Then, for a given (consistent) theory and all of its theorems, i.e.,
sentences finite and infinite in the language of the theory both
consistent and inconsistent, then they can't be separated, else there
would be a collection of one or the other. Now, the consistent ones
are theorems of the theory. They're facts. Being all the theorems
about the objects of the theory, there are not more theorems of the
objects of the theory. Any "new" theorems would be via the addition
of "new" properties of the objects. Then, given a collection of all
theorems, there aren't any "new" theorems without "new" properties of
those objects. If there are undecideable statements about those
objects, that means they have properties unaccounted for in the
original specification of the objects, "hidden properties."
So, there is thus an implicit non-logical axiom for any theory Goedel
analyzes: some anonymous axiom that purports to have a fact about the
objects. Otherwise the objects are fully specified. As there is thus
one for that and another ad infinitum, theories analyzed by Goedel are
no longer finitely axiomatized. Then, Goedel's results don't apply.
Then, as I was trying to explain to Larry before you decided to re-
enter the conversation, there are some reasonable justifications of
the rejection of Goedel's results, that there can be no complete and
consistent finitely-axiomatized theory strong enough to represent the
natural integers. Here, and nobody's ever told me this before, I
introduce the notion that for objects of a theory to have anonymous
properties that there are anonymous non-logical/proper axioms, and
when those are thus witnessed, then the finitely axiomatized theory
analyzed by Goedel has fleas ad infinitum.
Ah, that's a novel concept, in this discussion. Enjoy.
If there are uncountably many irrationals then the class function of
the transfinite recursion schema is defined up to some uncountable
ordinal, and then due the denseness properties of the rationals in the
reals ZFC is inconsistent.
Ross
--
Finlayson Consulting
Jonathan Hoyle
> 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.
Okay, let me restate what I think you are saying explicitly, and
correct me if you mean something else. Are you saying that you are
well ordering the irrationals? Okay, let's say so. Now this "index
set" X of your's, I presume that this is merely a subset of ordinals
large enough to contain the irrationals, is that right? In other
words, p_i is an irrational for any given ordinal i where i is an
ordinal less than the cardinality of the irrationals. Is this
correct?
> 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...
Whoa, whoa, whoa. I thought p_i was the i-th element. And how do you
know that p_i+1 is less than p_i? Nothing in the well ordering
requires this. You have not shown that there even exists an i such
that p_i+1 < p_i.
> where i+1 is the least element of the well-ordering X_i set minus i...
Huh? Isn't the i+1 simply the next ordinal after i? And what is this
X_i? I thought X was the index set; how do you define X_i?
Can you start again, explain each step here?
Thanks,
Jonathan Hoyle
http://www.jonhoyle.com
MoeBlee
> 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.
>
> Okay, let me restate what I think you are saying explicitly, and
> correct me if you mean something else. Are you saying that you are
> well ordering the irrationals? Okay, let's say so. Now this "index
> set" X of your's, I presume that this is merely a subset of ordinals
> large enough to contain the irrationals, is that right?
He won't say exactly, except that it is a "sufficiently large"
ordinal, whatever that means.
> In other
> words, p_i is an irrational for any given ordinal i where i is an
> ordinal less than the cardinality of the irrationals. Is this
> correct?
>
> > 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...
>
> Whoa, whoa, whoa. I thought p_i was the i-th element. And how do you
> know that p_i+1 is less than p_i? Nothing in the well ordering
> requires this. You have not shown that there even exists an i such
> that p_i+1 < p_i.
> > where i+1 is the least element of the well-ordering X_i set minus i...
>
> Huh? Isn't the i+1 simply the next ordinal after i? And what is this
> X_i? I thought X was the index set; how do you define X_i?
>
> Can you start again, explain each step here?
I tried to get him to be precise at each step, step by step. He won't
do it. He'll only explain in such a way as to come back around full
circle (or with added orbits of confusion) to the mishmosh
formulations he started with.
MoeBlee
MoeBlee
> Goedel has a theory about theories that no consistent theory strong
> enough to represent, say, the natural integers would be complete.
