Incompleteness of Cantor's enumeration of the rational numbers

Post by Jim Burns
There aren't any neighboring intervals.
Any two intervals have intervals between them.

That is wrong in geometry. The measure outside of the intervals is
infinite. Hence there exists at least one point outside. This point has
two nearest intervals

Regards, WM

Jim Burns

2024-11-04 14:37:30 UTC

Further there are never
two irrational numbers
without a rational number between them.
(Even the existence of neighbouring intervals
is problematic.)

There aren't any neighboring intervals.
Any two intervals have intervals between them.

That is wrong in geometry.
The measure outside of the intervals is infinite.
Hence there exists at least one point outside.
This point has two nearest intervals

This point,
which is on the boundary of two intervals,
is not two irrational points.

Further there are never
two irrational numbers
without an interval between them.

2024-11-04 17:32:10 UTC

Further there are never
two irrational numbers
without a rational number between them.
(Even the existence of neighbouring intervals
is problematic.)

There aren't any neighboring intervals.
Any two intervals have intervals between them.

That is wrong in geometry.
The measure outside of the intervals is infinite.
Hence there exists at least one point outside.
This point has two nearest intervals

This point,
which is on the boundary of two intervals,
is not two irrational points.

You are wrong. The intervals together cover a length of less than 3. The
whole length is infinite. Therefore there is plenty of space for a point
not in contact with any interval.

Post by Jim Burns
Further there are never
two irrational numbers
without an interval between them.

Not in reality. But in the used model.
The rationals are dense but the intervals are not. This proves that the
rationals are not countable.

Regards, WM

Jim Burns

2024-11-05 17:25:37 UTC

Further there are never
two irrational numbers
without a rational number between them.
(Even the existence of neighbouring intervals
is problematic.)

There aren't any neighboring intervals.
Any two intervals have intervals between them.

That is wrong in geometry.
The measure outside of the intervals is infinite.
Hence there exists at least one point outside.
This point has two nearest intervals

This point,
which is on the boundary of two intervals,
is not two irrational points.

You are wrong.
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.

⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.

I mean 'exterior' in the topological sense.

For a point x in the boundary ∂A of set A
each open set Oₓ which holds x
holds points in A and points not.in A

An interior point of A is
is in A and not.in ∂A

An exterior point of A is
not.in A and not.in ∂A

⎛ Assuming end.to.end.to.end intervals,
⎜ there are exterior points
⎝ a distance 10¹⁰⁰⁰⁰⁰ from any interval.

However,
the intervals aren't end.to.end.to.end.
Their midpoints are
the differences of ratios of countable.to,
and not any other points.

Each of {...,-3,-2,-1,0,1,2,3,...} is
the midpoint of an interval.
There can't be any exterior point
a distance 1 from any interval.

There can't be any exterior point
a distance ⅟2 from any interval.
Nor ⅟3. Nor ⅟4. Nor any positive distance.

An exterior point which is not
a positive distance from any interval
is not an exterior point.

Therefore,
in what is _almost_ your conclusion,
there are no exterior points.

Instead, there are boundary points.
For each x not.in the intervals,
each open set Oₓ which holds x
holds points in the intervals and
points not.in the intervals.
x is a boundary point.

All of the line except at most 2³ᐟ²⋅ε is
boundary of the intervals.

This is a figure.ground inversion of
how we are used to thinking about boundaries,
the expanse of a square's interior with
the boundary _line_ around it, ...

⎛ In the limit, there is no interior, too.
⎜ And no exterior.
⎝ It's only boundary.

However, ℚ is not a square, nor is it
close to anything else we could call a figure.

⎛ This is a nice example of the difference between
⎝ what is intuitive and what is true.

Post by Jim Burns
Further there are never
two irrational numbers
without an interval between them.

Not in reality. But in the used model.

What you're saying is:
⎛ I (WM) am not.talking about
⎝ what you all are talking about.

Which, in itself, is fine.
Billions of other people are not.talking about
what we all are talking about.

However, those people talking about
pop stars, or cosmology, or the rainy season
aren't imagining that they're talking about
what we all are talking about.
Which is what you are imagining, apparently.

Post by WM
The rationals are dense

Yes.
Each multi.point interval [x,x′] holds
rationals.

Post by WM
but the intervals are not.

No.
Each multi.point interval [x,x′] holds
ε.cover intervals.

Post by WM
This proves that
the rationals are not countable.

⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝ jₖ = k-iₖ
proves that
the rationals are countable.

Jim Burns

2024-11-05 18:15:59 UTC

[...]

⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝ jₖ = k-iₖ

jₖ = (iₖ+jₖ)-iₖ

Post by Jim Burns
proves that
the rationals are countable.

Ross Finlayson

2024-11-05 18:28:14 UTC

[...]

⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝ jₖ = k-iₖ

jₖ = (iₖ+jₖ)-iₖ

Post by Jim Burns
proves that
the rationals are countable.

Hausdorff even made for that all the
constructible is a countable union of countable.

Hausdorff was a pretty great geometer and
versed in set theory, along with Vitali he
has a lot going on with regards to "doubling
spaces" and "doubling measures", where there's
that Vitali made the first sort of example known
about "doubling measure", with splitting the
unit line segment into bits and re-composing
them length 2, then Vitali and Hausdorff also
made the geometric equi-decomposability of a ball.

Then, later, it's called Banach-Tarski for the
usual idea in measure theory that a ball can be
decomposed and recomposed equi-decomposable into
two identical copies, that it's a feature of
the measure theory and continuum mechanics actually.
Their results are ordinary-algebraic, though.

Then, it's said that von Neumann spent a lot
of examples in the equi-decomposable and the planar,
the 2D case, where Vitali wrote the 1D case and
Vitali and Hausdorff the 3D case, then I'd wonder
what sort of summary "von" Neumann, as he preferred
to be called, would make of "re-Vitali-ized"
measure theory.

There are also some modern theories about
"Rationals are HUGE" with regards to them
in various meaningful senses being much,
much larger than integers, among the integers.

Vitali and Hausdorff are considered great geometers,
and well versed in set theory. That's where
"non-measurable" in set theory comes from, because
Vitali and Hausdorff were more geometers than set theorists.

Ross Finlayson

2024-11-05 21:51:46 UTC

[...]

⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝ jₖ = k-iₖ

jₖ = (iₖ+jₖ)-iₖ

Post by Jim Burns
proves that
the rationals are countable.

Hausdorff even made for that all the
constructible is a countable union of countable.
Hausdorff was a pretty great geometer and
versed in set theory, along with Vitali he
has a lot going on with regards to "doubling
spaces" and "doubling measures", where there's
that Vitali made the first sort of example known
about "doubling measure", with splitting the
unit line segment into bits and re-composing
them length 2, then Vitali and Hausdorff also
made the geometric equi-decomposability of a ball.
Then, later, it's called Banach-Tarski for the
usual idea in measure theory that a ball can be
decomposed and recomposed equi-decomposable into
two identical copies, that it's a feature of
the measure theory and continuum mechanics actually.
Their results are ordinary-algebraic, though.
Then, it's said that von Neumann spent a lot
of examples in the equi-decomposable and the planar,
the 2D case, where Vitali wrote the 1D case and
Vitali and Hausdorff the 3D case, then I'd wonder
what sort of summary "von" Neumann, as he preferred
to be called, would make of "re-Vitali-ized"
measure theory.
There are also some modern theories about
"Rationals are HUGE" with regards to them
in various meaningful senses being much,
much larger than integers, among the integers.
Vitali and Hausdorff are considered great geometers,
and well versed in set theory. That's where
"non-measurable" in set theory comes from, because
Vitali and Hausdorff were more geometers than set theorists.

Of course "ye olde Pythagoreans" had all rational.

Chris M. Thomasson

2024-11-05 22:50:35 UTC

[...]

⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝ jₖ = k-iₖ

jₖ = (iₖ+jₖ)-iₖ

Post by Jim Burns
proves that
the rationals are countable.

Hausdorff even made for that all the
constructible is a countable union of countable.
Hausdorff was a pretty great geometer and
versed in set theory, along with Vitali he
has a lot going on with regards to "doubling
spaces" and "doubling measures", where there's
that Vitali made the first sort of example known
about "doubling measure", with splitting the
unit line segment into bits and re-composing
them length 2, then Vitali and Hausdorff also
made the geometric equi-decomposability of a ball.
Then, later, it's called Banach-Tarski for the
usual idea in measure theory that a ball can be
decomposed and recomposed equi-decomposable into
two identical copies, that it's a feature of
the measure theory and continuum mechanics actually.
Their results are ordinary-algebraic, though.
Then, it's said that von Neumann spent a lot
of examples in the equi-decomposable and the planar,
the 2D case, where Vitali wrote the 1D case and
Vitali and Hausdorff the 3D case, then I'd wonder
what sort of summary "von" Neumann, as he preferred
to be called, would make of "re-Vitali-ized"
measure theory.
There are also some modern theories about
"Rationals are HUGE" with regards to them
in various meaningful senses being much,
much larger than integers, among the integers.
Vitali and Hausdorff are considered great geometers,
and well versed in set theory. That's where
"non-measurable" in set theory comes from, because
Vitali and Hausdorff were more geometers than set theorists.

Of course "ye olde Pythagoreans" had all rational.

A fun animation about the Pythagoreans:

;^)

2024-11-06 10:35:36 UTC

Post by WM
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.

⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.

This plentiness does not change when the intervals are translated.

Post by Jim Burns
I mean 'exterior' in the topological sense.
For a point x in the boundary ∂A of set A
each open set Oₓ which holds x
holds points in A and points not.in A

The intervals are closed with irrational endpoints.

Post by Jim Burns
Each of {...,-3,-2,-1,0,1,2,3,...} is
the midpoint of an interval.
There can't be any exterior point
a distance 1 from any interval.
There can't be any exterior point
a distance ⅟2 from any interval.
Nor ⅟3. Nor ⅟4. Nor any positive distance.

Nice try. But there are points outside of intervals, and they are closer
to interval ends than to the interior, independent of the configuration
of the intervals. Note that only 3/oo of the points are inside.

Post by Jim Burns
An exterior point which is not
a positive distance from any interval
is not an exterior point.

Positive is what you can define, but there is much more in smaller
distance. Remember the infinitely many unit fractions within every eps >
0 that you can define.

Post by Jim Burns
Therefore,
in what is _almost_ your conclusion,
there are no exterior points.

There are 3/oo of all points exterior.

Post by Jim Burns
Instead, there are boundary points.
For each x not.in the intervals,
each open set Oₓ which holds x
holds points in the intervals and
points not.in the intervals.
x is a boundary point.

The intervals are closed

Post by WM
The rationals are dense

Yes.
Each multi.point interval [x,x′] holds
rationals.

Post by WM
but the intervals are not.

No.

Therefore not all rationals are enumerated.

Post by Jim Burns
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ+jₖ = ⌈(2⋅k+¼)¹ᐟ²+½⌉
⎜ iₖ = k-(iₖ+jₖ-1)(iₖ+jₖ-2)/2
⎝ jₖ = k-iₖ
proves that
the rationals are countable.

Contradiction. Something of your theory is inconsistent.

Regards, WM

Jim Burns

2024-11-06 14:22:05 UTC

Post by WM
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.

⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.

This plentiness does not change
when the intervals are translated.

⎛ When the intervals are end.to.end.to.end,
⎜ there are exterior points
⎝ a distance 10¹⁰⁰⁰⁰⁰ from any interval.

Are there points 10¹⁰⁰⁰⁰⁰ from any interval
when midpoints of intervals include
each of {...,-3,-2,-1,0,1,2,3,...} ?

Isn't that a plentiness which changes?

Post by Jim Burns
I mean 'exterior' in the topological sense.
For a point x in the boundary ∂A of set A
each open set Oₓ which holds x
holds points in A and points not.in A

The intervals are closed with irrational endpoints.

'Exterior' seems like a good way to say
'not in contact'.

It seems to me that you have a better argument
with open intervals instead of closed,
but let them be closed, if you like.

Either way,
there are no points 10¹⁰⁰⁰⁰⁰ from any interval.

Nice try.
But there are points outside of intervals,

Are any of these points.outside
⅟2 from any interval? ⅟3? ⅟4?

Post by WM
and they are closer to interval ends
than to the interior, independent of
the configuration of the intervals.

Shouldn't I be pointing that out
to you?

If there is no point with more.than.⅟2
between it and any midpoint,
shouldn't there be fewer.than.no points
with more.than.⅟2 between it and
any closer endpoint?

Post by WM
Note that
only 3/oo of the points are inside.

Yes, less than 2³ᐟ²⋅ε

If the intervals were open,
all of that would be "inside"
in the interior of their union.

Of the rest,
none of it is more.than.⅟2 from any interval.

Post by Jim Burns
An exterior point which is not
a positive distance from any interval
is not an exterior point.

Positive is what you can define,

Positive ℕ⁺ holds countable.to from.1
Positive ℚ⁺ holds ratios of elements of ℕ⁺
Positive ℝ⁺ holds points.between.splits of Q⁺

Post by WM
but there is much more in smaller distance.

Distances are positive or zero.
Two distinct points are a positive distance apart.

Post by WM
Remember the infinitely many unit fractions
within every eps > 0 that you can define.

For each of the infinitely.many unit fractions
there is no point a distance of that unit fraction
or more from any interval.

Post by Jim Burns
Therefore,
in what is _almost_ your conclusion,
there are no exterior points.

There are 3/oo of all points exterior.

Did you intend to write "interior"?

An exterior point is in
an open interval holding no rational.

There are no
open intervals holding no rational.

There are no exterior points.

However,
there are boundary points.
All but 2³ᐟ²⋅ε are boundary points.

