Discussion:
Incompleteness of Cantor's enumeration of the rational numbers
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WM
2024-11-03 08:38:01 UTC
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Apply Cantor's enumeration of the rational numbers q_n, n = 1, 2, 3, ...
Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
Let ε --> 0.
Then all intervals together have a measure m < 2ε*sqrt(2) --> 0.

By construction there are no rational numbers outside of the intervals.
Further there are never two irrational numbers without a rational number
between them. This however would be the case if an irrational number
existed between two intervals with irrational ends. (Even the existence
of neighbouring intervals is problematic.) Therefore there is nothing
between the intervals, and the complete real axis has measure 0.

This result is wrong but implied by the premise that Cantor's
enumeration is complete.

Regards, WM
Mikko
2024-11-03 13:57:37 UTC
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Post by WM
Apply Cantor's enumeration of the rational numbers q_n, n = 1, 2, 3,
... Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
Let ε --> 0.
Then all intervals together have a measure m < 2ε*sqrt(2) --> 0.
By construction there are no rational numbers outside of the intervals.
Further there are never two irrational numbers without a rational
number between them. This however would be the case if an irrational
number existed between two intervals with irrational ends.
No, it would not. Between any two distinct nubers, whether rational or
irrational, there are both rational and irrational numbers. There are
also intervals from the above specified set.
Post by WM
(Even the existence of neighbouring intervals is problematic.)
Not at all. Between any two non-interlapping intervals there is another
interval so there are not neighbouring intervals. Consequentely, all
these interval are lonely, and so are the rational numbers in their center.
Post by WM
Therefore there is nothing between the intervals, and the complete real
axis has measure 0.
As long as ε > 0 the intervals overlap so "between" is not well defined.
Anyway, there are real numbers that are not in any interval.
Post by WM
This result is wrong but implied by the premise that Cantor's
enumeration is complete.
Your result is wrong.

Cantor's enumeration is complete. Numbers not enumerated are not rational.
--
Mikko
WM
2024-11-03 16:40:01 UTC
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Post by WM
Apply Cantor's enumeration of the rational numbers q_n, n = 1, 2, 3,
... Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
Let ε --> 0.
Then all intervals together have a measure m < 2ε*sqrt(2) --> 0.
By construction there are no rational numbers outside of the
intervals. Further there are never two irrational numbers without a
rational number between them. This however would be the case if an
irrational number existed between two intervals with irrational ends.
No, it would not. Between any two distinct numbers, whether rational or
irrational, there are both rational and irrational numbers.
Not between two adjacent intervals. Such intervals must exist because
space between intervals must exist. Choose a point of this space and go
in both directions, find the adjacent intervals.
As long as ε > 0 the intervals overlap
Let ε = 1. If all intervals overlap and there is no space "between",
then the measure of the real line is less than 2*sqrt(2). Therefore not
all intervals overlap.
Anyway, there are real numbers that are not in any interval.
That is not possible because between two adjacent intervals there is no
rational number and hence no irrational number.

Regards, WM
Jim Burns
2024-11-03 22:18:01 UTC
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followups to sci.math sci.logic
Post by WM
Apply Cantor's enumeration of
the rational numbers q_n, n = 1, 2, 3, ...
Cover each q_n by the interval
ε[q_n - sqrt(2)/2^n, q_n + sqrt(2)/2^n].
= [xᵋₙ,xᵋₙ′]
xᵋₙ = qₙ - 2¹ᐟ²⁻ⁿ⋅ε
xᵋₙ′ = qₙ + 2¹ᐟ²⁻ⁿ⋅ε

μ[xᵋₙ,xᵋₙ′] = μ(xᵋₙ,xᵋₙ′) = 2³ᐟ²⁻ⁿ⋅ε
Post by WM
Let ε --> 0.
Then all intervals together have
a measure m < 2ε*sqrt(2) --> 0.
Yes.

For each ε > 0
⎛ there is an open ε.cover {(xᵋₙ,xᵋₙ′)ᴿ: n∈ℕ} of ℚ
⎜ ℚ ⊆ int.⋃{(xᵋₙ,xᵋₙ′)ᴿ:n∈ℕ}
⎜ and
⎜ 0 ≤ μ(ℚ) ≤
⎜ μ(⋃{(xᵋₙ,xᵋₙ′)ᴿ:n∈ℕ}) ≤¹
⎜ ∑ₙ2³ᐟ²⁻ⁿ⋅ε = 2³ᐟ²⋅ε
⎜
⎜ ¹ '≤' not '=' because the intervals (xᵋₙ,xᵋₙ′)ᴿ
⎝ are in.line, not in.sequence, and they overlap.

0 ≤ μ(ℚ) ≤
glb.{μ(⋃{(xᵋₙ,xᵋₙ′)ᴿ:n∈ℕ}): 0 < ε ∈ ℝ} ≤
glb.{2³ᐟ²⋅ε: 0 < ε ∈ ℝ} = 0

0 ≤ μ(ℚ) ≤ 0
Post by WM
By construction there are
no rational numbers outside of the intervals.
Further there are never
two irrational numbers without
a rational number between them.
This however would be the case
Maybe "This however would NOT be the case"
was intended?
Post by WM
This however would [not(?)] be the case
if an irrational number existed between
two intervals with irrational ends.
No.
It is still the case.
I admit that I find this difficult to picture,
but the mathematics is clear.

The intervals (xᵋₙ,xᵋₙ′)ᴿ of an ε.cover
are not strung out like a string of pearls.
They overlap, with each interval containing
a0.many smaller intervals.
Their total measure is < 2³ᐟ²⋅ε but they're smeared
like an infinitely.thin coat of butter on toast.
They're more cloud.like than necklace.like.

If you are imagining one.point "limit" intervals,
consider that
for each ε > 0
ℚ is in the interior of ⋃{(xᵋₙ,xᵋₙ′)ᴿ:n∈ℕ}
and the upper bound of that set's measure
is not problematic.

"In the limit",
ℚ is not in the interior, and,
if we naively say μ(ℚ) = ℵ₀⋅0
what, exactly, is that? 0? ℵ₀?
The naive method only trades questions
for other questions.
Post by WM
(Even the existence of neighbouring intervals
is problematic.)
There aren't any neighboring intervals.
Any two intervals have intervals between them.
Post by WM
Therefore
there is nothing between the intervals,
and the complete real axis has measure 0.
This result is wrong
but implied by the premise that
Cantor's enumeration is complete.
WM
2024-11-04 09:55:24 UTC
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Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals

Regards, WM
Mikko
2024-11-04 10:31:46 UTC
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Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't. Between that point an an interval there are rational
numbers and therefore other intervals, at least some of which do not
cover the point and don't overlap with the interval. Therefore the
point has no nearest interval.
--
Mikko
WM
2024-11-04 10:47:19 UTC
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Post by Mikko
Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Post by Mikko
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
Post by Mikko
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.

Regards, WM
Richard Damon
2024-11-04 12:28:50 UTC
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Post by WM
Post by Mikko
Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Post by Mikko
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
Unless "nearest" isn't a thing because things are dense.
Post by WM
Post by Mikko
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.

Your concept of "Nearest" is the unfounded assumption, based on
incorrect logic and thus is not accepted.
Post by WM
Regards, WM
WM
2024-11-04 16:55:35 UTC
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Post by Richard Damon
Post by WM
I said the nearest one. There is no interval nearer than the nearest one.
Unless "nearest" isn't a thing because things are dense.
The intervals cannot be dense because they have a length of less than 3
in an infinite space. The rationals are dense. This proves that they are
not countable.>>
Post by Richard Damon
Post by WM
Post by WM
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
It is proven in an inconsistent theory. I describe the true mathematics.

Regards, WM
Richard Damon
2024-11-05 03:12:18 UTC
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Post by WM
Post by Richard Damon
Post by WM
I said the nearest one. There is no interval nearer than the nearest one.
Unless "nearest" isn't a thing because things are dense.
The intervals cannot be dense because they have a length of less than 3
in an infinite space. The rationals are dense. This proves that they are
not countable.>>
Not at all. You just don't understand how infinity works.

Cantor showed how to count the Rationals in a countable infinity.

He showed that the Reals Could not be counted, not even a finite length
line of them.
Post by WM
Post by Richard Damon
Post by WM
Post by WM
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
It is proven in an inconsistent theory. I describe the true mathematics.
No, you describe an inconsisten mathematics, which has blown up in your
face and blineded you to the reality of the infinite numbers.
Post by WM
Regards, WM
WM
2024-11-05 08:45:57 UTC
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Post by Richard Damon
Cantor showed how to count the Rationals in a countable infinity.
No. Then the real axis would have measure zero.
Post by Richard Damon
He showed that the Reals Could not be counted, not even a finite length
line of them.
That is the nonsense believed by matheologians. He assumes a fixed set
ℕ. But if Hilbert's hotel was real, then always a diagonal number could
be enumerated by inserting it into the list at line no. 1.
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by WM
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
It is proven in an inconsistent theory. I describe the true mathematics.
No, you describe
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by the
believers in matheology.

Regards, WM
Richard Damon
2024-11-05 12:03:37 UTC
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Post by WM
Post by Richard Damon
Cantor showed how to count the Rationals in a countable infinity.
No. Then the real axis would have measure zero.
No, because the real axis is made of Real Numbers, not Rational Numbers.

Yes, the proportion of Real Numbers that are Rational is 0
Post by WM
Post by Richard Damon
He showed that the Reals Could not be counted, not even a finite
length line of them.
That is the nonsense believed by matheologians. He assumes a fixed set
ℕ. But if Hilbert's hotel was real, then always a diagonal number could
be enumerated by inserting it into the list at line no. 1.
A Fixed but COUNTABLY INFINITE set.

I don't think you understand what a diagonal proof is (I know you don't
as you have shown you don't).

The fact that Hilbert shows we can "add" new entries into the infinite
set of Natural Numbers by transforming them shows some of the properties
of infinite sets that finite sets do not have.

This isn't showing a "contradiction", it is showing that infinite sets
are diffferent than finite sets, and some of the propreties you are
insisting on, just don't work, so if you insist on them, you can't have
the infinite sets.

PERIOD.

Saying that infinite sets exist, and your properties still exist, is
like claiming that 1 == 2.

Your logic system just blows up and creates your "darkness" in the void
it leaves behind.
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by WM
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
It is proven in an inconsistent theory. I describe the true mathematics.
No, you describe
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by the
believers in matheology.
Regards, WM
You seem to be just babbling. If you mean that between to finite length
intervals, which include there endpoints, and a point that is between
those two intervals, then YES there exist two other intervals between
the point and those two intervals.

The key here is you need to be using CLOSED intervals that contain their
end points, and that just follows from the fact that the Real number
line is dense, and between any two points is an infinite number of other
points (and thus a finite interval).

