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Three incompatible facts require dark numbers
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Ganzhinterseher
Nov 16, 2019, 11:26:55 AM11/16/19
to
1) Every endsegment E(n) = {n, n+1, n+2, ...} of the set of natural numbers reduces the intersection of endsegments in natural order
E(1) ∩ E(2) ∩ E(3) ∩ ...
by one and only one natural number, namely n-1.
2) Every endsegment is infinite and leaves infinitely many numbers in the intersection:
∀n ∈ ℕ ∩{E(1), E(2), ..., E(n)} = E(n) =/= { } .
3) The intersection of all endsegments is empty
E(1) ∩ E(2) ∩ E(3) ∩ ... = { } .
If (3) is true, then (1) requires that endsegments are existing that reduce the intersection to a finite set of n, n-1, ..., 3, 2, 1 elements. But such endsegments cannot be identified because of (2).
Another proof of dark numbers is caused by the axiom of foundation according to which very descending chain of ordinals is finite. Counting back from ω by definable numbers we arive in a finite number of steps at zero. That implies that infinitely many natural numbers between 0 and ω are not touched because they are not available for counting.
Regards, WM
E(1) ∩ E(2) ∩ E(3) ∩ ...
by one and only one natural number, namely n-1.
2) Every endsegment is infinite and leaves infinitely many numbers in the intersection:
∀n ∈ ℕ ∩{E(1), E(2), ..., E(n)} = E(n) =/= { } .
3) The intersection of all endsegments is empty
E(1) ∩ E(2) ∩ E(3) ∩ ... = { } .
If (3) is true, then (1) requires that endsegments are existing that reduce the intersection to a finite set of n, n-1, ..., 3, 2, 1 elements. But such endsegments cannot be identified because of (2).
Another proof of dark numbers is caused by the axiom of foundation according to which very descending chain of ordinals is finite. Counting back from ω by definable numbers we arive in a finite number of steps at zero. That implies that infinitely many natural numbers between 0 and ω are not touched because they are not available for counting.
Regards, WM
FredJeffries
Nov 16, 2019, 4:00:18 PM11/16/19
to
On Saturday, November 16, 2019 at 8:26:55 AM UTC-8, Ganzhinterseher wrote:
> 1) Every endsegment E(n) = {n, n+1, n+2, ...} of the set of natural numbers reduces the intersection of endsegments in natural order
>
> E(1) ∩ E(2) ∩ E(3) ∩ ...
>
> by one and only one natural number, namely n-1.
Your 'E(1) ∩ E(2) ∩ E(3) ∩ ...' is complete and utter gibberish.
> 1) Every endsegment E(n) = {n, n+1, n+2, ...} of the set of natural numbers reduces the intersection of endsegments in natural order
>
> E(1) ∩ E(2) ∩ E(3) ∩ ...
>
> by one and only one natural number, namely n-1.
EOD
George Greene
Nov 17, 2019, 12:21:50 AM11/17/19
to
On Saturday, November 16, 2019 at 11:26:55 AM UTC-5, Ganzhinterseher wrote:
> 1) Every endsegment E(n) = {n, n+1, n+2, ...} of the set of natural numbers reduces the intersection of endsegments in natural order
The intersection operation in set theory IS COMMUTATIVE AND ASSOCIATIVE, YOU IDIOT. NO application of it happens IN *ANY* ORDER among the operands.
> 1) Every endsegment E(n) = {n, n+1, n+2, ...} of the set of natural numbers reduces the intersection of endsegments in natural order
Given a set of sets on which you are trying to compute an intersection,
you might as well think of ALL the intersections (despite the fact that
conjunction is generally thought of as binary) as happening IN PARALLEL
because THE ORDER *N*E*V*E*R* Matters.
Ganzhinterseher
Nov 17, 2019, 5:09:09 AM11/17/19
to
Am Sonntag, 17. November 2019 06:21:50 UTC+1 schrieb George Greene:
> On Saturday, November 16, 2019 at 11:26:55 AM UTC-5, Ganzhinterseher wrote:
> > 1) Every endsegment E(n) = {n, n+1, n+2, ...} of the set of natural numbers reduces the intersection of endsegments in natural order
>
> The intersection operation in set theory IS COMMUTATIVE AND ASSOCIATIVE^
> On Saturday, November 16, 2019 at 11:26:55 AM UTC-5, Ganzhinterseher wrote:
> > 1) Every endsegment E(n) = {n, n+1, n+2, ...} of the set of natural numbers reduces the intersection of endsegments in natural order
>
We can apply it in the given order:
E(1) ∩ E(2) ∩ E(3) ∩ ...
NO application of it happens IN *ANY* ORDER among the operands.
∀n ∈ ℕ ∩{E(1), E(2), E(3), ..., E(n)}
.
> Given a set of sets on which you are trying to compute an intersection,
> you might as well think of ALL the intersections (despite the fact that
> conjunction is generally thought of as binary) as happening IN PARALLEL
> because THE ORDER *N*E*V*E*R* Matters.
It does matter in mathematics. That's why you try to deny it. We have the sequence (a_n) with
> Given a set of sets on which you are trying to compute an intersection,
> you might as well think of ALL the intersections (despite the fact that
> conjunction is generally thought of as binary) as happening IN PARALLEL
> because THE ORDER *N*E*V*E*R* Matters.
a_n = ∩{E(1), E(2), E(3), ..., E(n)}.
Alas between the existing terms of this sequence and the final result
∩{E(1), E(2), E(3), ...} = { }
there exists a gap. If the final empty result is brought about by endsegments in mathematical fashion, i.e., removing one element each, then there must be last steps which cannot be enumerated although they must exist.
Regards, WM
Transfinite Numbers
Nov 17, 2019, 5:25:19 AM11/17/19
to
An intersection of end segments either results
in an end segment (finite many end segments were
intersected) or in the empty set (infinitely many
end segments were intersected). There is nothing
inbetween, even if when wish for it very hard,
it will not appear in ZFC.
in an end segment (finite many end segments were
intersected) or in the empty set (infinitely many
end segments were intersected). There is nothing
inbetween, even if when wish for it very hard,
it will not appear in ZFC.
Ganzhinterseher
Nov 17, 2019, 9:23:04 AM11/17/19
to
Am Sonntag, 17. November 2019 11:25:19 UTC+1 schrieb Transfinite Numbers:
> An intersection of end segments either results
> in an end segment (finite many end segments were
> intersected)
This holds for all endsegments that can be defined, i.e., for infinitely many endsegments:
> An intersection of end segments either results
> in an end segment (finite many end segments were
> intersected)
∀k ∈ ℕ ∩{E(1), E(2), ..., E(k)} = E(k) mit |E(k)| = ℵo . (*)
> or in the empty set (infinitely many
> end segments were intersected).
> There is nothing
> inbetween, even if when wish for it very hard,
> it will not appear in ZFC.
If set theory is the foundation of this mathematics, it is inconsistent. Otherwise it is useless.
Regards, WM
j4n bur53
Nov 17, 2019, 12:43:18 PM11/17/19
to
Since deMorgan juggling with Endsegments E(k) is the
same as juggling with Initialsegments A(k):
∩ E(k+1) = ∩ (N \ A(k))
k e N k e N
= N \ ∪ A(k)
k e N
= N \ N
So there is the following duality in your
annoying obsfuscation nonsense:
Endsegments Initialsegmemts
Notation E(k) A(k)
Operation ∩ ∪
Result when finitely many operands to operation:
E(k) A(k)
Result when infinitely many operands to operation:
{} N
WMs crank claim:
dark numbers dark numbers
WMs crank argument:
E(k)-1={} A(k)+1=omega
Once a crank, always a crank!
Ganzhinterseher schrieb:
same as juggling with Initialsegments A(k):
∩ E(k+1) = ∩ (N \ A(k))
k e N k e N
= N \ ∪ A(k)
k e N
= N \ N
So there is the following duality in your
annoying obsfuscation nonsense:
Endsegments Initialsegmemts
Notation E(k) A(k)
Operation ∩ ∪
Result when finitely many operands to operation:
E(k) A(k)
Result when infinitely many operands to operation:
{} N
WMs crank claim:
dark numbers dark numbers
WMs crank argument:
E(k)-1={} A(k)+1=omega
Once a crank, always a crank!
Ganzhinterseher schrieb:
j4n bur53
Nov 17, 2019, 12:45:12 PM11/17/19
to
https://en.wikipedia.org/wiki/De_Morgan's_laws
Thats basic hermeneutics, "Quod est superius,
est sicut (id) quod est inferius." If you now
count downwards, this doesn't improve your
nonsense, it doesn't open a magical door,
so that suddently your wishful thinking of
dark numbers comes true. sup and inf are
linked by deMorgan laws. That A(k) doesn't
reach omega, i.e. always A(k)+1=/=omega, cannot
be circumvented by a headstand and by operating
with E(k) instead. The picture of finite and
infinite is the same, even if you turn your
head 180° degree and work with the complement.
Even a crazy from Ausgburg Crank Institute,
who has only Volkswagen Omlette for brains,
should see that from the beginning.
But when there are no brain cells,
well then there are no brain cells.
j4n bur53 schrieb:
Thats basic hermeneutics, "Quod est superius,
est sicut (id) quod est inferius." If you now
count downwards, this doesn't improve your
nonsense, it doesn't open a magical door,
so that suddently your wishful thinking of
dark numbers comes true. sup and inf are
linked by deMorgan laws. That A(k) doesn't
reach omega, i.e. always A(k)+1=/=omega, cannot
be circumvented by a headstand and by operating
with E(k) instead. The picture of finite and
infinite is the same, even if you turn your
head 180° degree and work with the complement.
Even a crazy from Ausgburg Crank Institute,
who has only Volkswagen Omlette for brains,
should see that from the beginning.
But when there are no brain cells,
well then there are no brain cells.
j4n bur53 schrieb:
Ganzhinterseher
Nov 17, 2019, 12:52:23 PM11/17/19
to
Am Sonntag, 17. November 2019 18:43:18 UTC+1 schrieb Transfinite Numbers:
> Since deMorgan juggling with Endsegments E(k) is the
> same as juggling with Initialsegments A(k):
That is correct. It is as impossible to complete infinity as to empty it. But the chances are better that some unbiased reader will understand the latter:
> Since deMorgan juggling with Endsegments E(k) is the
> same as juggling with Initialsegments A(k):
When a set is emptied by a tool always subtracting only one element, then it cannot get from infinity to zero in less than infinitely many steps.
