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Infinite Induction and the Limits of Curves
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Mike Kelly
science forum Guru Wannabe


Joined: 30 Mar 2006
Posts: 119

PostPosted: Thu Jul 13, 2006 10:53 am    Post subject: Re: Infinite Induction and the Limits of Curves Reply with quote

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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Tony Orlow (aeo6)
science forum Guru


Joined: 24 Mar 2005
Posts: 4069

PostPosted: Mon Jul 17, 2006 4:56 pm    Post subject: Re: Infinite Induction and the Limits of Curves Reply with quote

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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David R Tribble
science forum Guru


Joined: 21 Jul 2005
Posts: 1005

PostPosted: Mon Jul 17, 2006 6:34 pm    Post subject: Re: Infinite Induction and the Limits of Curves Reply with quote

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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Virgil
science forum Guru


Joined: 24 Mar 2005
Posts: 5536

PostPosted: Mon Jul 17, 2006 7:00 pm    Post subject: Re: Infinite Induction and the Limits of Curves Reply with quote

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 pseudo-reals 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.

Non-responsive. 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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Mike Kelly
science forum Guru Wannabe


Joined: 30 Mar 2006
Posts: 119

PostPosted: Fri Jul 21, 2006 10:10 am    Post subject: Re: Infinite Induction and the Limits of Curves Reply with quote

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 non-negative 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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