No, you're leaving out that the theory must be formally axiomatizable.
We'd say recursively axiomatizable these days. The theory that is the
set of true sentences of arithmetic is complete and consistent; but
it's not recursively axiomatizable. And Godel doesn't have a "theory
about theories" so much as it would be more reasonable to say he
proved certain theorems about theories.
> So,
> is Goedel's theory incomplete?
What "Godel's theory"?
> Then, where that's the only statement
> of the theorem, for there to be any different property of his theory,
> it is that there is a complete theory. Otherwise, according to
> himself, it's inconsistent.
I have no idea what you're trying to say that makes sense.
> How else would you describe truth, in terms of the objects of a
> theory, except in terms of facts proven by theorems of the theory?
Truth of sentences in the language of a theory is defined by the
method of models.
> Is
> it not true that 1+1 = 2? Is it not a fact? What do you see as the
> difference between a truth and a fact?
As a formal sentence, it's true in any model of that sentence. And
it's true in the standard model of the language of arithmetic. Also,
it would seem to me that the sentence expresses a proposition that is
a basic finitary fact.
> Perhaps it's similar to the difference between a proper class, and a
> set, where each are defined by their elements.
Perhaps someday, before or after Doomsday, who knows, you'll actually
read a book on mathematical logic.
> If everything (each thing) in a set theory is a set, then why not
> everything (all things)?
You'd have to ask someone who thinks not.
> That the universe in ZFC is called a proper
> class in ZFC with classes instead of a set in ZFC,
I have no idea what you mean by "the universe in ZFC".
> a set theory,
> doesn't help quantifying over sets in the universe (domain of
> discourse, all sets) of ZFC. I certainly am not misattributing you,
> I'm not a liar, instead simply illustrating via simple plain language
> statements the inconsistency of a variety of your assumptions about
> (all) the objects of a theory to which you claim adherence (ZFC).
When you said, "[Moe] would deny in a set theory where everything is a
set that everything is a set", you said something untrue of me. But
i'll give you the benefit of the doubt now that you didn't intend to
lie.
> Where Goedel claims that no consistent theory strong enough to
> represent the natural integers could be complete, where a variety of
> trivial theories are obviously consistent and complete (eg x = x), the
> above is not an incorrect summarization of Goedel, just partial and in
> context.
Your argument after that point went on by missing the crucial fact
that what are incomplete are recursively axiomatized theories. Godel
never said that a theory that is not recursively axiomatized can't be
sufficient for arithmetic, consistent and complete.
> Then, for a given (consistent) theory and all of its theorems, i.e.,
> sentences finite and infinite in the language of the theory
We're not dealing with infinite sentences in the context of Godel's
incompleteness.
> both
> consistent and inconsistent, then they can't be separated, else there
> would be a collection of one or the other.
I don't know what you're trying to say. A sentence on itself is
inconsistent iff it's a self-contradiction. Anyway, it's a corollary
of Church's theorem that the set of self-contradictions is not
recursive. Maybe that's what you have in mind.
> Now, the consistent ones
> are theorems of the theory. They're facts. Being all the theorems
> about the objects of the theory, there are not more theorems of the
> objects of the theory. Any "new" theorems would be via the addition
> of "new" properties of the objects. Then, given a collection of all
> theorems, there aren't any "new" theorems without "new" properties of
> those objects. If there are undecideable statements about those
> objects, that means they have properties unaccounted for in the
> original specification of the objects, "hidden properties."
None of what you said generally holds for first order theories as
discussed in ordinary mathematical logic. So I really don't know what
theories you think you're talking about.
> So, there is thus an implicit non-logical axiom for any theory Goedel
> analyzes: some anonymous axiom that purports to have a fact about the
> objects. Otherwise the objects are fully specified. As there is thus
> one for that and another ad infinitum, theories analyzed by Goedel are
> no longer finitely axiomatized. Then, Goedel's results don't apply.
You really need to read a book on mathematical logic so that you can
write a post using the words you use but in a meaningful way.