Post by Jim Burns
Instead, there are boundary points.
  For each x not.in the intervals,
  each open set Oₓ which holds x
  holds points in the intervals and
  points not.in the intervals.
x is a boundary point.

The intervals are closed

We are only told
that Oₓ is an open set holding x
not that Oₓ is one of the ε.cover of ℚ
The question is whether x is a boundary point.

Post by WM
The rationals are dense

Yes.
Each multi.point interval [x,x′] holds
rationals.

Post by WM
but the intervals are not.

No.
Each multi.point interval [x,x′] holds
ε.cover intervals.

Therefore not all rationals are enumerated.

Explain why.

⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ 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 Jim Burns
proves that
the rationals are countable.

Contradiction.

It contradicts a non.empty exterior.
It doesn't contradict an almost.all boundary.

Post by WM
Something of your theory is inconsistent.

Your intuition is disturbed by
an almost.all boundary.

Disturbed intuitions and inconsistencies
are different.

2024-11-06 16:24:25 UTC

Post by WM
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.

⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.

This plentiness does not change
when the intervals are translated.

⎛ When the intervals are end.to.end.to.end,
⎜ there are exterior points
⎝ a distance 10¹⁰⁰⁰⁰⁰ from any interval.
Are there points 10¹⁰⁰⁰⁰⁰ from any interval
when midpoints of intervals include
each of {...,-3,-2,-1,0,1,2,3,...} ?

In fact, when ordered that way they include only intervals around the
positive integers because the natural numbers already are claimed to be
indexing all fractions. (Hence not more intervals are required.)

Post by Jim Burns
Isn't that a plentiness which changes?

No.

Post by WM
The intervals are closed with irrational endpoints.

'Exterior' seems like a good way to say
'not in contact'.

Every point outside is not an endpoint and is not in contact.

Post by Jim Burns
It seems to me that you have a better argument
with open intervals instead of closed,
but let them be closed, if you like.

Closed intervals with irrational endpoints prohibit any point outside.
Boundary or not! Of course there are points outside. This shows that the
rationals are not countable.

Post by Jim Burns
Either way,
there are no points 10¹⁰⁰⁰⁰⁰ from any interval.

Nice try.
But there are points outside of intervals,

Are any of these points.outside
⅟2 from any interval? ⅟3? ⅟4?

If not, then the measure of the real axis is less than 3.>

Post by Jim Burns
If there is no point with more.than.⅟2
between it and any midpoint,

In your first configuration, there are points with more than 1/3 between
it and a midpoint. If the intervals are translated, the distance may
become smaller for some points but necessarily becomes larger for
others. Shuffling does not increase the sum of the intervals.

Post by WM
There are 3/oo of all points exterior.

Did you intend to write "interior"?

Of course.

Post by Jim Burns
An exterior point is in
an open interval holding no rational.

and therefore no irrational either.

Post by Jim Burns
There are no
open intervals holding no rational.

That is true but shows that not all rationals are caught in intervals
because they are not countabel.

Post by Jim Burns
There are no exterior points.

If and only if all rationals could be enumerated!
That has been disproved.

Post by WM
Therefore not all rationals are enumerated.

Explain why.

There are rationals outside of all intervals in the infinite space
outside of the intervals covering less than 3 of the infinite space.

Post by WM
Contradiction.

It contradicts a non.empty exterior.
It doesn't contradict an almost.all boundary.

There is no difference between outside, exterior and you "boundary
points". The latter are only created by your inability to define small
enough intervals.

Post by WM
Something of your theory is inconsistent.

Your intuition is disturbed by
an almost.all boundary.

No. Your boundary is nonsense. If a point is outside of an interval,
then it is irrelevant whether you can construct an open interval not
covering points of the interior. It is outside.

Post by Jim Burns
Disturbed intuitions and inconsistencies
are different.

Tricks relating to your inability are not acceptable.

Regards, WM

Jim Burns

2024-11-06 19:08:58 UTC

Post by WM
The intervals are closed with irrational endpoints.

'Exterior' seems like a good way to say
'not in contact'.

Every point outside is not an endpoint

Yes.

Post by WM
and is not in contact.

You don't know that.

The _union_ of
arbitrarily.many _open_ sets
is an open set.

However,
the _union_ of
these infinitely.many _closed_ intervals
with irrational endpoints
is an open interval
with rational endpoints.

⋃{[⅟(k+√2),1-⅟(k+√2)]: k∈ℕ} = (0,1)

Recall that
0 = glb.{⅟(k+√2): k∈ℕ}
0 ∉ {⅟(k+√2): k∈ℕ}
1 = lub.{1-⅟(k+√2): k∈ℕ}
1 ∉ {1-⅟(k+√2): k∈ℕ}

Post by WM
Something of your theory is inconsistent.

Your intuition is disturbed by
an almost.all boundary.

No. Your boundary is nonsense.

"My" boundary is a definition.
It states what we take the term "boundary" to mean.
The open sets involved exist or not.exist,
independently of whether we use or not.use
the term "boundary".

The term "boundary" helps clarify
what I admit is a confusing situation.
_Of course_ you (WM) object to clarity.
Clarity is your nemesis.

https://en.wikipedia.org/wiki/Boundary_(topology)
⎛ It is the set of points p ∈ X such that
⎜ every neighborhood of p contains
⎜ at least one point of S and
⎝ at least one point not of S :

2024-11-06 19:50:23 UTC

Post by WM
The intervals are closed with irrational endpoints.

'Exterior' seems like a good way to say
'not in contact'.

Every point outside is not an endpoint

Yes.

Post by WM
and is not in contact.

You don't know that.

From every positive point we know that it is not 0 and not in contact
with (-oo, 0]. Same for every point not in an interval.

Post by Jim Burns
The _union_ of
arbitrarily.many _open_ sets
is an open set.
However,
the _union_ of
these infinitely.many _closed_ intervals
with irrational endpoints
is an open interval
with rational endpoints.

We use only intervals, not limits. A point is in an interval or it is not.

Post by Jim Burns
"My" boundary is a definition.

But it is irrelevant here like the offside rule in soccer.

Post by Jim Burns
The term "boundary" helps clarify
what I admit is a confusing situation.

No,it is not confusing at all. For every interval we can decide whether
a poin is inside or outside.

Post by Jim Burns
https://en.wikipedia.org/wiki/Boundary_(topology)
⎛ It is the set of points p ∈ X such that
⎜ every neighborhood of p contains
⎜ at least one point of S and

But there are no points outside of intervals because, if Cantor has
enumerated all rationals, then all rationals are caught and irrationals
cannot be outside. Therefore a boundary is excluded. Points p can only
exist insideof intervals.

Further: For every point x and for every interval we can find x - e
where e is the endpoint of an interval. No boundary is useful, in
particular because the infinite positive real axis except 3 unit
intervals cannot be covered by the blurred boundary.

Regards, WM

Jim Burns

2024-11-06 20:20:22 UTC

Post by WM
The intervals are closed with irrational endpoints.

'Exterior' seems like a good way to say
'not in contact'.

Every point outside is not an endpoint

Yes.

Post by WM
and is not in contact.

You don't know that.

From every positive point we know that
it is not 0 and
not in contact with (-oo, 0].
Same for every point not in an interval.

Is 0 "not in contact with" [-1,0) ⊆ ℝ
where
ℝ is points.between.splits of ℚ
ℚ is differences of ℚ⁺
ℚ⁺ is ratios of ℕ⁺
ℕ⁺ is countable.to from 1
?

We use only intervals, not limits.
A point is in an interval or it is not.

A point can be not.in the closure of each
of infinitely.many sets
and also in the closure of their union.

An ε.cover of ℚ holds infinitely.many sets.

2024-11-07 08:46:05 UTC

Post by WM
From every positive point we know that
it is not 0 and
not in contact with (-oo, 0].
Same for every point not in an interval.

Is 0 "not in contact with" [-1,0) ⊆ ℝ

0 is not a positive point.

Post by Jim Burns
A point can be not.in the closure of each
of infinitely.many sets
and also in the closure of their union.

These ideas are irrelevant because 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] (as soon as
the I(n) are smaller than 2/10).
These intervals (without splitting or modifying them) can be reordered,
to cover the whole positive axis except boundaries.
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.

Even using only intervals J(P) = [p - 1/10, p + 1/10] where p is a prime
number can accomplish the same.

Is this the power of infinity?
Or is it only the inertia of brains conquered by matheology?

Regards, WM

Jim Burns

2024-11-07 11:18:37 UTC

Post by WM
From every positive point we know that
it is not 0 and
not in contact with (-oo, 0].
Same for every point not in an interval.

Is 0 "not in contact with" [-1,0) ⊆ ℝ

0 is not a positive point.

I want to find out from you (WM)
what "not in contact with" means.

For a point
in the boundary but not in the set,
is it "not in contact with" the set?
Is it "in contact with" the set?

Point x′ is in the boundary of set S
iff
each interval [x,x″] such that
x′ is in its interior, x < x′ < x″,
holds points in S and points not.in S

For S = [-1,0) and x = 0
each [x,x″] such that x < 0 < x″
holds points in [-1,0) and points not.in [-1,0)

0 is in the boundary of [-1,0)
Is 0 "in contact with" [-1,0) ?

2024-11-07 13:03:23 UTC

Post by Jim Burns
I want to find out from you (WM)
what "not in contact with" means.
For a point
in the boundary but not in the set,
is it "not in contact with" the set?
Is it "in contact with" the set?

It is not in contact with the set. "For every eps" is not a valid
criterion because eps depends on what you can define, not on what
exists. The endpoint is in contact with the set.

Post by Jim Burns
0 is in the boundary of [-1,0) > Is 0 "in contact with" [-1,0) ?

I am not an expert on these things. I would say it is in contact with
the set because a point of the set is next to it. The closure of a set
is in contact with the set.

Regards, WM

Jim Burns

2024-11-07 15:29:03 UTC

It is not in contact with the set.
"For every eps" is not a valid criterion
because eps depends on
what you can define, not on what exists.
The endpoint is in contact with the set.

Post by Jim Burns
0 is in the boundary of [-1,0)
Is 0 "in contact with" [-1,0) ?

I am not an expert on these things.
I would say
it is in contact with the set
because a point of the set is next to it.
The closure of a set is in contact with the set.

That's what I thought you meant.
If, otherwise,
"not in contact with" meant "not in",
the easier, clearer way to say that is "not in".
Thank you for clarifying.

----
⎛ The boundary of a set S holds
⎜ those points x′ such that
⎜ each interval [x,x″] with
⎜ x′ in its interior, x < x′ < x″,
⎝ holds points in S and points not.in S

2024-11-07 17:59:02 UTC

Post by Jim Burns
⎛ The boundary of a set S holds
⎜ those points x′ such that
⎜ each interval [x,x″] with
⎜ x′ in its interior, x < x′ < x″,
⎝ holds points in S and points not.in S

Do you think you need the boundary in my last example?

When we cover the real axis by intervals
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
J(n) = [n - √2/10, n + √2/10]
and shuffle them in a clever way, then all rational numbers are
midpoints of intervals and no irrational number is outside of all intervals.
Do you believe this???

Regards, WM

Jim Burns

2024-11-07 20:32:20 UTC

Do you think you need the boundary in my last example?
When we cover the real axis by intervals
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
J(n) = [n - √2/10, n + √2/10]
and shuffle them in a clever way,
then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.

No irrational is not in contact with
the union of intervals.
That's different from
⎛ no irrational is not covered by
⎝ the union of intervals.
The first is true, the second is false.

The first is true because,
for each irrational x,
each interval of which x is in its interior
holds rationals, and
rationals are points in the union of intervals.

There is an enumeration of ℚ⁺
the set of ratios of ℕ⁺ countable.to from.1
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ/jₖ
⎜ 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

#x#ₖ is the -k.th digit of
the decimal representation of real number x

d is a Cantorian anti.diagonal of
the Cantorian rational.list ⟨iₖ/jₖ⟩
#d#ₖ = (#iₖ/jₖ#ₖ+5) mod 10

The closest that d and iₖ/jₖ could be
would be if, miraculously, all prior digits matched.
Even with that and the worst case for following digits,
|d-iₖ/jₖ| ≥ 4×10⁻ᵏ

Consider this ε.cover of
closed intervals with irrational endpoints.
⎛ ε.cover = {[x⁽ᵋₖ,xᵋ⁾ₖ]:k∈ℕ⁺}
⎜ x⁽ᵋₖ = iₖ/jₖ-2¹ᐟ²⋅10⁻ᵏ
⎝ xᵋ⁾ₖ = iₖ/jₖ+2¹ᐟ²⋅10⁻ᵏ

The infinite sum of measures = 2³ᐟ²/9
d is _not in contact with_ each interval.

However,
each interval of which d is in its interior
holds rationals, which are points in ⋃(ε.cover)
Thus, d is _in contact with_ ⋃(ε.cover)
but not with any of its intervals.

Post by WM
Do you believe this???

Don't you believe this???

2024-11-07 21:06:39 UTC

Post by WM
When we cover the real axis by intervals
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
J(n) = [n - √2/10, n + √2/10]
and shuffle them in a clever way,
then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.

No irrational is not in contact with
the union of intervals.

On the contrary, every irrational (and every rational) is in contact
with infinitely many intervals, because in the length of √2/5 there are
infinitely many rationals and therefore infinitely many midpoints of
intervals.

Post by Jim Burns
The first is true because,
for each irrational x,
each interval of which x is in its interior
holds rationals, and
rationals are points in the union of intervals.
There is an enumeration of ℚ⁺

If it is true, then the measure of the intervals of 3/10 of the real
axis grows to infinitely many times the real axis.