Your concept of "adjacent" points is the problem case which you just
showed can't exist, so apparently yours is the "mathology" that doesn't
have a firm foundation.
WM
2024-11-05 12:34:28 UTC
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Post by Richard Damon
Post by WM
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by the
believers in matheology.
If you mean that between to finite length
intervals, which include there endpoints, and a point that is between
those two intervals, then YES there exist two other intervals between
the point and those two intervals.
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.

Regards, WM
Richard Damon
2024-11-06 11:46:21 UTC
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Post by WM
Post by WM
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by
the believers in matheology.
If you mean that between to finite length intervals, which include
there endpoints, and a point that is between those two intervals, then
YES there exist two other intervals between the point and those two
intervals.
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Regards, WM
Or rational endpoints.

There is no "next to" on the dense line, which is your problem.

Between ANY two points are an infinite number of other points, so "Next
to" isn't an applicable term, and any logic based on it has exploded.
WM
2024-11-06 18:22:40 UTC
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Post by Richard Damon
Post by WM
Post by WM
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by
the believers in matheology.
If you mean that between to finite length intervals, which include
there endpoints, and a point that is between those two intervals,
then YES there exist two other intervals between the point and those
two intervals.
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Or rational endpoints.
No, the rationals are centres of their intervals.
Post by Richard Damon
There is no "next to" on the dense line
Every positive point is nearer to zero than to any negative point.
Of -x and 0 the latter is next to any positive x.

Regards, WM
Richard Damon
2024-11-07 00:51:03 UTC
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Post by WM
Post by Richard Damon
Post by WM
Post by WM
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by
the believers in matheology.
If you mean that between to finite length intervals, which include
there endpoints, and a point that is between those two intervals,
then YES there exist two other intervals between the point and those
two intervals.
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Or rational endpoints.
No, the rationals are centres of their intervals.
They can also be endpoints of intervals.
Post by WM
Post by Richard Damon
There is no "next to" on the dense line
Every positive point is nearer to zero than to any negative point.
Of -x and 0 the latter is next to any positive x.
But that isn't "next to".

There is no point that is NEXT TO any other point, as if X and Y were
"next to" each other, then the point (X+Y)/2 would be between them, and
thus they are not next to each other.

By the rules of arithmetic of Rational and Reals, that expresion is
ALWAYS a value, and will ALWAYS be between the two numbers.
Post by WM
Regards, WM
WM
2024-11-07 09:28:17 UTC
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Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Or rational endpoints.
No, the rationals are centres of their intervals.
They can also be endpoints of intervals.
Not of their own intervals.
Post by Richard Damon
Post by WM
Post by Richard Damon
There is no "next to" on the dense line
Every positive point is nearer to zero than to any negative point.
Of -x and 0 the latter is next to any positive x.
But that isn't "next to".
It is next to when between a point and the interval no further point
exists, like here:

Use the intervals J(n) = [n - 1/10, n + 1/10]. Without splitting or
modifying them they can be translated and reordered, to cover the whole
positive axis and every rational as a midpoint - if Cantor was right.

Regards, WM
Richard Damon
2024-11-07 12:04:09 UTC
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Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
These other intervals also have irrational endpoints. Every point
outside of an interval is next to some endpoint which is irrational.
Or rational endpoints.
No, the rationals are centres of their intervals.
They can also be endpoints of intervals.
Not of their own intervals.
But they can define intervals that way.

You don't seem to understand what thosse intervals do
Post by WM
Post by Richard Damon
Post by WM
Post by Richard Damon
There is no "next to" on the dense line
Every positive point is nearer to zero than to any negative point.
Of -x and 0 the latter is next to any positive x.
But that isn't "next to".
It is next to when between a point and the interval no further point
But that doesn't exist for closed intervals (intervals define with a
point on their end).
Post by WM
Use the intervals J(n) = [n - 1/10, n + 1/10]. Without splitting or
modifying them they can be translated and reordered, to cover the whole
positive axis and every rational as a midpoint - if Cantor was right.
I don't think you understand what Cantor was doing or trying to show.

But of course, you never seemed to have known what you were talking about.

IF you are talking about the thing I think you are, Cantor was just
showing that between any two irrational numbers will be a rational
number. (Not at its MID-point, just between them).
Post by WM
Regards, WM
WM
2024-11-05 08:46:30 UTC
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Post by Richard Damon
Cantor showed how to count the Rationals in a countable infinity.
No. Then the real axis would have measure zero.
Post by Richard Damon
He showed that the Reals Could not be counted, not even a finite length
line of them.
That is the nonsense believed by matheologians. He assumes a fixed set
ℕ. But if Hilbert's hotel was real, then always a diagonal number could
be enumerated by inserting it into the list at line no. 1.
Post by Richard Damon
Post by WM
Post by Richard Damon
Post by WM
Post by WM
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
No, it is a PROVEN statement, therefor true.
It is proven in an inconsistent theory. I describe the true mathematics.
No, you describe
I describe that a point between two finite intervals has two finite
intervals around it. Even this simple conclusion must be denied by the
believers in matheology.

Regards, WM
Mikko
2024-11-04 17:49:39 UTC
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Post by WM
Post by Mikko
Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
This discussion is about numbers, not geometry.
Post by WM
Post by Mikko
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
There is no nearesst one. There is always a nearer one.
Post by WM
Post by Mikko
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
It is not unfounded. Your conterclaim is unfounded (or at least its
foundation is not in anythhing relevant).
--
Mikko
WM
2024-11-04 18:12:55 UTC
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest
intervals
No, it hasn't.
In geometry it has.
This discussion is about numbers, not geometry.
Geometry is only another language for the same thing.
Post by Mikko
Post by WM
Post by Mikko
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
There is no nearesst one. There is always a nearer one.
Nonsense.
Post by Mikko
Post by WM
Post by Mikko
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
It is not unfounded.
Of course it is. It is the purest nonsense. Starting at a point outside
of the intervals we move and stop as soon as the first interval is
reached. If that is declared as impossible then your theory is irrelevant.

Regards, WM



Your conterclaim is unfounded (or at least its
Post by Mikko
foundation is not in anythhing relevant).
Mikko
2024-11-05 10:29:57 UTC
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Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
This discussion is about numbers, not geometry.
Geometry is only another language for the same thing.
Another language is an unnecessary complication that only reeasls
an intent to deceive.
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
There is no nearesst one. There is always a nearer one.
Nonsense.
No, the meaning is clear. Of course, because some intevals overlap,
you should have specified what exacly you mean by "nearer". But as
ε shriks the overlappings disappear and the distance between any
two intevals approaches the distance between their centers we may
define distance between the intervals as the distance between their
endpoints even wne ε > 0.
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
It is not unfounded.
Of course it is. It is the purest nonsense.
That you don't even try to support your clam to support your claim
indicates that you don't really believe it. Cantor's results are
conclusions of proofs and you have not shown any error in the proofs.
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't. Even if you will that
will not make the results unfounded. It only means that you want to
use a different foundation. Whether you can find one that you like
is your problem.
--
Mikko
WM
2024-11-05 11:26:58 UTC
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Post by Mikko
Post by Mikko
There is no nearesst one. There is always a nearer one.
And always the endpoint is irrational.
Post by Mikko
That you don't even try to support your clam to support your claim
indicates that you don't really believe it.
My claim says that every point outside of intervals has an irrational
interval end next to it. It does not matter how many intervals you claim
between the point and the nearest interval, because _all_ intervals have
irrational endpoints.

Cantor's results are
Post by Mikko
conclusions of proofs and you have not shown any error in the proofs.
I have. This example for instance proves that he did not enumerate all
rationals, because the rationals are dense, the intervals are not dense.
Post by Mikko
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't.
Cantor's bijections concern only potentially infinite sets, but are
assumed and claimed to concern the complete sets. That is the grave
mistake. His result says for all infinite "countable sets" that they are
infinite, nothing more. Mathematics proves that for all intervals (0,
2n) the ratio between even numbers and natural numbers is 1/2. The
sequence 1/2, 1/2, 1/2, ... has limit 1/2. That is the true result in
mathematics.

Regards, WM
Mikko
2024-11-06 14:59:43 UTC
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Post by WM
Post by Mikko
Post by Mikko
There is no nearesst one. There is always a nearer one.
And always the endpoint is irrational.
Post by Mikko
That you don't even try to support your clam to support your claim
indicates that you don't really believe it.
My claim says that every point outside of intervals has an irrational
interval end next to it. It does not matter how many intervals you
claim between the point and the nearest interval, because _all_
intervals have irrational endpoints.
Cantor's results are
Post by Mikko
conclusions of proofs and you have not shown any error in the proofs.
I have. This example for instance proves that he did not enumerate all
rationals, because the rationals are dense, the intervals are not dense.
You have not proven that. It is fairly easy to prove that there are
no positive rationals other than those enumerated by Cantor (if I
recall correctly he enumerated only positive rationals). To prove
that there are positive rationals that are not included in Cantor's
enumeration it suffices to show one but you have not shown any.
Post by WM
Post by Mikko
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't.
Cantor's bijections concern only potentially infinite sets, but are
assumed and claimed to concern the complete sets.
Everything Cantor said was about complete sets. He did neither deny the
possibility of potentially infinte sets nor said anything about them (as
far as I know and remember).
Post by WM
That is the grave mistake. His result says for all infinite "countable
sets" that they are infinite, nothing more.
He very clearly says and proves that all infinite sets are not equinumerous.
--
Mikko
WM
2024-11-06 17:52:01 UTC
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Post by Mikko
  Cantor's results are
Post by Mikko
conclusions of proofs and you have not shown any error in the proofs.
I have. This example for instance proves that he did not enumerate all
rationals, because the rationals are dense, the intervals are not dense.
You have not proven that. It is fairly easy to prove that there are
no positive rationals other than those enumerated by Cantor (if I
recall correctly he enumerated only positive rationals). To prove
that there are positive rationals that are not included in Cantor's
enumeration it suffices to show one but you have not shown any.
I have shown that without rational numbers outside of the intervals with
irrational endpoints covering 3 of infinitely many units the real axis
has measure 3. Completely independent of the form and configuration of
the intervals between them no real number could exist if all rationals
were included.
Post by Mikko
Post by Mikko
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't.
Cantor's bijections concern only potentially infinite sets, but are
assumed and claimed to concern the complete sets.
Everything Cantor said was about complete sets. He did neither deny the
possibility of potentially infinte sets nor said anything about them (as
far as I know and remember).
Georg Cantor did not get tired to explain the difference and the
importance of the actual infinite over and over again.

"In spite of significant difference between the notions of the potential
and actual infinite, where the former is a variable finite magnitude,
growing above all limits, the latter a constant quantity fixed in itself
but beyond all finite magnitudes, it happens deplorably often that the
one is confused with the other." [Cantor, p. 374]
Post by Mikko
That is the grave mistake. His result says for all infinite "countable
sets" that they are infinite, nothing more.
He very clearly says and proves that all infinite sets are not
equinumerous.
All countable infinite sets are equinumerous according to him, but not
in reality.