> Then, as I was trying to explain to Larry before you decided to re-
> enter the conversation, there are some reasonable justifications of
> the rejection of Goedel's results, that there can be no complete and
> consistent finitely-axiomatized theory strong enough to represent the
> natural integers. Here, and nobody's ever told me this before, I
> introduce the notion that for objects of a theory to have anonymous
> properties that there are anonymous non-logical/proper axioms, and
> when those are thus witnessed, then the finitely axiomatized theory
> analyzed by Goedel has fleas ad infinitum.
>
> Ah, that's a novel concept, in this discussion.
> Enjoy.
I enjoy various things. Your babbling is not one of them.
> If there are uncountably many irrationals then the class function of
> the transfinite recursion schema is defined up to some uncountable
> ordinal, and then due the denseness properties of the rationals in the
> reals ZFC is inconsistent.
Breathtaking.
MoeBlee
Math1723
>
> I tried to get him to be precise at each step, step by step. He
> won't do it. He'll only explain in such a way as to come back around
> full circle (or with added orbits of confusion) to the mishmosh
> formulations he started with.
I think he does that because subconsciously he knows that precision
would be his undoing. By speaking vaguely and sloppily, it allows him
to believe that his crankish ideas are true. Rigor would force him to
abandon his beloved crackpot theories. Psychologically, it's similar
to some people being afraid to go to the doctor, not wanting to hear
what they already know in their heart is true.
David R Tribble
>> but since there's no simple word to describe collectively
>> RF, TO, WM, and the others, I use the phrase "so-called."
>> opponent of mainstream mathematics, Archimedes Plutonium, and
>> he was very upset at my use of the word "crank" to describe him,
>> even with the tag "so-called." Therefore nothing is accomplished by
>> repeated name-calling (both sides are guilty of this).
>
MoeBlee wrote:
> For me, it's not a matter of name-calling. I don't begrudge anyone
> eschewing, rejecting, opposing, critiquing, or criticizing any
> particular mathematics, including ZFC. But if someone does that in a
> crank manner, then they're being a crank.
Exactly. It's not *what* they believe, it's *how* they believe.
Or, more precisely, how they present their beliefs to the
world.
There are several tell-tale indicators of crankiness; see:
http://en.wikipedia.org/wiki/Crank_%28person%29
http://tinyurl.com/8hah7
Primary among those traits is:
The true hallmark of the crank is not so much asserting
that the Earth is flat as making this assertion in the
face of all counterarguments and contrary evidence.
Certain authors who have studied the phenomenon of
crankery agree that this is the essential defining
characteristic of a crank: No argument or evidence
can ever be sufficient to make a crank abandon his belief.
I think we can safely say that this criteria applies to
several certain well-known posters to sci.math.
See also:
http://en.wikipedia.org/wiki/Duck_test
David R Tribble
> I don't ramble incoherently about mathematics, I ramble in a very
> concise and specific manner. I'm direct, dammit. That I saw reason
> to inspect the status quo's foundation of mathematics and find them
> lacking, is, I would agree, not a conventional perspective.
>
> Moe, you would deny in a set theory where everything is a set that
> everything is a set. Here's another one, consider for example G(zero)
> = G(alpha) = G(lambda), G(i) defined up to On, the "proper class" of
> ordinals, to be "the union restricted to i and i is an ordinal",
> there's Burali-Forti again, the paradox. You can't have the simple
> predicate x=x to define a set in ZFC. That is to say, one of the most
> direct truisms: tautology, identity, can't be used to apply to sets in
> ZFC, because there is no universe in ZFC. Yet, in theories containing
> the universe of ZFC, ZFC contains itself. When you read the above,
> does it not make sense?
I think most of us would agree that MoeBlee has a valid
point about Ross's writing style.
Jesse F. Hughes
> Goedel has a theory about theories that no consistent theory strong
> enough to represent, say, the natural integers would be complete. So,
> is Goedel's theory incomplete? Then, where that's the only statement
> of the theorem, for there to be any different property of his theory,
> it is that there is a complete theory. Otherwise, according to
> himself, it's inconsistent.
Man, you got his number. In fact, this is a way cooler criticism than
ol' elsiemelsi ever offered.