Regards, WM
j4n bur53
Nov 17, 2019, 1:06:49 PM11/17/19
to
Yes n+1=/=omega, you cannot complete infinity
by your finite intersection, we have forall n:
n {E(1),..,E(n)} =/= {}
On the other hand the infinite intersection is:
n {E(1),E(2),..} = {}
There is no Mückenschuss, and no dark numbers.
Ganzhinterseher schrieb:
by your finite intersection, we have forall n:
n {E(1),..,E(n)} =/= {}
On the other hand the infinite intersection is:
n {E(1),E(2),..} = {}
There is no Mückenschuss, and no dark numbers.
Ganzhinterseher schrieb:
Ganzhinterseher
Nov 17, 2019, 1:29:35 PM11/17/19
to
Am Sonntag, 17. November 2019 19:06:49 UTC+1 schrieb Transfinite Numbers:
> Yes n+1=/=omega, you cannot complete infinity
> by your finite intersection, we have forall n:
>
> n {E(1),..,E(n)} =/= {}
>
> On the other hand the infinite intersection is:
>
> n {E(1),E(2),..} = {}
Why do you use E(1)? Why do you use any definable ensegment E(k)? We know that the inclusion of all of them is useless. It is at most an attempt to deceive innocent newbees since
> Yes n+1=/=omega, you cannot complete infinity
> by your finite intersection, we have forall n:
>
> n {E(1),..,E(n)} =/= {}
>
> On the other hand the infinite intersection is:
>
> n {E(1),E(2),..} = {}
∀k ∈ ℕ ∩{E(1), E(2), ..., E(k)} = E(k) mit |E(k)| = ℵo .
Transfinite Numbers
Nov 17, 2019, 1:38:14 PM11/17/19
to
For the 1000-th time, lets recall:
Well the more general result was already posted many
times, at your disposal. But you never make use
of it. But the more general result is:
If M is finite then:
∩ E(k) =/= {}
k e M
If M is infinite then:
∩ E(k) = {}
k e M
This covers everything you can imagine of
removing from N, i.e. removing from N some A
such that N \ A = M.
Just check whether M is finite or infinite.
You need to train your memory better. You
always forget all results.
Well the more general result was already posted many
times, at your disposal. But you never make use
of it. But the more general result is:
If M is finite then:
∩ E(k) =/= {}
k e M
If M is infinite then:
∩ E(k) = {}
k e M
This covers everything you can imagine of
removing from N, i.e. removing from N some A
such that N \ A = M.
Just check whether M is finite or infinite.
You need to train your memory better. You
always forget all results.
Jim Burns
Nov 17, 2019, 2:30:15 PM11/17/19
to
On 11/17/2019 12:45 PM, j4n bur53 wrote:
> https://en.wikipedia.org/wiki/De_Morgan's_laws
>
> Thats basic hermeneutics, "Quod est superius,
> est sicut (id) quod est inferius."
I suspect that you want "hermetics" instead of
> https://en.wikipedia.org/wiki/De_Morgan's_laws
>
> Thats basic hermeneutics, "Quod est superius,
> est sicut (id) quod est inferius."
"hermeneutics".
https://en.wikipedia.org/wiki/As_above,_so_below
| "As above, so below" is an aphorism associated with
| sacred geometry, Hermeticism, and the Tarot.
|| Quod est inferius est sicut quod est superius.
|| Et quod est superius est sicut quod est inferius,
|| ad perpetranda miracula rei unius.
https://en.wiktionary.org/wiki/hermetic
(hermetic)
2. Pertaining to alchemy or occult practices;
magical, alchemical.
https://en.wiktionary.org/wiki/hermeneutics
(hermeneutics)
1. The study or theory of the methodical interpretation
of text, especially holy texts.
I'm not sure, but I think I've just been engaging
in hermeneutics.
----
I find "As above, so below" interesting to a surprising
degree. (Please excuse me for avoiding the Latin.)
"As above, so below" is basically what Isaac Newton
demonstrated, except that he did it better than Hermes
Trismegistus, Thrice-Great Hermes. Newton laid out laws
that were valid both above and below. _And they worked_
so we are still using them, or their descendants.
I can imagine the shadow of "As above, so below" in today's
grand unification projects, descendants of Newton's
unification of Heavenly motion and Earthly motion.
I would like to suggest that the difference between what
came of the teachings of T. Hermes and what came of the
teachings of I. Newton was largely a matter of _what to do_
_if we're wrong_ . I speculate that they started out in
pretty much the same place, but Newton and co. threw out
the bad stuff, while Hermes and co. obfuscated, layered
more symbolism and poetry on top of the old, and basically
chose _looking good_ over _finding truth_ .
Or, I could be over-generalizing from my experience of
certain posts. They just feel to me like they have an
alchemical quality to them. But that's just a gut feeling.
I haven't studied this.
Transfinite Numbers
Nov 17, 2019, 8:07:05 PM11/17/19
to
I guess its hermetics applied to hermeneutics,
Here is a nice picture of "As above, so below"
http://math.hawaii.edu/LatThy/
If a lattice has a complement operator ~: L->L,
and if its not only a pseudo complement, then
you have de Morgan. And this implies:
if a ≤ b then ~b ≤ ~a.
Proof:
a ≤ b iff a ∧ b = a iff a ∨ b = b.
Take the first definition, and complement both
sides you get ~a ∨ ~b = ~a, or which is the
same you get ~b ∨ ~a = ~a, and which is by
the second definition ~b ≤ ~a.
Disclaimer: Only a rough sketch. Warning deMorgan
alone does not lead to Boolean.
Here is a nice picture of "As above, so below"
http://math.hawaii.edu/LatThy/
If a lattice has a complement operator ~: L->L,
and if its not only a pseudo complement, then
you have de Morgan. And this implies:
if a ≤ b then ~b ≤ ~a.
Proof:
a ≤ b iff a ∧ b = a iff a ∨ b = b.
Take the first definition, and complement both
sides you get ~a ∨ ~b = ~a, or which is the
same you get ~b ∨ ~a = ~a, and which is by
the second definition ~b ≤ ~a.
Disclaimer: Only a rough sketch. Warning deMorgan
alone does not lead to Boolean.
Transfinite Numbers
Nov 17, 2019, 8:14:13 PM11/17/19
to
Maybe this guy can model WMs nonsense?
Axiomatizing Reasoning About Sets:
Cardinality, Mereology, and Decisiveness
Friday, January 17, 2020 - Wesley H. Holliday
http://jointmathematicsmeetings.org/meetings/national/jmm2020/2245_invspeakers#holliday
My assumption about Muckamatik, if N \ N =/= {},
the Muckamatik sets must be some kind of
porous sponge.
Axiomatizing Reasoning About Sets:
Cardinality, Mereology, and Decisiveness
Friday, January 17, 2020 - Wesley H. Holliday
http://jointmathematicsmeetings.org/meetings/national/jmm2020/2245_invspeakers#holliday
My assumption about Muckamatik, if N \ N =/= {},
the Muckamatik sets must be some kind of
porous sponge.
Ganzhinterseher
Nov 18, 2019, 5:15:25 AM11/18/19
to
Am Sonntag, 17. November 2019 19:38:14 UTC+1 schrieb Transfinite Numbers:
> Well the more general result was already posted many
> times, at your disposal. But you never make use
> of it. But the more general result is:
>
> If M is finite then:
>
> ∩ E(k) =/= {}
> k e M
>
> If M is infinite then:
>
> ∩ E(k) = {}
> k e M
>
> This covers everything you can imagine of
> removing from N
Alas it is in contradiction with mathematics:
> Well the more general result was already posted many
> times, at your disposal. But you never make use
> of it. But the more general result is:
>
> If M is finite then:
>
> ∩ E(k) =/= {}
> k e M
>
> If M is infinite then:
>
> ∩ E(k) = {}
> k e M
>
> This covers everything you can imagine of
> removing from N
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
if all endegments are definble.
Regards, WM
Transfinite Numbers
Nov 18, 2019, 6:13:15 AM11/18/19
to
Mostlikely there is an incompatibility of ZF
with Muckamatik. Beause in ZF, which is based on
FOL, there is no inference rule:
/* WMs Mückenüberdrussschuss */
f(A) = {} => exists B (f(B) = {1})
This one does also not exist in FOL:
/* WMs Mückenüberdrussschuss II */
f(A) = {} => f is making coffee
LoL
with Muckamatik. Beause in ZF, which is based on
FOL, there is no inference rule:
/* WMs Mückenüberdrussschuss */
f(A) = {} => exists B (f(B) = {1})
This one does also not exist in FOL:
/* WMs Mückenüberdrussschuss II */
f(A) = {} => f is making coffee
LoL
Ganzhinterseher
Nov 18, 2019, 11:57:49 AM11/18/19
to
Am Montag, 18. November 2019 12:13:15 UTC+1 schrieb Transfinite Numbers:
> Beause in ZF, which is based on
> FOL, there is no inference rule:
>
> Beause in ZF, which is based on
> FOL, there is no inference rule:
>
> f(A) = {} => exists B (f(B) = {1})
Maybe not in ZF, but in mathematics we have the inference rule: If a set is emptied in a one-by-one procedure then zero is preceded by 1. The one-by-one procedure is guaranteed by
∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
Regards, WM
Transfinite Numbers
Nov 18, 2019, 1:07:15 PM11/18/19
to
Now you show some clear signs of insanity.
LMAO!
LMAO!
Dan Christensen
Nov 18, 2019, 2:26:50 PM11/18/19
to
On Saturday, November 16, 2019 at 11:26:55 AM UTC-5, Ganzhinterseher wrote:
> 1) Every endsegment E(n) = {n, n+1, n+2, ...} of the set of natural numbers reduces the intersection of endsegments in natural order
>
> E(1) ∩ E(2) ∩ E(3) ∩ ...
>
> by one and only one natural number, namely n-1.
>
> 2) Every endsegment is infinite and leaves infinitely many numbers in the intersection:
>
> ∀n ∈ ℕ ∩{E(1), E(2), ..., E(n)} = E(n) =/= { } .
>
> 3) The intersection of all endsegments is empty
>
> E(1) ∩ E(2) ∩ E(3) ∩ ... = { } .
>
> If (3) is true, then (1) requires that endsegments are existing that reduce the intersection to a finite set of n, n-1, ..., 3, 2, 1 elements. But such endsegments cannot be identified because of (2).
>
So, you have the intersection of the finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
>
> E(1) ∩ E(2) ∩ E(3) ∩ ...