Post by Jim Burns
The infinite sum of measures = 2³ᐟ²/9
d is _not in contact with_ each interval.

The intervals are shifted such that every rational number is the
midpoint of an interval. Every point p on the real axis is covered by
infinitely many intervals, namely by all intervals having midpoint
rationals q with
d - √2/10 < q < d + √2/10.

Post by Jim Burns
However,
each interval of which d is in its interior
holds rationals, which are points in ⋃(ε.cover)
Thus, d is _in contact with_ ⋃(ε.cover)
but not with any of its intervals.

The intervals have length √2/5. There is no ε.

Post by WM
Do you believe this???

Don't you believe this???

No, it violates mathematics and logic when by reordering the measure of
a set of disjunct intervals grows.

Regards, WM

Chris M. Thomasson

2024-11-07 21:06:43 UTC

Do you think you need the boundary in my last example?
When we cover the real axis by intervals
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
J(n) = [n - √2/10, n + √2/10]
and shuffle them in a clever way, then all rational numbers are
midpoints of intervals and no irrational number is outside of all intervals.
Do you believe this???

Trying to relate it to a fractal wrt escape time. The fractal border is
infinitely complex. So we can iterate a point using the fractal
algorithm. If it stays within the boundary for how far it can go during
said iteration, then we say its in the set, otherwise its out of the
set. Some points can take billions of iterations to make that decision.

But this is a heck of a lot different that saying there must be a
largest natural number type shit... :^)

Jim Burns

2024-11-07 19:06:25 UTC

[...]

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".

Yes. Except boundaries.
⎛ The boundary of a set S holds
⎜ those points x′ such that
⎜ each interval [x,x″] with
⎜ x′ in its interior, x < x′ < x″,
⎝ holds points in S and points not.in S

ℝ is the set of points.between.splits of ℚ

Consider ℝ.points x < x″ with
their foresplits F F″ and hindsplits H H″ of ℚ
F ᵉᵃᶜʰ< x ≤ᵉᵃᶜʰ H and F∪H = ℚ
F″ ᵉᵃᶜʰ< x ≤ᵉᵃᶜʰ H″ and F″∪H″ = ℚ

F ≠⊂ F″
H ⊃≠ H″
F″∩H ≠ {}

F″∩H is
the set of rationals between x and x″
F″∩H is not empty, because x < x″

Each ℝ.interval [x,x″] holds rationals.

There is an enumeration of ℚ⁺
the set of ratios of ℕ⁺ countable.to from.1
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ/jₖ
⎜ 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

⎛ If you must have ℚ instead of ℚ⁺
⎜ I can add bells and whistles to the enumeration.
⎜ It won't significantly change the argument,
⎝ but it will give you more squiggles to look at.

There is an ε.cover of ℚ⁺
-- not scrunched together, but spread across ℚ⁺ --
of closed intervals with irrational endpoints
⎛ ε.cover = {[x⁽ᵋₖ,xᵋ⁾ₖ]:k∈ℕ⁺}
⎜ x⁽ᵋₖ = iₖ/jₖ-2ᵏ⁻¹ᐟ²⋅ε
⎝ xᵋ⁾ₖ = iₖ/jₖ+2ᵏ⁻¹ᐟ²⋅ε

Consider a point xᵒᵘᵗ not.in the ε.cover.
xᵒᵘᵗ ∉ ⋃(ε.cover)
For each interval [x,x″] such that x < xᵒᵘᵗ < x″
there are points in ℚ⁺ which are also in ⋃(ε.cover)
there are points not.in ⋃(ε.cover): xᵒᵘᵗ

xᵒᵘᵗ is in the boundary of ⋃(ε.cover)

Post by WM
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".

Yes.
In that clever re.ordering, not scrunched together,
the whole positive axis
is in the ε.cover or
in the boundary of the ε.cover.

A boundary so much more than its set
is not what we think of as "boundary"
but ℚ⁺ is not what we thing of as "interval".

2024-11-07 19:33:09 UTC

Post by WM
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".

Yes.
In that clever re.ordering, not scrunched together,
the whole positive axis
is in the ε.cover or
in the boundary of the ε.cover.

It is impossible however to cover the real axis (even many times) by the
intervals
J(n) = [n - 1/10, n + 1/10].
No boundaries are involved because every interval of length 1/5 contains
infinitely many rationals and therefore is essentially covered by
infinitely many intervals of length 1/5 - if Cantor is right.

Regards, WM

Jim Burns

2024-11-07 23:29:49 UTC

Post by WM
and shuffle them in a clever way,
then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.
That means by clever reordering them
you can cover the whole positive axis
except "boundaries".

"Except boundaries" is the key phrase.

Post by Jim Burns
Yes.
In that clever re.ordering, not scrunched together,
the whole positive axis
is in the ε.cover or
in the boundary of the ε.cover.

It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].

Those are not the cleverly.re.ordered intervals.

Post by Jim Burns
⎛ ε.cover = {[x⁽ᵋₖ,xᵋ⁾ₖ]:k∈ℕ⁺}
⎜ x⁽ᵋₖ = iₖ/jₖ-2¹ᐟ²⋅10⁻ᵏ
⎝ xᵋ⁾ₖ = iₖ/jₖ+2¹ᐟ²⋅10⁻ᵏ
There is an enumeration of ℚ⁺
the set of ratios of ℕ⁺ countable.to from.1
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ iₖ/jₖ
⎜ 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
d is a Cantorian anti.diagonal of
the Cantorian rational.list ⟨iₖ/jₖ⟩
#d#ₖ = (#iₖ/jₖ#ₖ+5) mod 10
#x#ₖ is the -k.th digit of
the decimal representation of real number x

For each interval [x⁽ᵋₖ,xᵋ⁾ₖ] in ε.cover
d is "not in contact with" [x⁽ᵋₖ,xᵋ⁾ₖ]

However,
d is "in contact with" ⋃(ε.cover)

d is in the closure of ⋃(ε.cover)
and not.in the closure of any of its intervals.

Post by WM
No boundaries are involved because
every interval of length 1/5 contains
infinitely many rationals and
therefore is essentially covered by
infinitely many intervals of length 1/5
- if Cantor is right.

I haven't claimed anything at all about
your all.1/5.length intervals.

2024-11-08 10:18:47 UTC

Post by WM
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].

Those are not the cleverly.re.ordered intervals.

They are the intervals that we start with.

Post by WM
No boundaries are involved because
every interval of length 1/5 contains infinitely many rationals and
therefore is essentially covered by infinitely many intervals of
length 1/5
- if Cantor is right.

I haven't claimed anything at all about
your all.1/5.length intervals.

Then consider the two only alternatives: Either by reordering (one after
the other or simultaneously) the measure of these intervals can grow
from 1/10 of the real axis to infinitely many times the real axis, or not.

My understanding of mathematics and geometry is that reordering cannot
increase the measure (only reduce it by overlapping). This is a basic
axiom which will certainly be agreed to by everybody not conditioned by
matheology. But there is also an analytical proof: Every reordering of
any finite set of intervals does not increase their measure. The limit
of a constant sequence is this constant however.

This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an interval.

Regards, WM

Richard Damon

2024-11-08 12:28:53 UTC

Post by WM
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].

Those are not the cleverly.re.ordered intervals.

They are the intervals that we start with.

Post by WM
No boundaries are involved because
every interval of length 1/5 contains infinitely many rationals and
therefore is essentially covered by infinitely many intervals of
length 1/5
- if Cantor is right.

I haven't claimed anything at all about
your all.1/5.length intervals.

Then consider the two only alternatives: Either by reordering (one after
the other or simultaneously) the measure of these intervals can grow
from 1/10 of the real axis to infinitely many times the real axis, or not.
My understanding of mathematics and geometry is that reordering cannot
increase the measure (only reduce it by overlapping). This is a basic
axiom which will certainly be agreed to by everybody not conditioned by
matheology. But there is also an analytical proof: Every reordering of
any finite set of intervals does not increase their measure. The limit
of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an interval.
Regards, WM

which makes the error that the properties of finite objects apply to the
infinite objects, which isn't true, and what just breaks your logic.

You take it as a given, but that just means that your logic is unable to
actually handle the infinite.

2024-11-08 16:43:15 UTC

Post by WM
My understanding of mathematics and geometry is that reordering cannot
increase the measure (only reduce it by overlapping). This is a basic
axiom which will certainly be agreed to by everybody not conditioned
by matheology. But there is also an analytical proof: Every reordering
of any finite set of intervals does not increase their measure. The
limit of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an interval.

which makes the error that the properties of finite objects apply to the
infinite objects, which isn't true, and what just breaks your logic.

The infinite of the real axis is a big supply but an as big drain.

Post by Richard Damon
You take it as a given, but that just means that your logic is unable to
actually handle the infinite.

I take it as evident that intervals of the measure 1/5 of the positive
real axis will not, by any shuffling, cover the real axis completely,
let alone infinitely often. I think who believes this is a deplorable
fanatic if not a fool.

Regards, WM

Richard Damon

2024-11-08 17:05:40 UTC

Post by WM
My understanding of mathematics and geometry is that reordering
cannot increase the measure (only reduce it by overlapping). This is
a basic axiom which will certainly be agreed to by everybody not
Every reordering of any finite set of intervals does not increase
their measure. The limit of a constant sequence is this constant
however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an interval.

which makes the error that the properties of finite objects apply to
the infinite objects, which isn't true, and what just breaks your logic.

The infinite of the real axis is a big supply but an as big drain.

What "drain", the numbers exist.

Post by Richard Damon
You take it as a given, but that just means that your logic is unable
to actually handle the infinite.

Since 1/5 of infinity isn't a finite measure, you can't use finite logic
to handle them.

You are just proving your use of broken logic.

2024-11-09 11:46:01 UTC

Post by WM
The infinite of the real axis is a big supply but an as big drain.

What "drain", the numbers exist.

But their density is everywhere the same.

Post by WM
I take it as evident that intervals of the measure 1/5 of the positive
real axis will not, by any shuffling, cover the real axis completely,
let alone infinitely often. I think who believes this is a deplorable
fanatic if not a fool.

Since 1/5 of infinity isn't a finite measure, you can't use finite logic
to handle them.

Since 1/5 is a finite number and for every finite set the covering is
1/5, the limit is 1/5 too.

Post by Richard Damon
You are just proving your use of broken logic.

Chuckle. Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
0--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
could grow in length or number to cover the whole real axis is a fool or
worse.

Regards, WM

joes

2024-11-08 19:05:04 UTC

which makes the error that the properties of finite objects apply to
the infinite objects, which isn't true, and what just breaks your
logic.

The infinite of the real axis is a big supply but an as big drain.

Post by Richard Damon
You take it as a given, but that just means that your logic is unable
to actually handle the infinite.

What is the measure you are using and what does it give for the real
axis?

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

Moebius

2024-11-08 20:35:09 UTC

which makes the error that the properties of finite objects apply to
the infinite objects, which isn't true, and what just breaks your
logic.

The infinite of the real axis is a big supply but an as big drain.

Post by Richard Damon
You take it as a given, but that just means that your logic is unable
to actually handle the infinite.

What is the measure you are using and what does it give for the real
axis?

Ob Du es nochmal schaffst, auf diesen saudummen Scheißdreck NICHT zu
antworten?

Moebius

2024-11-08 20:36:17 UTC

which makes the error that the properties of finite objects apply to
the infinite objects, which isn't true, and what just breaks your
logic.

The infinite of the real axis is a big supply but an as big drain.

Post by Richard Damon
You take it as a given, but that just means that your logic is unable
to actually handle the infinite.

What is the measure you are using and what does it give for the real
axis?

Ob Du es nochmal schaffst, auf diesen saudummen Scheißdreck NICHT zu
antworten?

Ich kann es jedenfalls nicht mehr sehen/lesen. Daher kommst Du jetzt
auch hier in mein Killfile. Tschüß!

2024-11-09 21:42:01 UTC

Post by joes
What is the measure you are using and what does it give for the real
axis?

Ob Du es nochmal schaffst, auf diesen saudummen Scheißdreck NICHT zu
antworten?

Ich kann es jedenfalls nicht mehr sehen/lesen.

Kein Wunder, weil es Deinen starken Glauben erschüttert.

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.

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 rationals are outside.
That implies that Cantor's above sequence does not contain all rationals.

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.

But note that Cantor's bijection between naturals and rationals does not
insert any non-natural number into ℕ. It confirms only that both sets
are very large. Therefore also the above sequence of intervals keeps the
same intervals and the same reality. And the same density 1/5 for every
finite interval and therefore also in the limit.

Regards, WM

Chris M. Thomasson

2024-11-09 21:45:28 UTC

Post by joes
What is the measure you are using and what does it give for the real
axis?

Ob Du es nochmal schaffst, auf diesen saudummen Scheißdreck NICHT zu
antworten?

Ich kann es jedenfalls nicht mehr sehen/lesen.

Kein Wunder, weil es Deinen starken Glauben erschüttert.
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.
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 rationals are outside.
That implies that Cantor's above sequence does not contain all rationals.
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.
But note that Cantor's bijection between naturals and rationals does not
insert any non-natural number into ℕ. It confirms only that both sets
are very large.

Cantor pairing can handle any natural number and convert into a unique
pair. Not very large... Infinite indeed. Cantor pairing can handle any
natural number.

Therefore also the above sequence of intervals keeps the

Post by WM
same intervals and the same reality. And the same density 1/5 for every
finite interval and therefore also in the limit.
Regards, WM

Moebius

2024-11-09 22:01:41 UTC

Post by WM
But note that Cantor's bijection between naturals and rationals does
not insert any non-natural number into ℕ.

Huh?!

Post by WM
It confirms only that [both] sets [have the same size, actually aleph_0].