Hey, I have an idea. Let's you and him fight.
--
Jesse F. Hughes
"Surround sound is going to be increasingly important in future
offices."
-- Microsoft marketing manager displays his keen insight
Ross A. Finlayson
David, to argue against (some) results of ZFC I present arguments as
to why, in my opinion, ZFC is inconsistent. To argue against
Goedelian results I present arguments as to why that there are no
strong, consistent, complete (and concrete) theories is inconsistent.
To argue against results of transfinite cardinals I present arguments
as to why they are not consistent.
So, I don't carry on in the face of obvious counterexamples and
reasonable theorems. Instead, I offer justifications as to why these
contentious issues (like there isn't a universe, true doesn't mean
provable, infinite sets lack elements and aren't infinite) may
conscientiously be nullified, because of their own inconsistencies.
_Then_ I feel comfortable in the consideration of various nonstandard
theories in mathematical foundations.
Consider the physical universe, and map mathematical objects to
physical objects. Then, (all) functions among those represent
physical objects, as do functions among those, ad infinitum. Then,
there are infinitely many objects in the universe, which
mathematically is its own powerset. Then, there is realizable
evidence ("contrary evidence") that it is reasonable to discredit the
powerset result.
I found my opinion around the use of mathematical proofs to illustrate
that what I say is so.
With regards to the duck test, proper classes are defined by their
elements, as are sets. Are not proper classes defined by their
elements, containing some specified elements, as are sets, similarly
walking, quacking, etcetera?
What's your opinion about the topic under discussion: is the
specified transfinite recursion schema not defined for ordinals up to
the cardinality of the irrationals? If it's not, the irrationals
aren't uncountable. If it is, they aren't either.
Ross
--
Finlayson Consulting
MoeBlee
> David, to argue against (some) results of ZFC I present arguments as
> to why, in my opinion, ZFC is inconsistent.
And your "arguments" as to your opinion that ZFC is inconsistent are
not proofs that ZFC is inconsistent.
> To argue against
> Goedelian results I present arguments as to why that there are no
> strong, consistent, complete (and concrete) theories is inconsistent.
Whatever "concrete" means, Godel didn't claim that there aren't
arithmetically sufficient, consistent, complete theories. Rather, he
(as improved by Rosser) proved that there are no RECURSIVELY
axiomatized, arithmetically sufficient, consistent, complete theories.
And whatever "arguments" you have about this matter are not proofs
that incompleteness is inconsistent with whatever you think it is
inconsistent with.
> To argue against results of transfinite cardinals I present arguments
> as to why they are not consistent.
Your arguments are not proofs.
> So, I don't carry on in the face of obvious counterexamples and
> reasonable theorems.
You carry on in face of the obvious fact that your word jumbles are
not proofs.
> Instead, I offer justifications as to why these
> contentious issues (like there isn't a universe, true doesn't mean
> provable, infinite sets lack elements and aren't infinite) may
> conscientiously be nullified, because of their own inconsistencies.
> _Then_ I feel comfortable in the consideration of various nonstandard
> theories in mathematical foundations.
Whatever that is supposed to be about, in none of it do we find proofs
of anything.
> Consider the physical universe, and map mathematical objects to
> physical objects. Then, (all) functions among those represent
> physical objects, as do functions among those, ad infinitum. Then,
> there are infinitely many objects in the universe, which
> mathematically is its own powerset. Then, there is realizable
> evidence ("contrary evidence") that it is reasonable to discredit the
> powerset result.
No proof of anything in the above.
> I found my opinion around the use of mathematical proofs to illustrate
> that what I say is so.
You don't use mathematical proof.
> With regards to the duck test, proper classes are defined by their
> elements, as are sets. Are not proper classes defined by their
> elements, containing some specified elements, as are sets, similarly
> walking, quacking, etcetera?
Objects are defined by having a property that is had only by that
object.
> What's your opinion about the topic under discussion: is the
> specified transfinite recursion schema not defined for ordinals up to
> the cardinality of the irrationals?
In ZFC there is no upper bound to the transfinite recursion schemata.