>
> by one and only one natural number, namely n-1.
>
> 2) Every endsegment is infinite and leaves infinitely many numbers in the intersection:
>
> ∀n ∈ ℕ ∩{E(1), E(2), ..., E(n)} = E(n) =/= { } .
>
> 3) The intersection of all endsegments is empty
>
> E(1) ∩ E(2) ∩ E(3) ∩ ... = { } .
>
> If (3) is true, then (1) requires that endsegments are existing that reduce the intersection to a finite set of n, n-1, ..., 3, 2, 1 elements. But such endsegments cannot be identified because of (2).
>
And the intersection of the infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all endsegments.
Both are easy to prove using basic set theory and arithmetic. No mysterious "dark numbers" are required.
Yet another of your "solutions" looking for a problem, like your alternative to Peano's Axioms with which you could not even prove 1=/=2? (Hee, hee!)
Dan
Download my DC Proof 2.0 freeware at http://www.dcproof.com
Visit my Math Blog at http://www.dcproof.wordpress.com
Python
Nov 18, 2019, 3:59:39 PM11/18/19
to
Crank Wolfgang Mueckenheim sockpuppet, Ganzhinterseher wrote:
> Am Montag, 18. November 2019 12:13:15 UTC+1 schrieb Transfinite Numbers:
>> Beause in ZF, which is based on
>> FOL, there is no inference rule:
>>
>> f(A) = {} => exists B (f(B) = {1})
>
> Maybe not in ZF, but in mathematics we have the inference rule: If a set is emptied in a one-by-one procedure then zero is preceded by 1.
Trick for Adjunct Lecturer Wolfgang Mueckenheim victims^H^H^H^H^H^H^H^H
> Am Montag, 18. November 2019 12:13:15 UTC+1 schrieb Transfinite Numbers:
>> Beause in ZF, which is based on
>> FOL, there is no inference rule:
>>
>> f(A) = {} => exists B (f(B) = {1})
>
> Maybe not in ZF, but in mathematics we have the inference rule: If a set is emptied in a one-by-one procedure then zero is preceded by 1.
unfortunate students at Hochschule Augsburg: frame this new silly
argument that a successor is actually a predecessor if you reverse
order so that you will pass your exam whatever grade you've got:
if 0 is the lowest grade it may be interpreted as the higher one.
This would be a good thing, as having good grades at Mueckenheim
shameful exams means to write down fallacies, lies and sophistries,
which is properly DISGUSTING as Hochschule Augsburg is a public
Academic German institution.
Ganzhinterseher
Nov 18, 2019, 4:16:18 PM11/18/19
to
Am Montag, 18. November 2019 20:26:50 UTC+1 schrieb Dan Christensen:
> On Saturday, November 16, 2019 at 11:26:55 AM UTC-5, Ganzhinterseher wrote:
> So, you have the intersection of the finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
>
> And the intersection of the infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all endsegments.
>
> Both are easy to prove using basic set theory
This shows that set theory is in contradiction with arithmetic.
> On Saturday, November 16, 2019 at 11:26:55 AM UTC-5, Ganzhinterseher wrote:
> So, you have the intersection of the finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
>
> And the intersection of the infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all endsegments.
>
> Both are easy to prove using basic set theory
> No mysterious "dark numbers" are required.
∀k ∈ ℕ ∩{E(1), E(2), ..., E(k)} = E(k) mit |E(k)| = ℵo
Regards, WM
Me
Nov 18, 2019, 4:22:03 PM11/18/19
to
Am 18.11.2019 um 22:00 schrieb Python:
>> If a set is emptied in a one-by-one procedure then <bla> [WM]
@Mückentroll: The intersection operation in not a "procedure" of any kind.
>> If a set is emptied in a one-by-one procedure then <bla> [WM]
@Mückentroll: The intersection operation in not a "procedure" of any kind.
Exfalso Quodlibet
Nov 18, 2019, 5:08:57 PM11/18/19
to
On Monday, November 18, 2019 at 11:57:49 AM UTC-5, Ganzhinterseher wrote:
> If a set is emptied in a one-by-one procedure
There's the source of your "incompatibility".
> If a set is emptied in a one-by-one procedure
Remove that, and there's no problem.
EFQ
Dan Christensen
Nov 18, 2019, 9:25:06 PM11/18/19
to
On Monday, November 18, 2019 at 4:16:18 PM UTC-5, Ganzhinterseher wrote:
> Am Montag, 18. November 2019 20:26:50 UTC+1 schrieb Dan Christensen:
> > On Saturday, November 16, 2019 at 11:26:55 AM UTC-5, Ganzhinterseher wrote:
>
>
> > So, you have the intersection of the finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
> >
> > And the intersection of the infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all endsegments.
> >
> > Both are easy to prove using basic set theory
>
> This shows that set theory is in contradiction with arithmetic.
>
Wrong again, Mucke.
> Am Montag, 18. November 2019 20:26:50 UTC+1 schrieb Dan Christensen:
> > On Saturday, November 16, 2019 at 11:26:55 AM UTC-5, Ganzhinterseher wrote:
>
>
> > So, you have the intersection of the finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
> >
> > And the intersection of the infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all endsegments.
> >
> > Both are easy to prove using basic set theory
>
> This shows that set theory is in contradiction with arithmetic.
>
> > No mysterious "dark numbers" are required.
>
> Only if you deliberately forbid to ask how the for every definable endsegment infinite set
>
> ∀k ∈ ℕ ∩{E(1), E(2), ..., E(k)} = E(k) mit |E(k)| = ℵo
>
> becomes empty in one step.
>
And going from a finite to an infinite set is really quite a "step," don't you think?
Ganzhinterseher
Nov 19, 2019, 7:01:54 AM11/19/19
to
f(k) = E(k+1) where f(k) = delete(k)
which has the same range as f(k) = q_k.
For details look here: https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
Regards, WM
Ganzhinterseher
Nov 19, 2019, 7:02:16 AM11/19/19
to
Am Montag, 18. November 2019 22:22:03 UTC+1 schrieb Me:
> Am 18.11.2019 um 22:00 schrieb Python:
>
> >> If a set is emptied in a one-by-one procedure then <bla> [WM]
>
> Am 18.11.2019 um 22:00 schrieb Python:
>
> >> If a set is emptied in a one-by-one procedure then <bla> [WM]
>
> The intersection operation in not a "procedure" of any kind.
But there is a bijection
f(k) = E(k+1) where f(k) = delete(k)
which has the same range as f(k) = q_k.
Regards, WM
Ganzhinterseher
Nov 19, 2019, 7:08:25 AM11/19/19
to
Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
> Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
>
> And going from a finite to an infinite set is really quite a "step," don't you think?
No, it is a step only for believers as fanatical as medieval monks and unwilling to see that reducing infinity to zero requires more than one step of the bijection delete(k) = E(k+1).
> And going from a finite to an infinite set is really quite a "step," don't you think?
Regards, WM
Exfalso Quodlibet
Nov 19, 2019, 8:51:13 AM11/19/19
to
On Tuesday, November 19, 2019 at 7:01:54 AM UTC-5, Ganzhinterseher wrote:
> Am Montag, 18. November 2019 23:08:57 UTC+1 schrieb Exfalso Quodlibet:
> > On Monday, November 18, 2019 at 11:57:49 AM UTC-5, Ganzhinterseher wrote:
> > > If a set is emptied in a one-by-one procedure
> >
> > There's the source of your "incompatibility".
> >
> > Remove that, and there's no problem.
> >
> Then we have to remove all infinite bijections.
Nope. Bijections are single objects, like f(n) = 2n. They are not "one-by-one procedures".
> Am Montag, 18. November 2019 23:08:57 UTC+1 schrieb Exfalso Quodlibet:
> > On Monday, November 18, 2019 at 11:57:49 AM UTC-5, Ganzhinterseher wrote:
> > > If a set is emptied in a one-by-one procedure
> >
> > There's the source of your "incompatibility".
> >
> > Remove that, and there's no problem.
> >
> Then we have to remove all infinite bijections.
EFQ
Dan Christensen
Nov 19, 2019, 9:43:00 AM11/19/19
to
On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
>
>
> > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
>
> No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
Researchgate? Really??? Couldn't get this bullshit past a peer review and publish in a reputable journal, eh, Mucke? What must your colleagues think? Or are they all cranks, too?
> Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
>
>
> > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
>
> No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
Post your "details" here, if you dare, Mucke. (Don't hold your breath, folks.)
> >
> > And going from a finite to an infinite set is really quite a "step," don't you think?
>
> No, it is a step only for believers as fanatical as medieval monks and unwilling to see that reducing infinity to zero requires more than one step of the bijection delete(k) = E(k+1).
>
You really don't have a clue, do you, Mucke? No one by crackpots like you talk about "reducing infinity to zero." With the possible exception of CS (is it really math?), there are are no "steps" in mathematics, no before and after, just facts that are always true.
Again, since you snipped it for obvious reasons:
You have the intersection of this finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
And the intersection of this infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all end segments.
Both are easy to prove using basic set theory and arithmetic. None of your mysterious "dark numbers" are required.
Yet another of your goofy "solutions" looking for a problem, like your alternative to Peano's Axioms with which you could not even prove 1=/=2? (HA, HA!)
R. Srinivasan
Nov 19, 2019, 12:10:21 PM11/19/19
to
On Tuesday, November 19, 2019 at 8:13:00 PM UTC+5:30, Dan Christensen wrote:
> On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> >
> >
> > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> >
> > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
>
> Researchgate? Really??? Couldn't get this bullshit past a peer review and publish in a reputable journal, eh, Mucke? What must your colleagues think? Or are they all cranks, too?
>
> Post your "details" here, if you dare, Mucke. (Don't hold your breath, folks.)
>
Can you point me to a "reputable" peer reviewed journal, say, in math or logic, whose editors and referees are honest and open-minded to dissenting points of view? In other words, people who are not interested in protecting turf. My impression is that mainstream mathematicians and logicians are undoubtedly smart, but are also part of a dishonest, unethical and intolerant cult that will put down dissent at any cost.
> On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> >
> >
> > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> >
> > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
>
> Researchgate? Really??? Couldn't get this bullshit past a peer review and publish in a reputable journal, eh, Mucke? What must your colleagues think? Or are they all cranks, too?
>
> Post your "details" here, if you dare, Mucke. (Don't hold your breath, folks.)
>
>
> > >
> > > And going from a finite to an infinite set is really quite a "step," don't you think?
