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Dave L. Renfro science forum Guru
Joined: 29 Apr 2005
Posts: 570

Posted: Mon Jul 03, 2006 7:59 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Tony Orlow wrote (in part):
Quote:  Thanks for your comment. I still am not sure what is
meant by "semicontinuous". The staircase is continuous
in that there are no gaps, locationwise.

Yes, each of the staircase functions are continuous.
Quote:  In the sense that the direction of the curve changes
instantaneously where riser meets tread, it is not
continuous.

This means the function fails to have a (finite) derivative
at one or more points. Continuity and differentiability are
different notions, although the latter does imply the
former.
My comment had to do with the function in which you
input a curve and output its length.
http://en.wikipedia.org/wiki/Semicontinuity
http://planetmath.org/encyclopedia/UpperSemiContinuous2.html
"Topology and Geometry in Polymer Science" by Stuart G. Whittington,
T. Lodge, and D. W. Sumners
Use 'Search in this book' = "semicontinuous", choose p. 71
http://books.google.com/books?vid=ISBN0387985808&id=FIPcAxs29ikC
"Geometric Tomography" by Richard J. Gardner
Use 'Search in this book' = "semicontinuous", choose p. 9
http://books.google.com/books?vid=ISBN0521451264&id=hwcKEZNLEmUC
"Elements of the Theory of Functions and Functional Analysis"
by A. N. Kolmogorov and S. V. Fomin
Use 'Search in this book' = "semicontinuous", choose p. 69
http://books.google.com/books?id=OyWeDwfQmeQC&vid=ISBN0486406830
"Selecta Mathematica: Volume 2" by Bert Schweizer, Abe Sklar,
and Karl Menger
Use 'Search in this book' = "semicontinuity", choose pp. 7780
http://books.google.com/books?vid=ISBN3211838341&id=kWVMlsqDYwoC
Dave L. Renfro 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Mon Jul 03, 2006 9:39 pm Post subject:
Re: Infinite Induction and the Limits of Curves



In article <MPG.1f131f9912a9289c98adaf@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Quote:  david petry said:
Tony Orlow wrote:
Hi All 
It has been my position that the method of inductive proof is
valid, not only for all finite natural n, but for the infinite
case as well, given certain precautions. An equality proven
inductively, such as f(n)=g(n), always holds in the infinite
case,
Hmm. I wonder if the following proof works in your mathematical
system. It doesn't work in ZFC.
Theorem: The rationals are equinumerous with the irrationals.
Hi David 
I'd be interested in the exact reason why this doesn't work in ZFC.

At least one good reason.
(1) It has been proved false in ZFC (and ZF, and NBG, as well)
Quote:  My thoughts on the rationals vs. the irrationals run as follows.
There are an infinite number of reals in the unit interval. Call this
number Big'un. Over the infinite real line, there are an equal number
of unit intervals (extending into infinite values).

In ZFC, and in NBG, there are no infinite values of R, so TO's argument
is cooked. 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Mon Jul 03, 2006 9:45 pm Post subject:
Re: Infinite Induction and the Limits of Curves



In article <MPG.1f1321d380f4fbce98adb0@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Quote:  Dave L. Renfro said:
Virgil wrote:
This is one of the cases in which two limit
processes do not commute, the limit of lengths
of approximating polygons as the number of segments
increases and the limiting set of points as the
number of steps increases. The result depends
on which limit is taken first.
David R Tribble wrote:
Yes, which was the whole point of Chas's example,
that you can't assume the limits work out the same.
For what it's worth, arc length, while not continuous,
is "halfway" continuous in this context:
sci.math, "Strange objcet...", 29 September 2002
http://groups.google.com/group/sci.math/msg/71efa9c214d3f721
Dave L. Renfro
Hi David 
Thanks for your comment. I still am not sure what is meant by
"semicontinuous".
The staircase is continuous
[garbage snipped]
the concept works, as far as I can tell.

Which shows how little TO can tell. 

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Mike Kelly science forum Guru Wannabe
Joined: 30 Mar 2006
Posts: 119

Posted: Mon Jul 03, 2006 10:01 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Tony Orlow wrote:
Quote:  david petry said:
Tony Orlow wrote:
Hi All 
It has been my position that the method of inductive proof is valid, not only
for all finite natural n, but for the infinite case as well, given certain
precautions. An equality proven inductively, such as f(n)=g(n), always holds in
the infinite case,
Hmm. I wonder if the following proof works in your mathematical
system. It doesn't work in ZFC.
Theorem: The rationals are equinumerous with the irrationals.
Hi David 
I'd be interested in the exact reason why this doesn't work in ZFC. Is there a
particular point where it breaks, or does it simply contradict the conclusion
of another proof in ZFC?

