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

Posted: Fri Jul 21, 2006 10:10 am Post subject:
Re: Infinite Induction and the Limits of Curves



Tony Orlow wrote:
Quote:  Mike Kelly wrote:
Hi Mike 
Sorry for the delay. Life is a big transition right now. So, let's see...

No worries.
Quote:  Tony Orlow wrote:
Mike Kelly said:
Tony Orlow wrote:
Mike Kelly wrote:
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.
What definition of value range are you using? To what sequences can it
be applied?
For any finite set of reals such as S_n for finite n, there is a least,
2, and a greatest, 2n. Clearly, the value range is the difference
between these two, and for every n, the difference between S_n and S_1
is twice the difference between T_n and T_1. So, it can be applied to
all finite sets of reals in this manner without controversy.

OK so the "value range" of a set of reals is the nonnegative real
number that is the maximum difference between two elements of the set,
if such a maximum eixsts? Is this a definition we can agree to use?
If we agree on it then I'd like to be able to use it from now on
without you claiming it means something else than what we agree on now.
If we don't agree on the definition then we're going to have to work on
it until we do agree otherwise all subsequent discussion is
meaningless.
....
Now, what does value range have to do with the number of elements in a
sequence? How does one use the value range in determining the number of
elements? Can you use it to determine the number of elements in any
sets or just some of them?
What happens for sequences that aren't monotonically increasing? Finite
sequences "up to" somewhere in the sequence {0, 1, 1/2, 3/4, 1/4, 7/8,
1/8, 5/8, 3/8, ... } all have the same value range : 1. And so does the
infinite sequence. Presumably you aren't going to tell me that {0, 1,
1/2 }, {0, 1, 1/2, 3/4, 1/4} and {0, 1, 1/2, 3/4, 1/4, 7/8, 1/8, 5/8,
3/8, ... } all have the same number of elements? Yet they all have the
same value range. So in what sense does value range determine the
number of elements?
Multiplying every element of a sequence by a constant will presumably
change the value range. Will it somehow change the number of elements?
Can you tell me honestly that it makes any sense at all that
multiplying each element of a set by a constant somehow changes the
size of the set? And yet this is exactly what you claim when you say
there are "more" integers than even integers. Does this make any sense?
Take each integer and multiply it by two. You now have half as many
numbers as you started with. Do you honestly think this is even vaguely
coherent?
Quote:  Where value range apparently becomes disputable is when it is applied to
the entire number line. I answered your question. Now, you answer mine.
Is the number line twice as long when we consider only the even
integers, as opposed to all integers?

I don't think "the number line" has a length so your question is
nonsensical to me.
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.
What does it mean to measure a set within a value range for the
elements?
Given a finite value range for a set, from the GLB to the LUB, and an
monotonically increasing algebraic formula mapping the naturals to the
elements of the set, we can calculate the number of elements in the set
using the Inverse Function Rule.

OK.... seems an utterly worthless invention but I don't see anything
*wrong* with it, other than its total lack of utility.
What happens if there aren't any monotonically increasing algebraic
functions involved, by the way? You can't say anything about the number
of elements in a set not "generated" by a monotonically increasing
algebraic function?
Quote:  Over the entire infinite set, we can
use Big'un as the LUB and express infinite sets formulaically in terms
of Big'un.

But Big'Un doesn't mean anything until you provide a definition. You've
never done that. (NB a definition isn't acceptable if it involves also
undefined terms). I have no way of evaluating your statement because it
is literally meaningless to me.
Once you've defined what BigUn is you should prove its existence in
some coherent system.
Then try explaining what your paragraph quoted above qctually means.
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.
What is the range of S_oo? What is the range of T_oo? In what sense is
one twice the other?
In no sense whatsoever. That's the point. They have the same value
range.

By the definition of value range I gave above, neither has a value
range at all. What definition are you using?
Quote:  For any S_n and T_m sharing the same value range, S_n has half
the number of elements (perhaps one less) than T_m. That is, n<=m/2.

OK.
Quote:  Now, is T_oo less than S_oo? If not, then oo<=oo/2.

By the definition of value range I gave above, neither has a value
range at all. What definition are you using?
Quote:  What does "half complete" mean?
It means T_oo only has half the range of S_oo.

