uncountability of the reals
Elaine Jackson
(1) the Cantor diagonal proof and (2) a proof in which, beginning with an
arbitrary sequence of reals, two subsequences are defined, one strictly
decreasing and one strictly increasing, both with the same limit; it is then
shown that this limit can't have been a term in the original, given
sequence. Does anybody know any other proofs?
spam...@nil.nil
Given a set A, P(A) (the power set of A, the set of all subsets of A)
has larger cardinality.
(if f: A --> P(A)
then the set {x in A: x is NOT in f(x)}
is not in the range of f - to prove it,
see what happens if f(y)=that_set -
is y in f(y) or not : contradictions in both cases)
(this can be considered, however, just another way of writing down
the Cantor diagonal proof - but it *looks* different)
To use it, you want to show that the reals are in a one to one
correspondence with the set of subsets of the natural numbers.
Let's just do it for the reals in [0,1).
Write any such real down in base three. Throw away anything that has
a "2" in its expansion (the only reason I am doing this is to avoid
things like .01012222222222222222.... which is also .0102). That set
is a subset of the reals in [0,1).
Now, for each such number (they are written as:
.01110101... etc., each "digit being 0 or 1) consider the subset of
the natural numbers corresponding to where the "digits" are one
(so for .01110101... this is the set {2,3,4,6,8,...} since the
second, third, fourth, sixth, eighth "digits" are one).
This gives a one-one (which is why I threw out things that had a "2"
in it) correspondence between this set and the subsets
of the natural numbers. But the set of subsets of the natural numbers
must have greater cardinality than the natural numbers themselves.
Elaine Jackson
<spam...@nil.nil> wrote in message news:3b3e36a5$1...@nntp2.nac.net...
Ross
Elaine Jackson wrote:
Can you please elaborate 2? This is the first that I have heard of that
method. It does not seem clear, for example, consider the arbitrary sequence of
those reals from zero to two, split into two subsequences (0,1) and (2,1), the
limit of each being one and that value existing in the original sequence.
Also, I am inclined to disbelieve its existence, as I disbelieve 1. I do not
accept the diagonal method, for that any number that would be the "diagonal
number" would actually be some other element of the presupposed enumeration of
the reals. I was wondering myself only earlier today what other possible reason
that there could be to that effect.
The unmappable powerset theorem says nothing about the reals.
--
contact Ross
company Apex Internet Software
email in...@apexinternetsoftware.com
website http://www.apexinternetsoftware.com/
Robert Glass
Elaine Jackson wrote:
Countable sets have Lebesgue measure zero.
Robert Glass
Ross wrote:
> Elaine Jackson wrote:
>
> > I'm curious about other proofs of this fact. The two I know about now are
> > (1) the Cantor diagonal proof and (2) a proof in which, beginning with an
> > arbitrary sequence of reals, two subsequences are defined, one strictly
> > decreasing and one strictly increasing, both with the same limit; it is then
> > shown that this limit can't have been a term in the original, given
> > sequence. Does anybody know any other proofs?
>
> Can you please elaborate 2? This is the first that I have heard of that
> method.
This proof was painstakingly explained to Mr Ross Finlayson in February on this
forum in the thread entitled "Cantor's First Proof".
http://forum.swarthmore.edu/epigone/sci.math/herdwixdee
http://groups.google.com/groups?hl=en&safe=off&ic=1&th=c41f2177ca043928,57&seekm=3A8BCCAB.41DF7EC8%40hot.rr.com#p
There is an indexing error in the original presentation of the proof which
was corrected in
http://forum.swarthmore.edu/epigone/sci.math/herdwixdee/3A8FD48A...@my-deja.com
> It does not seem clear, for example, consider the arbitrary sequence of
> those reals from zero to two, split into two subsequences (0,1) and (2,1), the
> limit of each being one and that value existing in the original sequence.
>
Mr Finlayson reveals the extent of his knowledge of the meaning of the term
"sequence".
>
> Also, I am inclined to disbelieve its existence, as I disbelieve 1.
Someone once asked Mark Twain whether he believed in infant baptism.
Twain replied: "Believe in it?! I've seen it!"
Ross
Robert Glass wrote:
To paraphrase myself, "don't try those tit-for-tat games with me, Glass."
Ross
Ross
Robert Glass wrote:
> Ross wrote:
>
> > Elaine Jackson wrote:
> >
> > > I'm curious about other proofs of this fact. The two I know about now are
> > > (1) the Cantor diagonal proof and (2) a proof in which, beginning with an
> > > arbitrary sequence of reals, two subsequences are defined, one strictly
> > > decreasing and one strictly increasing, both with the same limit; it is then
> > > shown that this limit can't have been a term in the original, given
> > > sequence. Does anybody know any other proofs?
> >
> > Can you please elaborate 2? This is the first that I have heard of that
> > method.
>
> This proof was painstakingly explained to Mr Ross Finlayson in February on this
> forum in the thread entitled "Cantor's First Proof".
> http://forum.swarthmore.edu/epigone/sci.math/herdwixdee
> http://groups.google.com/groups?hl=en&safe=off&ic=1&th=c41f2177ca043928,57&seekm=3A8BCCAB.41DF7EC8%40hot.rr.com#p
>
> There is an indexing error in the original presentation of the proof which
> was corrected in
> http://forum.swarthmore.edu/epigone/sci.math/herdwixdee/3A8FD48A...@my-deja.com
>
> > It does not seem clear, for example, consider the arbitrary sequence of
> > those reals from zero to two, split into two subsequences (0,1) and (2,1), the
> > limit of each being one and that value existing in the original sequence.
> >
>
> Mr Finlayson reveals the extent of his knowledge of the meaning of the term
> "sequence".
>
A sequence is an ordered list. That is the extent of its meaning.
>
> >
> > Also, I am inclined to disbelieve its existence, as I disbelieve 1.
>
> Someone once asked Mark Twain whether he believed in infant baptism.
> Twain replied: "Believe in it?! I've seen it!"
>
> > I do not
> > accept the diagonal method, for that any number that would be the "diagonal
> > number" would actually be some other element of the presupposed enumeration of
> > the reals. I was wondering myself only earlier today what other possible reason
> > that there could be to that effect.
> >
> > The unmappable powerset theorem says nothing about the reals.
> >
> > --
> > contact Ross
> > company Apex Internet Software
> > email in...@apexinternetsoftware.com
> > website http://www.apexinternetsoftware.com/
I copy here what we appear to be discussing, from that post which you mention:
> Suppose f is any map from N to R. We'll show that there is a real number a
> such that a is not in the range of f.
>
> Look at f(0) and f(1). If f(0)<f(1) consider the closed interval [f(0),f(1)],
> otherwise
> switch them. Call that interval U(1). Now for each n>1 we will construct a
> subinterval
> of U(n) of U(n-1) such that f(n) is not in U(n-1). (It may help to draw a picture.
Your function appears to be a function following what is often called the midpoint theorem, that is to say, at this
point f is necessarily increasing. That is because f(0)<f(1).
So, what you are saying is that for each n>1, that the superinterval U(n+1) is such that f(n+1) is not in U(n). That
implies that the length of U(n+1) is greater than that of U(n), as f is necessarily increasing, else that f(n) would
appear in some U(n-1).
> Also
> this should remind you of the "middle thirds" construction of the Cantor Set.)
>
> Consider U(n-1), with endpoints a(n-1) < b(n-1). If f(n) is outside U(n-1) let U(n)
> be the "middle third" of U(n-1), i.e. a(n) = a(n-1) + (b(n-1) - a(n-1))/3 and
> b(n) = a(n-1) + 2(b(n-1) - a(n-1))/3.
>
The lower bound a(n) is less than or equal to any a(m<n), that means for the interval U(n+1) to always be a
superinterval of U(n) that the lower bound a(n) can not be greater than that of a(n-1), and the upper bound b(n) can
not be less than b(n-1). What this says, among other things, is that a(n)<=f(0) and b(n) <= f(1), as well, b(n)
>=f(n), where f(n) >b(n-1) .
>...
> If f(n) is in U(n-1) and is to the left of the midpoint, let a(n) = f(n) and
> b(n) = a(n-1) + (b(n-1) - a(n-1))/2.
It says above that that can not be the case, that is to say, f(n) is not in U(n-1), up where it says "it may help to
draw a picture". Let us ignore it.
>If f(n) is to the right of the midpoint,
> let a(n) = a(n-1) + (b(n-1) - a(n-1))/2 and b(n) = f(n).
>
> In this way we have a nested sequence of closed intervals
> U(1) contains U(2) contains U(3) contains... where for each n in N f(n-1) is
> not an element of U(n).
>
Your description is of something opposite of that.
> Now by the nested interval property (see a real analysis book in the chapter
> on completeness, Bolzano-Weierstrass, Heine-Borel, etc if you need a reminder)
> the intersection of the U(i) is non-empty -- it is either a closed interval, in
> which
> case let a be any interior point (say the midpoint) or a single point, in which case
>
> let a be that point.
>
The intersection is (f(0), f(1)).
> Now a cannot be in the range of f because it is in ALL of the U(i) but for any
> k in N, k>1, f(k) is not in U(k+1) and f(0) is not in U(2).
The intersection is (f(0), f(1)).
Ross
Virgil
wrote:
> Elaine Jackson wrote:
>
> > I'm curious about other proofs of this fact. The two I know about now are
> > (1) the Cantor diagonal proof and (2) a proof in which, beginning with an
> > arbitrary sequence of reals, two subsequences are defined, one strictly
> > decreasing and one strictly increasing, both with the same limit; it is
> > then
> > shown that this limit can't have been a term in the original, given
> > sequence. Does anybody know any other proofs?
>
> Can you please elaborate 2? This is the first that I have heard of that
> method. It does not seem clear, for example, consider the arbitrary sequence
> of
> those reals from zero to two, split into two subsequences (0,1) and (2,1),
> the
> limit of each being one and that value existing in the original sequence.
If you assume the existence of a sequence containing all the real
numbers, it is easy to construct subsequences having the properties (2).
It was done in detail in a past post in one of the threads disproving
some of your ideas. But then perhaps you only remember the parts that
support your own beliefs.
Try to do it yourself, and you may discover how easy it is.
> Also, I am inclined to disbelieve its existence, as I disbelieve 1. I do not
> accept the diagonal method, for that any number that would be the "diagonal
> number" would actually be some other element of the presupposed enumeration
> of
> the reals. I was wondering myself only earlier today what other possible
> reason
> that there could be to that effect.
Your disbelief does not make it disappear.
>
> The unmappable powerset theorem says nothing about the reals.
>
It only says nothing about the reals to those who know little about the
reals to begin with. To those who know enough, it says quite a bit.
>
> --
> contact Ross
> company Apex Internet Software
> email in...@apexinternetsoftware.com
> website http://www.apexinternetsoftware.com/
>
>
Ross's beliefs, here as elsewhere, are too often supported only by an
unwarranted faith in the infallibility his own intuition and in the face
of the clearest of logic countradicting those same beliefs.
Virgil
wrote:
> To paraphrase myself, "don't try those tit-for-tat games with me, Glass."
>
> Ross
>
> --
> contact Ross
> company Apex Internet Software
> email in...@apexinternetsoftware.com
> website http://www.apexinternetsoftware.com/
>
>
I believe it was Oscar Wilde who said " I often quote myself. It
improves my conversation."
I don't think Ross has quite got the style of it, yet.
Ross
Virgil wrote:
> In article <3B3E67C0...@calpha.com>, Ross <ap...@calpha.com>
> wrote:
>
> > Elaine Jackson wrote:
> >
> > > I'm curious about other proofs of this fact. The two I know about now are
> > > (1) the Cantor diagonal proof and (2) a proof in which, beginning with an
> > > arbitrary sequence of reals, two subsequences are defined, one strictly
> > > decreasing and one strictly increasing, both with the same limit; it is
> > > then
> > > shown that this limit can't have been a term in the original, given
> > > sequence. Does anybody know any other proofs?
> >
> > Can you please elaborate 2? This is the first that I have heard of that
> > method. It does not seem clear, for example, consider the arbitrary sequence
> > of
> > those reals from zero to two, split into two subsequences (0,1) and (2,1),
> > the
> > limit of each being one and that value existing in the original sequence.
>
> If you assume the existence of a sequence containing all the real
> numbers, it is easy to construct subsequences having the properties (2).
>
It appears to me that if an ordered sequence of reals is split into two
subsequences where one is strictly increasing and the other strictly decreasing,
then where the original sequence has bounds a and b that the limit of any of the
subsequences could only be a, b, or some value between a and b. Where they are to
have the same limit, it would have to be somewhere between a and b, where that is
in the original sequence.
So, if it's easy, provide an example.
>
> It was done in detail in a past post in one of the threads disproving
> some of your ideas. But then perhaps you only remember the parts that
> support your own beliefs.
>
Provide an example.
>
> Try to do it yourself, and you may discover how easy it is.
>
> > Also, I am inclined to disbelieve its existence, as I disbelieve 1. I do not
> > accept the diagonal method, for that any number that would be the "diagonal
> > number" would actually be some other element of the presupposed enumeration
> > of
> > the reals. I was wondering myself only earlier today what other possible
> > reason
> > that there could be to that effect.
>
> Your disbelief does not make it disappear.
>
Hancher, this might seem off-topic, but I think you are a girl.
> >
> > The unmappable powerset theorem says nothing about the reals.
> >
>
> It only says nothing about the reals to those who know little about the
> reals to begin with. To those who know enough, it says quite a bit.
>
> Ross's beliefs, here as elsewhere, are too often supported only by an
> unwarranted faith in the infallibility his own intuition and in the face
> of the clearest of logic countradicting those same beliefs.
Hancher, about the reals, the reals are a subset of the closure of the
hyperintegers to division, where the rationals are the closure of the (finite)
integers to division. The reals are dense (and continuous) on the Euclidean number
line, where differences between them are marked in real amounts. About the
powerset of a set, it has 2^n elements where the set has n elements, constructing
some of the rationals from zero to one using base two shows that for a finite
string length n that there would 2^n possibly many.
Ross
Virgil
wrote:
> I copy here what we appear to be discussing, from that post which you
> mention:
>
> > Suppose f is any map from N to R. We'll show that there is a real number a
> > such that a is not in the range of f.
> >
> > Look at f(0) and f(1). If f(0)<f(1) consider the closed interval
> > [f(0),f(1)],
> > otherwise
> > switch them. Call that interval U(1). Now for each n>1 we will construct a
> > subinterval
> > of U(n) of U(n-1) such that f(n) is not in U(n-1). (It may help to draw a
> > picture.
>
> Your function appears to be a function following what is often called the
> midpoint theorem, that is to say, at this
> point f is necessarily increasing. That is because f(0)<f(1).
Wrong. The construction allows for either f(0) < f(1) or f(1) < f(0).
It did gloss over the case of equality:f(n) = f(m), n<>m. But this is
easily dealt with:
If f has only finitely many values, it clearly omits some reals, so this
case is no problem.
If it contains infinitely many different values, take a subseqence which
omits all repetitions of values after the first ocurence and the
subsequence will still cover all the values covered by the original.
>
> So, what you are saying is that for each n>1, that the superinterval U(n+1)
> is such that f(n+1) is not in U(n).
Not at all. It says f(n) is not in U(n+1). This is critical to the
proof, so of course, Ross messed it up.
> That
> implies that the length of U(n+1) is greater than that of U(n), as f is
> necessarily increasing, else that f(n) would
> appear in some U(n-1).