In ZFC, the rationals are countable and the irrationals are
uncountable. If you think that cardinality/bijection is a valid way to
determine equinumerosity then this is a proof that the sets are not
equinumerous (in ZFC).
David doesn't define what equinmerous means so he is attempting to
prove some undefined statement. I also have serious issues with his
claim that the limit of set S_n is the set of irrationals in the
interval.
His argument is, essentially, as you said "The number line alternates
between rationals and irrationals". Clearly this is bogus in ZFC as
"the number line" is not a sequence in ZFC. Moreover, in ZFC between
any two distinct reals are a countably infinite number of rationals and
an uncountably infinite number of irrationals.
<snip>
Quote:  But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n > oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED
Like I said, I think there is an assumption that each of these segments will
boil down to a single irrational number, though that's not provable in the
finite case. I am not sure it's provable in the infinite case either. There
seems to be something missing in that area. I think it's interesting, but I'm
not quite convinced it's right.

H a ha but you agree with the logic in that paragraph? Then you'll
agree with this logic :
T_n = { naturals up to n }
S_n = {even naturals up to n }
T_oo = { natural numbers }
S_oo = { even natural numbers }
For all n, the number of elements in T_n is exactly equal to the number
of elements in S_n, and hence, taking the limit as n > oo, we conclude
that the number of elements in T_oo equals the number of elements in
S_oo. Thus the naturals and the evens are equinumerous.

mike. 

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Tony Orlow (aeo6) science forum Guru
Joined: 24 Mar 2005
Posts: 4069

Posted: Wed Jul 05, 2006 6:56 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Dave L. Renfro said:
Quote:  Tony Orlow wrote (in part):
Thanks for your comment. I still am not sure what is
meant by "semicontinuous". The staircase is continuous
in that there are no gaps, locationwise.
Yes, each of the staircase functions are continuous.
In the sense that the direction of the curve changes
instantaneously where riser meets tread, it is not
continuous.
This means the function fails to have a (finite) derivative
at one or more points. Continuity and differentiability are
different notions, although the latter does imply the
former.

Yes, the derivative fails to be continuous.
I am not sure why you're citing these pages. Is either of these curves
semicontinuous? I don't see a formula such as you suggest. Maybe I am missing
something, since I am not familiar with semicontinuity.
Well, thanks for the references. I would be interested, though, in what you
thought of my solution using segmentwise definition of the curves which
demonstrates a clear difference between the diagonal and the staircase in the
limit.

Smiles,
Tony 

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Tony Orlow (aeo6) science forum Guru
Joined: 24 Mar 2005
Posts: 4069

Posted: Wed Jul 05, 2006 7:02 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Mike Kelly said:
Quote: 
Tony Orlow wrote:
david petry said:
Tony Orlow wrote:
Hi All 
It has been my position that the method of inductive proof is valid, not only
for all finite natural n, but for the infinite case as well, given certain
precautions. An equality proven inductively, such as f(n)=g(n), always holds in
the infinite case,
Hmm. I wonder if the following proof works in your mathematical
system. It doesn't work in ZFC.
Theorem: The rationals are equinumerous with the irrationals.
Hi David 
I'd be interested in the exact reason why this doesn't work in ZFC. Is there a
particular point where it breaks, or does it simply contradict the conclusion
of another proof in ZFC?
In ZFC, the rationals are countable and the irrationals are
uncountable. If you think that cardinality/bijection is a valid way to
determine equinumerosity then this is a proof that the sets are not
equinumerous (in ZFC).

In ZFC that's true. I wondered at what specific step the error occurs.
Quote: 
David doesn't define what equinmerous means so he is attempting to
prove some undefined statement. I also have serious issues with his
claim that the limit of set S_n is the set of irrationals in the
interval.
His argument is, essentially, as you said "The number line alternates
between rationals and irrationals". Clearly this is bogus in ZFC as
"the number line" is not a sequence in ZFC. Moreover, in ZFC between
any two distinct reals are a countably infinite number of rationals and
an uncountably infinite number of irrationals.

How do you know there are an uncountably infinite set of irrationals? Probably
because there is an uncountable set of reals, and if you only remove a
countable set of rationals, then you are left with an uncountable set. Is that
right? There is something a little fishy with that type of logic in my opinion,
but I don't disagree with the conclusion, particularly.
Quote: 
snip
But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n > oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED
Like I said, I think there is an assumption that each of these segments will
boil down to a single irrational number, though that's not provable in the
finite case. I am not sure it's provable in the infinite case either. There
seems to be something missing in that area. I think it's interesting, but I'm
not quite convinced it's right.
H a ha but you agree with the logic in that paragraph?

When I say "something's missing" you take that to mean I agree with the logic?
That's strange.
Quote:  Then you'll
agree with this logic :
T_n = { naturals up to n }
S_n = {even naturals up to n }
T_oo = { natural numbers }
S_oo = { even natural numbers }
For all n, the number of elements in T_n is exactly equal to the number
of elements in S_n, and hence, taking the limit as n > oo, we conclude
that the number of elements in T_oo equals the number of elements in
S_oo. Thus the naturals and the evens are equinumerous.