Regards. WM
Mikko
2024-11-07 09:24:32 UTC
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Post by WM
Post by Mikko
  Cantor's results are
Post by Mikko
conclusions of proofs and you have not shown any error in the proofs.
I have. This example for instance proves that he did not enumerate all
rationals, because the rationals are dense, the intervals are not dense.
You have not proven that. It is fairly easy to prove that there are
no positive rationals other than those enumerated by Cantor (if I
recall correctly he enumerated only positive rationals). To prove
that there are positive rationals that are not included in Cantor's
enumeration it suffices to show one but you have not shown any.
I have shown that without rational numbers outside of the intervals
with irrational endpoints covering 3 of infinitely many units the real
axis has measure 3.
No, you have not shown it, only said.
--
Mikko
Ross Finlayson
2024-11-06 02:48:44 UTC
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest
intervals
No, it hasn't.
In geometry it has.
This discussion is about numbers, not geometry.
Geometry is only another language for the same thing.
Another language is an unnecessary complication that only reeasls
an intent to deceive.
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
There is no nearesst one. There is always a nearer one.
Nonsense.
No, the meaning is clear. Of course, because some intevals overlap,
you should have specified what exacly you mean by "nearer". But as
ε shriks the overlappings disappear and the distance between any
two intevals approaches the distance between their centers we may
define distance between the intervals as the distance between their
endpoints even wne ε > 0.
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
It is not unfounded.
Of course it is. It is the purest nonsense.
That you don't even try to support your clam to support your claim
indicates that you don't really believe it. Cantor's results are
conclusions of proofs and you have not shown any error in the proofs.
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't. Even if you will that
will not make the results unfounded. It only means that you want to
use a different foundation. Whether you can find one that you like
is your problem.
Here what's considered an "opinion" of ZF is any axiom,
of the theory, what results "restriction of comprehension",
for example the Axiom of Regularity, or, the Axiom of (Regular)
Infinity.

Somebody like Mirimanoff, who introduced the plain "extra-ordinary",
then saw that as soon as Mirimanoff brought that up, then
ZF set theory had an axiom of regular/ordinary infinity added to it,
thus that Russell's "paradox" was put away, then for some
relatively simple things or the establishment of an inequality
the uncountability, to so follow.

The anti-diagonal argument as discovered by du Bois Reymond,
and nested intervals known since forever, the m-w proof as
is one of the number-theoretic proofs of uncountability,
another bit for continued fractions, these are the number-theoretic
results for uncountability, then there's the set-theoretic
bit or the powerset result.

So, the idea of providing an example to uncountability,
would be a 1-1 and onto function a bijection, between
countable domain and here most succinctly, the unit interval,
each of the points of the unit interval. This would be
with regards to the "number-theoretic" arguments.

Then, there would also need be a "set-theoretic" counter-example.


Here's that's provided by the "natural/unit equivalency function",
which falls out of the number-theoretic results un-contradicted,
and then some "ubiquitous ordinals" between ordering-theory and
set-theory, not unlike Cohen's forcing establishing the ndependence
of the Continuum Hypothesis, which one can also see as forestalling
what's a contradiction after ZF, since ordinals either would or
wouldn't live between cardinals with or without CH.

So, providing a counterexample and noting that
the "restrictions of comprehension" are _stipulations_
and thusly _non-logical_, makes for an inclusive take
on a foundation beneath _ordinary_ set theory: _extra-ordinary_
set theory. ("A theory of one relation: elt.")

In this way we can have extra-ordinary theory and plain
simple classical logical theory and plain ordinary regular
set theory, all quite thoroughly logical.
WM
2024-11-06 10:01:21 UTC
Reply
Permalink
Post by Mikko
Post by WM
Geometry is only another language for the same thing.
Another language is an unnecessary complication that only reeasls
an intent to deceive.
It is a clearer language.
Post by Mikko
No, the meaning is clear. Of course, because some intevals overlap,
you should have specified what exacly you mean by "nearer". But as
ε shriks the overlappings disappear and the distance between any
two intevals approaches the distance between their centers we may
define distance between the intervals as the distance between their
endpoints even wne ε > 0.
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.

Regards, WM
Mikko
2024-11-06 15:04:44 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by WM
Geometry is only another language for the same thing.
Another language is an unnecessary complication that only reeasls
an intent to deceive.
It is a clearer language.
No, what can be said about numbers can be stated at least as clearly
in the language of arithmetic.
Post by WM
Post by Mikko
No, the meaning is clear. Of course, because some intevals overlap,
you should have specified what exacly you mean by "nearer". But as
ε shriks the overlappings disappear and the distance between any
two intevals approaches the distance between their centers we may
define distance between the intervals as the distance between their
endpoints even wne ε > 0.
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
--
Mikko
WM
2024-11-06 17:55:15 UTC
Reply
Permalink
Post by Mikko
Post by WM
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according to
Cantor's results, 0. That shows that his results are wrong.
But the important question is also covered by ε = 1. The measure of the
real axis is, according to Cantor's results, less than 3. That shows
that his results are wrong.

Regards, WM
Mikko
2024-11-07 09:22:47 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by WM
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according to
Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
Post by WM
But the important question is also covered by ε = 1. The measure of the
real axis is, according to Cantor's results, less than 3. That shows
that his results are wrong.
No, that is not Cantor's result, so all we can say about it is that
you are wrong about Cantor's result.
--
Mikko
WM
2024-11-07 13:21:42 UTC
Reply
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according to
Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
But not between irrational points.
Post by Mikko
Post by WM
But the important question is also covered by ε = 1. The measure of
the real axis is, according to Cantor's results, less than 3. That
shows that his results are wrong.
No, that is not Cantor's result,
It is Cantor's result that all rationals are countable, hence inside my
intervals.

But we can use the following estimation that should convince everyone:

Use the intervals I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n]. Since n and
q_n can be in bijection, these intervals are sufficient to cover all
q_n. That means by clever reordering them you can cover the whole
positive axis except "boundaries".

And an even more suggestive approximation:
Replace the I(n) by intervals J(n) = [n - 1/10, n + 1/10].
These intervals (without splitting or modifying them) can be reordered,
to cover the whole positive axis except boundaries. Every rational is
the midpoint of an interval.That means the real axis is covered
infinitely often.
Reordering them again in an even cleverer way, they can be used to cover
the whole positive and negative real axes except boundaries. And
reordering them again, they can be used to cover 100 real axes in parallel.

That would be possible if Cantor was right.

Regards, WM
Mikko
2024-11-08 13:09:07 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according to
Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
But not between irrational points.
Real axis contains both real and irrational numbers and nothing else.
Between any two points of the real axis there are both rational and
irrational points.
Post by WM
Post by Mikko
Post by WM
But the important question is also covered by ε = 1. The measure of the
real axis is, according to Cantor's results, less than 3. That shows
that his results are wrong.
No, that is not Cantor's result,
It is Cantor's result that all rationals are countable, hence inside my
intervals.
That is but what you said above is not.
Post by WM
Use the intervals I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n]. Since n
and q_n can be in bijection, these intervals are sufficient to cover
all q_n. That means by clever reordering them you can cover the whole
positive axis except "boundaries".
Depends on the type of n.
Post by WM
Replace the I(n) by intervals J(n) = [n - 1/10, n + 1/10].
Likewise.
--
Mikko
WM
2024-11-08 16:30:23 UTC
Reply
Permalink
Post by Mikko
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
I leave ε = 1. No shrinking. Every point outside of the intervals
is nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according
to Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
But not between irrational points.
Real axis contains both real and irrational numbers and nothing else.
Between any two points of the real axis there are both rational and
irrational points.
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and none is
outside. Therefore also irrational numbers cannot be there. Of course
this is wrong. It proves that not all rational numbers are countable and
in the sequence.
Post by Mikko
Post by WM
It is Cantor's result that all rationals are countable, hence inside
my intervals.
That is but what you said above is not.
Post by WM
Use the intervals I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n]. Since n
and q_n can be in bijection, these intervals are sufficient to cover
all q_n. That means by clever reordering them you can cover the whole
positive axis except "boundaries".
Depends on the type of n.
The n are the natural numbers.
Post by Mikko
Post by WM
Replace the I(n) by intervals J(n) = [n - 1/10, n + 1/10].
Likewise.
Post by WM
J(n) = [n - 1/10, n + 1/10]
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
Only a very hard believer can believe that by shuffling the intervals
they could cover the real line infinitely often.

Regards, WM
Mikko
2024-11-09 14:03:30 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according to
Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
But not between irrational points.
Real axis contains both real and irrational numbers and nothing else.
Between any two points of the real axis there are both rational and
irrational points.
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and none is
outside.
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Post by WM
Therefore also irrational numbers cannot be there.
That is equally obvious.
Post by WM
Of course this is wrong.
You may call it wrong but that's the way they are.
Post by WM
It proves that not all rational numbers are countable and in the sequence.
Calling a truth wrong does not prove anything.
--
Mikko
WM
2024-11-09 21:30:47 UTC
Reply
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Post by Mikko
Post by WM
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... and none
is outside.
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Post by WM
Therefore also irrational numbers cannot be there.
That is equally obvious.
Post by WM
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller
than 3. If no irrationals are outside, then nothing is outside, then the
measure of the real axis is smaller than 3. That is wrong. Therefore
there are irrationals outside. That implies that rational are outside.
That implies that Cantor's above sequence does not contain all rationals.
Post by Mikko
Post by WM
It proves that not all rational numbers are countable and in the sequence.
Calling a truth wrong does not prove anything.
Proving that when Cantor is true the real axis has measure 3 proves that
Cantor is wrong.

Regards, WM
Mikko
2024-11-10 10:20:06 UTC
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Post by Mikko
Post by WM
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... and none is
outside.
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Post by WM
Therefore also irrational numbers cannot be there.
That is equally obvious.
Post by WM
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n. If all n is all reals then
the measure of their union is infinite.
--
Mikko
WM
2024-11-10 10:54:02 UTC
Reply
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Post by Mikko
Post by Mikko
Post by WM
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2,
4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... and
none is outside.
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Post by WM
Therefore also irrational numbers cannot be there.
That is equally obvious.
Post by WM
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
Post by Mikko
If all n is all reals then
the measure of their union is infinite.
But n is all reals as you could have found out yourself, by the measure < 3.

Regards, WM
Mikko
2024-11-11 11:15:56 UTC
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Post by WM
Post by Mikko
Post by Mikko
Post by WM
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1,  ... and none is
outside.
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Post by WM
Therefore also irrational numbers cannot be there.
That is equally obvious.
Post by WM
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
The measure of the interval J(n) is √2/5, which is roghly 0,28.
The measure of the set of all those intervals is infinite.
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
--
Mikko
WM
2024-11-11 11:33:52 UTC
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Post by Mikko
Post by WM
Post by Mikko
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
The measure of the interval J(n) is √2/5, which is roghly 0,28.
Agreed, I said smaller than 3.
Post by Mikko
The measure of the set of all those intervals is infinite.
The density or relative measure is √2/5. By shifting intervals this
density cannot grow. Therefore the intervals cannot cover the real axis,
let alone infinitely often.
Post by Mikko
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
Therefore infinitely many natural numbers must become centres of
intervals, if Cantor was right. But that is impossible.