> If it's not, the irrationals
> aren't uncountable. If it is, they aren't either.
You've not proven that.
MoeBlee
Ross A. Finlayson
The property that there remains an uncountable number or irrationals
left in the interval (p_lambda, 0) would only be so up to and
including any ordinal equivalent to the irrationals, were ZFC
consistent. For each element of an uncountable set there is a
distinct element of a countable set, then as there are 1-1 functions
either way there is a 1-1 and onto function, thus ZFC is inconsistent.
Do you just choose to disbelieve that part?
Arguments, a selection of supporting reasons sufficient to draw a
particular conclusion, are proofs of a sort. Proofs are simply
mathematical argument. That such sweeping arguments can be expressed
in so few words is a sign of their elegance not inapplicability.
Speaking of inapplicability, where there are nonstandard measure
theories, there aren't any applications of transfinite cardinals in
physics, say. Where as well via a simple argument the existence of a
physical universe along the lines of a mathematical universe
illustrates a stark counterexample (proof contrary) to the powerset
result, reasonable people may well find via a clear argument impetus
for the discovery of theories of the infinite more suitably evidenced
in reality. Concrete means real.
About the impromptu Goedel argument, that a recursively axiomatized
theorem strong enough to be Goedelianly incomplete thus has anonymous
axioms detailing properties of its objects, thus as well has another
ghost axiom further specifying that, as does that augmented set of
axioms etcetera, then a recursively axiomatized theory strong enough
to be Goedelianly incomplete isn't recursively axiomatized. In fact,
given the very first anonymous axiom, which presupposes properties
about the objects not otherwise specified in the tangible, explicitly
stated axioms, where those properties are unknown, then the theorem
isn't recursively axiomatizable, because there are as well elements of
the language unknown to any lector.
Proofs are arguments. Mathematical arguments are proofs, of varying
levels of rigor and explicitness.
Now, back to the argument at hand, it has to do with well-ordering a
subset of the irrationals. As described above using transfinite
induction, or a transfinite recursion schema indicating a well-
ordering of a subset of the irrationals, there is thus described in
ZFC using standard definitions of the irrational numbers a set of
irrationals. This set, which is constructible in ZFC via the well-
ordering principle, if it can't be uncountable, has that there aren't
uncountably many irrationals. Otherwise, it can be uncountable, in
ZFC, and is constructed so as to be. That done, it is illustrated
that due the denseness of the rationals in the reals, and reverse
normal ordering on the set, that for each element of the uncountable
set of irrationals there exists a particular, distinct rational
number. Otherwise the rationals aren't dense in the reals.
Then, via Cantor-Schroeder-Bernstein and a trivial injection the other
way, there is a bijection between the irrationals and rationals, due
to the denseness of each of the rationals and irrationals in the
reals.
So, then I've proven, some write proved, that given that the
irrationals are uncountable, in ZFC there exists (any of a wide
variety of) particular uncountable subsets of the irrationals such
that for each element there is a distinct rational, and that means ZFC
is inconsistent.
I have another proof of the existence of an injection from the
irrationals to rationals as referenced earlier in the thread, you
might consider that as well. Also, I argue that in theories with
expanded comprehension compared to ZFC, that ZFC's universe is the
Russell set, thus irregular, thus ZFC is inconsistent, as part of a
slate of arguments giving reasonable people justification to search
for novel mathematical foundations that aren't necessarily consistent
with the standard.
Address the arguments, as you haven't.
Ross
--
Finlayson Consulting
Brian Chandler
> On Sep 18, 3:21 pm, "Ross A. Finlayson" <r...@tiki-lounge.com> wrote:
<right, we'll skip this bit>
> Can you start again, explain each step here?
Ah, well, I _used_ to think that the definition of an "optimist" was
someone who doesn't look both ways before crossing a one-way street,
but now I've discovered Smirnoff.
Brian Chandler
http://imaginatorium.org
Robert Maas, see http://tinyurl.com/uh3t
> 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.
If you're claiming an uncountably-infinite set of disjoint open
intervals within the reals, I don't believe that's possible.
> ... well-ordering the rationals ...