The first sentence starts out with a false premise, yet you arrive at the
standard conclusion. Whatever. No, that's got things that SHOULD be missing.
Like, maybe the whole comment. :)

Smiles,
Tony 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Wed Jul 05, 2006 7:57 pm Post subject:
Re: Infinite Induction and the Limits of Curves



In article <MPG.1f15d6dae7d3da2098adc3@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Quote:  Mike Kelly said:
In ZFC, the rationals are countable and the irrationals are
uncountable. If you think that cardinality/bijection is a valid way to
determine equinumerosity then this is a proof that the sets are not
equinumerous (in ZFC).
In ZFC that's true. I wondered at what specific step the error occurs.

When TO adds one of his false assumptions.
Quote: 
David doesn't define what equinmerous means so he is attempting to
prove some undefined statement. I also have serious issues with his
claim that the limit of set S_n is the set of irrationals in the
interval.
His argument is, essentially, as you said "The number line alternates
between rationals and irrationals". Clearly this is bogus in ZFC as
"the number line" is not a sequence in ZFC. Moreover, in ZFC between
any two distinct reals are a countably infinite number of rationals and
an uncountably infinite number of irrationals.
How do you know there are an uncountably infinite set of irrationals?
Probably
because there is an uncountable set of reals, and if you only remove a
countable set of rationals, then you are left with an uncountable set. Is
that
right?

Mirabile dictu, TO got something right!!! 

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Dave L. Renfro science forum Guru
Joined: 29 Apr 2005
Posts: 570

Posted: Wed Jul 05, 2006 9:01 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Dave L. Renfro wrote:
Tony Orlow wrote:
Quote:  I am not sure why you're citing these pages. Is either of
these curves semicontinuous? I don't see a formula such as
you suggest. Maybe I am missing something, since I am not
familiar with semicontinuity.

Here's what I was talking about. Let X be the set of
curves (whose precise definition isn't very important
right now) and let Y be the set of nonnegative real numbers.
Then we can view "length" L as a function L: X > Y,
where L(C), for C in X, is the length of curve C. That is,
the function I'm talking about, which is semicontinuous but
not continuous, is the function that takes a curve C
to its length L(C).
Dave L. Renfro wrote (in part):
Tony Orlow wrote:
Quote:  Well, thanks for the references. I would be interested,
though, in what you thought of my solution using segmentwise
definition of the curves which demonstrates a clear difference
between the diagonal and the staircase in the limit.

I'll try to look through it (bit by bit) in my spare time
in the next day or two. Right now, from looking at the first
paragraph, I have two comments.
There are many notions of "infinity" in mathematics, and
you seem to be mixing unrelated notions. For example,
there's the idea of a limit and there is the idea of
cardinal numbers. These are entirely different ideas,
but you seem to be using elements of each type in
an incongruous ways.
You might want to look up transfinite induction. Roughly,
this is induction over "infinite lists" that can be
arbitrarily long (in the sense of cardinality). Specifically,
the induction is over statements indexed by ordinal
numbers, and ordinal numbers are sort of like taking
the natural number sequence and extending it by the
addition of another operation (besides the successor
operation), the "limit operation". Ordinals look
like this:
...... (...) ..... (...) .... (..) .. (..) . (.) (.) ... ((.)) ...
[The double parentheses is supposed to represent
a clustering of the parentheses, and hence a double
clustering of the points.]
More concretely, here's an example of points on the
real number line whose corresponding order type is w^2:
1, 1.9, 1.99, 1.999, ..., 2, 2.9, 2.99, 2.999, ..., 3,
3.9, 3.99, 3.999, ..., 4, ..., 5, ..., 6,
[skip forward a lot], 3386, 3386.9, 3386.99, 3386.999, ...,
and so on through all the natural numbers.
I don't think transfinite induction has any immediate
applications for what you seem to be looking at, but
it's something you might be interested in looking into.
Dave L. Renfro 

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Mike Kelly science forum Guru Wannabe
Joined: 30 Mar 2006
Posts: 119

Posted: Wed Jul 05, 2006 9:50 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Tony Orlow wrote:
Quote:  Mike Kelly said:
Tony Orlow wrote:
david petry said:
Tony Orlow wrote:
Hi All 
It has been my position that the method of inductive proof is valid, not only
for all finite natural n, but for the infinite case as well, given certain
precautions. An equality proven inductively, such as f(n)=g(n), always holds in
the infinite case,
Hmm. I wonder if the following proof works in your mathematical
system. It doesn't work in ZFC.
Theorem: The rationals are equinumerous with the irrationals.
Hi David 
I'd be interested in the exact reason why this doesn't work in ZFC. Is there a
particular point where it breaks, or does it simply contradict the conclusion
of another proof in ZFC?
In ZFC, the rationals are countable and the irrationals are
uncountable. If you think that cardinality/bijection is a valid way to
determine equinumerosity then this is a proof that the sets are not
equinumerous (in ZFC).
In ZFC that's true. I wondered at what specific step the error occurs.