> >
> > No, it is a step only for believers as fanatical as medieval monks and unwilling to see that reducing infinity to zero requires more than one step of the bijection delete(k) = E(k+1).
> >
>
> Does anyone take you seriously who isn't certifiably insane or an unfortunate student forced to take your math course of misinformation, Mucke? You can't even formally define these "dark numbers" of yours. Or any of your other bizarre notions. You have zero credibility.
>
I would rather be a student of Mueckenheim than that of a professor from the mainstream cult. While I would not accept everything that Mueckenheim says, at least I could be confident that he would not cut off my funding just because I am pursuing a credible anti-mainstream line of research. I do not wish to trapped in the thinking of the mainstream cult, from which there is no escape.
> > >
> > > And going from a finite to an infinite set is really quite a "step," don't you think?
> >
> > No, it is a step only for believers as fanatical as medieval monks and unwilling to see that reducing infinity to zero requires more than one step of the bijection delete(k) = E(k+1).
> >
>
> Does anyone take you seriously who isn't certifiably insane or an unfortunate student forced to take your math course of misinformation, Mucke? You can't even formally define these "dark numbers" of yours. Or any of your other bizarre notions. You have zero credibility.
>
>
> You really don't have a clue, do you, Mucke? No one by crackpots like you talk about "reducing infinity to zero." With the possible exception of CS (is it really math?), there are are no "steps" in mathematics, no before and after, just facts that are always true.
>
> Again, since you snipped it for obvious reasons:
>
> You have the intersection of this finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
>
> And the intersection of this infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all end segments.
>
> Both are easy to prove using basic set theory and arithmetic. None of your mysterious "dark numbers" are required.
>
Actually Mueckenheim does have a paradox here, and it can be made clear by looking at the following version of Zeno's dichotomy paradox.
> You really don't have a clue, do you, Mucke? No one by crackpots like you talk about "reducing infinity to zero." With the possible exception of CS (is it really math?), there are are no "steps" in mathematics, no before and after, just facts that are always true.
>
> Again, since you snipped it for obvious reasons:
>
> You have the intersection of this finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
>
> And the intersection of this infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all end segments.
>
> Both are easy to prove using basic set theory and arithmetic. None of your mysterious "dark numbers" are required.
>
Consider the sequence of nested half open real intervals
I(n) = (0, (1/2)**(n-1)], n=1,2,....
We may consider the real intervals I(n) to be coded by the infinite sets E(n) of Mueckenheim that you have defined above.
Clearly the infinite intersection of the nested intervals I(n) is the null set:
Intersect {I(1), I(2),....} = {}.
Here is the paradox. You have an infinite intersection of nested intervals, EACH of which is of nonzero length. Our strong intuition here is that the intersection should be an interval of nonzero length (can you see this?). This intuition corresponds to Zeno's dichotomy paradox, in the sense that if the infinite intersection is indeed an interval, that would mean that a runner starting from the point 1 will never reach the point 0 even after crossing an infinity of the intervals I(n). Yet we find that the intersection is empty and that is the paradox.
Now imagine the coding of I(n) by E(n) in the above paradox and we get Mueckenheim's argument.
I claim that these paradoxes arise essentially because of the existence of infinite sets and quantification over these. We need to restrict infinite classes to be proper classes and ban quantification over proper classes in order to avoid these paradoxes. This is done in my logic NAFL. Here is my recent preprint on Zeno's dichotomy paradox that I posted to the PhilSci archive:
http://philsci-archive.pitt.edu/15799/
This was discussed in sci.logic a few months back. I will try to publish it with the promised Part II, on the logic NAFL, which is in progress right now.
Regards, RS
Dan Christensen
Nov 19, 2019, 1:43:20 PM11/19/19
to
On Tuesday, November 19, 2019 at 12:10:21 PM UTC-5, R. Srinivasan wrote:
> On Tuesday, November 19, 2019 at 8:13:00 PM UTC+5:30, Dan Christensen wrote:
> > On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> > >
> > >
> > > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> > >
> > > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
> >
> > Researchgate? Really??? Couldn't get this bullshit past a peer review and publish in a reputable journal, eh, Mucke? What must your colleagues think? Or are they all cranks, too?
> >
> > Post your "details" here, if you dare, Mucke. (Don't hold your breath, folks.)
> >
> Can you point me to a "reputable" peer reviewed journal, say, in math or logic, whose editors and referees are honest and open-minded to dissenting points of view?
Ultra-finistists, constructivists and other dissenters from the mainstream somehow get published in peer reviewed journals. It's a question of quality of work.
> On Tuesday, November 19, 2019 at 8:13:00 PM UTC+5:30, Dan Christensen wrote:
> > On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> > >
> > >
> > > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> > >
> > > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
> >
> > Researchgate? Really??? Couldn't get this bullshit past a peer review and publish in a reputable journal, eh, Mucke? What must your colleagues think? Or are they all cranks, too?
> >
> > Post your "details" here, if you dare, Mucke. (Don't hold your breath, folks.)
> >
> Can you point me to a "reputable" peer reviewed journal, say, in math or logic, whose editors and referees are honest and open-minded to dissenting points of view?
WM repeatedly makes the same outrageous, unproven claims in this forums and others. Here is just a small sampling:
“In my system, two different numbers can have the same value.”
-- sci.math, 2014/10/16
“1+2 and 2+1 are different numbers.”
-- sci.math, 2014/10/20
“1/9 has no decimal representation.”
-- sci.math, 2015/09/22
"0.999... is not 1."
-- sci.logic 2015/11/25
“Axioms are rubbish!”
-- sci.math, 2014/11/19
“Formal definitions have lead to worthless crap like undefinable numbers.”
-- sci.math 2017/02/05
“No set is countable, not even |N.”
-- sci.logic, 2015/08/05
“Countable is an inconsistent notion.”
-- sci.math, 2015/12/05
“There is no actually infinite set |N.”
-- sci.math, 2015/10/26
“|N is not covered by the set of natural numbers.”
-- sci.math, 2015/10/26
“The set of all rationals can be shown not to exist.”
--sci.math, 2015/11/28
“Everything is in the list of everything and therefore everything belongs to a not uncountable set.”
-- sci.math, 2015/11/30
"'Not equal' and 'equal can mean the same.”
-- sci.math, 2016/06/09
“The set of numbers will get empty after all have numbers been used.”
-- sci.math, 2016/08/24
Deliberately ambiguous terminology, logical fallacies and mind-numbing repetition are his stock in trade. I suspect that his motives may be less than honourable.
Transfinite Numbers
Nov 19, 2019, 1:50:01 PM11/19/19
to
Your "intuition" is another form of Mückenschuss.
It reads, where (I_n) is a family of intervals:
/* R. Srinivasan, Mückenschuss */
forall n e N P(I_n) => P(n I_n)
Again some fallacy.
It reads, where (I_n) is a family of intervals:
/* R. Srinivasan, Mückenschuss */
forall n e N P(I_n) => P(n I_n)
Again some fallacy.
Dan Christensen
Nov 19, 2019, 2:01:29 PM11/19/19
to
On Tuesday, November 19, 2019 at 12:10:21 PM UTC-5, R. Srinivasan wrote:
> Actually Mueckenheim does have a paradox here, and it can be made clear by looking at the following version of Zeno's dichotomy paradox.
>
> Consider the sequence of nested half open real intervals
>
> I(n) = (0, (1/2)**(n-1)], n=1,2,....
>
> We may consider the real intervals I(n) to be coded by the infinite sets E(n) of Mueckenheim that you have defined above.
>
> Clearly the infinite intersection of the nested intervals I(n) is the null set:
>
> Intersect {I(1), I(2),....} = {}.
>
Yes.
>
> Consider the sequence of nested half open real intervals
>
> I(n) = (0, (1/2)**(n-1)], n=1,2,....
>
> We may consider the real intervals I(n) to be coded by the infinite sets E(n) of Mueckenheim that you have defined above.
>
> Clearly the infinite intersection of the nested intervals I(n) is the null set:
>
> Intersect {I(1), I(2),....} = {}.
>
> Here is the paradox. You have an infinite intersection of nested intervals, EACH of which is of nonzero length. Our strong intuition here is that the intersection should be an interval of nonzero length (can you see this?).
> This intuition corresponds to Zeno's dichotomy paradox, in the sense that if the infinite intersection is indeed an interval, that would mean that a runner starting from the point 1 will never reach the point 0 even after crossing an infinity of the intervals I(n). Yet we find that the intersection is empty and that is the paradox.
>
> Now imagine the coding of I(n) by E(n) in the above paradox and we get Mueckenheim's argument.
>
> I claim that these paradoxes arise essentially because of the existence of infinite sets and quantification over these.
> We need to restrict infinite classes to be proper classes and ban quantification over proper classes in order to avoid these paradoxes.
R. Srinivasan
Nov 19, 2019, 4:28:23 PM11/19/19
to
On Wednesday, November 20, 2019 at 12:20:01 AM UTC+5:30, Transfinite Numbers wrote:
> Your "intuition" is another form of Mückenschuss.
> It reads, where (I_n) is a family of intervals:
>
> /* R. Srinivasan, Mückenschuss */
> forall n e N P(I_n) => P(n I_n)
>
> Again some fallacy.
>
Can you explain your notation? What is P(I_n)? What is nI_n ?
> Your "intuition" is another form of Mückenschuss.
> It reads, where (I_n) is a family of intervals:
>
> /* R. Srinivasan, Mückenschuss */
> forall n e N P(I_n) => P(n I_n)
>
> Again some fallacy.
>
I hope you are aware of the age old philosophical objection to infinite summations, probably raised first by Zeno.
Which is, how can infinitely many positive intervals of length 2**(-n) have a finite sum. Here the intuition is that infinitely many times any positive magnitude, no matter how small, is always infinite.
This intuition is countered by the intersection argument I gave. That is, even though each of the intervals I(n) is of positive length, there is no smallest interval and secondly, the intersection of the infinitely many intervals I(n) is empty. That is, no positive magnitude gets multiplied infinitely many times, as our intuition tells us.
However, this argument only shifts the burden to the issue of how the intervals I(n) can intersect to the empty set. None of the upper end-points of these infinitely many nested intervals I(n) reaches infinitely close to zero, so how can the infinite intersection be empty? Note that we are considering the intervals I(n) as an infinite totality; there is no fallacy of "one at a time" here. This is a valid intuition and this is what Mueckenheim is essentially trying to express.