2024-11-10 09:18:35 UTC

It confirms only that both sets have the same size, actually aleph_0.

True but irrelevant. For covering of all rationals by all naturals there
must be the same reality, but that is not.

Regards, WM

Chris M. Thomasson

2024-11-10 20:34:53 UTC

It confirms only that both sets have the same size, actually aleph_0.

True but irrelevant. For covering of all rationals by all naturals there
must be the same reality, but that is not.

Did you know that Cantor pairing can turn any natural number into a
_unique_ pair of natural numbers? This unique pair can be turned back
into the original natural number. It's a mapping at its essence.
Actually, it can be used for hashing, anyway...

The original natural number is there. The pair of natural numbers is
also there. We can go back and forth between them? They are all there in
the infinite set of natural numbers!

2024-11-09 11:49:44 UTC

What is the measure you are using and what does it give for the real
axis?

The measure is the density 1/5 for every finite segment and its limit 1/5.

Regards, WM

Ross Finlayson

2024-11-08 16:55:39 UTC

Post by WM
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].

Those are not the cleverly.re.ordered intervals.

They are the intervals that we start with.

Post by WM
No boundaries are involved because
every interval of length 1/5 contains infinitely many rationals and
therefore is essentially covered by infinitely many intervals of
length 1/5
- if Cantor is right.

I haven't claimed anything at all about
your all.1/5.length intervals.

Then consider the two only alternatives: Either by reordering (one after
the other or simultaneously) the measure of these intervals can grow
from 1/10 of the real axis to infinitely many times the real axis, or not.
My understanding of mathematics and geometry is that reordering cannot
increase the measure (only reduce it by overlapping). This is a basic
axiom which will certainly be agreed to by everybody not conditioned by
matheology. But there is also an analytical proof: Every reordering of
any finite set of intervals does not increase their measure. The limit
of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an interval.
Regards, WM

Perhaps you've never heard of Vitali's doubling-space,
the Vitali and Hausdorff's what became Banach-Tarski
the equi-decomposability, the doubling in signal theory
according to Shannon and Nyquist, and as with regards to
the quasi-invariant measure theory, where: taking a
continuum apart and putting it back together doubles things.

It's part of continuum mechanics and as with regards to infinity.
(Mathematical infinity.)

Ross Finlayson

2024-11-08 17:03:03 UTC

Post by WM
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].

Those are not the cleverly.re.ordered intervals.

They are the intervals that we start with.

Post by WM
No boundaries are involved because
every interval of length 1/5 contains infinitely many rationals and
therefore is essentially covered by infinitely many intervals of
length 1/5
- if Cantor is right.

I haven't claimed anything at all about
your all.1/5.length intervals.

Then consider the two only alternatives: Either by reordering (one after
the other or simultaneously) the measure of these intervals can grow
from 1/10 of the real axis to infinitely many times the real axis, or not.
My understanding of mathematics and geometry is that reordering cannot
increase the measure (only reduce it by overlapping). This is a basic
axiom which will certainly be agreed to by everybody not conditioned by
matheology. But there is also an analytical proof: Every reordering of
any finite set of intervals does not increase their measure. The limit
of a constant sequence is this constant however.
This geometrical consequence of Cantor's theory has, to my knowledge,
never been discussed. By the way I got the idea after a posting of
yours: Each of {...,-3,-2,-1,0,1,2,3,...} is the midpoint of an interval.
Regards, WM

Perhaps you've never heard of Vitali's doubling-space,
the Vitali and Hausdorff's what became Banach-Tarski
the equi-decomposability, the doubling in signal theory
according to Shannon and Nyquist, and as with regards to
the quasi-invariant measure theory, where: taking a
continuum apart and putting it back together doubles things.
It's part of continuum mechanics and as with regards to infinity.
(Mathematical infinity.)

That's part of usually what's called "measure theory",
and even today physicists are quite agog about "the measure problem".

When Vitali showed his geometric and analytic result,
then set theory arrived at "non-measurable" or "un-measurable".
Yet, it's still good, and, shows that it's a real thing
that doubling and halving measures and spaces are mathematical.

So, the quantum theorists have got plenty problems
about what's continuous again. ("The measure problem.")

Then as well, this of course gets into the multiple law(s)
of large numbers, what's called "non-standard" probability
theory, analysis, models of integers, and so on.

It's quite sensible with an approach like mine with
the three definitions of continuous domains the line-reals,
field-reals, and signal-reals, and the doubling line-reals
and halving signal-reals, after of course an apologetics in
the mathematical foundations to explain the slate of
uncountability and logical paradoxes so that there are none,
my "slates", that I've written since a decade and decades.

So, it probably helps if you know Riemann and Lebesgue
and Cauchy and Seidel and all these things after Weierstrass
that Dirichlet and Poincare make all out as with regards
to Vitali and Hausdorff and the "re-Vitali-ization",
measure theory.

Jim Burns

2024-11-08 18:01:08 UTC

Post by WM
It is impossible however to cover
the real axis (even many times) by
the intervals
J(n) = [n - 1/10, n + 1/10].

Those are not the cleverly.re.ordered intervals.

They are the intervals that we start with.

Starting intervals
{[n-⅒,n+⅒]: n∈ℕ⁺}
with midpoints
{ 1, 2, 3, 4, 5, ... }
and cleverly.shifted intervals
{[iₙ/jₙ-⅒,iₙ/jₙ+⅒]: n∈ℕ⁺}
with midpoints
{ 1/1, 1/2, 2/1, 1/3, 2/2, ... }

⎛ n ↦ iₙ/jₙ
⎜ iₙ+jₙ = ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ iₙ = n-((iₙ+jₙ)-1)((iₙ+jₙ)-2)/2
⎜ jₙ = (iₙ+jₙ)-iₙ
⎝ (iₙ+jₙ-1)(iₙ+jₙ-2)/2+iₙ = n

I haven't claimed anything at all about
your all.1/5.length intervals.

Either by reordering
(one after the other or simultaneously)
the measure of these intervals

https://en.wikipedia.org/wiki/Measure_(mathematics)
⎛ Let X be a set and Σ a σ-algebra over X.
⎜ A set function μ from Σ to
⎜ the extended real number line [!]
⎜ is called a measure
⎜ if the following conditions hold:
⎜ • Non-negativity: For all E ∈ Σ, μ(E) ≥ 0.
⎜ • μ(∅) = 0.
⎜ • Countable additivity (or σ-additivity):
⎜ For all countable collections {Eₖ}ₖ⃛₌₁ of
⎜ pairwise disjoint sets in Σ,
⎝ μ(⋃ₖ⃛₌₁Eₖ) = ∑ₖ⃛₌₁μ(Eₖ).

https://en.wikipedia.org/wiki/Extended_real_number_line
⎛ In mathematics, the extended real number system
⎜ is obtained from the real number system ℝ
⎜ by adding two[!] elements denoted +∞ and −∞
⎜ that are respectively greater and lower than
⎝ every real number.

The value of a measure is 0 or +∞ or
a point (in ℝ⁺) between fore and hind of
a split (of ℚ⁺) of all ratios of
numbers (in ℕ⁺) the countable.to from.1
(AKA finite).

Post by WM
the measure of these intervals can grow
from 1/10 of the real axis

≠ 0
≠ a point between a split of
all ratios of the countable.to from.1
= +∞

Post by WM
to infinitely many times the real axis,

≠ 0
≠ a point between a split of
all ratios of the countable.to from.1
= +∞

Post by WM
or not.

The clever re.ordering does not
increase the measure.
It is +∞ before and +∞ after.

+∞ has properties different from
anything in (0,+∞)

In particular,
∀x ∈ (0,+∞): ∃x′ ∈ (0,x): 5⋅x′ = x
However,
+∞ ∉ (0,+∞)
¬∃x′ ∈ (0,+∞): 5⋅x′ = +∞

By
"everybody not conditioned by matheology"
you mean
"everybody who hasn't thought much about infinity"

Post by WM
Every reordering of
any finite set of intervals
does not increase their measure.
The limit of a constant sequence is
this constant however.

You are assuming that the measure of
the all.⅕.intervals at their starting positions
is some value outside the extended reals.
Otherwise,
for midpoints in {...,-2,-1,0,1,2,...}
the measure is in [0,+∞]
and not.in [0,+∞)
and thus equals +∞

Post by WM
This geometrical consequence of Cantor's theory
has, to my knowledge, never been discussed.
Each of {...,-3,-2,-1,0,1,2,3,...} is
the midpoint of an interval.

2024-11-09 11:45:05 UTC

By
"everybody not conditioned by matheology"
you mean
"everybody who hasn't thought much about infinity"

Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number to cover the whole real axis is a fool or
worse.

Regards, WM

Jim Burns

2024-11-09 23:27:42 UTC

By
"everybody not conditioned by matheology"
you mean
"everybody who hasn't thought much about infinity"

Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

Our sets do not change.

The set
{[n-⅒,n+⅒]: n∈ℕ⁺}
with the midpoints at
⟨ 1, 2, 3, 4, 5, ... ⟩
does not _change_ to the set
{[iₙ/jₙ-⅒,iₙ/jₙ+⅒]: n∈ℕ⁺}
with the midpoints at
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩

----
Either
all instances of a 𝗰𝗹𝗮𝗶𝗺 about a set
are _only_ true or _only_ false
or
a set changes.

In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which
has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.

Even though
we are _not_ physically able to check, for each number
in an infinite set of numbers,
that a 𝗰𝗹𝗮𝗶𝗺 is true about it,
we _are_ physically able to check, for each 𝗰𝗹𝗮𝗶𝗺
in a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀,
that it is not.first.false in that 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲.

Also, we already know some 𝗰𝗹𝗮𝗶𝗺𝘀 to be true.

Some finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲𝘀 of 𝗰𝗹𝗮𝗶𝗺𝘀 are
known to be only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀,
and thus known to be only true 𝗰𝗹𝗮𝗶𝗺𝘀.
Un.physically.checkable numbers do not
prevent us from knowing they're true 𝗰𝗹𝗮𝗶𝗺𝘀.

In the second case, with the changing sets,
who knows?
Perhaps something else could be done,
but not that.

In any case,
what.we.do is the first case, with
its not.changing sets and
its known.about infinity.

For that reason (and more, I suspect),
our sets do not change.

Post by WM
Everybody who believes that
the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

Our sets do not change.

Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".

Consider geometry.

Similar triangles have
corresponding sides in the same ratio.

Consider these points, line.segments, and triangles
in the ⟨x,y⟩.plane

A = ⟨0,-1⟩
B = ⟨0,0⟩
C = ⟨x,0⟩ with 0 < x < 1
D = ⟨1,0⟩
E = ⟨1,y⟩ with points A C E collinear.

△ABC and △EDC are similar
△ABC ∼ △EDC
μA͞B = 1
μB͞C = x
μE͞D = y
μD͞C = 1-x

Similar triangles.
μA͞B/μB͞C = μE͞D/μD͞C
1/x = y/(1-x)

y = 1/x - 1
x = 1/(y+1)

To each point C = ⟨x,0⟩ in (0,1)×{0}
there corresponds
exactly one point E = ⟨1,y⟩ in {1}×(0,+∞)
and vice versa.

(0,1)×{0} is not stretched over {1}×(0,+∞)
{1}×(0,+∞) is not shrunk to (0,1)×{0}
They both _are_
And their points correspond
by line A͞C͞E through point A.

Consider again the two sets of midpoints
⟨ 1, 2, 3, 4, 5, ... ⟩ and
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩

They both _are_
And their points correspond
by i/j ↦ n = (i+j-1)(i+j-2)/2+i

2024-11-10 09:35:40 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

It cannot do so because the reality of the rationals is much larger than
the reality of the naturals.

Post by Jim Burns
----
Either
all instances of a 𝗰𝗹𝗮𝗶𝗺 about a set
are _only_ true or _only_ false
or
a set changes.
In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which
has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.

But it will never complete an infinite set of claims. It will forever
remain in the status nascendi. Therefore irrelevant for actual or
completed infinity.

So yes, you can shift the intervals to midpoints of every finite initial
segment of 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, ...

But you must remove them from larger natural numbers. That will never
change.

Post by Jim Burns
In the second case, with the changing sets,
who knows?
Perhaps something else could be done,
but not that.

Certainly not. The intervals can neither grow in size nor in multitude.

Post by Jim Burns
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".

But they cannot become such sets.

Post by Jim Burns
Consider again the two sets of midpoints
⟨ 1, 2, 3, 4, 5, ... ⟩ and
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩
They both _are_
And their points correspond
by i/j ↦ n = (i+j-1)(i+j-2)/2+i

But they cannot be completely transformed into each other. That is
prohibited by geometry. It is possible for every finite initial segment
of the above sequence, but not possible to replace all the given
intervals to cover all rational midpoints.

Regards, WM

Ross Finlayson

2024-11-10 17:34:13 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

It cannot do so because the reality of the rationals is much larger than
the reality of the naturals.

But it will never complete an infinite set of claims. It will forever
remain in the status nascendi. Therefore irrelevant for actual or
completed infinity.
So yes, you can shift the intervals to midpoints of every finite initial
segment of 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, ...
But you must remove them from larger natural numbers. That will never
change.

Post by Jim Burns
In the second case, with the changing sets,
who knows?
Perhaps something else could be done,
but not that.

Certainly not. The intervals can neither grow in size nor in multitude.

Post by Jim Burns
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".

But they cannot become such sets.

Related rates again, McQuack?

That rationals are about on the order of 2^w,
and, reals are about on the order of 2^w,
is a thing, involved with that
"rationals are countable"
while
"rationals are huge"
and
"reals are uncountable"
while
"reals are constructible".