The proof doesn't define what "equinumerous" means or state anything
about what system it is working in so we don't have a definition of
"rational numbers" or "irrational numbers". So it fails to be a proof
of anything in particular before any steps at all have been taken.
Some other problems :
 Lack of definition of what a "piece of a number line" is. S_3 = {
(0, 1/3), (1/3, 1/2), (1/2, 1) } doesn't clear things up.
 Conflation of "line segment" with "number".
 The assertion that the limit of the sequence of S_n is the set of
irrational numbers.in the unit interval. Totally unjustified.
Quote:  David doesn't define what equinmerous means so he is attempting to
prove some undefined statement. I also have serious issues with his
claim that the limit of set S_n is the set of irrationals in the
interval.
His argument is, essentially, as you said "The number line alternates
between rationals and irrationals". Clearly this is bogus in ZFC as
"the number line" is not a sequence in ZFC. Moreover, in ZFC between
any two distinct reals are a countably infinite number of rationals and
an uncountably infinite number of irrationals.
How do you know there are an uncountably infinite set of irrationals?
Probably
because there is an uncountable set of reals, and if you only remove a
countable set of rationals, then you are left with an uncountable set. Is that
right? There is something a little fishy with that type of logic in my opinion,
but I don't disagree with the conclusion, particularly.

That's an easy way to prove it, yes.
1 The rationals are countable.
2 The reals are uncountable.
3 The set difference of an uncountable and a countable set is
uncountable.
4 The irrationals are the set difference of the reals and the
rationals.
==> The set of irrationals is uncountable
What's fishy about that? Could you list which of those premises you
agree and disagree with? Do you disagree with the implication from the
premises?
Quote:  snip
But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n > oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED
Like I said, I think there is an assumption that each of these segments will
boil down to a single irrational number, though that's not provable in the
finite case. I am not sure it's provable in the infinite case either. There
seems to be something missing in that area. I think it's interesting, but I'm
not quite convinced it's right.
H a ha but you agree with the logic in that paragraph?
When I say "something's missing" you take that to mean I agree with the logic?
That's strange.

Well, you didn't question the logic presented in the paragraph, you
questioned the assumption that the line segments "boil down to" an
irrational number. If you had a problem with the logic of the paragraph
then the assertion about the line segments would be a moot point anyway
because the conclusion still wouldn't follow. I might expect you, then,
to point out any problem you had with the logic in the paragraph. As
you didn't, I assumed you agreed with it. But I'm sure you know your
mind best.
Quote:  Then you'll
agree with this logic :
T_n = { naturals up to n }
S_n = {even naturals up to n }
T_oo = { natural numbers }
S_oo = { even natural numbers }
For all n, the number of elements in T_n is exactly equal to the number
of elements in S_n, and hence, taking the limit as n > oo, we conclude
that the number of elements in T_oo equals the number of elements in
S_oo. Thus the naturals and the evens are equinumerous.
The first sentence starts out with a false premise, yet you arrive at the
standard conclusion.

"For all n, the number of elements in T_n is exactly equal to the
number of elements in S_n"?
Choose any standard natural number n.
The number of elements in T_n is n.
The number of elements in S_n is n.
n = n
So the number of elements in T_n is exactly equal to the number of
elements in S_n.
What is wrong with that? Which steps do you agree with and which do you
disagree with?
Maybe the problem was that when I said "for all n" I should have
explicitly stated "for all standard natural numbers n"?
Quote:  Whatever. No, that's got things that SHOULD be missing.
Like, maybe the whole comment. :)

Smiles,
Tony 


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Tony Orlow (aeo6) science forum Guru
Joined: 24 Mar 2005
Posts: 4069

Posted: Mon Jul 10, 2006 6:56 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Dave L. Renfro wrote:
Quote:  Dave L. Renfro wrote:
http://en.wikipedia.org/wiki/Semicontinuity
http://planetmath.org/encyclopedia/UpperSemiContinuous2.html
Tony Orlow wrote:
I am not sure why you're citing these pages. Is either of
these curves semicontinuous? I don't see a formula such as
you suggest. Maybe I am missing something, since I am not
familiar with semicontinuity.
Here's what I was talking about. Let X be the set of
curves (whose precise definition isn't very important
right now) and let Y be the set of nonnegative real numbers.
Then we can view "length" L as a function L: X > Y,
where L(C), for C in X, is the length of curve C. That is,
the function I'm talking about, which is semicontinuous but
not continuous, is the function that takes a curve C
to its length L(C).
Dave L. Renfro wrote (in part):
"Topology and Geometry in Polymer Science" by Stuart G. Whittington,
T. Lodge, and D. W. Sumners
Use 'Search in this book' = "semicontinuous", choose p. 71
http://books.google.com/books?vid=ISBN0387985808&id=FIPcAxs29ikC
"Geometric Tomography" by Richard J. Gardner
Use 'Search in this book' = "semicontinuous", choose p. 9
http://books.google.com/books?vid=ISBN0521451264&id=hwcKEZNLEmUC
"Elements of the Theory of Functions and Functional Analysis"
by A. N. Kolmogorov and S. V. Fomin
Use 'Search in this book' = "semicontinuous", choose p. 69
http://books.google.com/books?id=OyWeDwfQmeQC&vid=ISBN0486406830
"Selecta Mathematica: Volume 2" by Bert Schweizer, Abe Sklar,
and Karl Menger
Use 'Search in this book' = "semicontinuity", choose pp. 7780
http://books.google.com/books?vid=ISBN3211838341&id=kWVMlsqDYwoC
Tony Orlow wrote:
Well, thanks for the references. I would be interested,
though, in what you thought of my solution using segmentwise
definition of the curves which demonstrates a clear difference
between the diagonal and the staircase in the limit.
I'll try to look through it (bit by bit) in my spare time
in the next day or two. Right now, from looking at the first
paragraph, I have two comments.