Regards, RS
FredJeffries
Nov 19, 2019, 4:51:25 PM11/19/19
to
On Tuesday, November 19, 2019 at 1:28:23 PM UTC-8, R. Srinivasan wrote:
> I hope you are aware of the age old philosophical objection to infinite summations, probably raised first by Zeno.
There are no 'infinite summations' in mathematics
> I hope you are aware of the age old philosophical objection to infinite summations, probably raised first by Zeno.
R. Srinivasan
Nov 19, 2019, 6:01:39 PM11/19/19
to
That does not make a difference to the counter-intuitive (paradoxical, but not necessarily contradictory) result that I have noted.
By the way, the currently accepted resolution of Zeno's dichotomy paradox is that the runner, while traversing the interval [0,1], crosses an actual infinity of points {0, 1/2, 3/4, ...}. If that is the case, the runner also crosses an actual infinity of intervals {[0,1/2],[1/2,3/4], ...} after which he reaches the end point 1. So isn't the runner physically performing an actually infinite summation of these intervals? In other words, the limit of the partial sums is physically realized (as claimed in my eprint that I cited).
Regards, RS
Transfinite Numbers
Nov 19, 2019, 6:05:15 PM11/19/19
to
P is some arbitrary predicate, I_n was explained,
n_n e N I_n is the infinite intersection. Assuming
such an "intuition" that allows you to form
an expectation, i.e. an inference rule:
/* R. Srinivasan, Mückenschuss */
forall n e N P(I_n) => P(∩_n e N I_n)
is a fallacy.
n_n e N I_n is the infinite intersection. Assuming
such an "intuition" that allows you to form
an expectation, i.e. an inference rule:
/* R. Srinivasan, Mückenschuss */
is a fallacy.
Transfinite Numbers
Nov 19, 2019, 6:07:15 PM11/19/19
to
A fallacy is the use of invalid or otherwise
faulty reasoning, or "wrong moves" in the
construction of an argument.
https://en.wikipedia.org/wiki/Fallacy
R. Srinivasan
Nov 19, 2019, 6:24:32 PM11/19/19
to
On Wednesday, November 20, 2019 at 4:35:15 AM UTC+5:30, Transfinite Numbers wrote:
> P is some arbitrary predicate, I_n was explained,
> n_n e N I_n is the infinite intersection. Assuming
>
> such an "intuition" that allows you to form
> an expectation, i.e. an inference rule:
>
> /* R. Srinivasan, Mückenschuss */
>
> forall n e N P(I_n) => P(∩_n e N I_n)
>
> is a fallacy.
>
OK, I get it. You are saying that a property of an interval I_n or for that matter, a finite intersection of the intervals I_n does not necessarily carry over to the infinite intersection.
> P is some arbitrary predicate, I_n was explained,
> n_n e N I_n is the infinite intersection. Assuming
>
> such an "intuition" that allows you to form
> an expectation, i.e. an inference rule:
>
> /* R. Srinivasan, Mückenschuss */
>
> forall n e N P(I_n) => P(∩_n e N I_n)
>
> is a fallacy.
>
You can always argue from the currently accepted axioms of set theory and formally defend counter-intuitive results. However, the purpose of raising such counter-intuitive results is to question the currently accepted axioms of set theory, in particular, the axiom of infinity.
In the case of Zeno's dichotomy paradox, I am claiming that these counter-intuitive results actually do lead to contradictions, albeit at the metamathematical or metatheoretical level, as noted in my cited preprint.
Transfinite Numbers
Nov 19, 2019, 7:34:12 PM11/19/19
to
Fallacies cannot produce credible contradictions. They
only produce non verifiable results. A proof that
uses a fallacious step is not a proof. A contradiction
based on a non-proof is a non-contradiction.
A meta theory might be indeed inconsistent, but for
FOL+ZFC, so far no crank could demonstrate an
inconsistent.
only produce non verifiable results. A proof that
uses a fallacious step is not a proof. A contradiction
based on a non-proof is a non-contradiction.
A meta theory might be indeed inconsistent, but for
FOL+ZFC, so far no crank could demonstrate an
inconsistent.
Transfinite Numbers
Nov 19, 2019, 7:37:17 PM11/19/19
to
If your meta-theory MT has a sentence A, with
MT |- A and MT |- ~A, then maybe you shouldn't
use this meta-theory MT, it can prove anything.
You could also prove "I have shown ZFC inconsistent".
LoL
MT |- A and MT |- ~A, then maybe you shouldn't
use this meta-theory MT, it can prove anything.
You could also prove "I have shown ZFC inconsistent".
LoL
Khong Dong
Nov 19, 2019, 9:48:50 PM11/19/19
to
On Tuesday, 19 November 2019 11:43:20 UTC-7, Dan Christensen wrote:
> On Tuesday, November 19, 2019 at 12:10:21 PM UTC-5, R. Srinivasan wrote:
> > On Tuesday, November 19, 2019 at 8:13:00 PM UTC+5:30, Dan Christensen wrote:
> > > On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > > > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> > > >
> > > >
> > > > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> > > >
> > > > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
> > >
> > > Researchgate? Really??? Couldn't get this bullshit past a peer review and publish in a reputable journal, eh, Mucke? What must your colleagues think? Or are they all cranks, too?
> > >
> > > Post your "details" here, if you dare, Mucke. (Don't hold your breath, folks.)
> > >
> > Can you point me to a "reputable" peer reviewed journal, say, in math or logic, whose editors and referees are honest and open-minded to dissenting points of view?
>
> Ultra-finistists, constructivists and other dissenters from the mainstream somehow get published in peer reviewed journals. It's a question of quality of work.
>
> WM repeatedly makes the same outrageous, unproven claims in this forums and others. Here is just a small sampling:
>
> “In my system, two different numbers can have the same value.”
It depends on what the underlying "system" be, what you'd mean by "number",
> On Tuesday, November 19, 2019 at 12:10:21 PM UTC-5, R. Srinivasan wrote:
> > On Tuesday, November 19, 2019 at 8:13:00 PM UTC+5:30, Dan Christensen wrote:
> > > On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > > > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> > > >
> > > >
> > > > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> > > >
> > > > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
> > >
> > > Researchgate? Really??? Couldn't get this bullshit past a peer review and publish in a reputable journal, eh, Mucke? What must your colleagues think? Or are they all cranks, too?
> > >
> > > Post your "details" here, if you dare, Mucke. (Don't hold your breath, folks.)
> > >
> > Can you point me to a "reputable" peer reviewed journal, say, in math or logic, whose editors and referees are honest and open-minded to dissenting points of view?
>
> Ultra-finistists, constructivists and other dissenters from the mainstream somehow get published in peer reviewed journals. It's a question of quality of work.
>
> WM repeatedly makes the same outrageous, unproven claims in this forums and others. Here is just a small sampling:
>
> “In my system, two different numbers can have the same value.”
and "value", etc... In the arithmetic system of the so called "standard
structure", two different prime elements - "numbers" - could have the same
"value" if you so choose.
Dan Christensen
Nov 19, 2019, 10:44:07 PM11/19/19
to
0 --> 3 --> 1 --> FluffyTheCat --> ...
Khong Dong
Nov 19, 2019, 10:53:09 PM11/19/19
to
So? No reason why you have to get frightened or whine about it.
Ganzhinterseher
Nov 20, 2019, 4:33:11 AM11/20/19
to
Am Dienstag, 19. November 2019 18:10:21 UTC+1 schrieb R. Srinivasan:
>
> I claim that these paradoxes arise essentially because of the existence of infinite sets and quantification over these.
Yes, that is the reason.
>
> I claim that these paradoxes arise essentially because of the existence of infinite sets and quantification over these.
> We need to restrict infinite classes to be proper classes and ban quantification over proper classes in order to avoid these paradoxes.
> This is done in my logic NAFL. Here is my recent preprint on Zeno's dichotomy paradox that I posted to the PhilSci archive:
>
> http://philsci-archive.pitt.edu/15799/
>
> This was discussed in sci.logic a few months back. I will try to publish it with the promised Part II, on the logic NAFL, which is in progress right now.
Regards, WM
Ganzhinterseher
Nov 20, 2019, 4:33:32 AM11/20/19
to
Am Dienstag, 19. November 2019 15:43:00 UTC+1 schrieb Dan Christensen:
> On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> >
> >
> > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> >
> > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
>
> Researchgate? Really??? Couldn't get this text past a peer review and publish in a reputable journal,
> On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> >
> >
> > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> >
> > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
>
These "reputable" journals are lead by fanatics who consider themselves experts and would no longer be considered experts if they had to publish the truth.
> You have the intersection of this finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
>
> And the intersection of this infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all end segments.
>
> Both are easy to prove using basic set theory and arithmetic. None of your mysterious "dark numbers" are required.
subtract({k}) = E(k+1)
between the infinite sets resulting from all finite E(n) and the empty set resulting from intersecting over all finite E(n).
Regards, WM
Ganzhinterseher
Nov 20, 2019, 4:35:16 AM11/20/19
to
The bijection
subtract({k}) = E(k+1)
Either there is no bijetcion or there are unidentifiable numbers.
Regards, WM
Ganzhinterseher
Nov 20, 2019, 4:44:52 AM11/20/19
to
Am Dienstag, 19. November 2019 22:28:23 UTC+1 schrieb R. Srinivasan:
> This intuition is countered by the intersection argument I gave. That is, even though each of the intervals I(n) is of positive length, there is no smallest interval and secondly, the intersection of the infinitely many intervals I(n) is empty. That is, no positive magnitude gets multiplied infinitely many times, as our intuition tells us.
>
> However, this argument only shifts the burden to the issue of how the intervals I(n) can intersect to the empty set.
They cannot. The simplest example is the sequenc (1/n) which has limit 0 but no term 0. The terms of the sequence can be considered as intersections of finite intervals [0, 1/n] and will not yield the degenerate interval [0]. Set theorists confuse the infinitely many terms with the limit. But here the mistake is not as obvious as with the endsgments.
> This intuition is countered by the intersection argument I gave. That is, even though each of the intervals I(n) is of positive length, there is no smallest interval and secondly, the intersection of the infinitely many intervals I(n) is empty. That is, no positive magnitude gets multiplied infinitely many times, as our intuition tells us.
>
> However, this argument only shifts the burden to the issue of how the intervals I(n) can intersect to the empty set.