Wrong again. The length of U(n+1) is either 1/ or 1/2 as great as that
of U(n).
>
> > Also
> > this should remind you of the "middle thirds" construction of the Cantor
> > Set.)
> >
>
> > Consider U(n-1), with endpoints a(n-1) < b(n-1). If f(n) is outside U(n-1)
> > let U(n)
> > be the "middle third" of U(n-1), i.e. a(n) = a(n-1) + (b(n-1) - a(n-1))/3
> > and
> > b(n) = a(n-1) + 2(b(n-1) - a(n-1))/3.
> >
>
> The lower bound a(n) is less than or equal to any a(m<n), that means for the
> interval U(n+1) to always be a
> superinterval of U(n) that the lower bound a(n) can not be greater than that
> of a(n-1), and the upper bound b(n) can
> not be less than b(n-1).
Wrong again. The a(n) form an increasing sequence, the b(n) form a
decreasing sequence, and a(n) < b(m) for all n,m in N.
> What this says, among other things, is that
> a(n)<=f(0) and b(n) <= f(1), as well, b(n)
> >=f(n), where f(n) >b(n-1) .
>
Wrong again.
What it actually says is a(n) >= min{f(0),f(1)} for n in N,
and b(n) <= max{f(0),f(1)}, for n in N.
> >...
> > If f(n) is in U(n-1) and is to the left of the midpoint, let a(n) = f(n)
> > and
> > b(n) = a(n-1) + (b(n-1) - a(n-1))/2.
>
> It says above that that can not be the case, that is to say, f(n) is not in
> U(n-1), up where it says "it may help to
> draw a picture". Let us ignore it.
>
> >If f(n) is to the right of the midpoint,
> > let a(n) = a(n-1) + (b(n-1) - a(n-1))/2 and b(n) = f(n).
> >
> > In this way we have a nested sequence of closed intervals
> > U(1) contains U(2) contains U(3) contains... where for each n in N f(n-1)
> > is
> > not an element of U(n).
> >
>
> Your description is of something opposite of that.
Your understanding may be opposite, but the words themselves say just
that.
>
> > Now by the nested interval property (see a real analysis book in the
> > chapter
> > on completeness, Bolzano-Weierstrass, Heine-Borel, etc if you need a
> > reminder)
> > the intersection of the U(i) is non-empty -- it is either a closed
> > interval, in
> > which
> > case let a be any interior point (say the midpoint) or a single point, in
> > which case
> >
> > let a be that point.
> >
>
> The intersection is (f(0), f(1)).
Wrong again. Since a(n) is an increasing sequence and is bounded above
by any of the b(n), it has a limit, say a(oo). Similarly b(n) is
decreasing and bounded below by any of the a(n), so ahs a limit which we
can denote by b(oo), with a(oo) <= b(oo).
Then the inersection of all the U(n) closed intervals is the non-empty
closed interval, U(oo) = [a(oo),b(oo)].
Since by construction, f(n) not in U(n+1), for each n, f(n) not in
U(oo). But U(oo) is not empty so none of its members are f(n)'s.
>
> > Now a cannot be in the range of f because it is in ALL of the U(i) but for
> > any
> > k in N, k>1, f(k) is not in U(k+1) and f(0) is not in U(2).
>
>
> The intersection is (f(0), f(1)).
Repeated wrongness. See above.
>
> Ross
> --
> contact Ross
> company Apex Internet Software
> email in...@apexinternetsoftware.com
> website http://www.apexinternetsoftware.com/
>
It is curious that Ross was unable to find the only true weakness in the
proof.
It transpires that Ross's critique of the proof is fatally flawed and
that the proof is valid once the matter of repeated values is dealt with.
Virgil
wrote:
Consider the sequence {f(0),f(1),f(2),...} of real numbers and assume
that there are infinitely many different values ( if there are only
finitely many values, clearly some reals have been left out). If there
are any repreated values delete all repetitions except the first, so
that the revised sequence is an injection from N = {0,1,2,...} to R =
the set of reals and is a surjection onto the range of the original
ssequence (no reals were omitted by deleting reptitions).
The following proves that the sequence cannot be a surjection.
Let U(1) = [a(1),b(1)] be the closed middle third of the interval from
min{f(0),f(1)} to max{f(0),f(1)}. Since f(0) <> f(1), U(1) has positive
length and contains neither f(0) nor f(1).
To construct U(n+1), divide U(n) into thirds and let U(n+1) be the
closure of whichever third is furthest from f(n), but if f(n) is the
midpoint of U(n), pick the upper third. thus f(n) is never in the closed
interval U(n+1).
The intersection of all the U(n), being a nested sequence of closed
intervals, is a non-empty set U.
Since, for all n, f(n) is not in U(n+1), then for all n, f(n) is also
not in U.
Every member of the non-empty set U is a real number not listed by f.
Thus f is not a surjection, as asserted.
Mike Oliver
>
> Elaine Jackson wrote:
>
>> Does anybody know any other proofs?
>
> Countable sets have Lebesgue measure zero.
To make that work, of course, you have to show that R does
*not* have measure zero, which is a certain amount of work.
The analogous observation wrt category (countable sets are
meager) is probably a little easier; then you just have to
prove the Baire Category Theorem. I haven't analyzed the question
in detail, but my guess is that comes down to the same thing
as Cantor's original proof, which is Elaine's proof (2) in the
original article.
Robert Glass
Ross wrote:
>
>
> Hancher, about the reals, the reals are a subset of the closure of the
> hyperintegers to division, where the rationals are the closure of the (finite)
> integers to division.
Lest any newcomer with genuine interest in the subject should be deceived,
this is another of Mr Ross Finlayson's errors which has been pointed out to
him on numerous occasions by various persons and explained to him
several times (with more or less patience), for example:
http://groups.google.com/groups?start=390&hl=en&safe=off&th=fcbef057c531d5bd,527&rnum=400&ic=1&selm=3AB2C236.A8EDF210%40hot.rr.com
Daniel Grubb
1) Baire Category Theorem: A complete metric space is not the union of
countably many nowhere dense sets. Let the sets involved be singletons.
2) The Lebesgue measure of [0,1] is 1, but the measure of any countable
set is 0.
3) Let {x_n} be a sequence. Construct a sequence of closed intervals
[a_n,b_n] such that x_n is not in [a_n, b_n] and b_n-a_n-->0. This is
achieved by taking an appropriate third of the previous interval at
each stage. The intersection of these intervals is non-empty and has
no element of the sequense {x_n}. (This may be just a restatement
of your second proof.)
---Dan Grubb
David C. Ullrich
>
>
>Elaine Jackson wrote:
>
>> I'm curious about other proofs of this fact. The two I know about now are
>> (1) the Cantor diagonal proof and (2) a proof in which, beginning with an
>> arbitrary sequence of reals, two subsequences are defined, one strictly
>> decreasing and one strictly increasing, both with the same limit; it is then
>> shown that this limit can't have been a term in the original, given
>> sequence. Does anybody know any other proofs?
>
>Can you please elaborate 2? This is the first that I have heard of that
>method. It does not seem clear, for example, consider the arbitrary sequence of
>those reals from zero to two, split into two subsequences (0,1) and (2,1), the
>limit of each being one and that value existing in the original sequence.
(0,1) and (2,1) are not _sequences_.
Can you state exactly what the word "sequence" means? If not you
should not use the word in a mathematical context - when you use
words and it's obvious you don't know what they mean it looks
bad. (And yes, whether you believe it or not, when you talk about
the sequence (0,1) it is obvious you do not know what the word
"sequence" means.)
>Also, I am inclined to disbelieve its existence, as I disbelieve 1. I do not
>accept the diagonal method, for that any number that would be the "diagonal
>number" would actually be some other element of the presupposed enumeration of
>the reals. I was wondering myself only earlier today what other possible reason
>that there could be to that effect.
>
>The unmappable powerset theorem says nothing about the reals.
>
>
>--
>contact Ross
>company Apex Internet Software
>email in...@apexinternetsoftware.com
>website http://www.apexinternetsoftware.com/
>
>
David C. Ullrich
*********************
"Sometimes you can have access violations all the
time and the program still works." (Michael Caracena,
comp.lang.pascal.delphi.misc 5/1/01)
Elaine Jackson
the description of (2). The quick answer is that the "arbitrary sequence of
reals" is actually supposed, by hypothesis, to exhaust all of the reals, and
you derive a contradiction from that. A slightly longer answer is that maybe
you'd rather start with an arbitrary sequence and adopt hypotheses about it
as you go through the proof, justifying each hypothesis with the fact that
the contrary assumption implies that the sequence already fails to exhaust
all of the reals. Then you end up proving that the sequence fails to exhaust
the reals anyway. It happens that I prefer thinking about it like that,
which is why I misstated it the way I did. The longest answer (and the most
informative) is simply to go through the proof:
Let {s[n]} be a sequence such that, for each real number x, there is at
least one n with s[n]=x. Let a[0] and b[0] satisfy a[0]<b[0], and for n>=0,
define as follows:
1) a[n+1]=s[p], where p is the smallest k with s[k] in the interval
(a[n],b[n])
2) b[n+1]=s[q], where q is the smallest k with s[k] in the interval
(a[n+1],b[n])
Then {a[n]}is strictly increasing and bounded above by b[0], so that {a[n]}
has a limit A, and {b[n]} is strictly decreasing and bounded below by a[0],
so that {b[n]} has a limit B. Since a[n]<b[n] for all n, A<=B. Hence there
is at least one real number x such that x is in (a[n],b[n]) for all n. Hence
there is an integer M>=0 with
a[n]<s[M]<b[n] for all n, and in particular a[M+1]<s[M]<b[M+1]. Let m be the
smallest M with a[M+1]<s[M]<b[M+1]. Since (a[m+1],b[m+1]) is a subset of
(a[m],b[m]), s[m] is in (a[m],b[m]). For any k<m, the minimality of m
implies that s[k] does not belong to (a[k+1],b[k+1]), and since m>=(k+1),
(a[m],b[m]) is a subset of (a[k+1],b[k+1]). Hence s[k] does not belong to
(a[m],b[m]) for all k<m, and so, by the definition, s[m]=a[n+1],
contradicting the hypothesis that a[m+1]<s[m]<b[m+1].
Ross <ap...@calpha.com> wrote in message news:3B3E67C0...@calpha.com...
Ross
Robert Glass wrote:
> Ross wrote:
>
> >
> >
> > Hancher, about the reals, the reals are a subset of the closure of the
> > hyperintegers to division, where the rationals are the closure of the (finite)
> > integers to division.
>
> Lest any newcomer with genuine interest in the subject should be deceived,
> this is another of Mr Ross Finlayson's errors which has been pointed out to
> him on numerous occasions by various persons and explained to him
> several times (with more or less patience), for example:
> http://groups.google.com/groups?start=390&hl=en&safe=off&th=fcbef057c531d5bd,527&rnum=400&ic=1&selm=3AB2C236.A8EDF210%40hot.rr.com
>
Bob, the rationals are the closure of the field of the finite integers to division, as well, the reals are a subset of the closure of
the hyperintegers to division. What's the error?
Have you yet repaired your proof again?
Ross
Ross
Elaine Jackson wrote:
> Right. I've already gotten flak in another newsgroup for the way I mangled
> the description of (2). The quick answer is that the "arbitrary sequence of
> reals" is actually supposed, by hypothesis, to exhaust all of the reals, and
> you derive a contradiction from that. A slightly longer answer is that maybe
> you'd rather start with an arbitrary sequence and adopt hypotheses about it
> as you go through the proof, justifying each hypothesis with the fact that
> the contrary assumption implies that the sequence already fails to exhaust
> all of the reals.
I don't know if that is a valid way. The first way, establishing contradiction,
would be a more valid way.
> Then you end up proving that the sequence fails to exhaust
> the reals anyway. It happens that I prefer thinking about it like that,
> which is why I misstated it the way I did. The longest answer (and the most
> informative) is simply to go through the proof:
>
> Let {s[n]} be a sequence such that, for each real number x, there is at
> least one n with s[n]=x. Let a[0] and b[0] satisfy a[0]<b[0], and for n>=0,
> define as follows:
>
> 1) a[n+1]=s[p], where p is the smallest k with s[k] in the interval
> (a[n],b[n])
> 2) b[n+1]=s[q], where q is the smallest k with s[k] in the interval
> (a[n+1],b[n])
>
> Then {a[n]}is strictly increasing and bounded above by b[0], so that {a[n]}
> has a limit A, and {b[n]} is strictly decreasing and bounded below by a[0],
> so that {b[n]} has a limit B. Since a[n]<b[n] for all n, A<=B. Hence there
> is at least one real number x such that x is in (a[n],b[n]) for all n. Hence
> there is an integer M>=0 with
> a[n]<s[M]<b[n] for all n, and in particular a[M+1]<s[M]<b[M+1]. Let m be the
> smallest M with a[M+1]<s[M]<b[M+1]. Since (a[m+1],b[m+1]) is a subset of
> (a[m],b[m]), s[m] is in (a[m],b[m]). For any k<m, the minimality of m
> implies that s[k] does not belong to (a[k+1],b[k+1]), and since m>=(k+1),
> (a[m],b[m]) is a subset of (a[k+1],b[k+1]). Hence s[k] does not belong to
> (a[m],b[m]) for all k<m, and so, by the definition, s[m]=a[n+1],
> contradicting the hypothesis that a[m+1]<s[m]<b[m+1].
>
A couple things, one is that where m>=(k+1), that m can equal k+1, thus that
(a[m],b[m])=(a[k+1],b[k+1]), as it is subset and not proper subset, thus that
s[k] can belong to (a[m],b[m]).
Another thing is about the minimality of M and whether it should be the
maximality of M.
Basically, the way I see it is that these two sequences each converge to some
value on (a[0], b[0]), where that value is tautologically in (a[0], b[0]).
Ross
Virgil
wrote:
> Elaine Jackson wrote:
>
> > Right. I've already gotten flak in another newsgroup for the way I mangled
> > the description of (2). The quick answer is that the "arbitrary sequence of
> > reals" is actually supposed, by hypothesis, to exhaust all of the reals,
> > and
> > you derive a contradiction from that. A slightly longer answer is that
> > maybe
> > you'd rather start with an arbitrary sequence and adopt hypotheses about it
> > as you go through the proof, justifying each hypothesis with the fact that
> > the contrary assumption implies that the sequence already fails to exhaust
> > all of the reals.
>
> I don't know if that is a valid way. The first way, establishing
> contradiction,
> would be a more valid way.
They are equally and totally valid.
Virgil
wrote:
> Bob, the rationals are the closure of the field of the finite integers to
> division, as well, the reals are a subset of the closure of
> the hyperintegers to division. What's the error?
The set of finite integers (with standard arithmetic) does not form a
field, it is the rationals that are the field.
You have been skipping your vocabulary lessons again!
Which formulation of hyperintegers are you using in which every standard
real is a quotient of hyperintegers?
Ross
Ross wrote:
> Elaine Jackson wrote:
>
> > Right. I've already gotten flak in another newsgroup for the way I mangled
> > the description of (2). The quick answer is that the "arbitrary sequence of
> > reals" is actually supposed, by hypothesis, to exhaust all of the reals, and
> > you derive a contradiction from that. A slightly longer answer is that maybe
> > you'd rather start with an arbitrary sequence and adopt hypotheses about it
> > as you go through the proof, justifying each hypothesis with the fact that
> > the contrary assumption implies that the sequence already fails to exhaust
> > all of the reals.