Okay. :)
Quote: 
There are many notions of "infinity" in mathematics, and
you seem to be mixing unrelated notions. For example,
there's the idea of a limit and there is the idea of
cardinal numbers. These are entirely different ideas,
but you seem to be using elements of each type in
an incongruous ways.

I can see how it would seem that way. My position on the matter is
apparently rather bizarre. A couple of decades ago I learned about
transfinite set theory, and did fine on the test, but never believed it
made sense. I suppose the most obviously objectionable aspect of its
conclusions was the idea that a proper subset could be the same size as
the proper superset. And so, I've basically rejected transfinite set
theory and cardinality as a proper measure of "size" for infinite sets,
and explored other notions over the years. So, I'm not talking about
transfinite cardinals or limit ordinals, but really, infinite naturals
and induction in the infinite case, and this does have a lot to do with
limits. If you drag in ordinals and cardinals it does look incongruous.
I make them stay in the yard. ;)
Quote: 
You might want to look up transfinite induction. Roughly,
this is induction over "infinite lists" that can be
arbitrarily long (in the sense of cardinality). Specifically,
the induction is over statements indexed by ordinal
numbers, and ordinal numbers are sort of like taking
the natural number sequence and extending it by the
addition of another operation (besides the successor
operation), the "limit operation". Ordinals look
like this:
..... (...) ..... (...) .... (..) .. (..) . (.) (.) ... ((.)) ...
[The double parentheses is supposed to represent
a clustering of the parentheses, and hence a double
clustering of the points.]

Yes, I understand. Every ordinal has a successor, but limit ordinals do
not have predecessors. In my mind, this violates the Peano ordering. I
agree with the conclusion of Robinson's Nonstandard Analaysis that there
is no smallest infinite, since the very same inductive argument made
against the largest finite given the ability to always increment applies
equally to the decrementing of any infinite. So, I cannot ascribe to the
notion of omega, etc.
Quote: 
More concretely, here's an example of points on the
real number line whose corresponding order type is w^2:
1, 1.9, 1.99, 1.999, ..., 2, 2.9, 2.99, 2.999, ..., 3,
3.9, 3.99, 3.999, ..., 4, ..., 5, ..., 6,
[skip forward a lot], 3386, 3386.9, 3386.99, 3386.999, ...,
and so on through all the natural numbers.
I don't think transfinite induction has any immediate
applications for what you seem to be looking at, but
it's something you might be interested in looking into.
Dave L. Renfro

Hi David, thanks for your comments. I don't think transfinite induction
has any applications for what I'm talking about either, but thanks for
the suggestion. It's common to think I'm just confused, when really, I'm
just stubborn in my convictions. Have a nice day! 

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Tony Orlow (aeo6) science forum Guru
Joined: 24 Mar 2005
Posts: 4069