Regards, WM
Mikko
2024-11-12 13:45:53 UTC
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Post by WM
Post by Mikko
Post by WM
Post by Mikko
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
The measure of the interval J(n) is √2/5, which is roghly 0,28.
Agreed, I said smaller than 3.
Post by Mikko
The measure of the set of all those intervals is infinite.
The density or relative measure is √2/5. By shifting intervals this
density cannot grow. Therefore the intervals cannot cover the real
axis, let alone infinitely often.
Post by Mikko
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
Therefore infinitely many natural numbers must become centres of
intervals, if Cantor was right. But that is impossible.
Where did Cantor say otherwise?
--
Mikko
WM
2024-11-12 13:59:24 UTC
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Post by Mikko
Post by WM
Post by Mikko
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
Therefore infinitely many natural numbers must become centres of
intervals, if Cantor was right. But that is impossible.
Where did Cantor say otherwise?
Cantor said that all rationals are within the sequence and hence within
all intervals. I prove that rationals are in the complement.

Regards, WM
Mikko
2024-11-13 10:39:39 UTC
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Post by WM
Post by Mikko
Post by WM
Post by Mikko
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
Therefore infinitely many natural numbers must become centres of
intervals, if Cantor was right. But that is impossible.
Where did Cantor say otherwise?
Cantor said that all rationals are within the sequence and hence within
all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of them
are in intervals that do not overlap with any of your J(n). You have not
proven that there is a rational that is not in any of Cantor's intervals.
Every rational is at the midpoint of one of Cantor's iterval.
--
Mikko
WM
2024-11-13 16:14:02 UTC
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Post by Mikko
Post by WM
Cantor said that all rationals are within the sequence and hence
within all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of them
are in intervals that do not overlap with any of your J(n).
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure 1/5
of ℝ+. By translating them to match Cantor's intervals they cover ℝ+
infinitely often. This is impossible. Therefore set theorists must
discard geometry.

Further all finitely many translations maintain the original relative
measure. The sequence 1/5, 1/5, 1/5, ... has limit 1/5 according to
analysis. Therefore set theorists must discard analysis.

Regards, WM
Mikko
2024-11-14 09:17:45 UTC
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Post by WM
Post by Mikko
Post by WM
Cantor said that all rationals are within the sequence and hence within
all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of them
are in intervals that do not overlap with any of your J(n).
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often. This is impossible. Therefore set theorists must
discard geometry.
The intervals J(n) are what they are. Translated intervals are not the same
intervals. The properties of the translated set dpend on how you translate.
For example, if you translate them to J'(n) = (n/100 - 1/10, n/100 + 1/10)
then the translated intervals J'(n) wholly cover the postive side of the
real line. Your "impossible" is false.
--
Mikko
WM
2024-11-14 10:34:52 UTC
Reply
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
Cantor said that all rationals are within the sequence and hence
within all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of them
are in intervals that do not overlap with any of your J(n).
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often. This is impossible. Therefore set theorists must
discard geometry.
The intervals J(n) are what they are. Translated intervals are not the same
intervals. The properties of the translated set depend on how you translate.
No. Covering by intervals is completely independent of their
individuality and therefore of their order. Therefore you can either
believe in set theory or in geometry. Both contradict each other.
Post by Mikko
For example, if you translate them to J'(n) = (n/100 - 1/10, n/100 + 1/10)
then the translated intervals J'(n) wholly cover the postive side of the
real line.
By shuffling the same set of intervals which do not cover ℝ+ in the
initial configuration, it is impossible to cover more. That's geometry.

Regards, WM
Mikko
2024-11-15 10:43:05 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
Cantor said that all rationals are within the sequence and hence within
all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of them
are in intervals that do not overlap with any of your J(n).
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often. This is impossible. Therefore set theorists must
discard geometry.
The intervals J(n) are what they are. Translated intervals are not the same
intervals. The properties of the translated set depend on how you translate.
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their
order but also their positions can be different as demonstrated by your
example and mine, too.
Post by WM
Therefore you can either believe in set theory or in geometry. Both
contradict each other.
Geometry cannot contradict set theory because there is nothing both
could say. But this discussion is about set theory so geometry is not
relevant.
Post by WM
Post by Mikko
For example, if you translate them to J'(n) = (n/100 - 1/10, n/100 + 1/10)
then the translated intervals J'(n) wholly cover the postive side of the
real line.
By shuffling the same set of intervals which do not cover ℝ+ in the
initial configuration, it is impossible to cover more. That's geometry.
So what part of ℝ+ is not covered by my J'?
--
Mikko
WM
2024-11-15 12:00:43 UTC
Reply
Permalink
Post by Mikko
Post by WM
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their
order but also their positions can be different as demonstrated by your
example and mine, too.
If the do not cover the whole figure in their initial order, then they
cannot do so in any other order.
Post by Mikko
Post by WM
Therefore you can either believe in set theory or in geometry. Both
contradict each other.
Geometry cannot contradict set theory because there is nothing both
could say. But this discussion is about set theory so geometry is not
relevant.
There is something both could say: Set theory claims that the intervals
are enough to cover every rational number by a midpoint.. i.e., to cover
ℝ+ infinitely often. Geometry denies this.
Post by Mikko
Post by WM
By shuffling the same set of intervals which do not cover ℝ+ in the
initial configuration, it is impossible to cover more. That's geometry.
So what part of ℝ+ is not covered by my J'?
Since according to Cantor's formula the smaller parts of ℝ+ are
frequently covered, in the larger parts much gets uncovered. Every
definable rational is covered. That is called potential infinity.

Regards, WM
Mikko
2024-11-16 09:21:02 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by WM
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their
order but also their positions can be different as demonstrated by your
example and mine, too.
If the do not cover the whole figure in their initial order, then they
cannot do so in any other order.
So you want to retract your claims that involve another order?
--
Mikko
WM
2024-11-16 19:42:22 UTC
Reply
Permalink
Post by Mikko
Post by Mikko
Post by WM
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their
order but also their positions can be different as demonstrated by your
example and mine, too.
If they do not cover the whole figure in their initial order, then they
cannot do so in any other order.
So you want to retract your claims that involve another order?
My claim is the obvious truth that the intervals [n - 1/10, n + 1/10] in
every order do not cover the positive real line, let alone infinitely often.

Regards, WM
Mikko
2024-11-17 08:55:55 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by Mikko
Post by WM
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their
order but also their positions can be different as demonstrated by your
example and mine, too.
If they do not cover the whole figure in their initial order, then they
cannot do so in any other order.
So you want to retract your claims that involve another order?
My claim is the obvious truth that the intervals [n - 1/10, n + 1/10]
in every order do not cover the positive real line, let alone
infinitely often.
Regards, WM
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often. This is impossible. Therefore set theorists must
discard geometry.
There you translate them so that not only their order but also their
positions change. But You also rejected my J' intervals without giving
any reason that does not reject your "translated" intervals.

So, you are arguing against your own words.
--
Mikko
WM
2024-11-17 10:29:31 UTC
Reply
Permalink
Post by WM
Post by WM
Post by Mikko
Post by Mikko
Post by WM
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their
order but also their positions can be different as demonstrated by your
example and mine, too.
If they do not cover the whole figure in their initial order, then
they cannot do so in any other order.
So you want to retract your claims that involve another order?
My claim is the obvious truth that the intervals [n - 1/10, n + 1/10]
in every order do not cover the positive real line, let alone
infinitely often.
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often.
That is the claim of set theory.
Post by WM
This is impossible.
This is my claim.
Post by WM
Post by WM
Therefore set theorists must
discard geometry.
There you translate them so that not only their order but also their
positions change. But You also rejected my J' intervals without giving
any reason that does not reject your "translated" intervals.
Your J'(n) = (n/100 - 1/10, n/100 + 1/10) are 100 times more than mine.
For every reordering of a finite subset of my intervals J(n) the
relative covering remains constant, namely 1/5.
The analytical limit proves that the constant sequence 1/5, 1/5, 1/5,
... has limit 1/5. This is the relative covering of the infinite set and
of every reordering.

Regards, WM
Mikko
2024-11-17 12:28:14 UTC
Reply
Permalink
Post by WM
Your J'(n) = (n/100 - 1/10, n/100 + 1/10) are 100 times more than mine.
For every reordering of a finite subset of my intervals J(n) the
relative covering remains constant, namely 1/5.
The analytical limit proves that the constant sequence 1/5, 1/5, 1/5,
... has limit 1/5. This is the relative covering of the infinite set
and of every reordering.
My J'(n) are your J(n) translated much as your translated J(n) except
that they are not re-ordered.

My J'(n) are as numerous as your J(n): there is one of each for every
natural number n.

Each my J'(n) has the same size as your corresponding J(n): 1/5.

One more similarity is that neither is relevant to the subject.
--
Mikko
WM
2024-11-17 12:46:29 UTC
Reply
Permalink
Post by Mikko
Post by WM
Your J'(n) = (n/100 - 1/10, n/100 + 1/10) are 100 times more than mine.
For every reordering of a finite subset of my intervals J(n) the
relative covering remains constant, namely 1/5.
The analytical limit proves that the constant sequence 1/5, 1/5, 1/5,
... has limit 1/5. This is the relative covering of the infinite set
and of every reordering.
My J'(n) are your J(n) translated much as your translated J(n) except
that they are not re-ordered.
My J'(n) are as numerous as your J(n): there is one of each for every
natural number n.
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are then
exhausted, yours are not.
Post by Mikko
Each my J'(n) has the same size as your corresponding J(n): 1/5.
One more similarity is that neither is relevant to the subject.
Only if you believe in matheology and resist mathematics.
Geometry says that your intervals cover the real line, my do not.
The sequence 1/5, 1/5, 1/5, has what limit?

Regards, WM
Mikko
2024-11-18 09:58:07 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by WM
Your J'(n) = (n/100 - 1/10, n/100 + 1/10) are 100 times more than mine.
For every reordering of a finite subset of my intervals J(n) the
relative covering remains constant, namely 1/5.
The analytical limit proves that the constant sequence 1/5, 1/5, 1/5,
... has limit 1/5. This is the relative covering of the infinite set
and of every reordering.
My J'(n) are your J(n) translated much as your translated J(n) except
that they are not re-ordered.
My J'(n) are as numerous as your J(n): there is one of each for every
natural number n.
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Irrelevant.
Post by WM
Post by Mikko
Each my J'(n) has the same size as your corresponding J(n): 1/5.
One more similarity is that neither is relevant to the subject.
Only if you believe in matheology and resist mathematics.
In mathematics unproven claims do not count.
Post by WM
Geometry says that your intervals cover the real line, my do not.
Geometry is mathematics so unproven claims do not count.
--
Mikko
WM
2024-11-18 14:29:40 UTC
Reply
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Post by Mikko
Post by WM
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Irrelevant.
Very relevant.
Post by Mikko
In mathematics unproven claims do not count.
Geometry is only another language of mathematics. The relative covering
1/5 of my intervals for every finite translation of every finite number
of intervals and the analytical limit of the constant sequence 1/5 is
mathematical proof that Cantor erred.
Don't you know analytical limits?