>
> I don't know if that is a valid way. The first way, establishing contradiction,
> would be a more valid way.
>
About that, what that means to say is that not(not A) does not imply A. In most
cases of intuitionistic logic it does, that is also called bivalent, Boolean, or
true/false logic as opposed to multivalent logic, but it doesn't always. For
example, in terms of cats and dogs, saying that X is not a cat does not imply that
it is a dog, where it might be something else, just not a cat.
I think I am unclear here. I reexamine your statement, I will here try to restate
your statement, thus that I can be sure to be discussing the same thing that you
have outlined.
First, here are the terms in this system:
n integer variable
x real variable
{s[n]} sequence of reals, sequence of all reals, for any real x, there exists
s(n)=x for function s, where the braces imply the sequence, thus that the
parenthetical notation is (s[0], s[1], ...), where that sequence is the reals.
s(n) function with domain of integer, range reals
a(n) function with domain of integers, range reals
b(n) function with domain integers, range reals
a(n+1) = s(p)
b(n+1) = s(q)
At this point, I will note that it appears that a and b can each be constant, where
a(0)=a(1)=a(2)... and b(0)=b(1)=b(2)=..., where p and q would each always be zero,
with the s(0) being in (a(0), b(0)).
Also, it would seem that the range of s is bounded by (a(0), b(0)). That is to say,
for a(n) to be less than arbitrary s(n) that a would be decreasing, similarly b
increasing, where they are having been described differently.
Thus, I would say that s(n) has a codomain a(0), b(0), for that to be otherwise.
It appears to me that the point of the theorem is that there are upper and lower
bounds b(0) and a(0) over an interval of reals, and that the function a(n) is
increasing, b(n) decreasing, where the limit of each function as n->oo is somewhere
on (a(0), b(0)).
I have an issue with this part: "a[n]<s[M]<b[n] for all n", where it might rather be
"a[n]<=s[M]<=b[n] for all n", where s(M) is the limit of a(n) and b(n).
I think another point to consider in light of the construction is that s should be
defined as increasing or decreasing, where it is not.
Now, as the limit of b(n)-a(n) goes to zero, with a(0)<b(0) and a increasing and b
decreasing, giving a(0) < (lim a(n)) = (lim b(n)) < b(0), where the limit of s(n)
also happens to be equal to lim a(n) and lim b(n), I don't see that as disqualifying
the sequence from also converging to lim s(n), where that was the original idea.
For example, consider the Cauchy sequence approximating the real value one-half
between zero and one as the a sequence, and the sequence of one minus each of those
values as the b sequence. Then, interleave them as (a(0), b(0), a(1), b(1), ...) to
form the s sequence. It converges to one-half.
Ross
Virgil wrote:
Have you yet repaired a proof again?
I happen to be using my own formulation of the hyperintegers, which are the
finite and infinite integers.
The rationals are a field, the integers are not a field. Inkpens are not a
field, nor are they integers.
The rationals are the field of the closure of the finite integers to division,
the reals are a field that is a subset of the closure of the hyperintegers to
division. The hyperreals is the field that is the closure of the hyperintegers
to division That is to say, defining the operation of division on the integers
induces the rationals, and defining the operation of division on the
hyperintegers induces the reals and hyperreals.
For example, 1/(non-finite hyperinteger X) evaluates to an infinitesimal
hyperreal, where 1/(finite hyperinteger X) evlauate to rational 1/X.
Robert Glass
Ross wrote:
>
> Bob, the rationals are the closure of the field of the finite integers to division, as well, the reals are a subset of the closure of
> the hyperintegers to division. What's the error?
>
The error is that your statement is wrong, which you would know if you ever
made an attempt to learn some mathematics instead of trolling around putting
random technical terms into your idiotic sentences.
The square root of two is not the ratio of two hyper-integers. The proof has been
known since the time of Pythagoras (Pythagoras just wasn't aware that the proof
worked for hyper-integers as well as integers.)
Ross
Robert Glass wrote:
The square root of two is not the ratio of any two finite integers, that's not a problem, nor does it contradict what I said. If there
exist any infinite hyperintegers, thus hyperintegers, then there exists infinite pairs of them for each irrational number thus that their
ratios are exactly equal to that irrational number.
The infinite hyperintegers have infinite precision, enough to represent the precision of an irrational number.
Robert Glass
Ross wrote:
> Virgil wrote:
>
> > In article <3B3F62CC...@calpha.com>, Ross <ap...@calpha.com>
> > wrote:
> >
> > > Bob, the rationals are the closure of the field of the finite integers to
> > > division, as well, the reals are a subset of the closure of
> > > the hyperintegers to division. What's the error?
> >
> > The set of finite integers (with standard arithmetic) does not form a
> > field, it is the rationals that are the field.
> >
> > You have been skipping your vocabulary lessons again!
> >
> > Which formulation of hyperintegers are you using in which every standard
> > real is a quotient of hyperintegers?
>
> Have you yet repaired a proof again?
>
> I happen to be using my own formulation of the hyperintegers, which are the
> finite and infinite integers.
>
Of course you have never given any definition of these hyperintegers. or
infinite integers for that matter. I know you have babbled about infinite
precision and "some hyperinteger starts with 314159
and contains all the digits of Pi, as it is infinite" but when asked for a proof
that hyper-integers can be represented in such a manner and asked how to
actually do this division of which your field of reals is (supposedly) the closure
of these hyperintegers, your reply was "No".
http://groups.google.com/groups?start=390&hl=en&safe=off&th=fcbef057c531d5bd,527&rnum=400&ic=1&selm=3AB2C236.A8EDF210%40hot.rr.com
Actually, that was probably the most intelligent answer anyone has ever
gotten from you.
>
> The rationals are a field, the integers are not a field.
So, the integers are a field when you write to me but not a field
when you write to Virgil.
Is that one of your axioms or is it a theorem in your system?
> Inkpens are not a
> field, nor are they integers.
>
> The rationals are the field of the closure of the finite integers to division,
> the reals are a field that is a subset of the closure of the hyperintegers to
> division.
Well, if you mean by "reals" what everybody else means by "rationals" you might
be almost right.
> The hyperreals is the field that is the closure of the hyperintegers
> to division That is to say, defining the operation of division on the integers
> induces the rationals, and defining the operation of division on the
> hyperintegers induces the reals and hyperreals.
>
So you are also calling "hyperreals" what everyone else calls "hyper-rationals".
Robert Glass
Ross wrote:
> Robert Glass wrote:
>
> > Ross wrote:
> >
> > >
> > > Bob, the rationals are the closure of the field of the finite integers to division, as well, the reals are a subset of the closure of
> > > the hyperintegers to division. What's the error?
> > >
> >
> > The error is that your statement is wrong, which you would know if you ever
> > made an attempt to learn some mathematics instead of trolling around putting
> > random technical terms into your idiotic sentences.
> >
> > The square root of two is not the ratio of two hyper-integers. The proof has been
> > known since the time of Pythagoras (Pythagoras just wasn't aware that the proof
> > worked for hyper-integers as well as integers.)
> >
> > >
> > > Have you yet repaired your proof again?
> > >
> > > Ross
>
> The square root of two is not the ratio of any two finite integers, that's not a problem, nor does it contradict what I said. If there
> exist any infinite hyperintegers, thus hyperintegers, then there exists infinite pairs of them for each irrational number thus that their
> ratios are exactly equal to that irrational number.
>
Prove it.
>
> The infinite hyperintegers have infinite precision, enough to represent the precision of an irrational number.
>
Prove it.
Robert Glass
Elaine Jackson wrote:
> Right. I've already gotten flak in another newsgroup for the way I mangled
> the description of (2).
Actually, Mr Finlayson did not accuse you of mangling your description. He said
that he didn't believe such a proof even existed. Congratulations! You have
been called a liar by Ross Finlayson -- that's kind of a badge of honor in this
group.
Ross
Robert Glass wrote:
I have discussed my reasoning why that I see that the infinite hyperintegers exist, and in having infinite precision why that their
ratios might represent any real number.
Those are some interesting comments, in my opinion. I don't know how to prove "that the the ring of hyper-integers can be represented
in this manner so that addition and multiplication can be meaningfully performed." I don't see it as necessary. I would like to know
how it is proven.
That's only about one thing.
>
> >
> > The rationals are a field, the integers are not a field.
>
> So, the integers are a field when you write to me but not a field
> when you write to Virgil.
>
> Is that one of your axioms or is it a theorem in your system?
>
Neither, it's a correction.
>
> > Inkpens are not a
> > field, nor are they integers.
> >
> > The rationals are the field of the closure of the finite integers to division,
> > the reals are a field that is a subset of the closure of the hyperintegers to
> > division.
>
> Well, if you mean by "reals" what everybody else means by "rationals" you might
> be almost right.
>
I think that the reals are a subset of the closure of the hyperintegers to division.
>
> > The hyperreals is the field that is the closure of the hyperintegers
> > to division That is to say, defining the operation of division on the integers
> > induces the rationals, and defining the operation of division on the
> > hyperintegers induces the reals and hyperreals.
> >
>
> So you are also calling "hyperreals" what everyone else calls "hyper-rationals".
>
I believe that any real can be represented as the ratio of two hyperintegers. Also, no.
>
> >
> > For example, 1/(non-finite hyperinteger X) evaluates to an infinitesimal
> > hyperreal, where 1/(finite hyperinteger X) evaluates to rational 1/X.
Ross
Robert Glass wrote:
> Elaine Jackson wrote:
>
> > Right. I've already gotten flak in another newsgroup for the way I mangled
> > the description of (2).
>
> Actually, Mr Finlayson did not accuse you of mangling your description. He said
> that he didn't believe such a proof even existed. Congratulations! You have
> been called a liar by Ross Finlayson -- that's kind of a badge of honor in this
> group.
>
That's not true.
>
> He said:
>
> >
> > >
> > > Can you please elaborate 2? This is the first that I have heard of that
> > > method. It does not seem clear, for example, consider the arbitrary
> > sequence of
> > > those reals from zero to two, split into two subsequences (0,1) and (2,1),
> > the
> > > limit of each being one and that value existing in the original sequence.
> > >
> > > Also, I am inclined to disbelieve its existence, as I disbelieve 1.
Bob, in other words, you had presented a reference (link) to what was described as
a similar method as 2, yet, I showed why it was meaningless, so, when I ask if it
had been repaired, that implies that I see it as broken.
Ross
Elaine Jackson
I was talking about? Can anybody give me the gist of the argument below? I
can't even make out who's talking to whom.
Ross <ap...@calpha.com> wrote in message news:3B3E8863...@calpha.com...
http://forum.swarthmore.edu/epigone/sci.math/herdwixdee/3A8FD48A.75C45998@my
Robert Glass
Ross wrote:
>
>
> I have discussed my reasoning why that I see that the infinite hyperintegers exist, and in having infinite precision why that their
> ratios might represent any real number.
>
> Those are some interesting comments, in my opinion. I don't know how to prove "that the the ring of hyper-integers can be represented
> in this manner so that addition and multiplication can be meaningfully performed." I don't see it as necessary.
That's revealing.
OK, here's an easier one:
You incessantly claim that in your system "for every n, n+1"
Let n be your number that starts 314159... and contains all the
digits of PI. What is n+1 ?
Robert Glass
Elaine Jackson wrote:
> Should I be trying to sort this out? Is this a different proof from the two
> I was talking about? Can anybody give me the gist of the argument below? I
> can't even make out who's talking to whom.
You have been trolled.
Ross Finlayson has been posting his nonsense to this group for at least 5 years
now.
He frequently states that his purpose is to get the longest threads.
In answer to your question, the proof I was talking about was the one you
labeled 2.
It was historically the first proof of the uncountability of the real numbers,
discovered by
Cantor in 1874. The "diagonalization proof" dates to 1891.
Robert Glass
Ross wrote:
> Bob, in other words, you had presented a reference (link) to what was described as
> a similar method as 2, yet, I showed why it was meaningless, so, when I ask if it
> had been repaired, that implies that I see it as broken.
All you showed was more evidence that you are an idiot, a lunatic and a troll. No one
cares what you see. We are all only doing this for our own entertainment and to warn
off unwary newcomers who might come across your gibberish and think that there
is some actual mathematics there.
Elaine Jackson
a[0]<b[0], you define a[1] as the "first" s[i] strictly between a[0] and
b[0], then define b[1] as the "first" s[i] strictly between a[1] and b[0],
and so on in similar fashion, where the "first" s[i] with a given property
is the one that has that property and whose index is smaller than the index
of any other s[i] that also has the property. This construction ensures that
a and b are subsequences of s (a point not actually mentioned in the proof
below) for reasons like the following: Suppose a[n]=s[k1] and a[n+1]=s[k2].
Then k1 and k2 are distinct because s[k2] is an interior point of the
interval (a[n],b[n]) whose left endpoint is s[k1]. Since s[k2] belongs to
(a[n],b[n]) and (a[n],b[n]) is a subset of
(a[n-1],b[n-1]), s[k2] belongs to (a[n-1],b[n-1]). Hence k2<k1 would
contradict the fact that, by definition, k1 is the *smallest* i for which
s[i] belongs to (a[n-1],b[n-1]). The only remaining possibility is that
k1<k2, as required.
I put in some comments below (IN CAPS, for readability), but in all honesty,
I can't say I really follow all of the points you're raising.
Ross <ap...@calpha.com> wrote in message news:3B3F83B2...@calpha.com...
p AND q ARE BOUND VARIABLES
> with the s(0) being in (a(0), b(0)).
>
> Also, it would seem that the range of s is bounded by (a(0), b(0)). That
is to say,
> for a(n) to be less than arbitrary s(n) that a would be decreasing,
similarly b
> increasing, where they are having been described differently.
>
> Thus, I would say that s(n) has a codomain a(0), b(0), for that to be
otherwise.
>
> It appears to me that the point of the theorem is that there are upper and
lower
> bounds b(0) and a(0) over an interval of reals, and that the function a(n)
is
> increasing, b(n) decreasing, where the limit of each function as n->oo is
somewhere
> on (a(0), b(0)).
>
> I have an issue with this part: "a[n]<s[M]<b[n] for all n", where it might
rather be
> "a[n]<=s[M]<=b[n] for all n", where s(M) is the limit of a(n) and b(n).
IF a[n]<=s[M]<=b[n] FOR ALL n, THEN ALSO a[n+1]<=s[M]<=b[n+1] FOR ALL n;
SINCE a[n]<a[n+1] AND b[n+1]<b[n], THE STRICT INEQUALITY FOLLOWS
> I think another point to consider in light of the construction is that s
should be
> defined as increasing or decreasing, where it is not.
>
> Now, as the limit of b(n)-a(n) goes to zero, with a(0)<b(0) and a
increasing and b
> decreasing, giving a(0) < (lim a(n)) = (lim b(n)) < b(0), where the limit
of s(n)
> also happens to be equal to lim a(n) and lim b(n), I don't see that as
disqualifying
> the sequence from also converging to lim s(n), where that was the original
idea.
>
> For example, consider the Cauchy sequence approximating the real value
one-half
> between zero and one as the a sequence, and the sequence of one minus each
of those
> values as the b sequence. Then, interleave them as (a(0), b(0), a(1),
b(1), ...) to
> form the s sequence. It converges to one-half.
>
> >
> > Another thing is about the minimality of M and whether it should be the
> > maximality of M.