By the definition of value range I gave above, neither has a value
range at all. What definition are you using?
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.
What does "the same entire value range" mean?
It means over the entire real line. I mean, come on. Pick ANY interval
on the real line. You have at least twice as many members of T as of S.
Over the entire real line, this proportion holds.

I mean, come on. Pick ANY subsequence of the sequences. You have
exactly the same number of elements in T as in S. Over the entire
sequence, this proportion holds.
Quote:  Consider sequences of the form
S_n = { 2, 4, 6, 8, ...., 2n}
R_n = { 1/3, 1/6, 1/9, 1/12, ... 1/3n }
Do these have the same element count up to the nth element? What about
in the infinite case? What does "value range" mean here?
Uh, yeah, that's what the "th" in "nth" means. Now consider that there
is no n that ends the sequence, and ask yourself if the evens really get
to travel twice as far along the line as the naturals.

Why should I consider your questions when you keep on ignoring mine?
What does value range mean when determining the number of elements in
R? R always has a value range no greater than 1/3. Does this mean it
never has more than 1/3 of an element? If you consider the interval
[0,1] you will find an infinite number of elements of R and none at all
of S. If you consider [1,2] you will find 1 from S and none from R. If
you consider [2,) then infinite from S, none from R. [0,)  infinite
from both So what?
Ask yourself whether, in general, the value range of a sequence has
anything at all to do with the number of elements of that sequence.

mike. 

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

Posted: Mon Jul 17, 2006 7:00 pm Post subject:
Re: Infinite Induction and the Limits of Curves



In article <e9gffn$2r0$1@ruby.cit.cornell.edu>,
Tony Orlow <aeo6@cornell.edu> wrote:
Quote:  Mike Kelly wrote:
Hi Mike 
Sorry for the delay. Life is a big transition right now. So, let's see...
Tony Orlow wrote:
Mike Kelly said:
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.
What definition of value range are you using? To what sequences can it
be applied?
For any finite set of reals such as S_n for finite n, there is a least,
2, and a greatest, 2n. Clearly, the value range is the difference
between these two, and for every n, the difference between S_n and S_1
is twice the difference between T_n and T_1. So, it can be applied to
all finite sets of reals in this manner without controversy.
Where value range apparently becomes disputable is when it is applied to
the entire number line. I answered your question. Now, you answer mine.
Is the number line twice as long when we consider only the even
integers, as opposed to all integers?
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.
What does it mean to measure a set within a value range for the
elements?
Given a finite value range for a set, from the GLB to the LUB, and an
monotonically increasing algebraic formula mapping the naturals to the
elements of the set, we can calculate the number of elements in the set
using the Inverse Function Rule. Over the entire infinite set, we can
use Big'un as the LUB and express infinite sets formulaically in terms
of Big'un.
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.
What is the range of S_oo? What is the range of T_oo? In what sense is
one twice the other?
In no sense whatsoever. That's the point. They have the same value
range. For any S_n and T_m sharing the same value range, S_n has half
the number of elements (perhaps one less) than T_m. That is, n<=m/2.
Now, is T_oo less than S_oo? If not, then oo<=oo/2.
What does "half complete" mean?
It means T_oo only has half the range of S_oo.

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.
What does "the same entire value range" mean?
It means over the entire real line. I mean, come on. Pick ANY interval
on the real line. You have at least twice as many members of T as of S.

Quote:  Over the entire real line, this proportion holds.

That is an assumption. And as none of those sets of naturals are
necessarily subsets of the set of reals, it does not follow.
In TO's world, the real spring complete with infinite pseudoreals from
TO's forehead, as if he were Zeus.
In fact the reals spring from the naturals but only after some
intermediate generations of integers and rationals. There are reals that
behave like grandchildren of the naturals, but the naturals are
progenitors of all other numbers.
Quote: 
Consider sequences of the form
S_n = { 2, 4, 6, 8, ...., 2n}
R_n = { 1/3, 1/6, 1/9, 1/12, ... 1/3n }
Do these have the same element count up to the nth element? What about
in the infinite case? What does "value range" mean here?
Uh, yeah, that's what the "th" in "nth" means.