Isn't that trivial, just use Cantor's zigzag mapping to the integers?
> The physical universe, as an infinite collection of items,
No such fact is in evidence. The total number of items in the
physical universe might be finite for all we know. The total number
of items in the *observable* universe almost surely is finite.
> mathematically is its own powerset,
You're a complete idiot!! Items, and sets of items, aren't the same
thing at all. Regardless of whether the number of items in the
Universe is finite or infinite, the number of subsets is two to
that power, which is more than the number of individual items.
> Ross
You wouldn't even make a decent intern, IMO.
Robert Maas, see http://tinyurl.com/uh3t
> irrationals.
Correct. x -> x+sqrt(2) will do nicely for example.
> Thus, with an injection either way, ...
What do you mean "either way"?? There is an injection in only one
direction, not the reverse.
> the Cantor-Bernstein theorem gives a bijection
Nope. To apply that theorem you would need injections both ways.
You have an injection only in one direction.
Robert Maas, see http://tinyurl.com/uh3t
> First Postulate: To draw a line from any point to any point.
> To me, this is a bit vague.
I agree. It's totally vague. It's not even a sentence. It's a
desideratum, a goal for a task, like you might include as a comment
in the preface to a software algorithm. For example, given two
positive integers, to compute their greatest common divisor.
But even as a goal, it's vague, as you explain.
> if between every two points there exists another point, then a
> line must be dense and consist of infinitely many points.
I agree about the infinite-many-points remark.
But the word "dense" is defined in a metric space, not in a
geometry, so what you've said so-far makes no sense, since all you
have is a geometry, not a metric space. Euclid didn't define any
metric as far as I know, since the concept hadn't yet been
conceived way back then.
Now the question arises whether the axioms of Euclid, plus the
choice of one particular line segment as a unit interval
(equivalently choosing two distinct points as unit distance from
each other), uniquely defines a metric space. I think so, but I'm
not sure. Can somebody verify this conjecture? If so, then it must
be the Euclidean metric, the usual metric on R^2 or R^3, and then
per that metric ... um ... you still have a problem, that dense is
a relation between two sets, not a property of a single set. One
set is dense in another. So what do you really mean?
I'm guessing that you mean that every point is a limit point per
that metric, i.e. for each point the greatest lower bound on
distances to other points is zero. Is that what you meant to say?
Now there's a concept of "nowhere dense", which seems to imply
there's a meaning of "dense" that applies to a single set, or to a
point of a set (the point is dense at that point), but I can't find
the definition via Google search. Maybe that meaning of "dense" is
what I defined in the previous paragraph, in which case I retract
my quibble. Can an expert on this jargon please find the definition
if it exists?
> But still, that doesn't imply that a line must have _uncountably_
> many points.
I think that's correct. Euclid's axioms seem to imply only
constructable points (from whatever points we start with to
generate a given geometry), and the constructable points are in
fact countably infinite if the starting points were either finite
or countably infinite.
I think we may interpret the axioms of Euclid to imply that for any
two distinct points there is a unique (closed) line segment between
the two points (and also the two half-open and also the one open
line segments too, by omitting one or both endpoints). With that
accepted, we can now define the word "convex" entirely in Euclidean
terms. (A set of points S is convex if for each pair of distinct
points in S p1,p2 the line segment p1,p2 is a subset of S.) Then a
single new axiom should suffice to generate uncountably infinite
points in any geometry that contains at least two points:
- Every convex subset of a line is either
- a line segment (which may be open or closed or half-open either way),
or
- a ray (which may extend in either direction from its starting
point, and may be open or closed at that starting point),
or
- the whole line.
That axiom basically establishes a bijection between points on a
line and Dedekind cuts, thereby giving us Cantor's uncountable reals
along any line.
Expert please, am I correct, or is that axiom not enough?
> (Of course not, since uncountability didn't exist in Euclid's day.)
But if a single new axiom, expressed in Euclid's own notational
conventions, is sufficient to generate Dedekind cuts hence reals,
then I think your remark is moot.
> 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.