M IS A BOUND VARIABLE
> >
> > Basically, the way I see it is that these two sequences each converge to
some
> > value on (a[0], b[0]), where that value is tautologically in (a[0],
b[0]).
YES
Virgil
wrote:
> The rationals are the field of the closure of the finite integers to
> division,
> the reals are a field that is a subset of the closure of the hyperintegers to
> division. The hyperreals is the field that is the closure of the
> hyperintegers
> to division That is to say, defining the operation of division on the
> integers
> induces the rationals, and defining the operation of division on the
> hyperintegers induces the reals and hyperreals.
>
> For example, 1/(non-finite hyperinteger X) evaluates to an infinitesimal
> hyperreal, where 1/(finite hyperinteger X) evlauate to rational 1/X.
More Rossaninity! Or is it Rossinanity?
Virgil
wrote:
> I think that the reals are a subset of the closure of the hyperintegers to
> division.
>
[snip]
>
> I believe that any real can be represented as the ratio of two hyperintegers.
> Also, no.
>
> >
> > >
> > > For example, 1/(non-finite hyperinteger X) evaluates to an infinitesimal
> > > hyperreal, where 1/(finite hyperinteger X) evaluates to rational 1/X.
> > >
> > > Ross
>
> --
> contact Ross
> company Apex Internet Software
> email in...@apexinternetsoftware.com
> website http://www.apexinternetsoftware.com/
As Ross refuses to let anyone else know what he means by "hyperinteger",
I suppose he feels free to attribute to them any properties that he
wishes.
On the other hand, until Ross provides some structure to his
hyperinteger theory, anyone else is quite as free to attribute different
properties to them.
The property that I attribute to Rossian hyperntegers is that using them
one can prove every statement true.
Thus every claim Ross makes about his hyperintegers is simultaneously
true and false. At least until Ross condescends to reveal what he means
by hyperintegers.
Virgil
wrote:
> Robert Glass wrote:
>
> > Elaine Jackson wrote:
> >
> > > Right. I've already gotten flak in another newsgroup for the way I
> > > mangled
> > > the description of (2).
> >
> > Actually, Mr Finlayson did not accuse you of mangling your description. He
> > said
> > that he didn't believe such a proof even existed. Congratulations! You have
> > been called a liar by Ross Finlayson -- that's kind of a badge of honor in
> > this
> > group.
> >
>
> That's not true.
Ah!. You can always tell when you are right when Ross says that.
>
> Bob, in other words, you had presented a reference (link) to what was
> described as
> a similar method as 2, yet, I showed why it was meaningless, so, when I ask
> if it
> had been repaired, that implies that I see it as broken.
Ross's "show" of meaninglessness was a congery of, possibly deliberate,
misreadings. His vision of it as broken is a problem with his vision and
not with the reference. It is the Rossian vision that is in need of
repair, at least in the matter of reading mathematics.
I and others have pointed out to him at least some of his errors of
interpretation, but we have only been acknowledged by his total silence
on the subject. In a way, that show us that we were right.
Virgil
wrote:
Rossian hyperintegers do not exist outsode Ross's Private Little
Universe until they are publicly defined.
And ratios of non-existent objects cannot exactly equal any irrational
number, or rational number for that matter.
Dave L. Renfro
[sci.math Sat, 30 Jun 2001 19:42:27 GMT]
<http://forum.swarthmore.edu/epigone/sci.math/clanfrendmel>
wrote
> I'm curious about other proofs of this fact. The two I know about
> now are (1) the Cantor diagonal proof and (2) a proof in which,
> beginning with an arbitrary sequence of reals, two subsequences
> are defined, one strictly decreasing and one strictly increasing,
> both with the same limit; it is then shown that this limit can't
> have been a term in the original, given sequence. Does anybody
> know any other proofs?
Well, there's Cantor's first proof, obtained in Dec. 1873 (we know
this because Cantor discussed it in some letters to other
mathematicians) and published in 1874. Actually, this is essentially
the same as your proof #2. Here's how it goes --->>
<< By "interval", I mean a connected subset of the reals
containing more than one point. >>
Given a countable set {x1, x2, ...}, we'll find a point in any
interval that doesn't belong to {x1, x2, ...}. Let J1 be a compact
interval in the given interval that doesn't contain x1. Pick a
compact interval J2 in J1 that doesn't contain x2. Continue in this
way. Then every point in the intersection of the Ji's is disjoint
from {x1, x2, ...}. [This intersection is nonempty by a theorem
which says that a decreasing sequence of compact sets has a nonempty
intersection. Actually, we're only applying this in the special case
of compact intervals, and it was in this special case that Cantor
and others knew this theorem. (In this special case, it's intimately
tied up with certain constructions of the real numbers that were
being worked on during this time.)]
If we recast this so that the following trivial property of compact
intervals is used, there is an interesting way to obtain a stronger
result.
LEMMA: Every interval contains two disjoint compact intervals.
<< Yes, yes . . . I know "compact" modifies "intervals" and
"disjoint" modifies the collection of these two intervals. >>
Here's how this lemma is used above. Say, for example, that we have
J1 and we want to show how to get J2. Using the lemma, there exist
two disjoint compact intervals in J1. By the pigeon-hole principle,
at least one of these compact intervals will not contain x2. Call
this interval J2.
Here's how to get a result stronger than the uncountability of
the reals. Note that the uncountability of the reals is equivalent
to the fact that the set of real numbers is not a countable union
of singleton sets.
LEMMA+: Every nonempty perfect set contains 2^(aleph_0) pairwise
disjoint compact perfect subsets.
Proof: We will use the fact that any perfect set can be
mapped continuously onto the unit square [0,1] x [0,1].
[First, map the perfect set continuously onto [0,1].
(For intervals this is trivial; for Cantor sets, use a
monotone increasing function similar to the Cantor
function.) Then follow this with a continuous mapping
of [0,1] onto the unit square (use a Peano curve).] Now
observe that the inverse images of the vertical fibers
{r} x [0,1], as r varies over [0,1], gives continuum many
pairwise disjoint perfect subsets of the given perfect set.
We can use Lemma+ to show that the reals cannot be written as a
countable union of sets each of which has cardinality less than
2^(aleph_0). Recall that this result is usually proved by Konig's
cardinal inequality result involving sums and products of cardinal
numbers.
Suppose E1, E2, ... are sets of real numbers and each has cardinality
less than 2^(aleph_0). We'll find a point in any interval that
doesn't belong to any of the Ei's. Let J be the given interval,
which we may assume is closed without loss of generality. Then by
Lemma+, J contains a collection of continuum many pairwise disjoint
compact subsets. By the pigeon-hole principle, at least one of these
perfect sets, call it P1, contains no points of E1. Now use Lemma+
in the same way to find a perfect subset P2 of P1 that contains no
points of E2. Continue in this way. Then the intersection of the
Pi's is nonempty and every point in this intersection does not
belong to any of the Ei's.
This neat topological proof of the fact that 2^(aleph_0) doesn't
have countable cofinality is due to Paul Mahlo (1913).
Incidentally, an expository paper I recently wrote that discusses
these and other issues is what led to the questions in my post
that began the sci.math thread titled <"How bad can 2^(Aleph_0) be?"
history question> at
<http://forum.swarthmore.edu/epigone/sci.math/stantaxfan>.
As far as I know, no one answered any of the questions I asked in
my initial post in this thread. However, I've since heard from
Robert Solovay. When I get a chance, I'll post some of what he had
to say about my questions.
<< I just moved, and so I'm still getting settled in. My post
yesterday at
<http://forum.swarthmore.edu/epigone/sci.math/pangsandsnax>
(thread titled "fallacy of diagonal argument; 0.9999... < 1")
was actually my first post in over a week. In fact, this marked
the first time in over a week that I've even looked at any
sci.math posts. Gee, I sure got a lot done this past week ... >>
Dave L. Renfro
Steve Leibel
"Elaine Jackson" <elaineja...@home.com> wrote:
> Should I be trying to sort this out? Is this a different proof from the two
> I was talking about? Can anybody give me the gist of the argument below? I
> can't even make out who's talking to whom.
>
That is the purpose of everything Ross posts. You just made his day.
D. Ross
| > Countable sets have Lebesgue measure zero.
|
| To make that work, of course, you have to show that R does
| *not* have measure zero, which is a certain amount of work.
Typically no more work than is involved already in the construction of
Lebesgue measure itself, which automatically gives intervals nonempty
measure. In fact, all one needs is the existence of *any* nontrivial Borel
measure on R.
The interesting thing in the context of this thread is that it changes the
nature of the reductio argument enough to bypass the most common
arguments levied against the diagonal argument by confused
students/hobbiests. In particular, if someone actually *believes* that R is
countable then it gives a *direct* proof that for any positive r, Leb(R)<r.
This might be more pedagogically convincing for some more stubborn minds.
Of course, for those who don't believe in Lebesgue measure, given a
countable subset A of R one can fairly easily construct for any r>0 a
continuous nonnegative function f with compact support such that f>1 on A
and the Riemann integral of f is <r. Now, if R is countable,...
- David R.
Jesse Hughes
> About that, what that means to say is that not(not A) does not imply
> A. In most cases of intuitionistic logic it does, that is also
> called bivalent, Boolean, or true/false logic as opposed to
> multivalent logic, but it doesn't always. For example, in terms of
> cats and dogs, saying that X is not a cat does not imply that it is
> a dog, where it might be something else, just not a cat.
And, yet, if X is not not a cat, then it is a cat. Golly,
intuitionism is hard.
Ross is as much a master of logic as he is of set theory. In fact,
his mastery of set theory is somehow related to his mastery of logic,
I dare say.
--
Jesse Hughes
"It's a dangerous place. It's a sacred place. They're going to live
as the Aborigines did." -Survivor producer Mark Burnett on the Outback
Dave Seaman
>| > Countable sets have Lebesgue measure zero.
>|
>| To make that work, of course, you have to show that R does
>| *not* have measure zero, which is a certain amount of work.
>Typically no more work than is involved already in the construction of
>Lebesgue measure itself, which automatically gives intervals nonempty
>measure. In fact, all one needs is the existence of *any* nontrivial Borel
>measure on R.
It's not automatic. In order to show that an interval has positive
Lebesgue measure, you need to show to things:
(1) an interval has positive outer measure, and
(2) an interval is a measurable set.
Since (2) is easy, let's focus on (1). By definition, the outer measure
of an interval is the inf over all countable open covers of that
interval, of the sums of the lengths in the intervals making up an open
cover.
How do you prove that the inf is the length of the interval?
Hint: It helps to make use of the Heine-Borel theorem, which says that
if A is a closed, bounded subset of R^n, then every open cover of A has a
finite subcover. Now apply the definition of outer measure, taking note
of the fact that you only need to consider the length sums over all
finite covers of the interval when taking the infimum. Therefore you can
use induction on the number of open sets making up the cover.
So, it's almost automatic if you take the Heine-Borel theorem as a given.
As Mike Oliver said, it's a certain amount of work.
--
Dave Seaman dse...@purdue.edu
Amnesty International calls for new trial for Mumia Abu-Jamal
<http://www.amnestyusa.org/abolish/reports/mumia/>
David C. Ullrich
<elaineja...@home.com> wrote:
[...]
>
>I put in some comments below (IN CAPS, for readability), but in all honesty,
>I can't say I really follow all of the points you[Ross]'re raising.
If you follow _any_ of them you're doing much better than the rest of
us.
David C. Ullrich
*********************
"Sometimes you can have access violations all the
time and the program still works." (Michael Caracena,
comp.lang.pascal.delphi.misc 5/1/01)
Randy Poe
>
> Virgil wrote:
>
> > In article <3B3E67C0...@calpha.com>, Ross <ap...@calpha.com>
> > wrote:
> >
> > > Elaine Jackson wrote:
> > >
> > > >(2) a proof in which, beginning with an
> > > > arbitrary sequence of reals, two subsequences are defined, one strictly
> > > > decreasing and one strictly increasing, both with the same limit; it is
> > > > then
> > > > shown that this limit can't have been a term in the original, given
> > > > sequence. Does anybody know any other proofs?
> >
> > numbers, it is easy to construct subsequences having the properties (2).
> >
>
> It appears to me that if an ordered sequence of reals is split into two
> subsequences where one is strictly increasing and the other strictly decreasing,
> then where the original sequence has bounds a and b that the limit of any of the
> subsequences could only be a, b, or some value between a and b.
As usual, step one of your argument is to make a completely
different statement than what was actually stated. Let me
show you where you left the thread, beginning with the word
"if":
> It appears to me that if an ordered sequence of reals
Oops. It said "arbitrary sequence of reals", not "ordered sequence
of reals".
>is split into two subsequences
Oops. The sequence contains two subsequences, it is not split
into two subsequences for any possible interpretation of the
word "split". "Split" implies that every element of the original
sequence is in one subsequence or the other. All we are saying
is that we are taking two subsets of the original. Some elements
may very well be left out of both.
> original sequence has bounds a and b
Oops. The original arbitrary sequence does not have to have any
bounds, upper or lower. In fact, if it is assumed to contain
all the real numbers, it has neither.
So other than no part of your premise having anything to do
with the discussion, your comments are entirely relevant.
- Randy
Jonathan Hoyle
Yes, that explains a lot, and I agree with that interpretation. In
any inconsistent theory (using the same rules of inference), all
statements (including their negations) are theorems, and thus Ross's
statements are both true and false in his theory. An inconsistent
system yielding a theory where truth and falsity have no meaning seems
to map well to Ross's style, since he has explicitly done this
himself. He has stated both that there are and there are not the same
of rationals are there are integers, usually within the same post. I
used to try to argue with him, but now I see that I merely lacked your
insight to his theory. Thanks, Virgil. :-)
Ross
Virgil wrote:
No, Hancher, if you had bothered to read it, you might have seen the truth that it
was syntactically incorrect, and thus invalid.
>
> I and others have pointed out to him at least some of his errors of
> interpretation, but we have only been acknowledged by his total silence
> on the subject. In a way, that show us that we were right.
Don't accept that, because it's not true.
First, Renfro has presented a better proof of this conversational piece than had
been presented. Where I argued against those non-proofs, I was correct. The
"diagonal argument" is no proof, it's only an argument. Also, I'm still thinking
about it. One problem with it is its presumption that the reals are not countable,
for if they were already countable (enumerable), then a list of them could contain
each of them, thus that the Lebesgue measure of such a set would be non-zero.
Thus, Renfro said "countable sets are not uncountable."
Second, there are many "uncountable sets" or "sets with non-negative Lebesgue
measure" thus that "compact set" or "decreasing interval pigeon-holing" shows the
existence of the intersection of some intervals on the reals as not containing any
element of a set with non-zero Lebegue measure that is not the complete set of
reals. That shows them not not uncountable, where that does not mean countable.
Third, I have said very many very true and valid things. Some of these things were
independently discovered by myself. Also, I am very good about proper references
and referencing.
Finally, if you read these recent posts, it is easily seen that I write clearly and
with logical comprehension.
So, in summary, I have made thousands of posts to sci.math, and they contain some
material that is strictly original and valid mathematics.
Bill Taylor
|> > Thus every claim Ross makes about his hyperintegers is simultaneously
|> > true and false.
|>
|> In any inconsistent theory (using the same rules of inference),
|> all statements (including their negations) are theorems, and thus
|> Ross's statements are both true and false in his theory.
So it would seem, in other words, that Ross (not to mention JH and many
others) are using "GENERALIZED MATHEMATICS".