Of course in mentioning real numbers the context is
to indicate when it means anything else "the complete
ordered field", where that's usually given to
"the real valued", when there are multiple models
of continuous domains here line-reals, field-reals, signal-reals.

So, enumerating rationals is on the order
of exponential, and what's on the order
of exponential also gets right to constructing
the indistinguishable according to 1/n, the 1/2^n.

Then, whence the rational field is constructed,
and numbers are dense in (among) the integers,
the precision of 1/n, is not 1/2^n.

So, book-keeping, and related-rates, and
asymptotics or after concrete mathematics,
then besides the algebra the words the language
of the things after axiomatics, words,
all go together just fine, then though
with neither your "unbounded bounded"
nor the usual "bounded unbounded".

Jim Burns

2024-11-10 17:49:56 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

It cannot do so because
the reality of the rationals is much larger than
the reality of the naturals.

The set
{3,4,5}
does not _change_ to the set
{6,7,8}
because
our sets do not change.

The word 'set' has many uses.
In some uses of 'set', sets change.
Break a plate, and your dinnerware set changes.

Not here.
In this use of 'set',
as in mathematics more generally,
sets do not change.

Add 3 to each of {3,4,5}.
You do not change {3,4,5}.
{3,4,5} continues to be {3,4,5}.
You get a different set, {6,7,8].
{6,7,8} has never been {3,4,5}.

The not.changing nature of our sets is a choice,
although it is such an ancient, universal, and
enormously.useful choice that it is easy
to overlook that a choice has been made.
But our sets could have been like dinnerware sets.

However,
our sets are not like dinnerware sets.
Our sets do not change.

⎛ I have argued for the correctness of that choice.
⎜ But, however good or bad my argument is,
⎜ that is the choice.
⎜ Pretending otherwise is like
⎝ pretending dogs are cats.

Post by Jim Burns
Either
all instances of a 𝗰𝗹𝗮𝗶𝗺 about a set
are _only_ true or _only_ false
or
a set changes.
In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which
has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.

But it

"It" refers to who or what?
Me? You? Chuck Norris? That finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲?

I will continue talking about that 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲.

Post by WM
But it will never complete
an infinite set of claims.

We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.

For that finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲, because 𝗶𝘁 is finite,
it is enough for us to see that
each 𝗰𝗹𝗮𝗶𝗺 in 𝗶𝘁 is true.or.not.first.false.
If we see that, we know that
each 𝗰𝗹𝗮𝗶𝗺 in 𝗶𝘁 is true.

Each 𝗰𝗹𝗮𝗶𝗺 in 𝗶𝘁 is true,
even if
a 𝗰𝗹𝗮𝗶𝗺 refers to one of infinitely.many.

There may be infinitely.many possible.referents,
but we know the 𝗰𝗹𝗮𝗶𝗺 is true
because
there are only finitely.many 𝗰𝗹𝗮𝗶𝗺𝘀.

Post by WM
It will forever remain in the status nascendi.
Therefore
irrelevant for actual or completed infinity.

A finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false,
will forever remain
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false.

Post by Jim Burns
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".

But they cannot become such sets.

Our sets do not change.

If matheologians would "concede" a claim which
they do not make and have never made,
would you (WM) stop
filling your students' heads with that gibberish?

But they cannot be completely transformed

< into each other.

Our sets do not change.

? That is prohibited by geometry.

Consider geometry.

Similar triangles have
corresponding sides in the same ratio.

Consider these points, line.segments, and triangles
in the ⟨x,y⟩.plane

A = ⟨0,-1⟩
B = ⟨0,0⟩
C = ⟨x,0⟩ with 0 < x < 1
D = ⟨1,0⟩
E = ⟨1,y⟩ with points A C E collinear.

△ABC and △EDC are similar
△ABC ∼ △EDC
μA͞B = 1
μB͞C = x
μE͞D = y
μD͞C = 1-x

Similar triangles.
μA͞B/μB͞C = μE͞D/μD͞C
1/x = y/(1-x)

y = 1/x - 1
x = 1/(y+1)

To each point C = ⟨x,0⟩ in (0,1)×{0}
there corresponds
exactly one point E = ⟨1,y⟩ in {1}×(0,+∞)
and vice versa.

(0,1)×{0} is not stretched over {1}×(0,+∞)
{1}×(0,+∞) is not shrunk to (0,1)×{0}
They both _are_
And their points correspond
by line A͞C͞E through point A.

Consider again the two sets of midpoints
⟨ 1, 2, 3, 4, 5, ... ⟩ and
⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩

They both _are_
And their points correspond
by i/j ↦ n = (i+j-1)(i+j-2)/2+i

2024-11-11 08:41:41 UTC

Post by Jim Burns
The set
{3,4,5}
does not _change_ to the set
{6,7,8}
because
our sets do not change.

But points or intervals in geometry can be translated on the real axis.

Post by Jim Burns
Our sets do not change.

But points or intervals in geometry can be translated on the real axis.

Post by Jim Burns
In the first case, with the not.changing sets,
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 which
has only true.or.not.first.false 𝗰𝗹𝗮𝗶𝗺𝘀
has only true 𝗰𝗹𝗮𝗶𝗺𝘀.

But it

"It" refers to who or what?

To that finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲.

Post by WM
But it will never complete
an infinite set of claims.

We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.

But you claim that _all_ fractions are in bijection with all natural
numbers, don't you?

Post by WM
It will forever remain in the status nascendi.
Therefore
irrelevant for actual or completed infinity.

Therefore such a sequence does not entitle you to claim infinite mappings.

Post by Jim Burns
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".

But they cannot become such sets.

Our sets do not change.

My intervals I(n) = [n - 1/10, n + 1/10] must be translated to all the
midpoints 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, ... if you want to
contradict my claim.

Post by WM
But they cannot be completely transformed

< into each other.
Our sets do not change.

But intervals can be shifted.

The first few terms do correspond or can be made c orresponding. That
can be proven by translating the due intervals. But the full claim is
nonsense because it is impossible to satisfy.

Regards, WM

Jim Burns

2024-11-11 18:23:49 UTC

But it

"It" refers to who or what?

To that finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲.

Post by WM
But it will never complete
an infinite set of claims.

We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.

But you claim that
_all_ fractions are in bijection with
all natural numbers,
don't you?

Yes, I claim that.
This is one 𝗰𝗹𝗮𝗶𝗺:
⎛ All fractions are in bijection with
⎝ all natural numbers.

We can peel that claim apart,
truthfully and finitely saying what it means, and
we can situate 𝗶𝘁 in a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀,
each 𝗰𝗹𝗮𝗶𝗺 of which is true.or.not.first.false.
Such a 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 has only true 𝗰𝗹𝗮𝗶𝗺𝘀.
That 𝗰𝗹𝗮𝗶𝗺, being in that 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲, is a true 𝗰𝗹𝗮𝗶𝗺.

⎛ I have been 𝗳𝗹𝗶𝗽𝗽𝗶𝗻𝗴 back 𝗮𝗻𝗱 𝗳𝗼𝗿𝘁𝗵 between 𝗳𝗼𝗻𝘁𝘀,
⎜ in an attempt to visibly mark a distinction between
⎜ claims as bearers of meaning and
⎜ 𝗰𝗹𝗮𝗶𝗺𝘀 as objects, as vibrations in the air,
⎜ smears of colored bear fat on a cave wall, bits, or pixels.
⎜
⎜ Finite sequences of smears of bear fat have
⎜ certain properties which we ("we" very broadly)
⎜ have learned to use in our exploration of infinity.
⎜
⎜ Note: that's _finite_ sequences.
⎜ Issues involving Scrooge McDuck and Disappearing Bob
⎝ do not come into play for these properties.

Post by WM
It will forever remain in the status nascendi.
Therefore
irrelevant for actual or completed infinity.

Therefore
such a sequence does not entitle you
to claim infinite mappings.

No.
Such a 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 does entitle me.

A finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.or.not.first.false,
is
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, each of which
is true.

Post by Jim Burns
Infinite sets can correspond to
other infinite sets which,
without much thought about infinity,
would seem to be a different "size".

But they cannot become such sets.

Our sets do not change.

My intervals I(n) = [n - 1/10, n + 1/10]
must be translated to all the midpoints
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, ...
if you want to contradict my claim.

Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".
If we are in the same discussion,
they are already contradicted at that point,

If we aren't in same discussion,
I don't see how to respond sensibly.
End.of.debate, I guess? If there ever was a debate?

2024-11-11 20:33:28 UTC

Post by WM
But it will never complete
an infinite set of claims.

We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.

But you claim that
_all_ fractions are in bijection with
all natural numbers,
don't you?

Yes, I claim that.

For that claim you need an infinite set of claims.

Post by Jim Burns
⎛ All fractions are in bijection with
⎝ all natural numbers.

It is wrong.

Post by WM
My intervals I(n) = [n - 1/10, n + 1/10]
must be translated to all the midpoints
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, ...
if you want to contradict my claim.

Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".

No, I claim that intervals can be translated. (The set of intervals
remains constant in size and multitude.) For every finite subset this is
possible.

Regards, WM

FromTheRafters

2024-11-11 21:50:07 UTC

Post by WM
But it will never complete
an infinite set of claims.

We do not need an infinite ???????? of ?????? completed.
We do not want an infinite ???????? of ?????? completed.

But you claim that
_all_ fractions are in bijection with
all natural numbers,
don't you?

Yes, I claim that.

For that claim you need an infinite set of claims.

Good thing we have such.

Post by Jim Burns
⎛ All fractions are in bijection with
⎝ all natural numbers.

It is wrong.

You 'think' it is wrong because you intuitively think it 'should' be
wrong. Saying it is wrong does not make it wrong no matter how many
times that you say it.

[...]

2024-11-12 08:58:12 UTC

Post by WM
For that claim you need an infinite set of claims.

Good thing we have such.

No.

Post by Jim Burns
⎛ All fractions are in bijection with
⎝ all natural numbers.

It is wrong.

You 'think' it is wrong because you intuitively think it 'should' be
wrong. Saying it is wrong does not make it wrong no matter how many
times that you say it.

Therefore I prove it. The set of intervals I(n) = [n - 1/10, n + 1/10]
cannot cover all fractions on the real axis.

Regards, WM

FromTheRafters

2024-11-12 10:10:52 UTC

Post by WM
For that claim you need an infinite set of claims.

Good thing we have such.

No.

see axiom vs. axiom schema.

joes

2024-11-12 11:23:47 UTC

Post by WM
For that claim you need an infinite set of claims.

Good thing we have such.

No.

Yes, we do claim something for each, every one and all of the
infinite set of (finite) natural numbers.

⎛ All fractions are in bijection with ⎝ all natural numbers.

It is wrong.

You 'think' it is wrong because you intuitively think it 'should' be
wrong. Saying it is wrong does not make it wrong no matter how many
times that you say it.

Therefore I prove it. The set of intervals I(n) = [n - 1/10, n + 1/10]
cannot cover all fractions on the real axis.

It absolutely does, for all fractions n.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

2024-11-12 14:30:25 UTC

Post by WM
Therefore I prove it. The set of intervals I(n) = [n - 1/10, n + 1/10]
cannot cover all fractions on the real axis.

It absolutely does, for all fractions n.

The claim however is for all fractions q, most of which are different
from n.

The density is provably 1/5 for all finite initial segments of the real
line. The sequence 1/5, 1/5, 1/5, ... has the limit 1/5. By translating
intervals neither their size nor their multitude changes. Therefore
never more than 1/5 of the real axis is covered. Most rationals remain
naked.

Regards, WM

Jim Burns

2024-11-12 05:00:07 UTC

Post by WM
But it will never complete
an infinite set of claims.

We do not need an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.
We do not want an infinite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀 completed.

But you claim that
_all_ fractions are in bijection with
all natural numbers,
don't you?

Yes, I claim that.

For that claim you need an infinite set of claims.

For that claim, I need
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, in which
each 𝗰𝗹𝗮𝗶𝗺 is true.or.not.first.false,
and in which that is one of the 𝗰𝗹𝗮𝗶𝗺𝘀.

Post by Jim Burns
⎛ All fractions are in bijection with
⎝ all natural numbers.

It is wrong.

It is one claim.

A conceivable argument that
claims.like.that are infinite and
therefore are not expressible finitely
fetches up hard against the bare fact of
it having been expressed, here, a few lines up.

I won't re.present here
a finite 𝘀𝗲𝗾𝘂𝗲𝗻𝗰𝗲 of 𝗰𝗹𝗮𝗶𝗺𝘀, in which
each 𝗰𝗹𝗮𝗶𝗺 is true.or.not.first.false,
and in which that is one of the 𝗰𝗹𝗮𝗶𝗺𝘀.
But I could.

Suppose I did.
Would you (WM) accept that 𝗰𝗹𝗮𝗶𝗺?

Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".

No, I claim that intervals can be translated.

Which means what?
Apparently, to you, it means that
sets change.

However,
our sets do not change.

Post by WM
(The set of intervals remains constant
in size and multitude.)

Our sets do not change.

Post by WM
For every finite subset this is possible.

Our sets do not change.

2024-11-12 14:24:32 UTC

Post by Jim Burns
⎛ All fractions are in bijection with
⎝ all natural numbers.

It is wrong.

It is one claim.

It is one wrong claim.

Post by Jim Burns
Apparently, to you, it means that
sets change.

It means that intervals I(n) = [n - 1/10, n + 1/10] can be shifted on
the real axis without getting larger and without getting more. The set
of intervals cannot change. The set of naturals differs from the set of
rationals. Therefore the rationals cannot be covered by the naturals.

Post by Jim Burns
Our sets do not change.