Posted: Mon Jul 10, 2006 7:17 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Mike Kelly wrote:
Quote:  Tony Orlow wrote:
Mike Kelly said:
Tony Orlow wrote:
david petry said:
Tony Orlow wrote:
Hi All 
It has been my position that the method of inductive proof is valid, not only
for all finite natural n, but for the infinite case as well, given certain
precautions. An equality proven inductively, such as f(n)=g(n), always holds in
the infinite case,
Hmm. I wonder if the following proof works in your mathematical
system. It doesn't work in ZFC.
Theorem: The rationals are equinumerous with the irrationals.
Hi David 
I'd be interested in the exact reason why this doesn't work in ZFC. Is there a
particular point where it breaks, or does it simply contradict the conclusion
of another proof in ZFC?
In ZFC, the rationals are countable and the irrationals are
uncountable. If you think that cardinality/bijection is a valid way to
determine equinumerosity then this is a proof that the sets are not
equinumerous (in ZFC).
In ZFC that's true. I wondered at what specific step the error occurs.
The proof doesn't define what "equinumerous" means or state anything
about what system it is working in so we don't have a definition of
"rational numbers" or "irrational numbers". So it fails to be a proof
of anything in particular before any steps at all have been taken.
Some other problems :
 Lack of definition of what a "piece of a number line" is. S_3 = {
(0, 1/3), (1/3, 1/2), (1/2, 1) } doesn't clear things up.
 Conflation of "line segment" with "number".
 The assertion that the limit of the sequence of S_n is the set of
irrational numbers.in the unit interval. Totally unjustified.
David doesn't define what equinmerous means so he is attempting to
prove some undefined statement. I also have serious issues with his
claim that the limit of set S_n is the set of irrationals in the
interval.
His argument is, essentially, as you said "The number line alternates
between rationals and irrationals". Clearly this is bogus in ZFC as
"the number line" is not a sequence in ZFC. Moreover, in ZFC between
any two distinct reals are a countably infinite number of rationals and
an uncountably infinite number of irrationals.
How do you know there are an uncountably infinite set of irrationals?
Probably
because there is an uncountable set of reals, and if you only remove a
countable set of rationals, then you are left with an uncountable set. Is that
right? There is something a little fishy with that type of logic in my opinion,
but I don't disagree with the conclusion, particularly.
That's an easy way to prove it, yes.
1 The rationals are countable.
2 The reals are uncountable.
3 The set difference of an uncountable and a countable set is
uncountable.
4 The irrationals are the set difference of the reals and the
rationals.
==> The set of irrationals is uncountable
What's fishy about that? Could you list which of those premises you
agree and disagree with? Do you disagree with the implication from the
premises?
snip
But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n > oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED
Like I said, I think there is an assumption that each of these segments will
boil down to a single irrational number, though that's not provable in the
finite case. I am not sure it's provable in the infinite case either. There
seems to be something missing in that area. I think it's interesting, but I'm
not quite convinced it's right.
H a ha but you agree with the logic in that paragraph?
When I say "something's missing" you take that to mean I agree with the logic?
That's strange.
Well, you didn't question the logic presented in the paragraph, you
questioned the assumption that the line segments "boil down to" an
irrational number. If you had a problem with the logic of the paragraph
then the assertion about the line segments would be a moot point anyway
because the conclusion still wouldn't follow. I might expect you, then,
to point out any problem you had with the logic in the paragraph. As
you didn't, I assumed you agreed with it. But I'm sure you know your
mind best.
Then you'll
agree with this logic :
T_n = { naturals up to n }
S_n = {even naturals up to n }
T_oo = { natural numbers }
S_oo = { even natural numbers }
For all n, the number of elements in T_n is exactly equal to the number
of elements in S_n, and hence, taking the limit as n > oo, we conclude
that the number of elements in T_oo equals the number of elements in
S_oo. Thus the naturals and the evens are equinumerous.
The first sentence starts out with a false premise, yet you arrive at the
standard conclusion.
"For all n, the number of elements in T_n is exactly equal to the
number of elements in S_n"?
Choose any standard natural number n.
The number of elements in T_n is n.
The number of elements in S_n is n.
n = n
So the number of elements in T_n is exactly equal to the number of
elements in S_n.

What are you on? For any n in N T_n has n elements and S_n has
floor(n/2) elements. You think there are as many even naturals between 1
and 10 as there are naturals between 1 and 10? Go back to kindergarten.
Quote: 
What is wrong with that? Which steps do you agree with and which do you
disagree with?
Maybe the problem was that when I said "for all n" I should have
explicitly stated "for all standard natural numbers n"?

Maybe you need to define what you mean by "up to n". That would
generally be read as some type of value range. Did you mean it as an
index? If so, say "the first n of the..." and it will be more clear.
Quote: 
Whatever. No, that's got things that SHOULD be missing.
Like, maybe the whole comment. :)

Smiles,
Tony
David Petry's original proof: 
Quote:  Theorem: The rationals are equinumerous with the irrationals.
Proof. Let U be [0,1) (i.e. the unit interval including 0
but not 1). I will show that the rationals in U are
equinumerous with the irrationals in U.
Let T be an enumeration of the rationals in U starting
with 0 (e.g. T = {0, 1/2, 1/3, 2/3, 1/4 ...} ) and let T_n
be the set containing the first 'n' elements of T.
Let S_n be the set of pieces (connected components) left over
after removing the elements of T_n from U. For example,
S_3 = { (0, 1/3), (1/3, 1/2), (1/2, 1) }
In the limit as n > oo, T_oo = {x  x in U and x is rational},
and S_oo = { {x}  x in U and x is irrational}
But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n > oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Tue Jul 11, 2006 1:47 am Post subject:
Re: Infinite Induction and the Limits of Curves



In article <e8u7tf$gmn$1@ruby.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Quote:  So, I'm not talking about
transfinite cardinals or limit ordinals, but really, infinite naturals
and induction in the infinite case,

But there is no system of axioms which allows TO's version of "infinite
naturals" or TO's version of "induction in the infinite case".
So TO resorts to handwaving. 