Nonresponsive. It is clear that TO has no legitimate answer.
Quote:  Now consider that there
is no n that ends the sequence

Then why does TO keep trying to make like there is an end??? 

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David R Tribble science forum Guru
Joined: 21 Jul 2005
Posts: 1005

Posted: Mon Jul 17, 2006 6:34 pm Post subject:
Re: Infinite Induction and the Limits of Curves



David Petry said:
Quote:  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}

Tony Orlow wrote:
Quote:  This part seems a little iffy. In all finite cases, the segment must include
rationals as well as irrationals, since between any two rationals is another
rational. Of course, between any two rationals must also lie an irrational, I
think. I think that uultimately the question becomes whether there must always
be a rational between any two irrationals. If you can prove this, as well as
there beign an irrational between any two rationals, then I think the case is
made, that the number line alternates between rationals and irrationals.
However, I don't think this is provable.

If between any two reals there is a rational, and between any two reals
there is an irrational, how do you conclude that the number line
alternates between rationals and irrationals? 

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

Posted: Mon Jul 17, 2006 4:56 pm Post subject:
Re: Infinite Induction and the Limits of Curves



Mike Kelly wrote:
Hi Mike 
Sorry for the delay. Life is a big transition right now. So, let's see...
Quote:  Tony Orlow wrote:
Mike Kelly said:
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.
What definition of value range are you using? To what sequences can it
be applied?

For any finite set of reals such as S_n for finite n, there is a least,
2, and a greatest, 2n. Clearly, the value range is the difference
between these two, and for every n, the difference between S_n and S_1
is twice the difference between T_n and T_1. So, it can be applied to
all finite sets of reals in this manner without controversy.
Where value range apparently becomes disputable is when it is applied to
the entire number line. I answered your question. Now, you answer mine.
Is the number line twice as long when we consider only the even
integers, as opposed to all integers?
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.
What does it mean to measure a set within a value range for the
elements?

Given a finite value range for a set, from the GLB to the LUB, and an
monotonically increasing algebraic formula mapping the naturals to the
elements of the set, we can calculate the number of elements in the set
using the Inverse Function Rule. Over the entire infinite set, we can
use Big'un as the LUB and express infinite sets formulaically in terms
of Big'un.
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.
What is the range of S_oo? What is the range of T_oo? In what sense is
one twice the other?

In no sense whatsoever. That's the point. They have the same value
range. For any S_n and T_m sharing the same value range, S_n has half
the number of elements (perhaps one less) than T_m. That is, n<=m/2.
Now, is T_oo less than S_oo? If not, then oo<=oo/2.
Quote: 
What does "half complete" mean?

It means T_oo only has half the range of S_oo.
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.
What does "the same entire value range" mean?

It means over the entire real line. I mean, come on. Pick ANY interval
on the real line. You have at least twice as many members of T as of S.
Over the entire real line, this proportion holds.
Quote: 
Consider sequences of the form
S_n = { 2, 4, 6, 8, ...., 2n}
R_n = { 1/3, 1/6, 1/9, 1/12, ... 1/3n }
Do these have the same element count up to the nth element? What about
in the infinite case? What does "value range" mean here?

Uh, yeah, that's what the "th" in "nth" means. Now consider that there
is no n that ends the sequence, and ask yourself if the evens really get
to travel twice as far along the line as the naturals. 

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

Posted: Thu Jul 13, 2006 10:53 am Post subject:
Re: Infinite Induction and the Limits of Curves



Tony Orlow wrote:
Quote:  Mike Kelly said:
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.

What definition of value range are you using? To what sequences can it
be applied?
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.

What does it mean to measure a set within a value range for the
elements?
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.

What is the range of S_oo? What is the range of T_oo? In what sense is
one twice the other?
What does "half complete" mean?
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.

What does "the same entire value range" mean?
Consider sequences of the form
S_n = { 2, 4, 6, 8, ...., 2n}
R_n = { 1/3, 1/6, 1/9, 1/12, ... 1/3n }
Do these have the same element count up to the nth element? What about
in the infinite case? What does "value range" mean here?

mike. 

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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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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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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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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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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

Like 

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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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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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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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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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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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