So you deny the application of set theory to geometry including
Banach-Tarski and therefore to algebra. What remains, what is it good for?

Regards, WM

Jim Burns

2024-11-12 15:58:29 UTC

Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".

No, I claim that intervals can be translated.

By which, you mean that translation changes intervals.

Intervals do not change.
"After" translation of [4-⅒,4+⅒] to [1/3-⅒,1/3+⅒]
[4-⅒,4+⅒] will continue being [4-⅒,4+⅒]
[1/3-⅒,1/3+⅒] will have never been [4-⅒,4+⅒]

Post by WM
(The set of intervals remains constant
in size and multitude.)

The set of intervals remains constant. Absolutely.
Sets do not change.
Intervals do not change.
Mathematical objects do not change.

⎛ But what about descriptions of change?
⎝ Matheologians are famous for those.

The description of a cannon ball's arc,
the description of the beat of a pendulum,
these are what matheologians are famous for.
But these are maps, not journeys.
The maps do not change.

Matheologians make true claims about
each one of infinitely.many journeys,
and then, they augment the description with
further true.or.not.first.false claims.
Without shooting cannons or swinging pendulums,
we know that the further claims must be true.

If the matheologians are clever enough or lucky enough,
then the further true claims eliminate
all but one description, and
we have a map (which has never changed).

The description and the further claims
are finite. Our finiteness is no barrier to saying them,
although we might not be clever or lucky enough
to eliminate all but one.

The description and the further claims
are always and everywhere _about_
the same infinitely.many, not.changing.

2024-11-12 16:43:32 UTC

Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".

No, I claim that intervals can be translated.

By which, you mean that translation changes intervals.

No, the intervals remain constant in size and multitude.

Post by WM
(The set of intervals remains constant
in size and multitude.)

The set of intervals remains constant. Absolutely.
Sets do not change.
Intervals do not change.
Mathematical objects do not change.

Therefore the intervals covering all naturals cannot cover more. But the
rationals are more in the sense that they include all naturals and 1/2.
By your argument Cantor has been falsified.

Regards, WM

Regards, WM

Jim Burns

2024-11-12 19:01:57 UTC

Your 𝗰𝗹𝗮𝗶𝗺𝘀 start with "Sets change".

No, I claim that intervals can be translated.

By which, you mean that translation changes intervals.

No, the intervals remain constant
in size and multitude.

Intervals which are constant _only_
in size and multitude
are not constant absolutely.

Post by WM
(The set of intervals remains constant
in size and multitude.)

The set of intervals remains constant. Absolutely.
Sets do not change.
Intervals do not change.
Mathematical objects do not change.

Therefore
the intervals covering all naturals
cannot cover more.

These intervals
{[n-⅒,n+⅒]: n∈ℕ⁺}
cover all naturals ℕ⁺ and
do not cover all fractions ℕ⁺/ℕ⁺

Post by WM
But the rationals are more in the sense that
they include all naturals and 1/2.

These intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
cover all fractions ℕ⁺/ℕ⁺

These intervals
{[n-⅒,n+⅒]: n∈ℕ⁺}
and these intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
are different intervals.

Our intervals do not change.
Our sets do not change.
Our mathematical objects do not change.

Post by WM
By your argument
Cantor has been falsified.

_What you understand_ as Cantor
has been falsified.

I'd say "misunderstand", but
I do not find it attractive to
dress up as Jacques Derrida and
deconstruct Georg Cantor.
Let someone else do it, if they choose.

That thing you think Cantor says,
whether or not Cantor says it,
is false.

Our sets and other things of their ilk
do not change.
Our finite sequences of claims which
are each true.or.not.first.false
are each true.

2024-11-12 21:38:50 UTC

Post by WM
No, the intervals remain constant
in size and multitude.

Intervals which are constant _only_
in size and multitude
are not constant absolutely.

They would suffer to cover all rationals completely if Cantor's
bijection was complete.

Post by Jim Burns
These intervals
{[n-⅒,n+⅒]: n∈ℕ⁺}
cover all naturals ℕ⁺ and
do not cover all fractions ℕ⁺/ℕ⁺

Right.

Post by WM
But the rationals are more in the sense that
they include all naturals and 1/2.

These intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
cover all fractions ℕ⁺/ℕ⁺

But these are more intervals.

Post by Jim Burns
These intervals
{[n-⅒,n+⅒]: n∈ℕ⁺}
and these intervals
{[i/j-⅒,i/j+⅒]: i/j∈ℕ⁺/ℕ⁺}
are different intervals.

In particular the second kind of intervals must be more. And this is the
solution: The identity of the intervals for the geometric covering is
irrelevant. I will elaborate o this in the next posting.

Regards, WM

Chris M. Thomasson

2024-11-10 20:36:09 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

It cannot do so because the reality of the rationals is much larger than
the reality of the naturals.[...]

Cantor pairing can create a unique pair of natural numbers from a single
natural number. Why do think of rationals at all!? Sigh.

2024-11-11 09:00:11 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

Our sets do not change.

But intervals on the real axis can be translated.

Post by Jim Burns
The set
   {[n-⅒,n+⅒]: n∈ℕ⁺}
with the midpoints at
   ⟨ 1, 2, 3, 4, 5, ... ⟩
does not _change_ to the set
   {[iₙ/jₙ-⅒,iₙ/jₙ+⅒]: n∈ℕ⁺}
with the midpoints at
   ⟨ 1/1, 1/2, 2/1, 1/3, 2/2, ... ⟩

It cannot do so because the reality of the rationals is much larger
than the reality of the naturals.[...]

Cantor pairing can create a unique pair of natural numbers from a single
natural number. Why do think of rationals at all!?

There it is easier to contradict Cantor, because naturals and rationals
can be interpreted as points on the real axis.
If pairing of naturals and rationals is possible for the complete set,
then all intervals with natural numbers as midpoints
I(n) = [n - 1/10, n + 1/10]
can be translated until all rational numbers
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, ...
are midpoints.

Obviously that is impossible because the density 1/5 of the intervals
can never increase. It is possible however to shift an arbitrarily large
(a potentially infinite) number of intervals to rational midpoints.

Regards, WM

Jim Burns

2024-11-11 18:38:56 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

Our sets do not change.

But intervals on the real axis can be translated.

The interval [4-⅒,4+⅒] can be translated to
the interval [1/3-⅒,1/3+⅒].

[4-⅒,4+⅒] does not change to [1/3-⅒,1/3+⅒]
[4-⅒,4+⅒] will continue being after translation.
[1/3-⅒,1/3+⅒] has never been [4-⅒,4+⅒].

Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

Ross Finlayson

2024-11-11 19:00:32 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

Our sets do not change.

But intervals on the real axis can be translated.

The interval [4-⅒,4+⅒] can be translated to
the interval [1/3-⅒,1/3+⅒].
[4-⅒,4+⅒] does not change to [1/3-⅒,1/3+⅒]
[4-⅒,4+⅒] will continue being after translation.
[1/3-⅒,1/3+⅒] has never been [4-⅒,4+⅒].
Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

How about Banach-Tarski equi-decomposability?

Ross Finlayson

2024-11-11 19:04:48 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

Our sets do not change.

But intervals on the real axis can be translated.

The interval [4-⅒,4+⅒] can be translated to
the interval [1/3-⅒,1/3+⅒].
[4-⅒,4+⅒] does not change to [1/3-⅒,1/3+⅒]
[4-⅒,4+⅒] will continue being after translation.
[1/3-⅒,1/3+⅒] has never been [4-⅒,4+⅒].
Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

How about Banach-Tarski equi-decomposability?

(We had a great long thread over on sci.logic
about Banach-Tarski and Vitali-Hausdorff, there's
quite a bit about the historical and technical arrival,
including references and links to Hausdorff's original.

Vitali's doubling-space reflects on "Zeno's graduation
course", where Zeno also has a doubling-space or
doubling-measure argument, since about 2300 years ago.

These are considered part of "mathematics", if
your project is wider than "bumbing- or dumbing-down W.M.".)

Jim Burns

2024-11-11 20:09:34 UTC

Post by Jim Burns
Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

How about Banach-Tarski equi-decomposability?

The parts do not change.

Post by Ross Finlayson
(We had a great long thread over on sci.logic
about Banach-Tarski and Vitali-Hausdorff, there's
quite a bit about the historical and technical arrival,
including references and links to Hausdorff's original.
Vitali's doubling-space reflects on "Zeno's graduation
course", where Zeno also has a doubling-space or
doubling-measure argument, since about 2300 years ago.
These are considered part of "mathematics", if
your project is wider than "bumbing- or dumbing-down W.M.".)

My project is potentially (dare I say it?) much wider
than explanations that sets do not change.

But what do I accomplish by talking about more,
if I say one thing, and a different thing is heard?

2024-11-11 20:40:37 UTC

Post by Jim Burns
Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

How about Banach-Tarski equi-decomposability?

The parts do not change.

Neither do my intervals [4-⅒,4+⅒] = [1/3-⅒,1/3+⅒].

Regards, WM

Jim Burns

2024-11-12 04:32:51 UTC

Post by Jim Burns
Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

How about Banach-Tarski equi-decomposability?

The parts do not change.

Neither do my intervals [4-⅒,4+⅒] = [1/3-⅒,1/3+⅒].

When I first read that,
I thought you meant [4-⅒,4+⅒] = [1/3-⅒,1/3+⅒]
Later,
I thought you meant [4-⅒,4+⅒] ≠ [1/3-⅒,1/3+⅒]

I feel that there's a good chance that
you mean one or the other.

Would you (WM) mind saying which?

2024-11-12 14:10:21 UTC

Post by Ross Finlayson
How about Banach-Tarski equi-decomposability?

The parts do not change.

Neither do my intervals [4-⅒,4+⅒] = [1/3-⅒,1/3+⅒].

When I first read that,
I thought you meant [4-⅒,4+⅒] = [1/3-⅒,1/3+⅒]
Later,
I thought you meant [4-⅒,4+⅒] ≠ [1/3-⅒,1/3+⅒]

Both intervals are one and the same, only shifted a bit. Or is it by
accident that you used n = 4 to cover q = 1/3?
1/1, 1/2, 2/1, 1/3, ...

Regards, WM

Jim Burns

2024-11-12 16:47:34 UTC

Post by Ross Finlayson
How about Banach-Tarski equi-decomposability?

The parts do not change.

Neither do my intervals [4-⅒,4+⅒] = [1/3-⅒,1/3+⅒].

When I first read that,
I thought you meant [4-⅒,4+⅒] = [1/3-⅒,1/3+⅒]
Later,
I thought you meant [4-⅒,4+⅒] ≠ [1/3-⅒,1/3+⅒]

Both intervals are one and the same,
only shifted a bit.

No.
[4-⅒,4+⅒] ≠ [1/3-⅒,1/3+⅒]

⎛ Assume otherwise.
⎜ 4 ∈ [4-⅒,4+⅒]
⎜ Translateᵂᴹ.
⎜ 4 ∉ [4-⅒,4+⅒] = [1/3-⅒,1/3+⅒]
⎜
⎜ ’Twas brillig, and the slithy toves
⎜ Did gyre and gimble in the wabe:
⎜ All mimsy were the borogoves,
⎝ And the mome raths outgrabe.

The borogove.claims are not.first.false.
It doesn't matter where your interval is,
one of those prior claims is false.

But
we can't accept mimsy borogroves and slithy toves.
They're nonsense, of course.
We must surrender our telescope into infinity:
finite sequences of claims which are
only true.or.not.first.false claims.

One can't have both
transmogrifying intervals and
finite sequences giving knowledge of infinity.

Post by WM
Or is it by accident that
you used n = 4 to cover q = 1/3?
1/1, 1/2, 2/1, 1/3, ...

No accident.
You got my point.

2024-11-12 18:06:33 UTC

Post by Jim Burns
[4-⅒,4+⅒] ≠ [1/3-⅒,1/3+⅒]

|[4-⅒,4+⅒]| = |[1/3-⅒,1/3+⅒]| and only that is important for my argument.

Post by WM
Or is it by accident that
you used n = 4 to cover q = 1/3?
1/1, 1/2, 2/1, 1/3, ...

No accident.
You got my point.

Then you will get my point, hopefully.

Regards, WM

Jim Burns

2024-11-12 20:03:14 UTC

Post by Jim Burns
[4-⅒,4+⅒] ≠ [1/3-⅒,1/3+⅒]

|[4-⅒,4+⅒]| = |[1/3-⅒,1/3+⅒]|
and only that is important for my argument.

Yes,
μ[4-⅒,4+⅒] = μ[1/3-⅒,1/3+⅒]

⎛ Also, |[0,1]| = |[0,2]|
⎝ so I think you mean 'measure', not 'cardinality'.

The value of a measure is defined to be
in the extended reals≥0.
μ⋃{ [n-⅒,n+⅒]:n∈ℕ⁺ } ∈ [0,+∞]

The extended reals≥0 are
the Archimedean reals≥0 [0,+∞)
plus one non.Archimedean point +∞
[0,+∞] = [0,+∞)∪{+∞}

The value of μ⋃{ [n-⅒,n+⅒]:n∈ℕ⁺ }
is not any Archimedean point, that is,
there is no finite m ∈ ℕ⁺ such that
μ⋃{ [n-⅒,n+⅒]:n∈ℕ⁺ } ≤ m

The value of μ⋃{ [n-⅒,n+⅒]:n∈ℕ⁺ }
is not Archimedean, so must be +∞
μ⋃{ [n-⅒,n+⅒]:n∈ℕ⁺ } = +∞

Also,
μ⋃{ [i/j-⅒,i/j+⅒]:i/j∈ℕ⁺/ℕ⁺ } = +∞

Post by WM
Or is it by accident that
you used n = 4 to cover q = 1/3?
1/1, 1/2, 2/1, 1/3, ...