Regards, WN
Mikko
2024-11-19 09:32:31 UTC
Reply
Permalink
Post by WM
Post by Mikko
Post by WM
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Irrelevant.
Very relevant.
It is not relevant if no relevancy is shown.
Post by WM
Post by Mikko
In mathematics unproven claims do not count.
Geometry is only another language of mathematics.
Therefore unproven claims don't count in geometry.
--
Mikko
WM
2024-11-19 11:04:08 UTC
Reply
Permalink
Post by Mikko
Post by WM
Post by Mikko
Post by WM
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Irrelevant.
Very relevant.
It is not relevant if no relevancy is shown.
But if relevancy is only deleted, it can show up again:

Every finite translation of any finite subset of intervals J(n)
maintains the relative covering 1/5. If the infinite set has the
relative covering 1 (or more), then you claim that the sequence 1/5,
1/5, 1/5, ... has limit 1 (or more).

So you deny analysis or / and geometry.

Regards, WM
Post by Mikko
Post by WM
Post by Mikko
In mathematics unproven claims do not count.
Geometry is only another language of mathematics.
Therefore unproven claims don't count in geometry.
Jim Burns
2024-11-16 19:30:21 UTC
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Post by WM
Post by Mikko
Translated intervals are not
the same as the original ones.
Not only their order
but also their positions can be different
as demonstrated by your example and mine, too.
If the do not cover the whole figure
in their initial order,
then they cannot do so in any other order.
Sets for which that is true
are finite sets.

Insisting on "for all sets"
does not change infinite sets into finite sets.

Insisting on "for all sets"
changes what.you.are.talking.about.
You will no longer be talking about infinite sets.
Post by WM
Since according to Cantor's formula
the smaller parts of ℝ+ are frequently covered,
in the larger parts much gets uncovered.
ℝ⁺ holds points.between.splits of ℚ⁺
ℚ⁺ holds ratios of numbers in ℕ⁺
ℕ⁺ holds numbers countable.to from.1

ℝ⁺ ℚ⁺ ℕ⁺ are infinite sets.

You (WM) are both talking.about and not.talking.about
infinite sets.
(Multi.tasking, I suppose.)
Post by WM
Every definable rational is covered.
countable.to from.1 ⟨i,j⟩ ↦ kᵢⱼ countable.to from.1
kᵢⱼ = (i+j-1)⋅(i+j-2)/2+i

For each countable.to from.1 k
⟨iₖ,jₖ⟩ is a pair of countable.to from.1

countable.to from.1 k ↦ ⟨iₖ,jₖ⟩ countable.to from.1
(iₖ+jₖ) = ⌈(2⋅k+¼)¹ᐟ²+½⌉
iₖ = k-((iₖ+jₖ)-1)⋅((iₖ+jₖ)-2)/2
jₖ := (iₖ+jₖ)-iₖ

(iₖ+jₖ-1)⋅(iₖ+jₖ-2)/2+iₖ = k
Post by WM
That is called potential infinity.
⎛ A finite set A can be ordered so that,
⎜ for each subset B of A,
⎜ either B holds first.in.B and last.in.B
⎜ or B is empty.
⎜
⎜ An infinite set is not finite.
⎜ An infinite set C can _only_ be ordered so that
⎜ there is a non.empty subset D of C, such that
⎜ either first.in.D or last.in.D or both don't exist,
⎜ not visibly and not darkly.
⎜
⎝ And our sets do not change.
WM
2024-11-16 19:54:41 UTC
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Post by Jim Burns
Post by Mikko
Translated intervals are not
the same as the original ones.
Not only their order
but also their positions can be different
as demonstrated by your example and mine, too.
If they do not cover the whole figure
 in their initial order,
then they cannot do so in any other order.
Sets for which that is true
are finite sets.
Sets for which that is true are sets which obey geometrical rules.
Post by Jim Burns
⎝ And our sets do not change.
Therefore the set of intervals cannot grow. The average density is and
remains 1/5, i.e., less than 1.

Regards, WM

Regards, WM
Jim Burns
2024-11-16 22:36:06 UTC
Reply
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Post by WM
Post by Jim Burns
Post by Mikko
Translated intervals are not
the same as the original ones.
Not only their order
but also their positions can be different
as demonstrated by your example and mine, too.
If they do not cover the whole figure
 in their initial order,
then they cannot do so in any other order.
Sets for which that is true
are finite sets.
Sets for which that is true are
sets which obey geometrical rules.
'Geometry' joins 'mathematics' and 'logic' among
words you (WM) invoke _in place of_ an argument.

Do you (WM) know whether, in your geometry,
triangles with equal angles (AKA, similar)
have corresponding sides in the same ratio?

If your geometry is like our geometry,
then they do,
and
there is a procedure in your geometry
(with similar triangles) which,
any one integral point determines
two integral points,
and
another procedure in your geometry
(with similar triangles) which,
from any two integral points determines
one integral point:
and
they are inverses of each other:
k ↦ ⟨i,j⟩ ↦ k
Post by WM
Post by Jim Burns
⎝ And our sets do not change.
Therefore
the set of intervals cannot grow.
An infinite set can match a proper superset
without growing.
Because it is infinite.
WM
2024-11-17 07:50:07 UTC
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Post by Jim Burns
Post by WM
Therefore
the set of intervals cannot grow.
An infinite set can match a proper superset
without growing.
But with shrinking. When it matches first itself and then a proper
subset, then it has decreased. The set of even numbers has fewer
elements than the set of integers.
Post by Jim Burns
Because it is infinite.
The interval [0, 1] is infinite because it can be split into infinitely
many subsets. But its measure remains constant. There is no reason
except naivety to believe that the intervals [n - 1/10, n + 1/10] could
cover the real line infinitely often.

Regards, WM
Jim Burns
2024-11-18 19:22:50 UTC
Reply
Permalink
Post by WM
Post by Jim Burns
Post by WM
Therefore
the set of intervals cannot grow.
An infinite set can match a proper superset
without growing.
But with shrinking.
No.
An infinite set
can match some proper supersets without growing and
can match some proper subsets without shrinking.

Sets which can't aren't infinite.
Post by WM
When it matches
first itself and then a proper subset,
then it has decreased.
There is no even number which,
after the evens matching the integers,
will not be an even number.

There is no even number which,
before the evens matching the integers,
was not an even number.

Even numbers do not change their evenicity.
Our sets do not change.
Post by WM
The set of even numbers has fewer elements
than the set of integers.
The set of even numbers is
a proper subset of the set of integers,
AND
the set of even numbers can match
the set of integers without either set changing.

Because it is infinite.

A _finite_ set can be ordered so that
each non.empty subset holds a first and a last.

An infinite set is _not.that_

However an infinite set is ordered,
some of its non.empty subsets
don't have a first or don't have a last,
not visibly and not darkly.
Post by WM
Post by Jim Burns
Because it is infinite.
The interval [0, 1] is infinite because
it can be split into infinitely many subsets.
But its measure remains constant.
There is no reason except naivety
to believe that the intervals [n - 1/10,  n + 1/10]
could cover the real line infinitely often.
There is no reason except
naivete and an almost fanatical devotion to the Pope.


No, wait! There is no reason except
naivete, an almost fanatical devotion to the Pope, and
⎛ k ↦ ⟨i,j⟩ ↦ k
⎜
⎜ (i+j) := ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ i := k-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = k
WM
2024-11-19 11:01:36 UTC
Reply
Permalink
Post by Jim Burns
The set of even numbers is
a proper subset of the set of integers,
AND
the set of even numbers can match
the set of integers without either set changing.
That implies that our well-known intervals can cover the real line or
reduce the average covering to 1/1000000000.

But every finite translation of any finite subset of intervals maintains
the relative covering 1/5. If the infinite set has the relative covering
1 (or more or less), then you claim that the sequence 1/5, 1/5, 1/5, ...
has limit 1 (or more or less).

So you deny analysis or / and geometry.

Regards, WM
Jim Burns
2024-11-19 16:26:43 UTC
Reply
Permalink
Post by WM
Post by Jim Burns
An infinite set
can match some proper supersets without growing and
can match some proper subsets without shrinking.
Sets which can't aren't infinite.
The set of even numbers is
a proper subset of the set of integers,
AND
the set of even numbers can match
the set of integers without either set changing.
That implies that
our well-known intervals
Sets with different intervals are different.
Our sets do not change.
Post by WM
our well-known intervals can
cover the real line or
reduce the average covering to 1/1000000000.
Sets of our well.known.intervals
can match some proper supersets without growing and
can match some proper subsets without shrinking.
So they're infinite.

Thank you for clearing that up.
So, it's time to move on, right?
Post by WM
But every finite translation of
any finite subset of intervals maintains
the relative covering 1/5.
Each finite subset is finite.
Some subsets of infinite sets aren't finite.
Post by WM
If the infinite set has
the relative covering 1 (or more or less),
then you claim that
the sequence 1/5, 1/5, 1/5, ... has
limit 1 (or more or less).
Relative covering isn't measure.
Each of those interval.unions has measure +∞

You haven't defined 'relative covering'.
Giving examples isn't a definition.
You (WM) don't know what you (WM) mean.

However,
let's keep going.

I claim that there are functions f:ℝ→ℝ
such that
⟨ f(⅟1) f(⅟2) f(⅟3) ... ⟩ =
⟨ ⅟5 ⅟5 ⅟5 ... ⟩
and f(0) = 1

lim.⟨ f(⅟1) f(⅟2) f(⅟3) ... ⟩ ≠
f(lim.⟨ ⅟1 ⅟2 ⅟3 ... ⟩)

f is discontinuous at 0
Yes, I claim there are functions like f
Post by WM
So you deny analysis or / and geometry.
I deny what you think analysis and geometry are.
I accept infinite sets
and discontinuous functions
and similar triangles in proportion.

What is it you (WM) accuse infinite sets of,
other than not being finite?

Note:
An infinite set
can match some proper supersets without growing and
can match some proper subsets without shrinking.

Sets which can't aren't infinite.
WM
2024-11-20 11:42:15 UTC
Reply
Permalink
Post by Jim Burns
Post by WM
That implies that
our well-known intervals
Sets with different intervals are different.
Our sets do not change.
The intervals before and after shifting are not different. Only their
positions are.