Regards, WM
Ganzhinterseher
Nov 20, 2019, 4:48:10 AM11/20/19
to
Regards, WM
Ganzhinterseher
Nov 20, 2019, 5:00:50 AM11/20/19
to
Regards, WM
Ganzhinterseher
Nov 20, 2019, 5:36:48 AM11/20/19
to
Am Mittwoch, 20. November 2019 00:05:15 UTC+1 schrieb Transfinite Numbers:
> P is some arbitrary predicate, I_n was explained,
> n_n e N I_n is the infinite intersection. Assuming
>
> such an "intuition" that allows you to form
> an expectation, i.e. an inference rule:
>
> /* R. Srinivasan, Mückenschuss */
>
> forall n e N P(I_n) => P(∩_n e N I_n)
>
> is a fallacy.
Instead of your silly claims find a way to name the natural numbers that reduce the infinite set
> P is some arbitrary predicate, I_n was explained,
> n_n e N I_n is the infinite intersection. Assuming
>
> such an "intuition" that allows you to form
> an expectation, i.e. an inference rule:
>
> /* R. Srinivasan, Mückenschuss */
>
> forall n e N P(I_n) => P(∩_n e N I_n)
>
> is a fallacy.
∀k ∈ ℕ ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
to the empty set
∩{E(1), E(2), E(3), ...} = { }
by the bijection
subtract({k}) = E(k+1) .
Or confess that there is no bijection resulting in the empty set. There is only the infinite jump applauded by believers in finished infinity.
Regards, WM
Exfalso Quodlibet
Nov 20, 2019, 8:01:08 AM11/20/19
to
On Wednesday, November 20, 2019 at 4:35:16 AM UTC-5, Ganzhinterseher wrote:
> Am Dienstag, 19. November 2019 14:51:13 UTC+1 schrieb Exfalso Quodlibet:
> > On Tuesday, November 19, 2019 at 7:01:54 AM UTC-5, Ganzhinterseher wrote:
> > > Am Montag, 18. November 2019 23:08:57 UTC+1 schrieb Exfalso Quodlibet:
> > > > On Monday, November 18, 2019 at 11:57:49 AM UTC-5, Ganzhinterseher wrote:
> > > > > If a set is emptied in a one-by-one procedure
> > > >
> > > > There's the source of your "incompatibility".
> > > >
> > > > Remove that, and there's no problem.
> > > >
> > > Then we have to remove all infinite bijections.
> >
> > Nope. Bijections are single objects, like f(n) = 2n. They are not "one-by-one procedures".
> >
> Why nope?
Because bijections are not "one-by-one procedures".
> Am Dienstag, 19. November 2019 14:51:13 UTC+1 schrieb Exfalso Quodlibet:
> > On Tuesday, November 19, 2019 at 7:01:54 AM UTC-5, Ganzhinterseher wrote:
> > > Am Montag, 18. November 2019 23:08:57 UTC+1 schrieb Exfalso Quodlibet:
> > > > On Monday, November 18, 2019 at 11:57:49 AM UTC-5, Ganzhinterseher wrote:
> > > > > If a set is emptied in a one-by-one procedure
> > > >
> > > > There's the source of your "incompatibility".
> > > >
> > > > Remove that, and there's no problem.
> > > >
> > > Then we have to remove all infinite bijections.
> >
> > Nope. Bijections are single objects, like f(n) = 2n. They are not "one-by-one procedures".
> >
> Why nope?
> Sets are single objcts, but they may have infinitela many elements.
> Infinite bijections with |N are such sets. Allegedly every pair (n, f(n)) of the bijection can be identified
> and every n is in a distance of one unit from its succesor and of n units from 0.
>
> The bijection
>
> subtract({k}) = E(k+1)
> results in the empty set
> (this can be proved) but it is impossible to find the natural numbers reducing the infinite set to the empty set.
EFQ
FredJeffries
Nov 20, 2019, 8:04:27 AM11/20/19
to
On Tuesday, November 19, 2019 at 3:01:39 PM UTC-8, R. Srinivasan wrote:
> By the way, the currently accepted resolution of Zeno's dichotomy paradox is that the runner, while traversing the interval [0,1], crosses an actual infinity of points {0, 1/2, 3/4, ...}.
Your argument devolves to your usage of vague, ambiguous, and undefined (undefinable?) terms 'runner', 'traversing', 'crosses', ...
> By the way, the currently accepted resolution of Zeno's dichotomy paradox is that the runner, while traversing the interval [0,1], crosses an actual infinity of points {0, 1/2, 3/4, ...}.
But, rather than examine your own premises, you would take the easy route and blame the (improperly applied) mathematics.
> If that is the case, the runner also crosses an actual infinity of intervals {[0,1/2],[1/2,3/4], ...} after which he reaches the end point 1. So isn't the runner physically performing an actually infinite summation of these intervals?
> In other words, the limit of the partial sums is physically realized (as claimed in my eprint that I cited).
R. Srinivasan
Nov 20, 2019, 8:11:50 AM11/20/19
to
Regards, RS
>
> Regards, WM
Transfinite Numbers
Nov 20, 2019, 9:05:40 AM11/20/19
to
As a proper class, this here is still empty:
∩_n e N I_n = { k e N | forall n (n e N => k e I_n) } = {}
If you ban the quatifier inside the infinite intersection,
the infinite intersection becomes kind of undefined.
= "cannot be expressed"
In either case Mückenschuss still does not hold, because neither
{} nor "cannot be expressed" is non-empty like I_n.
Does not solve the problem that your intuition is wrong.
∩_n e N I_n = { k e N | forall n (n e N => k e I_n) } = {}
If you ban the quatifier inside the infinite intersection,
the infinite intersection becomes kind of undefined.
= "cannot be expressed"
In either case Mückenschuss still does not hold, because neither
{} nor "cannot be expressed" is non-empty like I_n.
Does not solve the problem that your intuition is wrong.
Transfinite Numbers
Nov 20, 2019, 9:07:57 AM11/20/19
to
Corr.:
{} nor "cannot be expressed" is not non-empty like I_n.
{} nor "cannot be expressed" is not non-empty like I_n.
FredJeffries
Nov 20, 2019, 9:20:01 AM11/20/19
to
On Wednesday, November 20, 2019 at 6:05:40 AM UTC-8, Transfinite Numbers wrote:
> Does not solve the problem that your intuition is wrong.
Intuition is almost always wrong. That's why we do science and mathematics.
> Does not solve the problem that your intuition is wrong.
Dan Christensen
Nov 20, 2019, 9:43:23 AM11/20/19
to
On Wednesday, November 20, 2019 at 5:00:50 AM UTC-5, Ganzhinterseher wrote:
> Am Mittwoch, 20. November 2019 03:48:50 UTC+1 schrieb Khong Dong:
> > On Tuesday, 19 November 2019 11:43:20 UTC-7, Dan Christensen wrote:
>
>
> > > WM repeatedly makes the same outrageous, unproven claims in this forums and others. Here is just a small sampling:
> > >
> > > “In my system, two different numbers can have the same value.”
> >
> > It depends on what the underlying "system" be, what you'd mean by "number",
> > and "value", etc... In the arithmetic system of the so called "standard
> > structure", two different prime elements - "numbers" - could have the same
> > "value" if you so choose.
>
> Dan cannot understand that number and value need not always be synonymous. For example 1/2 and 2/4 are different fractions with the same value.
The same number, same value, different representations. See
> Am Mittwoch, 20. November 2019 03:48:50 UTC+1 schrieb Khong Dong:
> > On Tuesday, 19 November 2019 11:43:20 UTC-7, Dan Christensen wrote:
>
>
> > > WM repeatedly makes the same outrageous, unproven claims in this forums and others. Here is just a small sampling:
> > >
> > > “In my system, two different numbers can have the same value.”
> >
> > It depends on what the underlying "system" be, what you'd mean by "number",
> > and "value", etc... In the arithmetic system of the so called "standard
> > structure", two different prime elements - "numbers" - could have the same
> > "value" if you so choose.
>
> Dan cannot understand that number and value need not always be synonymous. For example 1/2 and 2/4 are different fractions with the same value.
https://ca.ixl.com/?partner=google&campaign=340146665&adGroup=22863574505&gclid=Cj0KCQiA5dPuBRCrARIsAJL7oehr_Wux1qEUrP8qpGbdijuJq-7vncoqNC8Qhp_JlLu9ElP0SjgH9mYaAlzVEALw_wcB
> But according to my experience it is useless to try to teach him mathematics or good behaviour or to stop his insults.
>
Dan
Dan Christensen
Nov 20, 2019, 9:55:01 AM11/20/19
to
On Wednesday, November 20, 2019 at 4:33:32 AM UTC-5, Ganzhinterseher wrote:
> Am Dienstag, 19. November 2019 15:43:00 UTC+1 schrieb Dan Christensen:
> > On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> > >
> > >
> > > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> > >
> > > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
> >
> > Researchgate? Really??? Couldn't get this text past a peer review and publish in a reputable journal,
>
> These "reputable" journals are lead by fanatics who consider themselves experts and would no longer be considered experts if they had to publish the truth.
>
You mean "truths", such as you have posted here:
> Am Dienstag, 19. November 2019 15:43:00 UTC+1 schrieb Dan Christensen:
> > On Tuesday, November 19, 2019 at 7:08:25 AM UTC-5, Ganzhinterseher wrote:
> > > Am Dienstag, 19. November 2019 03:25:06 UTC+1 schrieb Dan Christensen:
> > >
> > >
> > > > Every end segment is definable as follows: For all n in N: For all x: [x in E(n) <=> x in N & x>=n]
> > >
> > > No. For details see https://www.researchgate.net/publication/336220780_Dark_natural_numbers_in_set_theory
> >
> > Researchgate? Really??? Couldn't get this text past a peer review and publish in a reputable journal,
>
> These "reputable" journals are lead by fanatics who consider themselves experts and would no longer be considered experts if they had to publish the truth.
>
“In my system, two different numbers can have the same value.”
“1+2 and 2+1 are different numbers.”
-- sci.math, 2014/10/20
“1/9 has no decimal representation.”
-- sci.math, 2015/09/22
"0.999... is not 1."
-- sci.logic 2015/11/25
“Axioms are rubbish!”
-- sci.math, 2014/11/19
“Formal definitions have lead to worthless crap like undefinable numbers.”
-- sci.math 2017/02/05
“No set is countable, not even |N.”
-- sci.logic, 2015/08/05
“Countable is an inconsistent notion.”
-- sci.math, 2015/12/05
“There is no actually infinite set |N.”
-- sci.math, 2015/10/26
“|N is not covered by the set of natural numbers.”