No accident.
You got my point.

Then you will get my point, hopefully.

Your point is that
μ⋃{ [n-⅒,n+⅒]:n∈ℕ⁺ }
isn't in the extended reals.

I get it.
Getting it and
agreeing with your changes to our definitions,
changes which turn the discussion into gibberish,
are different.

2024-11-12 21:38:51 UTC

Post by Jim Burns
[4-⅒,4+⅒] ≠ [1/3-⅒,1/3+⅒]

|[4-⅒,4+⅒]| = |[1/3-⅒,1/3+⅒]|
and only that is important for my argument.

Yes,
μ[4-⅒,4+⅒] = μ[1/3-⅒,1/3+⅒]
⎛ Also, |[0,1]| = |[0,2]|
⎝ so I think you mean 'measure', not 'cardinality'.

Yes.

Post by Jim Burns
Your point is that
μ⋃{ [n-⅒,n+⅒]:n∈ℕ⁺ }
isn't in the extended reals.
I get it.

Here is the final solution of the problem:

1) If Cantor was right, then the initial intervals would be sufficient
to cover all rationals in the order 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,
... (these rationals are then centres of intervals).

2) If all rationals could be covered, then this could be accomplished in
arbitrary order, not only along Cantors sequence. Because only the
measure is important, not the identity of an interval.

3) Then we could first cover all naturals and then all halves and then
all quarters and so on. But we know that already after covering all
naturals no further intervals are available.

Regards, WM

joes

2024-11-12 22:03:06 UTC

Post by WM
3) Then we could first cover all naturals and then all halves and then
all quarters and so on. But we know that already after covering all
naturals no further intervals are available.

Well, because this order has the type
omega + omega + … omega = omega*omega = omega^2.
This amounts to saying that the naturals are a subset of the rationals.
It goes back to the lines (or columns) of your tired matrix.
That is not a bijection between N and Q. That doesn’t prove there is none.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.

Ross Finlayson

2024-11-11 20:59:59 UTC

Post by Jim Burns
Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

How about Banach-Tarski equi-decomposability?

The parts do not change.

My project is potentially (dare I say it?) much wider
than explanations that sets do not change.
But what do I accomplish by talking about more,
if I say one thing, and a different thing is heard?

Well, you see, the story goes, that a mathematical ball,
with a unit volume, is according to measure, that the
theory of measure, indicates that any manner of partitioning
said ball or its decomposition, would result in whatever
re-composition, a volume, the same.

Yet, according to the theory, the Banach-Tarski "equi-decomposability"
arrives instead at making two identical copies.

This is quite after Hausdorff established this geometrically,
and quite after Vitali made a unit line segment, partitioned,
by "not-equi-decomposed", re-composed, to a length L where 1 < L < 3,
or that exactly, L = 2, which is exactly twice, double, the L = 1.

Then, it's said that von Neumann in one of his phases
established quite a variety of similar results on the plane.
Though, I imagine he had his influences or who really
came up with these ideas.

Yet, it's to Vitali to whom this great and important aspect
of mathematics is assigned, that broke standard measure theory
and made "non-measurable" out of things.

While that is so, we can point at "Zeno's graduation course"
or the many runners doing wind-sprints on the field, and
how it's so that he shows it add up to twice, when otherwise
these days the bumblebee flying back and forth between
the oncoming trains is simply derived the distance from the time.

So, "mathematics" is rather an inclusive endeavor,
and it includes these things.

Then, given that WM is a nuisance soft-ball straw-man,
then, what are we to make of this observation that
"intervals could grow in length", under what transformations?

It's called doubling-spaces and doubling-measure and
with regards to halving-spaces and halving-measure,
and it's a feature of continuum and, as it's called,
infinitesimal analysis.

The, ..., "re-Vitali-ization", or, "quasi-invariant",
measure theory, ....

It's a thing, ....

Look back on the discussion on sci.logic "Banach-Tarski",
you'll find learnings.

Ross Finlayson

2024-11-12 17:40:21 UTC

Post by Jim Burns
Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

How about Banach-Tarski equi-decomposability?

The parts do not change.

My project is potentially (dare I say it?) much wider
than explanations that sets do not change.
But what do I accomplish by talking about more,
if I say one thing, and a different thing is heard?

Well, you see, the story goes, that a mathematical ball,
with a unit volume, is according to measure, that the
theory of measure, indicates that any manner of partitioning
said ball or its decomposition, would result in whatever
re-composition, a volume, the same.
Yet, according to the theory, the Banach-Tarski "equi-decomposability"
arrives instead at making two identical copies.
This is quite after Hausdorff established this geometrically,
and quite after Vitali made a unit line segment, partitioned,
by "not-equi-decomposed", re-composed, to a length L where 1 < L < 3,
or that exactly, L = 2, which is exactly twice, double, the L = 1.
Then, it's said that von Neumann in one of his phases
established quite a variety of similar results on the plane.
Though, I imagine he had his influences or who really
came up with these ideas.
Yet, it's to Vitali to whom this great and important aspect
of mathematics is assigned, that broke standard measure theory
and made "non-measurable" out of things.
While that is so, we can point at "Zeno's graduation course"
or the many runners doing wind-sprints on the field, and
how it's so that he shows it add up to twice, when otherwise
these days the bumblebee flying back and forth between
the oncoming trains is simply derived the distance from the time.
So, "mathematics" is rather an inclusive endeavor,
and it includes these things.
Then, given that WM is a nuisance soft-ball straw-man,
then, what are we to make of this observation that
"intervals could grow in length", under what transformations?
It's called doubling-spaces and doubling-measure and
with regards to halving-spaces and halving-measure,
and it's a feature of continuum and, as it's called,
infinitesimal analysis.
The, ..., "re-Vitali-ization", or, "quasi-invariant",
measure theory, ....
It's a thing, ....
Look back on the discussion on sci.logic "Banach-Tarski",
you'll find learnings.

So, do you reject the existence of these?

Mathematics doesn't, ....

Jim Burns

2024-11-12 21:36:08 UTC

Post by Jim Burns
Our sets do not change.
Everybody who believes that
intervals could grow in length or number
is deeply mistaken about
what our whole project is.

How about Banach-Tarski equi-decomposability?

The parts do not change.

any manner of partitioning said ball or its decomposition,
would result in whatever re-composition,
a volume, the same.

So, do you reject the existence of these?

No.

What I mean by "The parts do not change" might be
too.obvious for you to think useful.to.state.
Keep in mind with whom I am primarily in discussion.
I am of the strong opinion that
"too obvious" is not possible, here.

Finitely.many pieces of the ball.before are
associated.by.rigid.rotations.and.translations to
finitely.many pieces of two same.volumed balls.after.

They are associated pieces.
They are not the same pieces.

Galileo found it paradoxical that
each natural number can be associated with
its square, which is also a natural number.
But 137 is associated with 137²
137 isn't 137²

I don't mean anything more than that.
I hope you agree.

Post by Ross Finlayson
Mathematics doesn't, ....

Mathematics thinks 137 ≠ 137²

2024-11-11 20:37:15 UTC

Post by WM
But intervals on the real axis can be translated.

The interval [4-⅒,4+⅒] can be translated to
the interval [1/3-⅒,1/3+⅒].

Obviously.

Post by Jim Burns
[4-⅒,4+⅒] does not change to [1/3-⅒,1/3+⅒]
[4-⅒,4+⅒] will continue being after translation.
[1/3-⅒,1/3+⅒] has never been [4-⅒,4+⅒].

That denies geometry. I use geometry. Your set theory must shy away from
it. Geometry can be expressed by algebra. What remains for your set theory?

Regards, WM

Chris M. Thomasson

2024-11-11 21:42:51 UTC

Post by WM
Everybody who believes that the intervals
I(n) = [n - 1/10, n + 1/10]
could grow in length or number
to cover the whole real axis
is a fool or worse.

Our sets do not change.

But intervals on the real axis can be translated.

It cannot do so because the reality of the rationals is much larger
than the reality of the naturals.[...]

Cantor pairing can create a unique pair of natural numbers from a
single natural number. Why do think of rationals at all!?

There it is easier to contradict Cantor, because naturals and rationals
can be interpreted as points on the real axis.
If pairing of naturals and rationals is possible for the complete set,
then all intervals with natural numbers as midpoints
I(n) = [n - 1/10, n + 1/10]
can be translated until all rational numbers
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, ...
are midpoints.
Obviously that is impossible because the density 1/5 of the intervals
can never increase. It is possible however to shift an arbitrarily large
(a potentially infinite) number of intervals to rational midpoints.

I don't think you know how to take any natural number and turn it into a
unique pair, and then back again via Cantor pairing.

2024-11-12 14:05:18 UTC

Post by WM
I(n) = [n - 1/10, n + 1/10]
can be translated until all rational numbers
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, ...
are midpoints.
Obviously that is impossible because the density 1/5 of the intervals
can never increase. It is possible however to shift an arbitrarily
large (a potentially infinite) number of intervals to rational midpoints.

I don't think you know how to take any natural number and turn it into a
unique pair, and then back again via Cantor pairing.

I don't think that you understand what's going on here.

Regards, WM

Chris M. Thomasson

2024-11-12 22:00:44 UTC

I don't think you know how to take any natural number and turn it into
a unique pair, and then back again via Cantor pairing.

I don't think that you understand what's going on here.

I understand Cantor pairing quite well. You are the one with the weird
problem.

Ross Finlayson

2024-11-06 17:31:44 UTC

Post by WM
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.

⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.

This plentiness does not change
when the intervals are translated.

⎛ When the intervals are end.to.end.to.end,
⎜ there are exterior points
⎝ a distance 10¹⁰⁰⁰⁰⁰ from any interval.
Are there points 10¹⁰⁰⁰⁰⁰ from any interval
when midpoints of intervals include
each of {...,-3,-2,-1,0,1,2,3,...} ?
Isn't that a plentiness which changes?

Post by Jim Burns
I mean 'exterior' in the topological sense.
For a point x in the boundary ∂A of set A
each open set Oₓ which holds x
holds points in A and points not.in A

The intervals are closed with irrational endpoints.

'Exterior' seems like a good way to say
'not in contact'.
It seems to me that you have a better argument
with open intervals instead of closed,
but let them be closed, if you like.
Either way,
there are no points 10¹⁰⁰⁰⁰⁰ from any interval.

Nice try.
But there are points outside of intervals,

Are any of these points.outside
⅟2 from any interval? ⅟3? ⅟4?

Post by WM
and they are closer to interval ends
than to the interior, independent of
the configuration of the intervals.

Shouldn't I be pointing that out
to you?
If there is no point with more.than.⅟2
between it and any midpoint,
shouldn't there be fewer.than.no points
with more.than.⅟2 between it and
any closer endpoint?

Post by WM
Note that
only 3/oo of the points are inside.

Yes, less than 2³ᐟ²⋅ε
If the intervals were open,
all of that would be "inside"
in the interior of their union.
Of the rest,
none of it is more.than.⅟2 from any interval.

Post by Jim Burns
An exterior point which is not
a positive distance from any interval
is not an exterior point.

Positive is what you can define,

Positive ℕ⁺ holds countable.to from.1
Positive ℚ⁺ holds ratios of elements of ℕ⁺
Positive ℝ⁺ holds points.between.splits of Q⁺

Post by WM
but there is much more in smaller distance.

Distances are positive or zero.
Two distinct points are a positive distance apart.

Post by WM
Remember the infinitely many unit fractions
within every eps > 0 that you can define.

For each of the infinitely.many unit fractions
there is no point a distance of that unit fraction
or more from any interval.

Post by Jim Burns
Therefore,
in what is _almost_ your conclusion,
there are no exterior points.

There are 3/oo of all points exterior.

Did you intend to write "interior"?
An exterior point is in
an open interval holding no rational.
There are no
open intervals holding no rational.
There are no exterior points.
However,
there are boundary points.
All but 2³ᐟ²⋅ε are boundary points.

The intervals are closed

We are only told
that Oₓ is an open set holding x
not that Oₓ is one of the ε.cover of ℚ
The question is whether x is a boundary point.

Post by WM
The rationals are dense

Yes.
Each multi.point interval [x,x′] holds
rationals.

Post by WM
but the intervals are not.

No.
Each multi.point interval [x,x′] holds
ε.cover intervals.

Therefore not all rationals are enumerated.

Explain why.
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ 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 Jim Burns
proves that
the rationals are countable.

Contradiction.

It contradicts a non.empty exterior.
It doesn't contradict an almost.all boundary.

Post by WM
Something of your theory is inconsistent.

Your intuition is disturbed by
an almost.all boundary.
Disturbed intuitions and inconsistencies
are different.

The definitions of "functions", and "topologies",
vary, and while each may be considered as their
own sort of theory, for relations and continua or wholes,
as well, in various theories like "ZF" or "when
the topology is defined as the open topology",
with regards to metric spaces and so on, it can help
to understand that in the usual account of descriptive
set theory, they are _opinions_ and there are _others_.

So, WM's "disturbed intuitions" here reflect variously
on unstated assumptions, or un-necessarily opinionated
or simply oppositely opinionated, with regards to the
theories, each their own, then as with regards to most
common aspects of:

geometry
number theory

arithmetic
algebra

function theory
topology

operator calculus

that it comes back around to geometry, that each these
has their theory where they're primary, i.e. elementary
in the theory and all else derived from them or defined
after them: a sort of heno-theory, with regards to
foundations from the many perspectives and the many
perspectives of foundations.