Is the set {1} different from the set {1} because they have different
positions? Is the set {1} in 1, 2, 3, ... different from the set {1} in
-oo, ..., -1, 0, 1,... oo?
Post by Jim Burns
Sets of our well.known.intervals
can match some proper supersets without growing
They cannot match the rational numbers without covering the whole
positive real line. That means the relative covering has increased from
1/5 to 1.
Post by Jim Burns
Relative covering isn't measure.
It is a measure! For every finite interval between natural numbers n and
m the covered part is 1/5.
Post by Jim Burns
You haven't defined 'relative covering'.
Giving examples isn't a definition.
If you are really too stupid to understand relative covering for finite
intervals, then I will help you. But I can't believe that it is
worthwhile. Your only reason of not knowing it is to defend set theory
which has been destroyed by my argument.
Post by Jim Burns
I claim that there are functions f:ℝ→ℝ
such that
⟨ f(⅟1) f(⅟2) f(⅟3) ... ⟩  =
⟨ ⅟5    ⅟5    ⅟5    ... ⟩
and  f(0)  =  1
Not in case of geometric shifting. All definable intervals fail in all
definable positions.
Post by Jim Burns
Post by WM
So you deny analysis or / and geometry.
I deny what you think analysis and geometry are.
I accept infinite sets
and discontinuous functions
Discontinuity is not acceptable in the geometry of shifting intervals.
Post by Jim Burns
What is it you (WM) accuse infinite sets of,
other than not being finite?
Nothing against infinite sets. I accuse matheologians to try to deceive.
Post by Jim Burns
An infinite set
can match some proper supersets without growing
I have proven that this is nonsense.

Regards, WM
Mikko
2024-11-21 09:16:33 UTC
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Post by WM
Post by Jim Burns
Post by WM
That implies that
our well-known intervals
Sets with different intervals are different.
Our sets do not change.
The intervals before and after shifting are not different. Only their
positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
--
Mikko
WM
2024-11-21 10:21:40 UTC
Reply
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Post by Mikko
Post by WM
The intervals before and after shifting are not different. Only their
positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
Consider this simplified argument. Let every unit interval after a
natural number n which is divisible by 10 be coloured black: (10n,
10n+1]. All others are white. Is it possible to shift the black
intervals so that the whole real axis becomes black?

No. Although there are infinitely many black intervals, the white
intervals will remain in the majority. For every finite distance (0,
10n) the relative covering is precisely 1/10, whether or not the
intervals have been moved or remain at their original sites. That means
the function decribing this, 1/10, 1/10, 1/10, ... has limit 1/10. That
is the quotient of the infinity of black intervals and the infinity of
all intervals.

Regards, WM
Mikko
2024-11-21 10:59:04 UTC
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Post by WM
Post by Mikko
Post by WM
The intervals before and after shifting are not different. Only their
positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
Consider this simplified argument. Let every unit interval after a
natural number n which is divisible by 10 be coloured black: (10n,
10n+1]. All others are white. Is it possible to shift the black
intervals so that the whole real axis becomes black?
Yes. Shift the interval (10n, 10n+1) to (n/2, n/2+1).
--
Mikko
WM
2024-11-21 11:03:28 UTC
Reply
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
The intervals before and after shifting are not different. Only
their positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
Consider this simplified argument. Let every unit interval after a
natural number n which is divisible by 10 be coloured black: (10n,
10n+1]. All others are white. Is it possible to shift the black
intervals so that the whole real axis becomes black?
Yes. Shift the interval (10n, 10n+1) to (n/2, n/2+1).
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.

Regards, WM
Mikko
2024-11-22 08:42:09 UTC
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Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
The intervals before and after shifting are not different. Only their
positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
Consider this simplified argument. Let every unit interval after a
natural number n which is divisible by 10 be coloured black: (10n,
10n+1]. All others are white. Is it possible to shift the black
intervals so that the whole real axis becomes black?
Yes. Shift the interval (10n, 10n+1) to (n/2, n/2+1).
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.
That is irrelevant to your question whether the whole interval becomes
black if the shifted intervals (n/2, n/2+1) are painted black.

And that question is irrelevant to the topic specified on the subject line.
--
Mikko
WM
2024-11-22 10:53:32 UTC
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Post by Mikko
Post by WM
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.
That is irrelevant to your question whether the whole interval becomes
black if the shifted intervals (n/2, n/2+1) are painted black.
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the
additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has
left, the white must remain such that the whole real axis can never
become black.

These facts prevent the Cantor-bijection for different sets of natural
numbers.
Post by Mikko
And that question is irrelevant to the topic specified on the subject line.
If different sets of natural numbers already cannot be in bijection,
then the rationals are also excluded.

Regards, WM
Mikko
2024-11-23 08:07:51 UTC
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Post by WM
Post by Mikko
Post by WM
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.
That is irrelevant to your question whether the whole interval becomes
black if the shifted intervals (n/2, n/2+1) are painted black.
1) The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the
additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has
left, the white must remain such that the whole real axis can never
become black.
You say that it is relevant but you don't show how that is relevant
to the fact that there is no real number between the intervals (n/2, n/2+1)
that is not a part of at least one of those intervals.

And you have not shown how the fact that there is no real number between
the intervals (n/2, n/2+1) is relevant to anything mentioned on the
subject line.
--
Mikko
WM
2024-11-23 08:49:18 UTC
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.
That is irrelevant to your question whether the whole interval becomes
black if the shifted intervals (n/2, n/2+1) are painted black.
1) The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the
additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has
left, the white must remain such that the whole real axis can never
become black.
You say that it is relevant but you don't show how that is relevant
to the fact that there is no real number between the intervals (n/2, n/2+1)
that is not a part of at least one of those intervals.
Because that has nothing to do with the topic under discussion. See
points 1, 2, and 3. They are to be discussed.

Regards, WM
Jim Burns
2024-11-21 15:39:36 UTC
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Post by WM
Post by Mikko
Post by WM
Post by Jim Burns
Post by WM
That implies that
our well-known intervals
Sets with different intervals are different.
Our sets do not change.
The intervals before and after shifting are not different.
Only their positions are.
Intervals with different points are different.
Our sets do not change.
Post by WM
Post by Mikko
The intervals are different.
A shifted interval contains a different set of numbers.
Post by WM
The intervals before and after shifting are not different.
Only their positions are.
The intervals are different.
A shifted interval contains a different set of numbers.
Consider this simplified argument.
Let every unit interval after
a natural number n which is divisible by 10
be coloured black: (10n, 10n+1].
All others are white.
Is it possible to shift the black intervals
No,
it is not possible to shift the black intervals.
Because our sets do not change.

Is it possible to match the black intervals
to the proper superset of all the intervals?

Yes.
10⋅n ↦ n ↦ 10⋅n
The two sets, of black and of all, are infinite.
Post by WM
Is it possible to shift the black intervals
so that the whole real axis becomes black?
No.
Although there are infinitely many black intervals,
the white intervals will remain in the majority.
Here, you have used 'majority' in a way which
includes 10⋅n ↦ n ↦ 10⋅n
That makes your answer irrelevant to the question.
Post by WM
For every finite distance (0, 10n)
Each, unlike the whole, finite.
Post by WM
 the relative covering is precisely 1/10,
The measure of black and the measure of all are
unbounded by any countable.to number, thus are +∞

+∞/+∞ is undefined.
Post by WM
whether or not the intervals have been moved or
remain at their original sites.
That means the function decribing this,
1/10, 1/10, 1/10, ...
has limit 1/10.
That is the quotient of
the infinity of black intervals and
the infinity of all intervals.
The Paradox of the Discontinuous Function
(not a paradox):

lim.⟨ rc(1), rc(2), rc(3), ... ⟩ ≠
rc( lim.⟨ 1, 2, 3, ... ⟩ )

You (WM) do not "believe in"
proper.superset.matching sets
discontinuous functions
similar triangles in proportion

And you (WM) feel that
those feelings should be enough.
WM
2024-11-21 16:24:51 UTC
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Post by Jim Burns
That means the function describing this,
1/10, 1/10, 1/10, ...
has limit 1/10.
That is the quotient of
the infinity of black intervals and
the infinity of all intervals.
The Paradox of the Discontinuous Function
It is a paradox that only 1/10 of the real line is covered for every
finite interval (0, n] but all is covered completely in the limit. By
what is it covered, after all n have been proved unable?
Post by Jim Burns
lim.⟨ rc(1), rc(2), rc(3), ... ⟩  ≠
rc( lim.⟨ 1, 2, 3, ... ⟩ )
You (WM) do not "believe in"
proper.superset.matching sets
discontinuous functions
There is no reason to believe in magic. But if you do, then all
Cantor-bijections can fail as well "in the infinite". Then mathematics
is insufficient to determine limits.

Regards, WM
Jim Burns
2024-11-21 18:54:57 UTC
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Post by WM
Post by Jim Burns
That means the function describing this,
1/10, 1/10, 1/10, ...
has limit 1/10.
That is the quotient of
the infinity of black intervals and
the infinity of all intervals.
The Paradox of the Discontinuous Function
It is a paradox
that only 1/10 of the real line is covered
for every finite interval (0, n]
but all is covered completely in the limit.
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
Post by WM
Post by Jim Burns
lim.⟨ rc(1), rc(2), rc(3), ... ⟩  ≠
rc( lim.⟨ 1, 2, 3, ... ⟩ )
You (WM) do not "believe in"
proper.superset.matching sets
discontinuous functions
There is no reason to believe in magic.
⎛ Arthur C Clarke's Third Law
⎜
⎜ Any sufficiently advanced technology
⎝ is indistinguishable from magic.

What is sufficiently advanced for you (WM)?
Arithmetic.
Post by WM
But if you do, then
all Cantor-bijections can fail as well
"in the infinite".
Then mathematics is insufficient
to determine limits.
I am not enough of a scholar to know
that this is true of _all_ mathematics, but
I know that much knowledge of infinity,
including what I'm most familiar with,
is grounded in the _finite_

Here, I DON'T refer to finite numbers, etc.
I refer to finite sequences of CLAIMS,
each of which is true.or.not.first.false.

Because the sequence of CLAIMS is finite,
it is well.ordered.in.both.directions.

Because its subset of false CLAIMS
cannot hold a first false claim,
then, by well.order, that subset is empty.

In a sequence with no false claim,
each claim is true.
Even if a claim refers to
an indefinite one of infinitely many,
that claim is true for
that indefinite one of infinitely.many.

NOT because magically
we can check infinitely.many numbers
Because we can see FINITELY.many claims
and, for some sequences,
SEE that they are each true.or.not.false.
WM
2024-11-21 19:21:06 UTC
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Post by Jim Burns
Post by WM
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
That is not an answer. Further it is only valid for the first numbers
which are followed by almost all numbers. Never completed.
Post by Jim Burns
Post by WM
There is no reason to believe in magic.
But if you do, then
all Cantor-bijections can fail as well
"in the infinite".
Then mathematics is insufficient
to determine limits.
I am not enough of a scholar to know
that this is true of _all_ mathematics, but
I know that much knowledge of infinity,
including what I'm most familiar with,
is grounded in the _finite_
Either limits can be calculated from the finite, or not. If not, then
Cantor's attempts are in vain from the scratch. If yes, then Cantor's
attempts have been contradicted.
Post by Jim Burns
Here, I DON'T refer to finite numbers, etc.
I refer to finite sequences of CLAIMS,
each of which is true.or.not.first.false.
That is the sequence of claims that limits can be calculated from the
finite and never the real axis is coloured black.