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Mike Kelly science forum Guru Wannabe
Joined: 30 Mar 2006
Posts: 119

Posted: Tue Jul 11, 2006 9:33 am Post subject:
Re: Infinite Induction and the Limits of Curves



Tony Orlow wrote:
Quote:  Mike Kelly wrote:
Tony Orlow wrote:
Mike Kelly said:
Then you'll
agree with this logic :
T_n = { naturals up to n }
S_n = {even naturals up to n }
T_oo = { natural numbers }
S_oo = { even natural numbers }
For all n, the number of elements in T_n is exactly equal to the number
of elements in S_n, and hence, taking the limit as n > oo, we conclude
that the number of elements in T_oo equals the number of elements in
S_oo. Thus the naturals and the evens are equinumerous.
The first sentence starts out with a false premise, yet you arrive at the
standard conclusion.
"For all n, the number of elements in T_n is exactly equal to the
number of elements in S_n"?
Choose any standard natural number n.
The number of elements in T_n is n.
The number of elements in S_n is n.
n = n
So the number of elements in T_n is exactly equal to the number of
elements in S_n.
What are you on? For any n in N T_n has n elements and S_n has
floor(n/2) elements. You think there are as many even naturals between 1
and 10 as there are naturals between 1 and 10? Go back to kindergarten.
What is wrong with that? Which steps do you agree with and which do you
disagree with?
Maybe the problem was that when I said "for all n" I should have
explicitly stated "for all standard natural numbers n"?
Maybe you need to define what you mean by "up to n". That would
generally be read as some type of value range. Did you mean it as an
index? If so, say "the first n of the..." and it will be more clear.

I should have said "Up to the nth". Sorry if that was unclear. You seem
to have figured out what I intended to say, though...
Let's start again :
T_n = { naturals up to the nth natural }
S_n = { even naturals up to the nth even natural }
T_3 = {1, 2, 3}  3 elements
S_3 = {2, 4, 6}  3 elements
T_n = {1, 2, 3, 4, ..., n}  n elements
S_n = {2, 4, 6, 8, ..., 2n}  n elements
T_oo = { All naturals }
S_oo = { All even naturals }
Now do you agree that the inductively proven equality between element
count of the set T and the element count of the set S which holds for
all n also holds in the infinite case? If not, why not?
For all n, the number of elements in T_n is exactly equal to the number
of elements in S_n, and hence, taking the limit as n > oo, we conclude
that the number of elements in T_oo equals the number of elements in
S_oo. Thus the naturals and the evens are equinumerous.

What about the sets
X_n = { 2 * 1, 2 * 2, 2 * 3, 2 * 4, ..., 2 * n }  n elements
Y_n = { 17 * 1, 17 * 2, 17 * 3, 17 * 4, .... 17 * n }  n elements
A_n = { "1", "2", "3", "4", ..., "n" }  n elements
B_n = { "2", "4", "6", "8", ..., "2n" }  n elements
X_3 = { 2, 4, 6 }  3 elements
Y_n = { 17, 34, 51 }  3 elements
A_n = { "1", "2", "3" }  3 elements
B_n = { "2", "4", "6", }  3 elements
X_oo = { All even naturals }
Y_oo = { All naturals divisible by 17 }
A_oo = { All strings which represent naturals }
B_oo = { All strings which represent even naturals }
Do these all have the same element count up to the nth element? What
about in the infinite case?

mike.
Quote:  David Petry's original proof:
Theorem: The rationals are equinumerous with the irrationals.
Proof. Let U be [0,1) (i.e. the unit interval including 0
but not 1). I will show that the rationals in U are
equinumerous with the irrationals in U.
Let T be an enumeration of the rationals in U starting
with 0 (e.g. T = {0, 1/2, 1/3, 2/3, 1/4 ...} ) and let T_n
be the set containing the first 'n' elements of T.
Let S_n be the set of pieces (connected components) left over
after removing the elements of T_n from U. For example,
S_3 = { (0, 1/3), (1/3, 1/2), (1/2, 1) }
In the limit as n > oo, T_oo = {x  x in U and x is rational},
and S_oo = { {x}  x in U and x is irrational}
But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n > oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED 


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Tony Orlow (aeo6) science forum Guru
Joined: 24 Mar 2005
Posts: 4069

Posted: Tue Jul 11, 2006 4:32 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Mike Kelly said:
Quote: 
Tony Orlow wrote:
Mike Kelly wrote:
Tony Orlow wrote:
Mike Kelly said:
Then you'll
agree with this logic :
T_n = { naturals up to n }
S_n = {even naturals up to n }
T_oo = { natural numbers }
S_oo = { even natural numbers }
For all n, the number of elements in T_n is exactly equal to the number
of elements in S_n, and hence, taking the limit as n > oo, we conclude
that the number of elements in T_oo equals the number of elements in
S_oo. Thus the naturals and the evens are equinumerous.
The first sentence starts out with a false premise, yet you arrive at the
standard conclusion.
"For all n, the number of elements in T_n is exactly equal to the
number of elements in S_n"?
Choose any standard natural number n.
The number of elements in T_n is n.
The number of elements in S_n is n.
n = n
So the number of elements in T_n is exactly equal to the number of
elements in S_n.
What are you on? For any n in N T_n has n elements and S_n has
floor(n/2) elements. You think there are as many even naturals between 1
and 10 as there are naturals between 1 and 10? Go back to kindergarten.
What is wrong with that? Which steps do you agree with and which do you
disagree with?
Maybe the problem was that when I said "for all n" I should have
explicitly stated "for all standard natural numbers n"?
Maybe you need to define what you mean by "up to n". That would
generally be read as some type of value range. Did you mean it as an
index? If so, say "the first n of the..." and it will be more clear.
I should have said "Up to the nth". Sorry if that was unclear. You seem
to have figured out what I intended to say, though...
Let's start again :
T_n = { naturals up to the nth natural }
S_n = { even naturals up to the nth even natural }
T_3 = {1, 2, 3}  3 elements
S_3 = {2, 4, 6}  3 elements
T_n = {1, 2, 3, 4, ..., n}  n elements
S_n = {2, 4, 6, 8, ..., 2n}  n elements
T_oo = { All naturals }
S_oo = { All even naturals }