Then, when fields make their own restrictions of comprehension,
definitions being stipulations and from them are derivable
contradictions, then those are no longer foundations, instead
"practical theories", of sorts, where for example "functions
aren't necessarily ZF's Cartesian functions" and "topologies
aren't necessarily the standard open topology", with respect
to otherwise the central and primary objects of the universe
of mathematics, which according to theories of types, for
example, according to model theory, are either equi-interpretable,
or, aren't, AND, making a deconstructive account into assumptions
or stipulations, DOES make a wider and fuller dialectic, which
MAY thusly point out that thusly non-logical assumptions or
stipulations may be negated, logically.

Or, "ZF set theory with respect to number theory or geometry,
is, at best: incomplete, and in some theories, there are
counterexamples to otherwise its results, according to each
non-logical stipulation, or restriction of comprehension".

Ross Finlayson

2024-11-06 17:53:37 UTC

Post by WM
The intervals together cover a length of less than 3.
The whole length is infinite.
Therefore there is plenty of space for
a point not in contact with any interval.

⎛ Assuming the covering intervals are translated
⎜ to where they are end.to.end.to.end,
⎜ there is plenty of space for
⎝ not.in.contact exterior points.

This plentiness does not change
when the intervals are translated.

⎛ When the intervals are end.to.end.to.end,
⎜ there are exterior points
⎝ a distance 10¹⁰⁰⁰⁰⁰ from any interval.
Are there points 10¹⁰⁰⁰⁰⁰ from any interval
when midpoints of intervals include
each of {...,-3,-2,-1,0,1,2,3,...} ?
Isn't that a plentiness which changes?

Post by Jim Burns
I mean 'exterior' in the topological sense.
For a point x in the boundary ∂A of set A
each open set Oₓ which holds x
holds points in A and points not.in A

The intervals are closed with irrational endpoints.

'Exterior' seems like a good way to say
'not in contact'.
It seems to me that you have a better argument
with open intervals instead of closed,
but let them be closed, if you like.
Either way,
there are no points 10¹⁰⁰⁰⁰⁰ from any interval.

Nice try.
But there are points outside of intervals,

Are any of these points.outside
⅟2 from any interval? ⅟3? ⅟4?

Post by WM
and they are closer to interval ends
than to the interior, independent of
the configuration of the intervals.

Shouldn't I be pointing that out
to you?
If there is no point with more.than.⅟2
between it and any midpoint,
shouldn't there be fewer.than.no points
with more.than.⅟2 between it and
any closer endpoint?

Post by WM
Note that
only 3/oo of the points are inside.

Yes, less than 2³ᐟ²⋅ε
If the intervals were open,
all of that would be "inside"
in the interior of their union.
Of the rest,
none of it is more.than.⅟2 from any interval.

Post by Jim Burns
An exterior point which is not
a positive distance from any interval
is not an exterior point.

Positive is what you can define,

Positive ℕ⁺ holds countable.to from.1
Positive ℚ⁺ holds ratios of elements of ℕ⁺
Positive ℝ⁺ holds points.between.splits of Q⁺

Post by WM
but there is much more in smaller distance.

Distances are positive or zero.
Two distinct points are a positive distance apart.

Post by WM
Remember the infinitely many unit fractions
within every eps > 0 that you can define.

For each of the infinitely.many unit fractions
there is no point a distance of that unit fraction
or more from any interval.

Post by Jim Burns
Therefore,
in what is _almost_ your conclusion,
there are no exterior points.

There are 3/oo of all points exterior.

Did you intend to write "interior"?
An exterior point is in
an open interval holding no rational.
There are no
open intervals holding no rational.
There are no exterior points.
However,
there are boundary points.
All but 2³ᐟ²⋅ε are boundary points.

The intervals are closed

We are only told
that Oₓ is an open set holding x
not that Oₓ is one of the ε.cover of ℚ
The question is whether x is a boundary point.

Post by WM
The rationals are dense

Yes.
Each multi.point interval [x,x′] holds
rationals.

Post by WM
but the intervals are not.

No.
Each multi.point interval [x,x′] holds
ε.cover intervals.

Therefore not all rationals are enumerated.

Explain why.
⎛ i/j ↦ kᵢⱼ = (i+j-1)(i+j-2)/2+i
⎜ k ↦ 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 Jim Burns
proves that
the rationals are countable.

Contradiction.

It contradicts a non.empty exterior.
It doesn't contradict an almost.all boundary.

Post by WM
Something of your theory is inconsistent.

Your intuition is disturbed by
an almost.all boundary.
Disturbed intuitions and inconsistencies
are different.

The definitions of "functions", and "topologies",
vary, and while each may be considered as their
own sort of theory, for relations and continua or wholes,
as well, in various theories like "ZF" or "when
the topology is defined as the open topology",
with regards to metric spaces and so on, it can help
to understand that in the usual account of descriptive
set theory, they are _opinions_ and there are _others_.
So, WM's "disturbed intuitions" here reflect variously
on unstated assumptions, or un-necessarily opinionated
or simply oppositely opinionated, with regards to the
theories, each their own, then as with regards to most
geometry
number theory
arithmetic
algebra
function theory
topology
operator calculus
that it comes back around to geometry, that each these
has their theory where they're primary, i.e. elementary
in the theory and all else derived from them or defined
after them: a sort of heno-theory, with regards to
foundations from the many perspectives and the many
perspectives of foundations.
Then, when fields make their own restrictions of comprehension,
definitions being stipulations and from them are derivable
contradictions, then those are no longer foundations, instead
"practical theories", of sorts, where for example "functions
aren't necessarily ZF's Cartesian functions" and "topologies
aren't necessarily the standard open topology", with respect
to otherwise the central and primary objects of the universe
of mathematics, which according to theories of types, for
example, according to model theory, are either equi-interpretable,
or, aren't, AND, making a deconstructive account into assumptions
or stipulations, DOES make a wider and fuller dialectic, which
MAY thusly point out that thusly non-logical assumptions or
stipulations may be negated, logically.
Or, "ZF set theory with respect to number theory or geometry,
is, at best: incomplete, and in some theories, there are
counterexamples to otherwise its results, according to each
non-logical stipulation, or restriction of comprehension".

Or, either of yours otherwise non-logical, stipulated exceptions.

In "Moment and Motion: ideals and foundations", Sep. 29 2024,

there's considered "complementary duals" and as with
regards to absolutes and ideals, what would of course
be ultimately _relevant_, in foundations and mathematical theory.

2024-11-06 16:36:43 UTC

For every definable x we can decide whether it is inside the interval
including its endpoints or outside. No open intervals are necesseary or
useful. Your trick is cunning bot not accepted.

Regards, WM

Chris M. Thomasson

2024-11-06 20:59:27 UTC

Post by Jim Burns
Instead, there are boundary points.
  For each x not.in the intervals,
  each open set Oₓ which holds x
  holds points in the intervals and
  points not.in the intervals.
x is a boundary point.

For every definable x we can decide whether it is inside the interval
including its endpoints or outside. No open intervals are necesseary or
useful. Your trick is cunning bot not accepted.

For any z we can decide if its inside or outside of an escape time fractal.

Moebius

2024-11-07 00:00:11 UTC

Post by Jim Burns
Instead, there are boundary points.
  For each x not.in the intervals,
  each open set Oₓ which holds x
  holds points in the intervals and
  points not.in the intervals.
x is a boundary point.

For every definable x we can decide whether it is inside the interval
including its endpoints or outside. No open intervals are necesseary
or useful. Your trick is cunning bot not accepted.

For any z we can decide if its inside or outside of <whatever>

Hint: In contrast to WM's psychomath we do not have to "decide" if a
number z is inside or outside of a certain interval. It IS EITHER inside
of the interval OR NOT. (No "decision" necessary.)

2024-11-07 09:04:42 UTC

Hint: we do not have to "decide" if a
number z is inside or outside of a certain interval. It IS EITHER inside
of the interval OR NOT. (No "decision" necessary.)

And when we cover the real axis by intervals
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
J(n) = [n - √2/10, n + √2/10]
in a clever way, then all rational numbers are midpoints of intervals
and no irrational number is outside of all intervals.
That is the power of infinity!!!

Regards, WM

Chris M. Thomasson

2024-11-07 21:02:41 UTC

Instead, there are boundary points.
  For each x not.in the intervals,
  each open set Oₓ which holds x
  holds points in the intervals and
  points not.in the intervals.
x is a boundary point.

For every definable x we can decide whether it is inside the interval
including its endpoints or outside. No open intervals are necesseary
or useful. Your trick is cunning bot not accepted.

For any z we can decide if its inside or outside of <whatever>

Hint: In contrast to WM's psychomath we do not have to "decide" if a
number z is inside or outside of a certain interval. It IS EITHER inside
of the interval OR NOT. (No "decision" necessary.)

Indeed!

Chris M. Thomasson

2024-11-04 20:54:26 UTC

[...]

Further there are never
two irrational numbers
without an interval between them.

Unless they equal each other? ;^)

Jim Burns

2024-11-04 22:02:47 UTC

Further there are never
two irrational numbers
without an interval between them.

Unless they equal each other? ;^)

And thus are one, not two, points?

Chris M. Thomasson

2024-11-04 23:08:25 UTC

Further there are never
two irrational numbers
without an interval between them.

Unless they equal each other? ;^)

And thus are one, not two, points?

Well, yeah in a sense.

A = (-1, .5, 3, 7)
B = (-1, .5, 3, 7)

A = B = true

;^)

Moebius

2024-11-05 00:07:42 UTC

Further there are never
two irrational numbers
without an interval between them.

Unless they equal each other? ;^)

And thus are one, not two, points?

Well, yeah in a sense.
A = (-1, .5, 3, 7)
B = (-1, .5, 3, 7)
A = B = true

Well, rather:

"A = B" is true.

After all, (-1, .5, 3, 7) =/= true (I'd say). :-P

__________________________________________________

Actually, some authors would prefer to write:

"there are never two different irrational numbers without an [nonempty]
interval between them".

It's a matter of "style" or "precision", if you like.

If we talk about "two" real numbers x, y [in math], usually x = y is NOT
excluded.

For example,

for any two real numbers x, y ... bla bla

USUALLY does not exclude x = y. [Might look strange, but is just the
usual math lingo.]

.
.
.

Moebius

2024-11-05 00:18:58 UTC

Post by Moebius
For example,
for any two real numbers x, y ... bla bla
USUALLY does not exclude x = y. [Might look strange, but is just the
usual math lingo.]

Hint: Archimedes Plutonium hat some problems with the convention. Seems
that Jim wants to follow his lead.

Post by Moebius
.
.
.

Ross Finlayson

2024-11-06 01:49:47 UTC

Post by Moebius
For example,
for any two real numbers x, y ... bla bla
USUALLY does not exclude x = y. [Might look strange, but is just the
usual math lingo.]

Hint: Archimedes Plutonium hat some problems with the convention. Seems
that Jim wants to follow his lead.

Post by Moebius
.
.
.

It's fair to say that Burns is a great example of dogmatic correctness.

There are others, there's a fuller milieu of "the theory",
as to what constitutes "greater dogmatic correctness".

Burns, or Professor Burns as I sometimes refer to him,
to which he has modestly demurred, is quite great and
here among the best of both foils and straight-men,
even were he a bot, in constancy and a continued
development, and a personal style helping write
first-class logic, in natural language.

Trichotomy, the property for two real numbers that
exactly one of x = y or x < y or x > y, and that
being a transitive property, is about a most fundamental
aspect of "order" or "ordering" theory, where "orderings"
are the fundamental elements of the theory, much like
in "set" theory were sets are fundamental, and elementary.

Then, with regards to ordering and lack thereof, there
is a notion about sampling from trichotomous domains,
what results a "vague fugue", as with regards to that
it's not so much guaranteed that two samples, after
error in precision, close enough together, are ordered.

Then, "sampling, observer, and measurement effects",
generally are seen to include that this makes for
what's called "interval arithmetic" with regards to
the representation of real-valued variables in
fixed or for that matter arbitrary precision machine
numbers.

AP, if you don't recall about 30 years ago, was pretty
great, for a couple notions including "universal atomism"
and "atomic universalism", then though it's largely agreed
that his working or practical theory, if that's not too
generous an appelation, was of insufficient probity
for a candidate in foundations.

So, Burns then is of certainly a high measure of reliability,
then as with regards to a certain perceived inflexibility,
it's not uncommon with regards to the "inductive impasse"
and the role of deductive analysis and the fuller dialectic
in the "analytical bridges (analytische bruecken)", that
the linear curriculum barely has enough time to get one
formalism in place, then that the wider and fuller curriculum,
is largely for more thoroughly logical and widely-mindful
students of _all_ the theory, vis-a-vis, practical working theory.

Even if that's a bot, ..., of which there are plenty.

Moebius

2024-11-05 00:19:50 UTC

Post by Moebius
For example,
for any two real numbers x, y ... bla bla
USUALLY does not exclude x = y. [Might look strange, but is just the
usual math lingo.]

Hint: Archimedes Plutonium had some problems with the convention. Seems
that Jim wants to follow his lead.

Post by Moebius
.
.
.

Moebius

2024-11-05 00:20:19 UTC

Post by Moebius
For example,
for any two real numbers x, y ... bla bla
USUALLY does not exclude x = y. [Might look strange, but is just the
usual math lingo.]

Hint: Archimedes Plutonium had some problems with that convention. Seems
that Jim wants to follow his lead.

Post by Moebius
.
.
.

Moebius

2024-11-05 02:27:35 UTC

Post by Moebius
Hint: Archimedes Plutonium had some problems with that convention. Seems
that Jim wants to follow his lead.

Actually, I'd recommend not to follow JB, FtA and/or RD blindly.
Sometimes they just talk nonsense.

Moebius

2024-11-05 02:31:13 UTC