Regards, WM
Jim Burns
2024-11-21 21:46:17 UTC
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Post by WM
Post by Jim Burns
Post by WM
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
That is not an answer.
You (WM) see it as "indistinguishable from magic".
That's a shame for your students.

⎛ Arthur C Clarke's Third Law
⎜
⎜ Any sufficiently advanced technology
⎝ is indistinguishable from magic.
Post by WM
Further it is only valid for
the first numbers which are followed by
almost all numbers.
ℝ⁺ points between splits of ℚ⁺
ℚ⁺ ratios of numbers in ℕ⁺
ℕ⁺ countable.to from.1

n ↦ ⟨i,j⟩ ↦ n
is valid for all of ℕ⁺ and all of ℕ⁺×ℕ⁺
ℕ⁺×ℕ⁺ → onto ℚ⁺
Post by WM
Never completed.
The _description_ is completed.
It's right there.

Finite sequences of claims, each of claim of which
is true.or.not.first.false
are completed.
WM
2024-11-21 21:57:52 UTC
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Post by Jim Burns
Post by WM
Post by Jim Burns
Post by WM
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
That is not an answer.
You (WM) see it as "indistinguishable from magic".
That's a shame for your students.
To believe that the intervals can colour the real axis black is a shame
for all members of humankind who do so.
Post by Jim Burns
The _description_ is completed.
It's right there.
The description of the set not of all its elements.

Either limits can be calculated from the finite, or not. If not, then
Cantor's attempts are in vain from the scratch. If yes, then Cantor's
attempts have been contradicted.

Regards, WM
Jim Burns
2024-11-22 18:51:06 UTC
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Permalink
Post by WM
Post by Jim Burns
Post by WM
Post by Jim Burns
Post by WM
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
That is not an answer.
Further it is only valid for
the first numbers which are followed by
almost all numbers.
Never completed.
The _description_ is completed.
It's right there.
The description of the set
not of all its elements.
The description is sufficient in order to
finitely.investigate infinitely.many.

Therefore, no,
the description is an answer.

----
We have spent a lot of pixels discussing FISONs,
finite initial segments of naturals.

However,
here I consider FISOCs,
finite initial segments of claims.

FISOCs share a useful property with FISONs,
they are well.ordered.
If any claim has a property,
then some claim has that property first.
If any claim is written in Comic Sans,
then some claim is in Comic Sans first.
If any claim is false,
then some claim is false first.

Consider a specific FISOC with
a description of what.we.are.considering, broadly.
⎛⎛ ℕ⁺ holds numbers countable.to from.1
⎜⎜ ℚ⁺ holds ratios of numbers in ℕ⁺
⎜⎝ ℝ⁺ holds points between splits of ℚ⁺
⎜ Further claims about elements of ℕ⁺ ℚ⁺ ℝ⁺
⎝ which are each true.or.not.first.false

Broadly speaking,
claims can be true and can be false.

Broadly speaking,
the initial ℕ⁺.ℚ⁺.ℝ⁺ claims can be false
about some three sets or other.

In those broader after.false instances,
the following not.first.false claims
are not.first.false
whether they are true or they are false.

In the broader after.false instances,
the following not.first.false claims
are NOT an answer.

However,
more narrowly,
for what.we.are.considering,
the initial ℕ⁺.ℚ⁺.ℝ⁺ claims are true.

More narrowly,
for what.we.are.considering,
no claim in that sequence of
true.or.not.first.false claims
is first.false,
so,
by the finiteness of the sequence,
no claim in that sequence of
true.or.not.first.false claims
is false.

More narrowly,
for what.we.are.considering,
the following not.first.false claims
ARE an answer.
WM
2024-11-22 21:30:02 UTC
Reply
Permalink
Post by Jim Burns
Post by WM
Post by Jim Burns
The _description_ is completed.
It's right there.
The description of the set
not of all its elements.
The description is sufficient in order to
finitely.investigate infinitely.many.
For instance, not all indices of the endsegments can be counted to.
Remember: The intersection of all endsegments is empty, but the
intersection of endsegments which can be counted to is infinite.
Note that every endsegment loses only one number. Therefore there must
exist infinitely many finite endsegments.
Post by Jim Burns
----
We have spent a lot of pixels discussing FISONs,
finite initial segments of naturals.
However,
here I consider FISOCs,
finite initial segments of claims.
FISOCs share a useful property with FISONs,
 they are well.ordered.
If any claim has a property,
 then some claim has that property first.
If any claim is written in Comic Sans,
 then some claim is in Comic Sans first.
If any claim is false,
 then some claim is false first.
Consider a specific FISOC with
a description of what.we.are.considering, broadly.
⎛⎛ ℕ⁺ holds numbers countable.to from.1
⎜⎜ ℚ⁺ holds ratios of numbers in ℕ⁺
⎜⎝ ℝ⁺ holds points between splits of ℚ⁺
⎜ Further claims about elements of ℕ⁺ ℚ⁺ ℝ⁺
⎝ which are each true.or.not.first.false
Broadly speaking,
claims can be true and can be false.
Broadly speaking,
the initial ℕ⁺.ℚ⁺.ℝ⁺ claims can be false
about some three sets or other.
In those broader after.false instances,
the following not.first.false claims
are not.first.false
whether they are true or they are false.
In the broader after.false instances,
the following not.first.false claims
are NOT an answer.
However,
more narrowly,
for what.we.are.considering,
the initial ℕ⁺.ℚ⁺.ℝ⁺ claims are true.
They are true for a potential infinity. Consider the claim: Every
endsegment is infinite. This claim is true for the potentially infinite
sequence of infinite endsegments. It is not true for finite endsegments.
But without finite endsegments there is no empty intersection of
endsegments possible.

Regards, WM
Jim Burns
2024-11-22 22:56:28 UTC
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Post by WM
Post by WM
Post by WM
Post by Jim Burns
The _description_ is completed.
It's right there.
The description of the set
not of all its elements.
The description
⎛ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎝ (i+j-1)⋅(i+j-2)/2+i = n
Post by WM
Post by WM
is sufficient in order to
finitely.investigate infinitely.many.
For instance,
not all indices of the endsegments
can be counted to.
You aren't referring to
the set ℕ⁺ of countable.to from.1

For n,i,j countable.to from.1
n ↦ ⟨i,j⟩ ↦ n: one.to.one

ℕ⁺ covers ℕ⁺×ℕ⁺ and ℕ⁺×ℕ⁺ covers ℕ⁺
ℕ⁺ and ℕ⁺×ℕ⁺ are infinite sets.
Post by WM
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.
No one should "remember" that.
It is incorrect.
Post by WM
Note that every endsegment loses only one number.
Finite cardinalities can lose one number.

After all finite cardinalites,
there are the infinite cardinalities,
which aren't finite,
and can't lose one number.
Post by WM
Therefore there must
exist infinitely many finite endsegments.
In ℕ⁺
each foresegment is finite.

If any endsegment is finite,
the union of endsegment and its foresegment is finite.

The union of any endsegment and its foresegment
is ℕ⁺

ℕ⁺ isn't finite,
and no endsegment is finite.
WM
2024-11-23 08:54:36 UTC
Reply
Permalink
Post by Jim Burns
Post by WM
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.
No one should "remember" that.
It is incorrect.
∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k), |E(k)| = ℵ₀
∩{E(1), E(2), ...} = { }.
Post by Jim Burns
Post by WM
Note that every endsegment loses only one number.
Finite cardinalities can lose one number.
For all endsegments:
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1
Post by Jim Burns
Post by WM
Therefore there must
exist infinitely many finite endsegments.
Regards, WM
Jim Burns
2024-11-23 15:33:46 UTC
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Permalink
Post by WM
Post by Jim Burns
Post by WM
[...]
[...]
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
|E(k+1)| = |E(k)| - 1
You (WM) mostly don't disagree with yourself
in the same post.

Perhaps what you intend to say is
⎛ For all post.definable end.segments
⎜ ∀k ∈ (ℕ\ℕ_def):
⎝ |E(k+1)| = |E(k)| - 1

Perhaps what you intend to say is
⎛ There are post.definable end.segments
⎝ which have finite cardinalities.

I nearly agreed with that,
because our ℕ = ℕ_def
so that says nothing,
we have no such k

However,
your mention of 'E(k+1)' implies
⎛ ∀k ∈ (ℕ\ℕ_def):
⎝ (ℕ\ℕ_def) ∋ k+1

For _your_ end.segments,
k ↦ k+1 : one.to.one
E(k) → E(k+1) : one.to.one
|E(k)| ≤ |E(k+1)|

Also,
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|

|E(k)| = |E(k+1)|

Thus,
actually,
even your post.definable end.segments have
cardinalities which don't change by 1
and thus are infinite cardinalities.
Post by WM
Post by Jim Burns
Post by WM
Therefore there must
exist infinitely many finite endsegments.
Your rabbit.out.of.the.hat:
more end segments than numbers in them.

----
Post by WM
Post by Jim Burns
Post by WM
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.
No one should "remember" that.
It is incorrect.
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
∩{E(1), E(2), ...} = { }.
⎛ ℕ_def is
⎜ ℕ_def ∋ 0 ∧ ∀k ∈ ℕ_def: ℕ_def ∋ k+1
⎜ ℕ_def ⊆ S ⇐ S ∋ 0 ∧ ∀k ∈ S: S ∋ k+1
⎜
⎜ ∀k ∈ ℕ_def:
⎜ |ℕ_def\E(k)| < |ℕ_def\E(k+1)| < |ℕ_def| = ℵ₀
⎜
⎝ Each end.segment of ℕ_def is countable.to.

⋂{E(k):k∈ℕ_def} = {}

because
∀j ∈ ℕ_def:
j ∉ E(j+1) ∈ {E(k):k∈ℕ_def}
j ∉ ⋂{E(k):k∈ℕ_def}

The end.segment.intersection is empty because
each end.segment of ℕ_def is countable.past.
WM
2024-11-23 17:23:45 UTC
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Post by WM
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
|E(k+1)| = |E(k)| - 1
Perhaps what you intend to say  is
precisely what I wrote.
⎛ For all post.definable end.segments
⎝ |E(k+1)| = |E(k)| - 1
|E(k+1)| = |E(k)| - 1 is true for _all_ endsegments.
Cardinality ℵo of infinite endsegements cannot describe this because
ℵo - 1 = ℵo.
Perhaps what you intend to say  is
⎛ There are post.definable end.segments
⎝ which have finite cardinalities.
Otherwise the intersection is infinite.
I nearly agreed with that,
because our ℕ = ℕ_def
The intersection of all definable endsegements is infinite.
The intersection of all endsegments is empty.
However,
your mention of 'E(k+1)' implies
⎝ (ℕ\ℕ_def) ∋ k+1
For _your_ end.segments,
k ↦ k+1 : one.to.one
E(k) → E(k+1) : one.to.one
|E(k)| ≤ |E(k+1)|
No. |E(k)| ≥ |E(k+1)| = |E(k)| - 1.
Also,
Only!
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
True.
|E(k)| = |E(k+1)|
Wrong. True only when cardinality is used because
ℵo - 1 = ℵo.
Post by WM
Post by Jim Burns
Post by WM
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.
No one should "remember" that.
It is incorrect.
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
∩{E(1), E(2), ...} = { }.
⎜ ℕ_def ∋ 0
No.