I presume you've heard of this? Though as hidebound reactionary academeaucrats
of very limited imagination, it's highly posssible you haven't.
It was doing the rounds informally a while ago. I don't recall the full
details, (anyone?), but it had a most memorable and impressive start line...
=======================================================================
In orthodox mathematics, we prove true results, by using valid methods.
But in Generalized Mathematics, both of these restrictions are dropped!
=======================================================================
Simple, really!
--------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
--------------------------------------------------------------------------------
Math and science require reasoning, the fuzzy subjects just require scholarship.
-- R. A. Heinlien.
--------------------------------------------------------------------------------
Virgil
wrote:
Whatever errors of syntax may have occurred in the presentation of
Cantor's original proof of the uncountbility of the reals, any person
who claims the mathematical talents that Ross claims should have been
able to repair them.
But Ross made such elementary errors as misreading the direction of an
inequality. Ross then argues the invalidity of the proof based on his
own errors of misreading.
>
> >
> > I and others have pointed out to him at least some of his errors of
> > interpretation, but we have only been acknowledged by his total silence
> > on the subject. In a way, that show us that we were right.
>
> Don't accept that, because it's not true.
Oh?
>
> First, Renfro has presented a better proof of this conversational piece than
> had
> been presented. Where I argued against those non-proofs, I was correct. The
> "diagonal argument" is no proof, it's only an argument.
Does Ross now suggest that a valid argument is no proof? Then exactly
what does he recognize as a proof?
> Also, I'm still
> thinking
> about it.
That hasn't worked in the past.
Why should it work now, or in the future?
> One problem with it is its presumption that the reals are not
> countable,
Neither the Cantor diagonal proof nor the earlier Cantor nested closed
intervals proof needs to presume that the reals are uncountable. It
fact, they work better when no assumption is made except that one starts
with a sequence of reals f:N->R. Clearly the image of f is countable,
but there is no need for an initial assumption that f is onto R.
> for if they were already countable (enumerable), then a list of them could
> contain
> each of them, thus that the Lebesgue measure of such a set would be non-zero.
> Thus, Renfro said "countable sets are not uncountable."
Does Ross actually know of a way to define a Lebegue_like measure for
which countable sets have positive measure? This is a notable discovery
, if true.
>
> Second, there are many "uncountable sets" or "sets with non-negative Lebesgue
> measure" thus that "compact set" or "decreasing interval pigeon-holing" shows
> the
> existence of the intersection of some intervals on the reals as not
> containing any
> element of a set with non-zero Lebegue measure that is not the complete set
> of
> reals. That shows them not not uncountable, where that does not mean
> countable.
Don't step in the oongah.
>
> Third, I have said very many very true and valid things. Some of these
> things were
> independently discovered by myself. Also, I am very good about proper
> references
> and referencing.
>
> Finally, if you read these recent posts, it is easily seen that I write
> clearly and
> with logical comprehension.
The Rossaninity view of "clearly and with logical comprehension" seems
to differ strongly from everyone else's view.
>
> So, in summary, I have made thousands of posts to sci.math, and they contain
> some
> material that is strictly original and valid mathematics.
While I do not recall any bits of Ross's material which were both
original and valid mathematics, but among so many thousands of posts, I
may have missed one or two.
Robert Glass
Ross wrote:
>
> Second, there are many "uncountable sets" or "sets with non-negative Lebesgue
> measure" thus that "compact set" or "decreasing interval pigeon-holing" shows the
> existence of the intersection of some intervals on the reals as not containing any
> element of a set with non-zero Lebegue measure that is not the complete set of
> reals. That shows them not not uncountable, where that does not mean countable.
Excuse me, I'm a bit lost.
Is not not uncountable a dog or a cat?
Ross
Bill Taylor wrote:
> jonh...@mac.com (Jonathan Hoyle) writes:
>
> |> > Thus every claim Ross makes about his hyperintegers is simultaneously
> |> > true and false.
> |>
> |> Yes, that explains a lot, and I agree with that interpretation.
> |> In any inconsistent theory (using the same rules of inference),
> |> all statements (including their negations) are theorems, and thus
> |> Ross's statements are both true and false in his theory.
>
> So it would seem, in other words, that Ross (not to mention JH and many
> others) are using "GENERALIZED MATHEMATICS".
>
It does not seem that way to me, and also, I am one of those people who is largely
ready to use valid mathematics.
So, you're mistaken. I speak only for myself.
>
> I presume you've heard of this? Though as hidebound reactionary academeaucrats
> of very limited imagination, it's highly posssible you haven't.
>
You want the reactionaries down the hall. I use valid mathematics, and am prepared
to do so in creative ways.
>
> It was doing the rounds informally a while ago. I don't recall the full
> details, (anyone?), but it had a most memorable and impressive start line...
>
> =======================================================================
> In orthodox mathematics, we prove true results, by using valid methods.
> But in Generalized Mathematics, both of these restrictions are dropped!
> =======================================================================
>
> Simple, really!
>
> --------------------------------------------------------------------------------
> Bill Taylor W.Ta...@math.canterbury.ac.nz
> --------------------------------------------------------------------------------
> Math and science require reasoning, the fuzzy subjects just require scholarship.
>
> -- R. A. Heinlien.
> --------------------------------------------------------------------------------
Heinlein is spelled Heinlein.
Ross
Virgil wrote:
No, that's false, Hancher.
>
> >
> > >
> > > I and others have pointed out to him at least some of his errors of
> > > interpretation, but we have only been acknowledged by his total silence
> > > on the subject. In a way, that show us that we were right.
> >
> > Don't accept that, because it's not true.
>
> Oh?
>
Yes, quite. Don't be coy, ugly.
>
> >
> > First, Renfro has presented a better proof of this conversational piece than
> > had
> > been presented. Where I argued against those non-proofs, I was correct. The
> > "diagonal argument" is no proof, it's only an argument.
>
> Does Ross now suggest that a valid argument is no proof? Then exactly
> what does he recognize as a proof?
>
Hancher, that argument is only one that fails. It is not a proof. A proof is a
specific set of consistent sentences about a theorem that show indubitably that the
theorem is true or false.
The diagonal argument is a shell game, not a proof.
>
> > Also, I'm still
> > thinking
> > about it.
>
> That hasn't worked in the past.
> Why should it work now, or in the future?
>
Again, Hancher, you are wrong.
For example, if a(n) is increasing, b(n) decreasing, and lim a(n)=lim(b(n), then
there is no interval (a(n), b(n)), only the point a(n)=b(n).
>
> > One problem with it is its presumption that the reals are not
> > countable,
>
> Neither the Cantor diagonal proof nor the earlier Cantor nested closed
> intervals proof needs to presume that the reals are uncountable. It
> fact, they work better when no assumption is made except that one starts
> with a sequence of reals f:N->R. Clearly the image of f is countable,
> but there is no need for an initial assumption that f is onto R.
>
>
Too bad for you.
>
> > for if they were already countable (enumerable), then a list of them could
> > contain
> > each of them, thus that the Lebesgue measure of such a set would be non-zero.
> > Thus, Renfro said "countable sets are not uncountable."
>
> Does Ross actually know of a way to define a Lebegue_like measure for
> which countable sets have positive measure? This is a notable discovery
> , if true.
>
I did not say anything like that. They might have infinitesimal Lebesgue like
measure, where the only definition of countable is having Lebesgue measure zero,
where infinitesimals are considered zero.
>
> >
> > Second, there are many "uncountable sets" or "sets with non-negative Lebesgue
> > measure" thus that "compact set" or "decreasing interval pigeon-holing" shows
> > the
> > existence of the intersection of some intervals on the reals as not
> > containing any
> > element of a set with non-zero Lebegue measure that is not the complete set
> > of
> > reals. That shows them not not uncountable, where that does not mean
> > countable.
>
> Don't step in the oongah.
>
Virgil, you'll have to explain that one yourself.
>
> >
> > Third, I have said very many very true and valid things. Some of these
> > things were
> > independently discovered by myself. Also, I am very good about proper
> > references
> > and referencing.
> >
> > Finally, if you read these recent posts, it is easily seen that I write
> > clearly and
> > with logical comprehension.
>
> The Rossaninity view of "clearly and with logical comprehension" seems
> to differ strongly from everyone else's view.
>
That is false. Only few points are disputed. For you to make blanket statements
to that effect is irresponsible, and detrimental to your believability.
> >
> > So, in summary, I have made thousands of posts to sci.math, and they contain
> > some
> > material that is strictly original and valid mathematics.
>
> While I do not recall any bits of Ross's material which were both
> original and valid mathematics, but among so many thousands of posts, I
> may have missed one or two.
>
Hancher, you don't get a say.
Ross
Robert Glass wrote:
Neither.
One thing I was thinking was that these sets with non-negative Lebesgue measure are
having different scalar values of Lbesgue measure. For example, a constant unit
function from zero to one has a Lebesgue measure of about one half that of the measure
of that function defined on zero to two.
Virgil
wrote:
> Virgil wrote:
>
> > In article <3B40DE92...@calpha.com>, Ross <ap...@calpha.com>
> > wrote:
> >
[snipped multilevel quotes]
> > >
> > > No, Hancher, if you had bothered to read it, you might have seen the
> > > truth
> > > that it
> > > was syntactically incorrect, and thus invalid.
> >
> > Whatever errors of syntax may have occurred in the presentation of
> > Cantor's original proof of the uncountbility of the reals, any person
> > who claims the mathematical talents that Ross claims should have been
> > able to repair them.
> >
> > But Ross made such elementary errors as misreading the direction of an
> > inequality. Ross then argues the invalidity of the proof based on his
> > own errors of misreading.
> >
>
> No, that's false, Hancher.
Ah, the confirmation of my correctness! Thanks, Ross.
[snip]
> > > The
> > > "diagonal argument" is no proof, it's only an argument.
> >
> > Does Ross now suggest that a valid argument is no proof? Then exactly
> > what does he recognize as a proof?
> >
>
> Hancher, that argument is only one that fails. It is not a proof. A proof
> is a
> specific set of consistent sentences about a theorem that show indubitably
> that the
> theorem is true or false.
I do not need any proof that a conjecture is indubitably true or false,
I need proof to determine which.
Ross has this way of not answering a question directly, but sort of
sneaking around it. My last question above is best answered either with
a "yes" or a "no". Ross answers by suggesting that the proof of a
"theorem" may in fact be something that disproves it.
Does Ross mean to suggest that his above definition of a proof excludes
valid arguments? Possibly.
>
> The diagonal argument is a shell game, not a proof.
It is a game at which Ross can never win, since it is so constructed
that Cantor never loses.
>
>
> >
> > > Also, I'm still
> > > thinking
> > > about it.
> >
> > That hasn't worked in the past.
> > Why should it work now, or in the future?
> >
>
> Again, Hancher, you are wrong.
Again, you confirm my rightness.
>
> For example, if a(n) is increasing, b(n) decreasing, and lim a(n)=lim(b(n),
> then
> there is no interval (a(n), b(n)), only the point a(n)=b(n).
In the above, if a(n) or b(n) is _strictly_ monotone, you never have
a(n) = b(n), and you need not have it when neither is strictly monotone.
>
> >
> > > One problem with it is its presumption that the reals are not
> > > countable,
> >
> > Neither the Cantor diagonal proof nor the earlier Cantor nested closed
> > intervals proof needs to presume that the reals are uncountable. It
> > fact, they work better when no assumption is made except that one starts
> > with a sequence of reals f:N->R. Clearly the image of f is countable,
> > but there is no need for an initial assumption that f is onto R.
> >
> >
>
> Too bad for you.
It works for me, but it blows your "disproof" all to blazes.
>
> >
> > > for if they were already countable (enumerable), then a list of them
> > > could
> > > contain
> > > each of them, thus that the Lebesgue measure of such a set would be
> > > non-zero.
> > > Thus, Renfro said "countable sets are not uncountable."
> >
> > Does Ross actually know of a way to define a Lebegue_like measure for
> > which countable sets have positive measure? This is a notable discovery
> > , if true.
> >
>
> I did not say anything like that.
If Ross believes this, Ross cannot even understand his own statements
correctly.
> They might have infinitesimal Lebesgue
> like
> measure, where the only definition of countable is having Lebesgue measure
> zero,
> where infinitesimals are considered zero.
>
> >
> > >
> > > Second, there are many "uncountable sets" or "sets with non-negative
> > > Lebesgue
> > > measure" thus that "compact set" or "decreasing interval pigeon-holing"
> > > shows
> > > the
> > > existence of the intersection of some intervals on the reals as not
> > > containing any
> > > element of a set with non-zero Lebegue measure that is not the complete
> > > set
> > > of
> > > reals. That shows them not not uncountable, where that does not mean
> > > countable.
> >
> > Don't step in the oongah.
> >
>
> Virgil, you'll have to explain that one yourself.
There should be enough who understand me to suit me.
>
> >
> > >
> > > Third, I have said very many very true and valid things. Some of these
> > > things were
> > > independently discovered by myself. Also, I am very good about proper
> > > references
> > > and referencing.
> > >
> > > Finally, if you read these recent posts, it is easily seen that I write
> > > clearly and
> > > with logical comprehension.
> >
> > The Rossaninity view of "clearly and with logical comprehension" seems
> > to differ strongly from everyone else's view.
> >
>
> That is false. Only few points are disputed. For you to make blanket
> statements
> to that effect is irresponsible, and detrimental to your believability.
In the past fortnight or so, there has been posted a groundswell of
support for the opinion that Rossian logic and comprehension differ from
that of the majority. One expects Ross to dispute the majority view.
>
> > >
> > > So, in summary, I have made thousands of posts to sci.math, and they
> > > contain
> > > some
> > > material that is strictly original and valid mathematics.
> >
> > While I do not recall any bits of Ross's material which were both
> > original and valid mathematics, but among so many thousands of posts, I
> > may have missed one or two.
> >
>
> Hancher, you don't get a say.
I seem to be getting a say despite Ross's objections.
Virgil
wrote:
> Bill Taylor wrote:
>
> > jonh...@mac.com (Jonathan Hoyle) writes:
> >
> > |> > Thus every claim Ross makes about his hyperintegers is simultaneously
> > |> > true and false.
> > |>
> > |> Yes, that explains a lot, and I agree with that interpretation.
> > |> In any inconsistent theory (using the same rules of inference),
> > |> all statements (including their negations) are theorems, and thus
> > |> Ross's statements are both true and false in his theory.
> >
> > So it would seem, in other words, that Ross (not to mention JH and many
> > others) are using "GENERALIZED MATHEMATICS".
> >
>
> It does not seem that way to me, and also, I am one of those people who is
> largely
> ready to use valid mathematics.
You may think yourself ready, but our experience with your mathematical
postings is that you are not able.
>
> So, you're mistaken. I speak only for myself.
For which much thanks.
>
> >
> > I presume you've heard of this? Though as hidebound reactionary
> > academeaucrats
> > of very limited imagination, it's highly posssible you haven't.
> >
>
> You want the reactionaries down the hall. I use valid mathematics, and am
> prepared
> to do so in creative ways.
You may use valid mathematics elsewhere, but what you have displayed
here has not been valid mathematics.
>
> >
> > It was doing the rounds informally a while ago. I don't recall the full
> > details, (anyone?), but it had a most memorable and impressive start
> > line...
> >
> > =======================================================================
> > In orthodox mathematics, we prove true results, by using valid methods.