Agreed, but see how easy it is to add just one more axiom, which
would have seemed "obvious" (even moreso than the parallel
postulate), to yield Cantor's uncountable reals! So I believe the
geometry that Euclid really believed in was in fact Cantor's
system.
Is there anyone here who takes the opposite position, that Euclud
would have accepted a partition of a line into two convex "rays"
for which there was no point whatsoever between them?
> Fifth Axiom: The whole is greater than the part.
Clearly he's talking about measure (length) of line segments, not
number of points in line segments.
After all he devised an elementary construction to produce a
bijection between any two (closed) line segments.
> So where does one get the idea that there exist uncountably many
> points anyway?
From my axiom about convex subsets of a line.
Ask any schoolkid: My axiom is obviously true about realworld lines, right?
> The Ruler Postulate clearly states that there exists a bijection
> between the set of real numbers and the set of points on a line,
I think my convex-subset postulate is more intuitively obvious.
Back to Euclid without my extra postulate:
> 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).
Indeed, in that system, a *curve* of arc-length pi does exist,
but a *line-segment* of (arc-)length pi does not exist,
assuming you start with just one origin point plus one extra point
unit distance away for each dimension of the space, to generate the
geometry via construction. (Of course if you *start* with three
points, 0 1 and pi along a single line, then you already have a
line segment of exactly pi, <bronxCheer/>)
Ross A. Finlayson
wrote:
Excuse me, you seem to be somewhat out of touch with the current
discussion. Using ZFC an injection is constructed from an uncountable
set of irrationals to a subset of the rationals.
Ross
--
Finlayson Consulting
David R Tribble
>> There are several tell-tale indicators of crankiness; see:
>> http://en.wikipedia.org/wiki/Crank_%28person%29
>> http://tinyurl.com/8hah7
>>
>> http://en.wikipedia.org/wiki/Duck_test
>
Ross A. Finlayson wrote:
> With regards to the duck test, proper classes are defined by their
> elements, as are sets. Are not proper classes defined by their
> elements, containing some specified elements, as are sets, similarly
> walking, quacking, etcetera?
I meant the Duck Test to be applied to people, not to sets or
classes.
David R Tribble
> What's your opinion about the topic under discussion: is the
> specified transfinite recursion schema not defined for ordinals up to
> the cardinality of the irrationals? If it's not, the irrationals
> aren't uncountable. If it is, they aren't either.
Your attempted proof that the irrationals are uncountable
is wrong. From what I know and what I've seen so far,
I agree with MoeBlee that you are not properly defining
the limit ordinals upon which you base your theorem.
You use an ordinal X that is "sufficiently large". So how
is X defined, and what is it exactly? You still have not
told us.
David R Tribble
>> There are several tell-tale indicators of crankiness; see:
>> http://en.wikipedia.org/wiki/Crank_%28person%29
>> http://tinyurl.com/8hah7
>
Ross A. Finlayson wrote:
> Consider the physical universe, and map mathematical objects to
> physical objects. Then, (all) functions among those represent
> physical objects, as do functions among those, ad infinitum.
> Then, there are infinitely many objects in the universe, ...
How do you know the physical universe is infinite?
Or not? Do you have a mathematical argument for this,
or a physical argument? Your "realizable evidence" is
no such thing.
> ... which mathematically is its own powerset.
> Then, there is realizable evidence ("contrary evidence")
> that it is reasonable to discredit the powerset result.
In order for an infinite set to be its own powerset, it
would have to contain members that are themselves
sets (permutations) of objects already in the set.
How is this possible using physical objects?
How is an infinite set of objects its own powerset?
How does this work, since it contradicts the elementary
proof that no set bijects with its own powerset?
Are you assuming some set theory other than ZFC?
MoeBlee
I suggested that he might as well take X to be the cardinality of the
set of real numbers. So, suppose he does that or suppose he just says
X is an arbitrary uncountable ordinal. Fine. But then his next step is
to introduce F and C, which he "defines" by "defining" each in terms
of the other. If he can define one of them and then the other, then
fine, his argument can then continue. But his argument is stopped when
he can't define his F and C without "defining" them in terms of each
other.
MoeBlee