∀k ∈ ℕ_def: ℕ_def ∋ k+1

Yes.
⎜ ℕ_def ⊆ S  ⇐  S ∋ 0 ∧ ∀k ∈ S: S ∋ k+1
There is no S.
⎝ Each end.segment of ℕ_def is countable.to.
Yes.
⋂{E(k):k∈ℕ_def} = {}
No.
because
j ∉ E(j+1) ∈ {E(k):k∈ℕ_def}
j ∉ ⋂{E(k):k∈ℕ_def}
The end.segment.intersection is empty because
No, the intersection contains all dark numbers ℕ\ℕ_def.
each end.segment of ℕ_def is countable.past.
What is countable.past?

Regards, WM
Jim Burns
2024-11-23 21:20:25 UTC
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Post by WM
Post by Jim Burns
For _your_ end.segments,
k ↦ k+1 : one.to.one
E(k) → E(k+1) : one.to.one
|E(k)| ≤ |E(k+1)|
No.
|E(k)| ≥ |E(k+1)| = |E(k)| - 1.
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
doesn't contradict
|E(k)| ≤ |E(k+1)|

Together,
|E(k)| = |E(k+1)|
and
|E(k+1)| doesn't lose one number.

Finite cardinalities can lose one number.

After all the finite cardinalities,
there are the infinite cardinalities,
which aren't finite
and can't lose one number.

|E(k)| = |E(k+1)| is infinite.

----
Do you (WM) object to
k ↦ k+1 : one.to.one

Do you (WM) object to
E(k) → E(k+1) : one.to.one

If you object, why?

If you don't object,
one consequence is
|E(k)| ≤ |E(k+1)|
and
|E(k)| = |E(k+1)|
and
|E(k)| = |E(k+1)| is infinite.
WM
2024-11-23 21:39:11 UTC
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Post by Jim Burns
Post by WM
|E(k)| ≥ |E(k+1)| = |E(k)| - 1.
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
 doesn't contradict
|E(k)| ≤ |E(k+1)|
It does.
Post by Jim Burns
Together,
|E(k)| = |E(k+1)|
and
|E(k+1)| doesn't lose one number.
Spare your nonsense.
Post by Jim Burns
|E(k)| = |E(k+1)| is infinite.
The cardinality is an unsharp measure.
Post by Jim Burns
----
Do you (WM) object to
 k ↦ k+1 : one.to.one
I don't know what that waffle should mean.

Regards, WM

WM
2024-11-11 11:44:51 UTC
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Post by Mikko
The measure of the set of all those intervals is infinite.
The density or relative measure over the complete real axis is √2/5. By
shifting intervals this density cannot grow. Therefore the intervals
cannot cover the real axis, let alone infinitely often.
Post by Mikko
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
Therefore infinitely many natural numbers must become centres of
intervals, if Cantor was right. But that is impossible.

Regards, WM
WM
2024-11-10 10:56:26 UTC
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Post by Mikko
Post by Mikko
Post by WM
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2,
4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and
none is outside.
Post by Mikko
Post by Mikko
All positive rationals quite obviously are in the sequence.
Non-positive
Post by Mikko
Post by Mikko
rationals are not.
Post by WM
Therefore also irrational numbers cannot be there.
That is equally obvious.
Post by WM
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
Post by Mikko
If all n is all reals then
the measure of their union is infinite.
But n is all naturals as you could have found out yourself, by the
measure < 3.

Regards, WM
Mikko
2024-11-11 11:17:52 UTC
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2,
4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and
none is outside.
Post by Mikko
Post by WM
Post by Mikko
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Post by WM
Therefore also irrational numbers cannot be there.
That is equally obvious.
Post by WM
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is
smaller than 3.
Post by Mikko
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
Post by Mikko
If all n is all reals then
the measure of their union is infinite.
But n is all naturals as you could have found out yourself, by the measure < 3.
That is not something one can find out. The symbol n means whatever you
say it means, and in this case you didn't say.
--
Mikko
Ross Finlayson
2024-11-07 22:31:46 UTC
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Post by WM
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according to
Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
Post by WM
But the important question is also covered by ε = 1. The measure of
the real axis is, according to Cantor's results, less than 3. That
shows that his results are wrong.
No, that is not Cantor's result, so all we can say about it is that
you are wrong about Cantor's result.
How about results _after_ Cantor's result in the extra-ordinary?
Ross Finlayson
2024-11-06 21:48:59 UTC
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Post by WM
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Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest
intervals
No, it hasn't.
In geometry it has.
This discussion is about numbers, not geometry.
Geometry is only another language for the same thing.
Another language is an unnecessary complication that only reeasls
an intent to deceive.
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
There is no nearesst one. There is always a nearer one.
Nonsense.
No, the meaning is clear. Of course, because some intevals overlap,
you should have specified what exacly you mean by "nearer". But as
ε shriks the overlappings disappear and the distance between any
two intevals approaches the distance between their centers we may
define distance between the intervals as the distance between their
endpoints even wne ε > 0.
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.
It is not unfounded.
Of course it is. It is the purest nonsense.
That you don't even try to support your clam to support your claim
indicates that you don't really believe it. Cantor's results are
conclusions of proofs and you have not shown any error in the proofs.
You are free to deny one of more of the assumptions that constitue
the foudations of the results but you havn't. Even if you will that
will not make the results unfounded. It only means that you want to
use a different foundation. Whether you can find one that you like
is your problem.
Here what's considered an "opinion" of ZF is any axiom,
of the theory, what results "restriction of comprehension",
for example the Axiom of Regularity, or, the Axiom of (Regular)
Infinity.
Somebody like Mirimanoff, who introduced the plain "extra-ordinary",
then saw that as soon as Mirimanoff brought that up, then
ZF set theory had an axiom of regular/ordinary infinity added to it,
thus that Russell's "paradox" was put away, then for some
relatively simple things or the establishment of an inequality
the uncountability, to so follow.
The anti-diagonal argument as discovered by du Bois Reymond,
and nested intervals known since forever, the m-w proof as
is one of the number-theoretic proofs of uncountability,
another bit for continued fractions, these are the number-theoretic
results for uncountability, then there's the set-theoretic
bit or the powerset result.
So, the idea of providing an example to uncountability,
would be a 1-1 and onto function a bijection, between
countable domain and here most succinctly, the unit interval,
each of the points of the unit interval. This would be
with regards to the "number-theoretic" arguments.
Then, there would also need be a "set-theoretic" counter-example.
Here's that's provided by the "natural/unit equivalency function",
which falls out of the number-theoretic results un-contradicted,
and then some "ubiquitous ordinals" between ordering-theory and
set-theory, not unlike Cohen's forcing establishing the ndependence
of the Continuum Hypothesis, which one can also see as forestalling
what's a contradiction after ZF, since ordinals either would or
wouldn't live between cardinals with or without CH.
So, providing a counterexample and noting that
the "restrictions of comprehension" are _stipulations_
and thusly _non-logical_, makes for an inclusive take
on a foundation beneath _ordinary_ set theory: _extra-ordinary_
set theory. ("A theory of one relation: elt.")
In this way we can have extra-ordinary theory and plain
simple classical logical theory and plain ordinary regular
set theory, all quite thoroughly logical.
See, it's possible to make a thorough, reasoned, deconstructive
account of ordinary set theory, with regards to many sorts
under-served yet perfectly apropos models of continuity,
with regards to infinity.
WM
2024-11-04 10:47:39 UTC
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Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Post by Mikko
Between that point an an interval there are rational
numbers and therefore other intervals
I said the nearest one. There is no interval nearer than the nearest one.
Post by Mikko
Therefore the
point has no nearest interval.
That is an unfounded assertions and therefore not accepted.

Regards, WM
Mikko
2024-11-21 09:21:10 UTC
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Post by Mikko
Post by WM
Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Depends on the set of intervals. There is no nearest from Cantor's set.
And "interval" is not a term of geometry.
--
Mikko
WM
2024-11-21 10:50:32 UTC
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Post by WM
Post by Mikko
Post by WM
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest
intervals
No, it hasn't.
In geometry it has.
Depends on the set of intervals.
No. Every point in the complement is closer to the end of an interval
than to its contents of rationals.

Regards, WM
Mikko
2024-11-21 11:01:43 UTC
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Depends on the set of intervals.
No. Every point in the complement is closer to the end of an interval
than to its contents of rationals.
True but irrelevant because it may be even closer to the end of
another interval. In particular with Cantor's set of intervals
where there is no nearest interval.
--
Mikko
WM
2024-11-21 19:22:39 UTC
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
That is wrong. The measure outside of the intervals is infinite.
Hence there exists a point outside. This point has two nearest
intervals
No, it hasn't.
In geometry it has.
Depends on the set of intervals.
No. Every point in the complement is closer to the end of an interval
than to its contents of rationals.
True but irrelevant because it may be even closer to the end of
another interval.
Every end of any interval is irrational.
Post by Mikko
In particular with Cantor's set of intervals
where there is no nearest interval.
For every point of the complement, every interval has irrational ends.

Regards, WM
Mikko
2024-11-22 08:52:16 UTC
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Post by Mikko
Post by WM
Post by Mikko
Post by WM
Post by Mikko
Post by WM
That is wrong. The measure outside of the intervals is infinite. Hence
there exists a point outside. This point has two nearest intervals
No, it hasn't.
In geometry it has.
Depends on the set of intervals.
No. Every point in the complement is closer to the end of an interval
than to its contents of rationals.
True but irrelevant because it may be even closer to the end of
another interval.
Every end of any interval is irrational.
Irrelevant to the claim that a point outside of all intervals has two
nearest intervals and also to the claim that there is a point outside
of all intervals and also to the topic specified on the subjec line.
All those claims are true for some sets of intervals and false for others.
--
Mikko
WM
2024-11-22 11:05:58 UTC
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Post by Mikko
True but irrelevant because it may be even closer to the end of
another interval.
Every end of any interval is irrational.
Irrelevant to the claim that a point outside of all intervals has two
nearest intervals
What else should it have?
Post by Mikko
and also to the claim that there is a point outside
of all intervals
That is proven by the measure of the intervals being less than 3.
Post by Mikko
and also to the topic specified on the subjec line.
That is concerned by the fact that in the complement of measure oo - 3
rational numbers must exist.
Post by Mikko
All those claims are true for some sets of intervals and false for others.
Let me know what else than an irrational number could be next to a point
in the complement, if there were no rational numbers.

Regards, WM
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