-- sci.math, 2015/10/26
“The set of all rationals can be shown not to exist.”
--sci.math, 2015/11/28
“Everything is in the list of everything and therefore everything belongs to a not uncountable set.”
-- sci.math, 2015/11/30
"'Not equal' and 'equal can mean the same.”
-- sci.math, 2016/06/09
“The set of numbers will get empty after all have numbers been used.”
-- sci.math, 2016/08/24
>
> > You have the intersection of this finite set {E(1),E(2), E(3), ... E(n)} = E(n) which is itself always an infinite set.
> >
> > And the intersection of this infinite set {E(1),E(2), E(3), ... } = { } since no number is common to all end segments.
> >
> > Both are easy to prove using basic set theory and arithmetic. None of your mysterious "dark numbers" are required.
>
> But it is impossible to investigate the bijection
>
> subtract({k}) = E(k+1)
>
> between the infinite sets resulting from all finite E(n) and the empty set resulting from intersecting over all finite E(n).
>
R. Srinivasan
Nov 20, 2019, 11:26:32 AM11/20/19
to
On Wednesday, November 20, 2019 at 7:35:40 PM UTC+5:30, Transfinite Numbers wrote:
> As a proper class, this here is still empty:
>
> ∩_n e N I_n = { k e N | forall n (n e N => k e I_n) } = {}
>
> If you ban the quatifier inside the infinite intersection,
> the infinite intersection becomes kind of undefined.
>
> = "cannot be expressed"
>
> In either case Mückenschuss still does not hold, because neither
> {} nor "cannot be expressed" is non-empty like I_n.
>
> Does not solve the problem that your intuition is wrong.
>
If we insist that infinite classes are proper classes and ban quantification over proper classes, we need to first of all work out how real analysis can be carried out in that framework. A real number itself becomes a proper class and then how do you define intervals of real numbers without quantification? Looks daunting, but there is a way, which I will describe in my paper. It turns out that we can only define closed intervals of real numbers and half open intervals like I_n will not exist. So a whole lot of stuff needs to be worked out and I will do this in my forthcoming paper.
> As a proper class, this here is still empty:
>
> ∩_n e N I_n = { k e N | forall n (n e N => k e I_n) } = {}
>
> If you ban the quatifier inside the infinite intersection,
> the infinite intersection becomes kind of undefined.
>
> = "cannot be expressed"
>
> In either case Mückenschuss still does not hold, because neither
> {} nor "cannot be expressed" is non-empty like I_n.
>
> Does not solve the problem that your intuition is wrong.
>
Regards, RS
Ganzhinterseher
Nov 20, 2019, 2:25:09 PM11/20/19
to
Am Mittwoch, 20. November 2019 15:20:01 UTC+1 schrieb FredJeffries:
> On Wednesday, November 20, 2019 at 6:05:40 AM UTC-8, Transfinite Numbers wrote:
>
> > Does not solve the problem that your intuition is wrong.
>
> Intuition is almost always wrong.
No, it is only wrong in dirty mathematics. We will debate, experiment, prove and conjecture until some picture emerges that satisfies this wonderful taskmaster that is our intuition.
> On Wednesday, November 20, 2019 at 6:05:40 AM UTC-8, Transfinite Numbers wrote:
>
> > Does not solve the problem that your intuition is wrong.
>
> Intuition is almost always wrong.
> That's why we do science and mathematics.
subtract({k}) = E(k+1)
∀k ∈ ℕ ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
∩{E(1), E(2), E(3), ...} = { }
Regards, WM
Ganzhinterseher
Nov 20, 2019, 2:31:18 PM11/20/19
to
Am Mittwoch, 20. November 2019 14:01:08 UTC+1 schrieb Exfalso Quodlibet:
> On Wednesday, November 20, 2019 at 4:35:16 AM UTC-5, Ganzhinterseher wrote:
> Because bijections are not "one-by-one procedures".
They are one-to-one mappings. If natural numbers are involved, then they are one-by-one procedures because of the definition of |N.
> On Wednesday, November 20, 2019 at 4:35:16 AM UTC-5, Ganzhinterseher wrote:
> Because bijections are not "one-by-one procedures".
>
> > Sets are single objcts, but they may have infinitela many elements.
>
> Having multiple elements, infinitely many or otherwise, doesn't magically require any set to be a "one-by-one procedure".
Anyhow, it is clear that there is a sequence of endsegments and that each one has precisely one natural number less than its predecessor. If the result is empty, then it cannot have been accomplished by the bijection
> > Sets are single objcts, but they may have infinitela many elements.
>
> Having multiple elements, infinitely many or otherwise, doesn't magically require any set to be a "one-by-one procedure".
subtract({k}) = E(k+1) .
> > Infinite bijections with |N are such sets. Allegedly every pair (n, f(n)) of the bijection can be identified
>
> Define "identified".
>
> > and every n is in a distance of one unit from its succesor and of n units from 0.
> >
> > The bijection
> >
> > subtract({k}) = E(k+1)
>
> ...is probably irrelevant.
is the mathematical foundation of the sequence of endsegments.
> > and every n is in a distance of one unit from its succesor and of n units from 0.
> >
> > The bijection
> >
> > subtract({k}) = E(k+1)
>
> ...is probably irrelevant.
>
> > results in the empty set
>
> "Results" how? If you perform some sort of "one-by-one procedure"? Those aren't relevant.
They are relevant for proving equicardinality.
> > results in the empty set
>
> "Results" how? If you perform some sort of "one-by-one procedure"? Those aren't relevant.
>
> > (this can be proved) but it is impossible to find the natural numbers reducing the infinite set to the empty set.
>
> "Reducing" how? If you perform some sort of "one-by-one procedure"? Those aren't relevant.
Set theory is based on it. Of course set theory is not relevant but if we assume it, then we have no other way to prove equicardinality of countable sets.
> > (this can be proved) but it is impossible to find the natural numbers reducing the infinite set to the empty set.
>
> "Reducing" how? If you perform some sort of "one-by-one procedure"? Those aren't relevant.
Regards, WM
Transfinite Numbers
Nov 20, 2019, 3:54:24 PM11/20/19
to
You don't need to use reals here. The intervals
can also be rational numbers. In Zenos paradoxon they
are rational numbers what concerns the end points,
and you can also fill them with rational numbers.
There is not much loss. Here you see the end points
a,b of the intervals [a,b] are rational:
https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
Also the above geometry gives a fine intuition how
an infinite intersection can result into a point.
Thats just the point where the two straight
lines cross. I really wonder where you get your
intuition from, and intuition that even contradicts
geometry?
can also be rational numbers. In Zenos paradoxon they
are rational numbers what concerns the end points,
and you can also fill them with rational numbers.
There is not much loss. Here you see the end points
a,b of the intervals [a,b] are rational:
https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
Also the above geometry gives a fine intuition how
an infinite intersection can result into a point.
Thats just the point where the two straight
lines cross. I really wonder where you get your
intuition from, and intuition that even contradicts
geometry?
Transfinite Numbers
Nov 20, 2019, 3:56:52 PM11/20/19
to
Dan Christensen is the opinion that the Greeks
failed in Zenos paradoxon, and it needed Galileo
to sort things out. My opinion is rather that
Zenos paradoxon is a fine tool to sharpen
intuition, similar like Russels Set Paradox.
Every paradox is like an Egyptian fable,
that wants to teach you something.
failed in Zenos paradoxon, and it needed Galileo
to sort things out. My opinion is rather that
Zenos paradoxon is a fine tool to sharpen
intuition, similar like Russels Set Paradox.
Every paradox is like an Egyptian fable,
that wants to teach you something.
Exfalso Quodlibet
Nov 20, 2019, 6:03:29 PM11/20/19
to
On Wednesday, November 20, 2019 at 2:31:18 PM UTC-5, Ganzhinterseher wrote:
> Am Mittwoch, 20. November 2019 14:01:08 UTC+1 schrieb Exfalso Quodlibet:
> > On Wednesday, November 20, 2019 at 4:35:16 AM UTC-5, Ganzhinterseher wrote:
>
>
> > Because bijections are not "one-by-one procedures".
>
> They are one-to-one mappings.
"one-to-one mapping" is not at all the same thing as a "one-by-one procedure".
> Am Mittwoch, 20. November 2019 14:01:08 UTC+1 schrieb Exfalso Quodlibet:
> > On Wednesday, November 20, 2019 at 4:35:16 AM UTC-5, Ganzhinterseher wrote:
>
>
> > Because bijections are not "one-by-one procedures".
>
> They are one-to-one mappings.
> > Having multiple elements, infinitely many or otherwise, doesn't magically require any set to be a "one-by-one procedure".
>
> Anyhow, it is clear that there is a sequence of endsegments and that each one has precisely one natural number less than its predecessor. If the result is empty, then it cannot have been accomplished by the bijection
>
> subtract({k}) = E(k+1) .
>
>
> > > Infinite bijections with |N are such sets. Allegedly every pair (n, f(n)) of the bijection can be identified
> >
> > Define "identified".
>
> Define "define".
EFQ
Ganzhinterseher
Nov 21, 2019, 5:07:08 AM11/21/19
to
Am Donnerstag, 21. November 2019 00:03:29 UTC+1 schrieb Exfalso Quodlibet:
> On Wednesday, November 20, 2019 at 2:31:18 PM UTC-5, Ganzhinterseher wrote:
> > Am Mittwoch, 20. November 2019 14:01:08 UTC+1 schrieb Exfalso Quodlibet:
> > > On Wednesday, November 20, 2019 at 4:35:16 AM UTC-5, Ganzhinterseher wrote:
> >
> >
> > > Because bijections are not "one-by-one procedures".
> >
> > They are one-to-one mappings.
>
> "one-to-one mapping" is not at all the same thing as a "one-by-one procedure".
In cases where the natural numbers are involved it is a one-by-one procedure because the set ℕ is defined by a one-by-one procedure.
> On Wednesday, November 20, 2019 at 2:31:18 PM UTC-5, Ganzhinterseher wrote:
> > Am Mittwoch, 20. November 2019 14:01:08 UTC+1 schrieb Exfalso Quodlibet:
> > > On Wednesday, November 20, 2019 at 4:35:16 AM UTC-5, Ganzhinterseher wrote:
> >
> >
> > > Because bijections are not "one-by-one procedures".
> >
> > They are one-to-one mappings.
>
> "one-to-one mapping" is not at all the same thing as a "one-by-one procedure".