> > But in Generalized Mathematics, both of these restrictions are dropped!
> > =======================================================================
> >
> > Simple, really!
> >
> > ----------------------------------------------------------------------------
> > ----
> > Bill Taylor W.Ta...@math.canterbury.ac.nz
> > ----------------------------------------------------------------------------
> > ----
> > Math and science require reasoning, the fuzzy subjects just require
> > scholarship.
> >
> > -- R. A.
> > Heinlien.
> > ----------------------------------------------------------------------------
> > ----
>
> Heinlein is spelled Heinlein.
Something correct! Congratulations.
Ross
> there is no interval (a(n), b(n)), only the point a(n)=b(n).
Please disregard that part.
Ross Presser
post under Ross Presser, and I have very little training in maths. I
often make erroneous statements, but I seldom defend them vigorously.
:)
--
Ross Presser * ross_p...@imtek.com
"A free-range shoggoth is a happy shoggoth, and a happy shoggoth is
generally less inclined to eat all of you at once." - Tim Morgan
Ross
Virgil wrote:
> In article <3B420940...@calpha.com>, Ross <ap...@calpha.com>
> wrote:
>
> > Virgil wrote:
> >
> > > In article <3B40DE92...@calpha.com>, Ross <ap...@calpha.com>
> > > wrote:
> > >
> [snipped multilevel quotes]
>
> > > >
> > > > No, Hancher, if you had bothered to read it, you might have seen the
> > > > truth
> > > > that it
> > > > was syntactically incorrect, and thus invalid.
> > >
> > > Whatever errors of syntax may have occurred in the presentation of
> > > Cantor's original proof of the uncountbility of the reals, any person
> > > who claims the mathematical talents that Ross claims should have been
> > > able to repair them.
> > >
> > > But Ross made such elementary errors as misreading the direction of an
> > > inequality. Ross then argues the invalidity of the proof based on his
> > > own errors of misreading.
> > >
> >
> > No, that's false, Hancher.
>
> Ah, the confirmation of my correctness! Thanks, Ross.
>
Virgil, no, it's actual confirmation of your incorrectness.
For you to say something so blatantly untrue as that whgen I say something that
the opposite is true, that is a sign of stupidness, as it is a stupid thing to
do.
>
> [snip]
>
> > > > The
> > > > "diagonal argument" is no proof, it's only an argument.
> > >
> > > Does Ross now suggest that a valid argument is no proof? Then exactly
> > > what does he recognize as a proof?
> > >
> >
> > Hancher, that argument is only one that fails. It is not a proof. A proof
> > is a
> > specific set of consistent sentences about a theorem that show indubitably
> > that the
> > theorem is true or false.
>
> I do not need any proof that a conjecture is indubitably true or false,
> I need proof to determine which.
>
The proof is showing is showing that a theorem is true or that it is false, the
"proof by counterexample."
>
> Ross has this way of not answering a question directly, but sort of
> sneaking around it. My last question above is best answered either with
> a "yes" or a "no". Ross answers by suggesting that the proof of a
> "theorem" may in fact be something that disproves it.
>
First, I say the diagonal argument is not a valid argument. Second, I described
what an actual proof would be, where a theorem is proven or disproven. Third, I
am quite direct.
>
> Does Ross mean to suggest that his above definition of a proof excludes
> valid arguments? Possibly.
>
No.
>
>
> >
> > The diagonal argument is a shell game, not a proof.
>
> It is a game at which Ross can never win, since it is so constructed
> that Cantor never loses.
>
That is actually not the case. For example, it is quite simple to construe it
thus that that argument fails. For example, when there is supposed to be some
number of the list not in the list, it is actually in the list.
> >
> >
> > >
> > > > Also, I'm still
> > > > thinking
> > > > about it.
> > >
> > > That hasn't worked in the past.
> > > Why should it work now, or in the future?
> > >
> >
> > Again, Hancher, you are wrong.
>
> Again, you confirm my rightness.
>
No, not any time recently.
> >
> > For example, if a(n) is increasing, b(n) decreasing, and lim a(n)=lim(b(n),
> > then
> > there is no interval (a(n), b(n)), only the point a(n)=b(n).
>
> In the above, if a(n) or b(n) is _strictly_ monotone, you never have
> a(n) = b(n), and you need not have it when neither is strictly monotone.
>
Well, a(n)= .5 for any n as does b(n), then you're again wrong, which actually
does mean that you are actually wrong, Virgil.
> >
> > >
> > > > One problem with it is its presumption that the reals are not
> > > > countable,
> > >
> > > Neither the Cantor diagonal proof nor the earlier Cantor nested closed
> > > intervals proof needs to presume that the reals are uncountable. It
> > > fact, they work better when no assumption is made except that one starts
> > > with a sequence of reals f:N->R. Clearly the image of f is countable,
> > > but there is no need for an initial assumption that f is onto R.
> > >
> > >
> >
> > Too bad for you.
>
> It works for me, but it blows your "disproof" all to blazes.
>
Which one?
That statement of yours implies to me that you have ideas of arguments that I
have raised, where my arguments are effective to the point of being acceptable,
and in this case, implicit.
So, which one?
> >
> > >
> > > > for if they were already countable (enumerable), then a list of them
> > > > could
> > > > contain
> > > > each of them, thus that the Lebesgue measure of such a set would be
> > > > non-zero.
> > > > Thus, Renfro said "countable sets are not uncountable."
> > >
> > > Does Ross actually know of a way to define a Lebegue_like measure for
> > > which countable sets have positive measure? This is a notable discovery
> > > , if true.
> > >
> >
> > I did not say anything like that.
>
> If Ross believes this, Ross cannot even understand his own statements
> correctly.
>
Hancher, I did not say anything about any "countable" set having a non-zero
Lebesgue or "Lebesgue-like" measure, because in this context countable is having
zero Lebesgue measure.
I am definitely not affiliated with Hancher, I want nothing to do with Hancher,
because Hancher has made untrue statements and lies about me.
>
> > They might have infinitesimal Lebesgue
> > like
> > measure, where the only definition of countable is having Lebesgue measure
> > zero,
> > where infinitesimals are considered zero.
> >
> > >
> > > >
> > > > Second, there are many "uncountable sets" or "sets with non-negative
> > > > Lebesgue
> > > > measure" thus that "compact set" or "decreasing interval pigeon-holing"
> > > > shows
> > > > the
> > > > existence of the intersection of some intervals on the reals as not
> > > > containing any
> > > > element of a set with non-zero Lebegue measure that is not the complete
> > > > set
> > > > of
> > > > reals. That shows them not not uncountable, where that does not mean
> > > > countable.
> > >
> > > Don't step in the oongah.
> > >
> >
> > Virgil, you'll have to explain that one yourself.
>
> There should be enough who understand me to suit me.
>
Does it not stink?
> >
> > >
> > > >
> > > > Third, I have said very many very true and valid things. Some of these
> > > > things were
> > > > independently discovered by myself. Also, I am very good about proper
> > > > references
> > > > and referencing.
> > > >
> > > > Finally, if you read these recent posts, it is easily seen that I write
> > > > clearly and
> > > > with logical comprehension.
> > >
> > > The Rossaninity view of "clearly and with logical comprehension" seems
> > > to differ strongly from everyone else's view.
> > >
> >
> > That is false. Only few points are disputed. For you to make blanket
> > statements
> > to that effect is irresponsible, and detrimental to your believability.
>
> In the past fortnight or so, there has been posted a groundswell of
> support for the opinion that Rossian logic and comprehension differ from
> that of the majority. One expects Ross to dispute the majority view.
>
Actually, I am not caring much about that.
For example, I have discussed several simple proofs that there are more rationals
than integers, and I don't care if others call it an argument, conjecture, or
anything else, because as time passes, the statements are already there. I have
confidence that a purely objective proof checker will read them easily.
> > > >
> > > > So, in summary, I have made thousands of posts to sci.math, and they
> > > > contain
> > > > some
> > > > material that is strictly original and valid mathematics.
> > >
> > > While I do not recall any bits of Ross's material which were both
> > > original and valid mathematics, but among so many thousands of posts, I
> > > may have missed one or two.
> > >
> >
> > Hancher, you don't get a say.
>
> I seem to be getting a say despite Ross's objections.
> >
> >
My point is that my original statements are already there, and nothing that you
can do can change that. Also, there are many witnesses.
Ross
Ross
quite clear that I have a quite clear enough comprehension of what I say that when I
say it that it is correct, also, when I make a mistake, I readily and ably
understand why that is so, and correct it, mostly on my own.
Is someone says I am wrong and I disagree, I have clear reasoning why. Also, I
defend my reasoning, and I defend it effectively.
Another thing that is quite clear is that I have made some statements that
contradict some practices.
A thing that is quite true is that the large part of things that I say that I do not
correct myself, that they are true.
Virgil wrote:
> In article <3B420710...@calpha.com>, Ross <ap...@calpha.com>
> wrote:
>
> > Bill Taylor wrote:
> >
> > > jonh...@mac.com (Jonathan Hoyle) writes:
> > >
> > > |> > Thus every claim Ross makes about his hyperintegers is simultaneously
> > > |> > true and false.
> > > |>
No, they are true, only.
>
> > > |> Yes, that explains a lot, and I agree with that interpretation.
> > > |> In any inconsistent theory (using the same rules of inference),
> > > |> all statements (including their negations) are theorems, and thus
> > > |> Ross's statements are both true and false in his theory.
> > >
> > > So it would seem, in other words, that Ross (not to mention JH and many
> > > others) are using "GENERALIZED MATHEMATICS".
> > >
> >
> > It does not seem that way to me, and also, I am one of those people who is
> > largely
> > ready to use valid mathematics.
>
> You may think yourself ready, but our experience with your mathematical
> postings is that you are not able.
I disagree, in terms of ability. The displayed logical comprehension is very real
and effective.
I also think that you should not attempt to speak for the community. They are each
able to speak for themselves.
I am able to speak for myself, and I say sometimes controversial things, because
there is no point in saying accepted things, as they are accepted. The point of
saying controversial things like "there are more rationals than integers" is that I
consider the things I say true.
>
> >
> > So, you're mistaken. I speak only for myself.
>
> For which much thanks.
Speak for yourself.
>
> >
> > >
> > > I presume you've heard of this? Though as hidebound reactionary
> > > academeaucrats
> > > of very limited imagination, it's highly posssible you haven't.
> > >
> >
> > You want the reactionaries down the hall. I use valid mathematics, and am
> > prepared
> > to do so in creative ways.
>
> You may use valid mathematics elsewhere, but what you have displayed
> here has not been valid mathematics.
That is not true. Hancher, that is statement by you is not true. I have shown
plentiful use of valid mathematics.
>
> >
> > >
> > > It was doing the rounds informally a while ago. I don't recall the full
> > > details, (anyone?), but it had a most memorable and impressive start
> > > line...
> > >
> > > =======================================================================
> > > In orthodox mathematics, we prove true results, by using valid methods.
> > > But in Generalized Mathematics, both of these restrictions are dropped!
> > > =======================================================================
> > >
> > > Simple, really!
> > >
> > > ----------------------------------------------------------------------------
> > > ----
> > > Bill Taylor W.Ta...@math.canterbury.ac.nz
> > > ----------------------------------------------------------------------------
> > > ----
> > > Math and science require reasoning, the fuzzy subjects just require
> > > scholarship.
> > >
> > > -- R. A.
> > > Heinlien.
> > > ----------------------------------------------------------------------------
> > > ----
> >
> > Heinlein is spelled Heinlein.
>
> Something correct! Congratulations.
There are enough of my things that are right that I am right.
The spelling of Heinlein's thing, it is a relatively trivial thing, as it is
accepted how he actually spells his name.
Then, there are other things that we have argued here that are non-trivial with
regards to the practice of valid mathematics. For example, I make "controversial"
statements about the treatments of infinite sets, but not about calculus or
trigonometry, because I accept, for example, that sin^2(x)+cos^2(x)=1, and that
d2x/dx = 2. Those are accepted, valid mathematical concepts, where when there are
others that are argued against by me and many others, then that is a reason to
consider why that is the case. So, when I weigh in about these controversial
treatments of the infinite sets, it is as others do, and some reason behind my
actions is that I see some valid truths that contradict the entrenched philosophy.
So, when I talk about mathematical things that contradict mathematical practice, you
can be quite sure that, rationally, I am only going to do so as I have firm
mathematical ground upon which to stand. You'll have to find another to argue
against the real, empirical results of geometry or calculus, but when it comes to
actually arguing for things that I see as true, I will and do.
Elaine Jackson
Robert Glass <bo...@my-deja.com> wrote in message
news:3B3FCE55...@my-deja.com...
>
>
> Elaine Jackson wrote:
>
> > Should I be trying to sort this out? Is this a different proof from the
two
> > I was talking about? Can anybody give me the gist of the argument below?
I
> > can't even make out who's talking to whom.
>
> You have been trolled.
>
> Ross Finlayson has been posting his nonsense to this group for at least 5
years
> now.
> He frequently states that his purpose is to get the longest threads.
>
> In answer to your question, the proof I was talking about was the one you
> labeled 2.
> It was historically the first proof of the uncountability of the real
numbers,
> discovered by
> Cantor in 1874. The "diagonalization proof" dates to 1891.
>
Ross
Elaine Jackson wrote:
> Good to know, but what is "trolled"?
>
I am not a troll, I am sincere.
>
> Robert Glass <bo...@my-deja.com> wrote in message
> news:3B3FCE55...@my-deja.com...
> >
> >
> > Elaine Jackson wrote:
> >
> > > Should I be trying to sort this out? Is this a different proof from the
> two
> > > I was talking about? Can anybody give me the gist of the argument below?
I discussed that given your wording that there were sequences (of intervals)
that converged to a point, not an interval.
>
> I
> > > can't even make out who's talking to whom.
> >
> > You have been trolled.
> >
> > Ross Finlayson has been posting his nonsense to this group for at least 5
> years
> > now.
That is false, for several reasons.
>
> > He frequently states that his purpose is to get the longest threads.
> >
That is also false. That has never been my stated purpose, for what I might
have accomplished it.
Virgil
wrote:
> Virgil wrote:
>
[snipped multilevel quotes]
> > Ah, the confirmation of my correctness! Thanks, Ross.
> >
>
> Virgil, no, it's actual confirmation of your incorrectness.
In your universe, maybe, but not in mine, which I seem to be sharing
with most posters.
>
> For you to say something so blatantly untrue as that whgen I say something
> that
> the opposite is true, that is a sign of stupidness, as it is a stupid thing
> to
> do.
It would be stupid of me to expect you to understand.
[snip]
> First, I say the diagonal argument is not a valid argument. Second, I
> described
> what an actual proof would be, where a theorem is proven or disproven.
> Third, I
> am quite direct.
You have certainly claimed that the diagonal argument is not valid, but
your claim lacks the "actual proof" you say you need, at least in the
judgement of all those responding to your claim.
What is the point of reiterating a claim which everyone who responds
rejects?
>
> >
> > Does Ross mean to suggest that his above definition of a proof excludes
> > valid arguments? Possibly.
> >
>
> No.
Finally a direct answer. Now. Ross, was that so painful?
>
> >
> >
> > >
> > > The diagonal argument is a shell game, not a proof.
> >
> > It is a game at which Ross can never win, since it is so constructed
> > that Cantor never loses.
> >
> That is actually not the case. For example, it is quite simple to construe
> it
> thus that that argument fails. For example, when there is supposed to be
> some
> number of the list not in the list, it is actually in the list.