Note that the value range for every S_n is twice that for the corresponding
T_n. When you include ALL of them, to the elements of S_n still have twice the
value range of T_n, or do they both now share the same value range, that of N?
This is where standard theory errs.
Quote: 
Now do you agree that the inductively proven equality between element
count of the set T and the element count of the set S which holds for
all n also holds in the infinite case? If not, why not?

If it does, then the value ranges are also different for T_oo and S_oo, but
both are taken to cover the range of all naturals. If you measure the sets
within any value range for the elements, T_n will have twice as many elements
as S_n.
Quote: 
For all n, the number of elements in T_n is exactly equal to the number
of elements in S_n, and hence, taking the limit as n > oo, we conclude
that the number of elements in T_oo equals the number of elements in
S_oo. Thus the naturals and the evens are equinumerous.

That would be a nice proof if you also included the part about S_oo having
twice the range of T_oo, but then you'd run into the contradiction that T_oo is
only half complete.
Quote: 

What about the sets
X_n = { 2 * 1, 2 * 2, 2 * 3, 2 * 4, ..., 2 * n }  n elements
Y_n = { 17 * 1, 17 * 2, 17 * 3, 17 * 4, .... 17 * n }  n elements
A_n = { "1", "2", "3", "4", ..., "n" }  n elements
B_n = { "2", "4", "6", "8", ..., "2n" }  n elements
X_3 = { 2, 4, 6 }  3 elements
Y_n = { 17, 34, 51 }  3 elements
A_n = { "1", "2", "3" }  3 elements
B_n = { "2", "4", "6", }  3 elements
X_oo = { All even naturals }
Y_oo = { All naturals divisible by 17 }
A_oo = { All strings which represent naturals }
B_oo = { All strings which represent even naturals }
Do these all have the same element count up to the nth element? What
about in the infinite case?

Up to the nth element, yes, but over the same entire value range, no.
Quote: 

mike.
David Petry's original proof:
Theorem: The rationals are equinumerous with the irrationals.
Proof. Let U be [0,1) (i.e. the unit interval including 0
but not 1). I will show that the rationals in U are
equinumerous with the irrationals in U.
Let T be an enumeration of the rationals in U starting
with 0 (e.g. T = {0, 1/2, 1/3, 2/3, 1/4 ...} ) and let T_n
be the set containing the first 'n' elements of T.
Let S_n be the set of pieces (connected components) left over
after removing the elements of T_n from U. For example,
S_3 = { (0, 1/3), (1/3, 1/2), (1/2, 1) }
In the limit as n > oo, T_oo = {x  x in U and x is rational},
and S_oo = { {x}  x in U and x is irrational}
But for all n, the number of elements in T_n is exactly
equal to the number of elements in S_n, and hence, taking
the limit as n > oo, we conclude that the number of elements
in T_oo equals the number of elements in S_oo QED


Smiles,
Tony 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Tue Jul 11, 2006 4:46 pm Post subject:
Re: Infinite Induction and the Limits of Curves



In article <MPG.1f1d9c9bca84ceeb98adc4@newsstand.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Quote:  Mike Kelly said:
T_oo = { All naturals } S_oo = { All even naturals }
Note that the value range for every S_n is twice that for the
corresponding T_n. When you include ALL of them, to the elements of
S_n still have twice the value range of T_n, or do they both now
share the same value range, that of N? This is where standard theory
errs.

Since when one includes all of either any "value range" ceases to exist
as a number, this is where TO errs in assuming properties of the
nonexistent.
Quote: 
Now do you agree that the inductively proven equality between
element count of the set T and the element count of the set S which
holds for all n also holds in the infinite case? If not, why not?
If it does, then the value ranges are also different for T_oo and
S_oo

How can one tell whether nonexistent "numbers" are equal or not
Quote: 
For all n, the number of elements in T_n is exactly equal to the
number of elements in S_n, and hence, taking the limit as n > oo,
we conclude that the number of elements in T_oo equals the number
of elements in S_oo. Thus the naturals and the evens are
equinumerous.
That would be a nice proof if you also included the part about S_oo
having twice the range of T_oo

It is just as validly half the range, or other ratio one wants to dream
up, as the ration of nonexistent to nonexistent is itself nonexistent. 

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