There does not have to be any number of the list which is also not in
the list. To say that there must be such a number is to create and
demolish a straw man, and then claim to have domolished the real man.
It is as reasonable as saying that no birds fly because ostrichs don't
fly.
>
> > >
> > > For example, if a(n) is increasing, b(n) decreasing, and lim
> > > a(n)=lim(b(n),
> > > then
> > > there is no interval (a(n), b(n)), only the point a(n)=b(n).
> >
> > In the above, if a(n) or b(n) is _strictly_ monotone, you never have
> > a(n) = b(n), and you need not have it when neither is strictly monotone.
> >
>
> Well, a(n)= .5 for any n as does b(n), then you're again wrong, which
> actually
> does mean that you are actually wrong, Virgil.
No, it means that you are wrong again.
What it means is that Ross, typically, did not understand the
consquences of what I said.
To wit:
A _strictly_ monotone sequence, {a(n)}, is an injection, so a(n) = a(m)
requires n = m.
Also if the strictly decreasing sequence,{b(n)}, and the strictly
increasing sequence,{a(n)} have the same limit then a(n) < b(m) for all
n and m.
> >
> > It works for me, but it blows your "disproof" all to blazes.
> >
>
> Which one?
The one you were trying, at the moment, to establish:your phony counter
to the Cantor diagonal proof.
> I am definitely not affiliated with Hancher, I want nothing to do with
> Hancher,
> because Hancher has made untrue statements and lies about me.
>
Thank you for the disclaimer. I would not wish anyone to suppose an
affiliation between us.
On the other hand, the truth of my statements about Ross and his
attempts at mathematics I leave to the judgement of those other than
Ross. I suspect Ross is too biased to give a fair assessment of them.
[snip]
> >
> > In the past fortnight or so, there has been posted a groundswell of
> > support for the opinion that Rossian logic and comprehension differ from
> > that of the majority. One expects Ross to dispute the majority view.
> >
>
> Actually, I am not caring much about that.
>
> For example, I have discussed several simple proofs that there are more
> rationals
> than integers, and I don't care if others call it an argument, conjecture, or
> anything else, because as time passes, the statements are already there. I
> have
> confidence that a purely objective proof checker will read them easily.
>
Actually, one does not have to be an objective proof checker to read
them. The question is who will agree with them. So far, no one but Ross.
[snip]
>
> My point is that my original statements are already there, and nothing that
> you
> can do can change that. Also, there are many witnesses.
>
> Ross
>
I wouldn't change them for anything. They are sui generis, a monument to
something I do not choose to name, but it isn't mathematucs.
Virgil
Ross Presser <rpre...@NOSPAMimtek.com.invalid> wrote:
> By the way, *this* Ross is not the same Ross as *that* Ross. I always
> post under Ross Presser, and I have very little training in maths. I
> often make erroneous statements, but I seldom defend them vigorously.
> :)
There have been other Ross's that "that" Ross who have, unfortunately,
been confused with him. Our apologies.
Virgil
wrote:
> I am angry because of the concerted efforts to discredit me. I think it
> should be
> quite clear that I have a quite clear enough comprehension of what I say that
> when I
> say it that it is correct, also, when I make a mistake, I readily and ably
> understand why that is so, and correct it, mostly on my own.
>
> Is someone says I am wrong and I disagree, I have clear reasoning why. Also,
> I
> defend my reasoning, and I defend it effectively.
>
> Another thing that is quite clear is that I have made some statements that
> contradict some practices.
>
> A thing that is quite true is that the large part of things that I say that I
> do not
> correct myself, that they are true.
Ross, as a "mathematician" posting "mathematics" in a newsgroup viewed
by many mathematicians considerably more skilled than you are, you are
like a lamb among wolves.
You have not the skills to persuade anyone that you are one of the
wolves yourself, and the approach you have been using in sci.math is
unlikely to gain you any of those skills.
One of the first things to learn is not to be too dogmatic outside your
own area of expertise. Even as clever a mathematician as Perrti Lounesto
looks a little lamb-like when he overlooks this rule.
You are not nearly expert enough in set theory to be wolfish there.
If you ever wish to be one of the wolves, you need to sharpen your fangs
by learning a litle real mathematics. Unfortunately, this newsgroup is
not a good school for that purpose.
Ross
Virgil wrote:
Mostly, it is not a problem, in that to some extent, we don't care. Also, I
have corrected obvious errors with regards to who is being who. I'm not
going to speak for the Virgil's, I've met several other people named Ross,
but never any Virgils. I happen to readily and correctly use my own first
name, and am quite aware that it is a common name shared with many others.
Also, I see myself as being honorable and as well as good for the name.
So, Ross is certainly his own man, as are the other Ross', and other Ross
Finlayson's, and the other Ross Andrew Finlayson's.
To the extent that I personally am the poster who posts here and is often
referred to as Ross in the context of sci.math, I can say that I am right
too often to allow Hancher's character attacks to go unanswered, although as
far as I can tell, that is all that he has.
Ross
Virgil wrote:
Hancher, I think your analogy is stupid. Instead of being a lamb, by now I have
been the commisioner, pope, grand troll, and "the Ross" of sci.math, where those
are appellations of others not regularly used by myself.
About mathematicians and mathematics, mathematics is huge. I have made comments
about several areas of mathematics over the last three or four years on sci.math
and sci.logic:
Foundational vis-a-vis Anti-foundational set theory
The Absurdity of Paradox
Regularity and Null
The Equivalency of Infinite Sets
Multivalent Logics
The Synthesis of Numbers
x-Lebesgue Integrability
Scalar Treatments of Infinities
Existence and the Logic of Reality
...
There are more concepts that we have discussed, that's just a short reminder of
some of the reasons why I think the study of mathematics and logic is an
interesting pursuit. As well, they show me to be productive.
Also, if I am at this caliber, and you are at yours, then considering that
supposedly you have been studying a much longer time than me, then maybe it shows
that I have somewhat of a higher level of capacity than Virgil.
Virgil
wrote:
> Hancher, I think your analogy is stupid. Instead of being a lamb, by now I
> have
> been the commisioner, pope, grand troll, and "the Ross" of sci.math, where
> those
> are appellations of others not regularly used by myself.
If, Ross, you choose to be judged mutton instead of lamb, I will not
gainsay you.
David C. Ullrich
>I am angry because of the concerted efforts to discredit me.
Then stop posting nonsense. This has been explained to you
many times.
> I think it should be
>quite clear that I have a quite clear enough comprehension of what I say that when I
>say it that it is correct, also, when I make a mistake, I readily and ably
>understand why that is so, and correct it, mostly on my own.
No, just the opposite of _all_ of that is clear to everyone but you.
Honest.
>Is someone says I am wrong and I disagree, I have clear reasoning why. Also, I
>defend my reasoning, and I defend it effectively.
I've never seen you do that.
David C. Ullrich
Name one person who has agreed that you are often right. All I see
is _everyone_ explaining your errors, or trying to.
>Ross
>--
>contact Ross
>company Apex Internet Software
>email in...@apexinternetsoftware.com
>website http://www.apexinternetsoftware.com/
>
>
David C. Ullrich
Ross
David C. Ullrich wrote:
Privately, I have received notice of agreement on some key points, by
contemporary, published, contemporarily published mathematicians.
Ross
David C. Ullrich wrote:
> On Tue, 03 Jul 2001 15:45:43 -0400, Ross <ap...@calpha.com> wrote:
>
> >I am angry because of the concerted efforts to discredit me.
>
> Then stop posting nonsense. This has been explained to you
> many times.
>
I post some valid reason, from time to time.
As I see fit to post, I shall, because I have no necessity to follow mob reason.
>
> > I think it should be
> >quite clear that I have a quite clear enough comprehension of what I say that when I
> >say it that it is correct, also, when I make a mistake, I readily and ably
> >understand why that is so, and correct it, mostly on my own.
>
> No, just the opposite of _all_ of that is clear to everyone but you.
> Honest.
>
Your blanket statement is invalid for its overgenerality.
>
> >Is someone says I am wrong and I disagree, I have clear reasoning why. Also, I
> >defend my reasoning, and I defend it effectively.
>
> I've never seen you do that.
>
I do not care.
>
> David C. Ullrich
> *********************
> "Sometimes you can have access violations all the
> time and the program still works." (Michael Caracena,
> comp.lang.pascal.delphi.misc 5/1/01)
Sometimes, Ullrich is wrong. Sometimes, I am wrong. Sometimes, I am right.
David C. Ullrich
Privately I've recieved assurance that I'm the king of France.
I take it you're admitting that nobody here on sci.math has
_posted_ anything that could be regarded as agreement with
anything you've ever said.
_Name_ someone who's agreed. (What key points?)
Lee Rudolph
>Privately I've recieved assurance that I'm the king of France.
Oh, good! So we can finally settle at least *one* of the
logical points raised by Bertrand Russell. Are you bald
or not?
Lee Rudolph
Virgil
wrote:
> Privately, I have received notice of agreement on some key points, by
> contemporary, published, contemporarily published mathematicians.
>
> Ross
Then perhaps they are too ashamed to agree publicly.
Virgil
wrote:
> Sometimes, Ullrich is wrong. Sometimes, I am wrong. Sometimes, I am right.
In a random selection of one statement from Dave's posts versus a random
selection of one staatement from Ross's posts, I would give odds in
favor of Dave being correct and odds against Ross being correct.
Ross
Virgil wrote:
He didn't say that. The guy, I have not seen him post to sci.math, he
wrote a doctoral thesis for his doctorate that says there are more elements
in a proper subset than the set. Then, he discusses treatments of the
sets. So, I guess that's more or less public.
Ross
Ross wrote:
> Virgil wrote:
>
> > In article <3B430BD6...@calpha.com>, Ross <ap...@calpha.com>
> > wrote:
> >
> > > Privately, I have received notice of agreement on some key points, by
> > > contemporary, published, contemporarily published mathematicians.
> > >
> > > Ross
> >
> > Then perhaps they are too ashamed to agree publicly.
>
> He didn't say that. The guy, I have not seen him post to sci.math, he
> wrote a doctoral thesis for his doctorate that says there are more elements
> in a proper subset than the set. Then, he discusses treatments of the
> sets. So, I guess that's more or less public.
>
A correction, he wrote that there are less elements in the proper subset than
the set. There are more elements in the proper superset than the set.
Jesse Hughes
> Virgil wrote:
>
> > In article <3B430BD6...@calpha.com>, Ross <ap...@calpha.com>
> > wrote:
> >
> > > Privately, I have received notice of agreement on some key points, by
> > > contemporary, published, contemporarily published mathematicians.
> > >
> > > Ross
> >
> > Then perhaps they are too ashamed to agree publicly.
>
> He didn't say that. The guy, I have not seen him post to sci.math, he
> wrote a doctoral thesis for his doctorate that says there are more elements
> in a proper subset than the set. Then, he discusses treatments of the
> sets. So, I guess that's more or less public.
Then, how about a name and affiliation? As you say, dissertations are
publications, so surely he wouldn't mind a citation.
--
Jesse Hughes
"The only 'intuitive' interface is the nipple. After that, it's all
learned." -Bruce Ediger on X interfaces
"Nipples are intuitive, my ass!" -Quincy Prescott Hughes
Ross
Ross wrote:
> Ross wrote:
>
> > Virgil wrote:
> >
> > > In article <3B430BD6...@calpha.com>, Ross <ap...@calpha.com>
> > > wrote:
> > >
> > > > Privately, I have received notice of agreement on some key points, by
> > > > contemporary, published, contemporarily published mathematicians.
> > > >
> > > > Ross
> > >
> > > Then perhaps they are too ashamed to agree publicly.
> >
> > He didn't say that. The guy, I have not seen him post to sci.math, he
> > wrote a doctoral thesis for his doctorate that says there are more elements
> > in a proper subset than the set. Then, he discusses treatments of the
> > sets. So, I guess that's more or less public.
> >
>
> A correction, he wrote that there are less elements in the proper subset than
> the set. There are more elements in the proper superset than the set.
Some people consider that the count of elements in the powerset is a natural
limit of some related processes. For example to compare how many reals there are
compared to how many integers or rationals, then the cardinal number describes a
whole class of infinite sets. Yet, among the sets with similar denities to each
other on the intervals, those sets fall into either the "countable" or
"uncountable" types, yet among them and between them precise arithmetical
determinations can be made.
For example, the set of continuous reals on the interval on the real number line,
it's a function of the range of the interval. There is reason to surmise that an
interval with twice the length another interval has has twice as many of the most
critical elements. Also, the cardinal numbers are telling us definite things
about the topology of that set compared to the entire complex system and domains
of that signal.
It's mostly about a synthetic system.
Virgil
wrote:
> For example, the set of continuous reals on the interval on the real number
> line,
> it's a function of the range of the interval.
One must presume that "set" is the antecedent of "function", but sets
are not, generally, functions of anything.
The "set of continuous reals on an interval" has no discernible
mathematical meaning if one uses the standard mathematical meanings of
all the words in it.
Intervals , at least finite ones, have lengths, functions have ranges.
If Ross wants to write about mathematics, and be understood, much less
accepted, he must either use words in their generally understood
mathematical meanings or he must provide definitions explaning his
meanings.
Mathematicians like to be sure of meanings, not to have to guess at them.
> There is reason to surmise
> that an
> interval with twice the length another interval has has twice as many of the
> most
> critical elements.
The most critical elements of an interval are its endpoints.
Does Ross "have reason to surmise" that an interval twice as long as
another has twice as many endpoints?
> Also, the cardinal numbers are telling us definite things
> about the topology of that set compared to the entire complex system and
> domains
> of that signal.
There are a number of obscure points in the above.
Which cardinal numbers?
Which set?
Which topology on that set?
What "entire complex system"?
What domain?
What signal?
When those questions are resolved, we may then consider whether the
quote means anything.
Ross
Virgil wrote:
We've already talked about that, or, I've talked about that here in the context of
this kind of reason, and it's talked about elsewhere because it is enough of an
obvious question that it would almost be irregular to not consider it.
What it is is about applying the set of facts that are known about these things
without utter concern that it not contradict exactly a practice is that.
>
> > Also, the cardinal numbers are telling us definite things
> > about the topology of that set compared to the entire complex system and
> > domains
> > of that signal.
>
> There are a number of obscure points in the above.
> Which cardinal numbers?
> Which set?
> Which topology on that set?
> What "entire complex system"?
> What domain?
> What signal?
>
They are not obscure if they are understood.
>
> When those questions are resolved, we may then consider whether the
> quote means anything.
Have at it.
D. Ross
but any construction which does not include an argument that it gives the
right measure to intervals can hardly be deemed a complete construction of
Lebesgue measure.
Moreover, one can (and often does) _start_ with the length function on the
algebra of finite unions of intervals (which is a finitely additive measure)
and then apply the Caratheodory theorem to extend to Lebesgue measure. Now,
you are right that something like Heine-Borel is used to show that the
length function satisfies the Caratheodory condition, but no *extra* work is
required to get that it gives the right measure to intervals, as that really
is automatic.
One is moved to mention as well Fisher's nice nonstandard construction
(which is trivial modulo an ultrafilter).
In any event, the main point - in the context of this thread - is that if
one believes in the existence of _any_ nontrivial Borel measure on R
(probably I should require sigma-finite here too), then uncountability of R
follows immediately, without resorting to the diagonalization argument which
seems to offend some people.
- David R.
a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
| In article <3b4025dc.11770129@localhost>,
| D. Ross <ro...@math.hawaii.NOSPAM.edu> wrote:
| >Mike Oliver <oli...@math.ucla.edu> wrote:
|
| >| > Countable sets have Lebesgue measure zero.
| >|
| >| To make that work, of course, you have to show that R does
| >| *not* have measure zero, which is a certain amount of work.
|
| >Typically no more work than is involved already in the construction of
| >Lebesgue measure itself, which automatically gives intervals nonempty
| >measure. In fact, all one needs is the existence of *any* nontrivial Borel
| >measure on R.
|
| It's not automatic. In order to show that an interval has positive
| Lebesgue measure, you need to show to things:
|
| (1) an interval has positive outer measure, and
| (2) an interval is a measurable set.
|
| Since (2) is easy, let's focus on (1). By definition, the outer measure
| of an interval is the inf over all countable open covers of that
| interval, of the sums of the lengths in the intervals making up an open
| cover.
|
| How do you prove that the inf is the length of the interval?
|
| Hint: It helps to make use of the Heine-Borel theorem, which says that
| if A is a closed, bounded subset of R^n, then every open cover of A has a
| finite subcover. Now apply the definition of outer measure, taking note
| of the fact that you only need to consider the length sums over all
| finite covers of the interval when taking the infimum. Therefore you can
| use induction on the number of open sets making up the cover.
|
| So, it's almost automatic if you take the Heine-Borel theorem as a given.
| As Mike Oliver said, it's a certain amount of work.
|
| --
| Dave Seaman dse...@purdue.edu
| Amnesty International calls for new trial for Mumia Abu-Jamal
| <http://www.amnestyusa.org/abolish/reports/mumia/>
D. Ross
Dave Seaman
D. Ross <ro...@math.hawaii.NOSPAM.edu> wrote:
>Oh good grief. There is certainly work in constructing Lebesgue Measure,
>but any construction which does not include an argument that it gives the
>right measure to intervals can hardly be deemed a complete construction of
>Lebesgue measure.
My point exactly. I gave a sketch of one approach that is commonly used
to define Lebesgue measure, together with an explanation of why the
measure so defined gives the right measure to intervals. What, exactly,
are you complaining about?
>Moreover, one can (and often does) _start_ with the length function on the
>algebra of finite unions of intervals (which is a finitely additive measure)
>and then apply the Caratheodory theorem to extend to Lebesgue measure. Now,
>you are right that something like Heine-Borel is used to show that the
>length function satisfies the Caratheodory condition, but no *extra* work is
>required to get that it gives the right measure to intervals, as that really
>is automatic.
I am not convinced that proving the Caratheodory theorem is any simpler
than the approach I took using compactness to show that a simple
induction suffices.
>One is moved to mention as well Fisher's nice nonstandard construction
>(which is trivial modulo an ultrafilter).
And I suppose you are claiming that ultrafilters are simpler than the
approach via compactness and induction over finite covers?
>In any event, the main point - in the context of this thread - is that if
>one believes in the existence of _any_ nontrivial Borel measure on R
>(probably I should require sigma-finite here too), then uncountability of R
>follows immediately, without resorting to the diagonalization argument which
>seems to offend some people.
Again, I claim it is simper and more direct to talk about Lebesgue
measure in particular than it is to introduce Borel measures in general
and to deduce the relevant properties. I don't buy your argument. The
method I sketched is "a bit of work," as claimed. Your suggested methods
are no less so.
Virgil
wrote:
[SNIP]
> > If Ross wants to write about mathematics, and be understood, much less
> > accepted, he must either use words in their generally understood
> > mathematical meanings or he must provide definitions explaning his
> > meanings.
> >
> > Mathematicians like to be sure of meanings, not to have to guess at them.
> >
> > > There is reason to surmise
> > > that an
> > > interval with twice the length another interval has has twice as many of
> > > the
> > > most
> > > critical elements.
> >
> > The most critical elements of an interval are its endpoints.
> > Does Ross "have reason to surmise" that an interval twice as long as
> > another has twice as many endpoints?
> >
>
> We've already talked about that, or, I've talked about that here in the
> context of
> this kind of reason, and it's talked about elsewhere because it is enough of
> an
> obvious question that it would almost be irregular to not consider it.
Does this "answer" relate in any way to the question preceding it?
>
> What it is is about applying the set of facts that are known about these
> things
> without utter concern that it not contradict exactly a practice is that.
>
The preceding terminates with a full stop, but doesn't seem to be a
sentence. Can anyone, except Finlayson, guess what it is supposed to
mean? I can't.
> >
> > > Also, the cardinal numbers are telling us definite things
> > > about the topology of that set compared to the entire complex system and
> > > domains
> > > of that signal.
> >
> > There are a number of obscure points in the above.
> > Which cardinal numbers?
> > Which set?
> > Which topology on that set?
> > What "entire complex system"?
> > What domain?
> > What signal?
> >
>
> They are not obscure if they are understood.
Does anyone, except Finlayson, know the aswers to these questions?
I suspect the answers are known only by Finlayson (to avoid the
confusion of multiple Rosses), if indeed by him.
The Scarlet Manuka
news:3B430BD6...@calpha.com...
>
> Privately, I have received notice of agreement on some key points, by
> contemporary, published, contemporarily published mathematicians.
"The lurkers support me in email!"
Traditional last cry of the troll who is caught out making a fool of
himself.
--
The Scarlet Manuka
D. Ross
| In article <3b42da40.5019829@localhost>,
| D. Ross <ro...@math.hawaii.NOSPAM.edu> wrote:
| >Oh good grief. There is certainly work in constructing Lebesgue Measure,
| >but any construction which does not include an argument that it gives the
| >right measure to intervals can hardly be deemed a complete construction of
| >Lebesgue measure.
|
| My point exactly. I gave a sketch of one approach that is commonly used
| to define Lebesgue measure, together with an explanation of why the
| measure so defined gives the right measure to intervals. What, exactly,
| are you complaining about?
My only complaint is that by pointing out (correctly) that Robert Glass's
proof is not elementary, you and Mike are intimating (I believe not
correctly) that it is not complete. I think you have misread the intent of
my posts.
Robert Glass answered the original poster's question with the nice
1-sentence proof for uncountability of R: "Countable sets have Lebesgue
measure zero." Mike Oliver suggested that this was not trivial in that it
can be hard to prove that R has nonzero measure. While I acknowledge that
this nontriviality can be the case - depending on the proof one uses - to me
this does not seem to me to negate the fact that Glass's answer is complete,
in that once we *have* Lebesgue measure, we certainly know that R has
nonzero measure; this is an essential part of the meaning of the term
"Lebesgue measure".
The other constructions I mentioned are secondary to this argument, and I
raised them simply to suggest that the locus of nontriviality in the
construction of Lebegue measure can be shifted away from the proof that the
measure is nontrivial, and on to other things (eg, confirmation of
sigma-additivity). I certainly do not suggest that the overall construction
is trivial.
| I am not convinced that proving the Caratheodory theorem is any simpler
| than the approach I took using compactness to show that a simple
| induction suffices.
I am not arguing as to the relative difficulty of any of these
constructions.
| >One is moved to mention as well Fisher's nice nonstandard construction
| >(which is trivial modulo an ultrafilter).
|
| And I suppose you are claiming that ultrafilters are simpler than the
| approach via compactness and induction over finite covers?
As it happens I do, but again simplicity was never my point.
- David R.
Robert Glass
> Good to know, but what is "trolled"?
>
From: http://www.jargon.net/jargonfile/t/troll.html
To utter a posting on Usenet designed to attract predictable responses or
flames.
Derives from the phrase "trolling for newbies" which in turn comes from
mainstream
"trolling", a style of fishing in which one trails bait through a likely spot
hoping for a bite
>
> Robert Glass <bo...@my-deja.com> wrote in message
> news:3B3FCE55...@my-deja.com...
> >
> >
> > Elaine Jackson wrote:
> >
> > > Should I be trying to sort this out? Is this a different proof from the
> two
> > > I was talking about? Can anybody give me the gist of the argument below?
> I
> > > can't even make out who's talking to whom.
> >
> > You have been trolled.
> >
> > Ross Finlayson has been posting his nonsense to this group for at least 5
> years
> > now.
> > He frequently states that his purpose is to get the longest threads.
Dave Seaman
D. Ross <ro...@math.hawaii.NOSPAM.edu> wrote:
>My only complaint is that by pointing out (correctly) that Robert Glass's
>proof is not elementary, you and Mike are intimating (I believe not
>correctly) that it is not complete. I think you have misread the intent of
>my posts.
I certainly didn't say the proof was not complete. But you have to
remember the context. If this proof is supposed to convince someone who
is struggling to understand the Cantor diagonal proof, then it looks like
a lost cause. I know of no proof based on measure theory that is any
simpler to understand than the diagonal proof.
Randy Poe
>
> David C. Ullrich wrote:
>
> >
> > is _everyone_ explaining your errors, or trying to.
>
> contemporary, published, contemporarily published mathematicians.
>
(To the tune of "My Bonny lies over the ocean")
The Lurkers support me in email
They all think I'm great don't you know.
You posters just don't understand me
But soon you will reap what you sow.
(excerpt from the song "The Lurkers support
me in e-mail", by Jo Walton,
http://www.bluejo.demon.co.uk/poetry/interstichia/lurkers.htm
The author's website requests that it not be reproduced
without permission).
Could you respond to David's question and name sombody
who agrees that you are often right, or even ever right?
- Randy
Randy Poe
> > He didn't say that. The guy, I have not seen him post to sci.math, he
> > wrote a doctoral thesis for his doctorate
the usual intent of a doctoral thesis...
> > that says there are more elements
> > in a proper subset than the set. Then, he discusses treatments of the
> > sets. So, I guess that's more or less public.
> >
>
> A correction, he wrote that there are less elements in the proper subset than
> the set. There are more elements in the proper superset than the set.
Are you saying that there exists a dissertation in mathematics
that contains the assertion that a proper subset OF AN INFINITE
SET A has lower cardinality than A?
Dissertations are publicly available, so I would be very interested
in the citation for this dissertation. Name, school, year, etc.
- Randy
Mike Oliver
> Are you saying that there exists a dissertation in mathematics
> that contains the assertion that a proper subset OF AN INFINITE
> SET A has lower cardinality than A?
Well, that's a true assertion. Depending, of course, on
what the meaning of the word "a" is.
Ross
Mike Oliver wrote:
Moreso, it's about cardinality. The premise of cardinality is a simple
feature of the infinite sets, there are many features of the infinite
sets. For example, you might compare "countable" and "uncountable" as
you might "surreal" and "positive".
The problem with cardinality is that because it does not have
multiplication or addition upon the infinite sets, that it is an
inadequate rule to determine the truth of the comparisons of infinite
sets. So, when people are expressing the addition and multiplication of
infinite sets using set theory, it's valid, where other people say that
it is impossible in terms of cardinality.
In terms of set theory, here we are having the cardinality as being
categorical about only one thing: that the powerset has more elements
than the set. Yet, it does not say what the powerset is. It's not
simple to say whether powerset is characteristic of the reals or the
rationals or other sets, or where it is adequately characteristic. It
is simple to write an acronym of that statement, where huge meaning is
compressed for a context, (not GCH), for something that is not the
Generalized Continuum Hypothesis.
Virgil
Robert Glass <bo...@my-deja.com> wrote:
> Elaine Jackson wrote:
>
> > Good to know, but what is "trolled"?
> >
>
> From: http://www.jargon.net/jargonfile/t/troll.html
>
> To utter a posting on Usenet designed to attract predictable responses or
> flames.
> Derives from the phrase "trolling for newbies" which in turn comes from
> mainstream
> "trolling", a style of fishing in which one trails bait through a likely spot
> hoping for a bite
Also, the fact that the folklore creature called a troll is reputed to
be of monumental ugliness contributes to the intended meaning.
Randy Poe
I meant to clarify after I posted that, but decided that the
effort of clarifying for Ross' benefit is an exercise in
futility.
- Randy
Virgil
wrote:
> Mike Oliver wrote:
>
> > Randy Poe wrote:
> >
> > > Are you saying that there exists a dissertation in mathematics
> > > that contains the assertion that a proper subset OF AN INFINITE
> > > SET A has lower cardinality than A?
> >
> > Well, that's a true assertion. Depending, of course, on
> > what the meaning of the word "a" is.
>
> Moreso, it's about cardinality. The premise of cardinality is a simple
> feature of the infinite sets, there are many features of the infinite
> sets. For example, you might compare "countable" and "uncountable" as
> you might "surreal" and "positive".
What Finlayson cannot seem to comprehend is that if A is a countably
infinite set, then only finite subsets of it have lower cardinality.
Infinite subsets, whether proper or not, have the same cardinality, by
which everyone but Finlayson understands that they can be put into
one-to-one correspondence.
The only relevant comparison I find between "surreal" and "positive" is
that Finlayson tends to be both when neither is warranted.
>
> The problem with cardinality is that because it does not have
> multiplication or addition upon the infinite sets, that it is an
> inadequate rule to determine the truth of the comparisons of infinite
> sets. So, when people are expressing the addition and multiplication of
> infinite sets using set theory, it's valid, where other people say that
> it is impossible in terms of cardinality.
>
> In terms of set theory, here we are having the cardinality as being
> categorical about only one thing: that the powerset has more elements
> than the set. Yet, it does not say what the powerset is.
Cardinality does not say what the Pope is either, even though he is
elected by cardinals.
Cardinality does not say what sets are, so why should it say what a
particular set is? Once sets, including powersets are given, cardinality
can say something about them.
> It's not
> simple to say whether powerset is characteristic of the reals or the
> rationals or other sets, or where it is adequately characteristic.
It is very simple to say that the powerset of any given set is the class
of all subsets of the given set. And that completely characterizes the
"powerset" operation.
> It
> is simple to write an acronym of that statement, where huge meaning is
> compressed for a context, (not GCH), for something that is not the
> Generalized Continuum Hypothesis.
Perhaps Finlayson would be happier making up acronyms and avoiding math.
Ross
Randy Poe wrote:
Hi. I appear to be making many of the same coherent statements that I
was before.
So, the term cardinality and the cardinal number system has its specific
meaning in its own namespace, but there are infinitely many other ways
to consider the comparisons of infinite sets. Many of them are very
useful and relevant, for example where the infinite set theory is used
to consider the logical calculus.
For example, there are rational discussions about the infinitesimals and
infinite ordinals, as there have been for thousands of years.
D. Ross
| I certainly didn't say the proof was not complete.
OK.
| But you have to
| remember the context. If this proof is supposed to convince someone who
| is struggling to understand the Cantor diagonal proof, then it looks like
| a lost cause.
Strictly speaking, the original context was that Elaine Jackson was looking
for an alternate proof that R is uncountable - in this context, Robert
Glass's post is perfect.
As for the second point, you are right, but I myself find the silence here
of the "R-is-countable" crowd encouraging. If someone asserts that "R is
countable because the diagonalization proof is wrong", and if one can
silence them up by offering a completely different 1-line proof...surely
this is a good thing?
| I know of no proof based on measure theory that is any
| simpler to understand than the diagonal proof.
I agree if the phrase "simpler to understand" is replaced by "elementary".
Glass's proof is certainly simple to understand as long as one doesn't have
to actually construct Lebesgue measure, or explain why it is (and should be)
countably additive.
- David R.