Axiom VII. The domain contains at least a set Z which contains the null-set as an element and is such that each of its elements a is related to another element of the form {a}, or which with each of its elements a contains also the related set {a} as an element. (Axiom of the infinite.)
>
> > > Having multiple elements, infinitely many or otherwise, doesn't magically require any set to be a "one-by-one procedure".
> >
> > Anyhow, it is clear that there is a sequence of endsegments and that each one has precisely one natural number less than its predecessor. If the result is empty, then it cannot have been accomplished by the bijection
> >
> > subtract({k}) = E(k+1) .
> >
> >
> > > > Infinite bijections with |N are such sets. Allegedly every pair (n, f(n)) of the bijection can be identified
> > >
> > > Define "identified".
> >
> > Define "define".
>
> Specify the meaning of "identified" with sufficient rigour and precision so that any two arbitrarily chosen rational people working with that definition will reliably and unavoidably reach the same conclusions about whether "every pair (n, f(n)) of the bijection can be identified".
>
And here is what all such people can understand:
> > > Having multiple elements, infinitely many or otherwise, doesn't magically require any set to be a "one-by-one procedure".
> >
> > Anyhow, it is clear that there is a sequence of endsegments and that each one has precisely one natural number less than its predecessor. If the result is empty, then it cannot have been accomplished by the bijection
> >
> > subtract({k}) = E(k+1) .
> >
> >
> > > > Infinite bijections with |N are such sets. Allegedly every pair (n, f(n)) of the bijection can be identified
> > >
> > > Define "identified".
> >
> > Define "define".
>
> Specify the meaning of "identified" with sufficient rigour and precision so that any two arbitrarily chosen rational people working with that definition will reliably and unavoidably reach the same conclusions about whether "every pair (n, f(n)) of the bijection can be identified".
>
A natnumber is called definable if it can be expressed by digits.
But I can give also a less restricted definition:
Every endsegment indexed by an identifiable natnumber belongs to
∀k ∈ ℕ ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo .
Every endsegment indexed by an identifiable natnumber can be removed from
∩{E(1), E(2), E(3), ...} = { }
Regards, WM
Transfinite Numbers
Nov 21, 2019, 6:30:21 AM11/21/19
to
You can see the infinite here, nothing mysterious:
https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
R. Srinivasan
Nov 21, 2019, 9:15:00 AM11/21/19
to
On Thursday, November 21, 2019 at 2:24:24 AM UTC+5:30, Transfinite Numbers wrote:
> You don't need to use reals here. The intervals
> can also be rational numbers. In Zenos paradoxon they
> are rational numbers what concerns the end points,
>
> and you can also fill them with rational numbers.
> There is not much loss. Here you see the end points
> a,b of the intervals [a,b] are rational:
>
In order to state Zeno's dichotomy paradox, you need the concept of space and time, which are continua, meaning real numbers are required. For example, the velocity v = dx/dt is a real number.
> You don't need to use reals here. The intervals
> can also be rational numbers. In Zenos paradoxon they
> are rational numbers what concerns the end points,
>
> and you can also fill them with rational numbers.
> There is not much loss. Here you see the end points
> a,b of the intervals [a,b] are rational:
>
> https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
>
> Also the above geometry gives a fine intuition how
> an infinite intersection can result into a point.
> Thats just the point where the two straight
>
> lines cross. I really wonder where you get your
> intuition from, and intuition that even contradicts
> geometry?
>
The geometric analogy that you are making here is not correct. The geometric representation of the infinity of intervals that Achilles crosses *necessarily* includes the end-point where Achilles catches up with the tortoise, at least in a diagrammatic sense. Show me how you can geometrically represent a countable infinity of intervals, e.g.,
>
> Also the above geometry gives a fine intuition how
> an infinite intersection can result into a point.
> Thats just the point where the two straight
>
> lines cross. I really wonder where you get your
> intuition from, and intuition that even contradicts
> geometry?
>
{[0,1/2],[1/2,3/4],[3/4,7/8],...},
that does NOT include the end-point 1. Not possible, at least in a diagrammatic sense.
For that matter, there is no geometric representation of the half open interval [0,1}. You cannot diagrammatically stretch a line segment from 0 through all points less than 1 without reaching the point 1.
Interesting that you bring up geometry and accuse me of contradicting it. On the contrary, the real analysis that I am proposing in my logic NAFL actually tacitly accepts (diagrammatic) Euclidean geometry as consistent and true. Euclidean geometry is unavoidably programmed into the human brain and we cannot reason without it. In the logic NAFL, non-Euclidean geometries (and therefore relativity theory) are inconsistent. The basic ideas are contained in this eprint of mine written in 2005:
https://arxiv.org/abs/math/0506475
I am now working on giving a clear description of the logic NAFL and how real analysis works in NAFL and my forthcoming paper will be better written than the above.
To give you a very brief idea, in NAFL, infinite sets do not exist and quantification over proper (infinite) classes is banned. Real numbers are represented by sequences of rationals. A real interval, say, [0,1] is represented by a sequence of rationals that has precisely [0,1] as its limit points. You can immediately see that there is no sequence of rationals that has only [0,1} as limit points because any such sequence must necessarily include the end point 1 as a limit point (according to a standard theorem of classical real analysis). This is the algebraic equivalent of the evident geometric fact that you cannot draw a line segment [0,1) that excludes the end-point 1.
You can also do calculus without open intervals if you accept that dy/dx IS 0/0. In classical real analysis 0/0 is undefined (can be anything). In the NAFL version of real analysis, every time you algebraically define a derivative like dy/dx, all you are doing is giving a definition of 0/0. Since there is no unique definition of 0/0, which is not a real number, this definition of the derivative does not contradict anything.
In the NAFL version of real analysis you cannot even state Zeno's dichotomy paradox, because a countable infinity of real numbers that excludes its limit points does not exist.
I am eager to write all this up unhurriedly and clearly, now that I am retired and nobody is harassing me with petty nonsense (as was the case in 2005, when my research funding was cut off by IBM). From 2005 to 2015 (when I retired) I could not pursue my research work at IBM.
Transfinite Numbers
Nov 21, 2019, 11:11:35 AM11/21/19
to
If you look closely:
https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
The two crossing straight lines, without the intersection
point, are certainly what you are looking for.
IBM would agree, that the upper bound of the sequence
(0,10),(0,15), (0,17 1/2), etc.. is 20. Even IBM
Switzerland would agree, and that 20 is not an upper
end of one of the intervalls in the sequence.
Its the same as:
0.999... = 1
https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
The two crossing straight lines, without the intersection
point, are certainly what you are looking for.
IBM would agree, that the upper bound of the sequence
(0,10),(0,15), (0,17 1/2), etc.. is 20. Even IBM
Switzerland would agree, and that 20 is not an upper
end of one of the intervalls in the sequence.
Its the same as:
0.999... = 1
R. Srinivasan
Nov 21, 2019, 11:43:07 AM11/21/19
to
On Thursday, November 21, 2019 at 9:41:35 PM UTC+5:30, Transfinite Numbers wrote:
> If you look closely:
> https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
>
> The two crossing straight lines, without the intersection
> point,
>
Hold it right there. Give me a *geometric* (meaning diagrammatic) construction of "the two crossing straight lines, without the intersection point".
> If you look closely:
> https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
>
> The two crossing straight lines, without the intersection
> point,
>
Not possible to diagrammatically remove one point from a geometric construction of a line segment. The problem is you are mixing algebra and geometry here. You have tacitly accepted the existence of open sets of real numbers in an algebraic sense and are inserting your belief into a geometric construction.
>
>are certainly what you are looking for.
> IBM would agree, that the upper bound of the sequence
>
> (0,10),(0,15), (0,17 1/2), etc.. is 20. Even IBM
> Switzerland would agree, and that 20 is not an upper
> end of one of the intervalls in the sequence.
>
> Its the same as:
>
> 0.999... = 1
>
Again you are making the mistake of tacitly accepting the existence of the infinite sequence of intervals in question in an algebraic sense and inserting that belief into a geometric construction. I am saying that there is no geometric (which, for me, is diagrammatic) construction of such a sequence of intervals that excludes the stated upper bound 20 (which can be thought of as an interval of zero length). And in my logic NAFL, the diagrammatic construction is supreme and the algebraic coding of real numbers and superclasses of real numbers by rational sequences follows the diagrammatic construction faithfully. Note that if a real number is a proper class, classes of real numbers do not exist. Hence I use the word "superclass" for a collection of real numbers coded by a sequence of rationals.
>are certainly what you are looking for.
> IBM would agree, that the upper bound of the sequence
>
> (0,10),(0,15), (0,17 1/2), etc.. is 20. Even IBM
> Switzerland would agree, and that 20 is not an upper
> end of one of the intervalls in the sequence.
>
> Its the same as:
>
> 0.999... = 1
>
Regards, RS
R. Srinivasan
Nov 21, 2019, 11:58:36 AM11/21/19
to
On Thursday, November 21, 2019 at 10:13:07 PM UTC+5:30, R. Srinivasan wrote:
> On Thursday, November 21, 2019 at 9:41:35 PM UTC+5:30, Transfinite Numbers wrote:
> > If you look closely:
> > https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
> >
> > The two crossing straight lines, without the intersection
> > point,
> >
> Hold it right there. Give me a *geometric* (meaning diagrammatic) construction of "the two crossing straight lines, without the intersection point".
>
> Not possible to diagrammatically remove one point from a geometric construction of a line segment. The problem is you are mixing algebra and geometry here. You have tacitly accepted the existence of open sets of real numbers in an algebraic sense and are inserting your belief into a geometric construction.
> >
> >are certainly what you are looking for.
> > IBM would agree, that the upper bound of the sequence
> >
> > (0,10),(0,15), (0,17 1/2), etc.. is 20. Even IBM
> > Switzerland would agree, and that 20 is not an upper
> > end of one of the intervalls in the sequence.
> >
> > Its the same as:
> >
> > 0.999... = 1
> >
> Again you are making the mistake of tacitly accepting the existence of the infinite sequence of intervals in question in an algebraic sense and inserting that belief into a geometric construction. I am saying that there is no geometric (which, for me, is diagrammatic) construction of such a sequence of intervals that excludes the stated upper bound 20 (which can be thought of as an interval of zero length).
>
What I mean here is that you cannot avoid constructing [0,20] with any geometric (diagrammatic) construction of all the intervals
> On Thursday, November 21, 2019 at 9:41:35 PM UTC+5:30, Transfinite Numbers wrote:
> > If you look closely:
> > https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
> >
> > The two crossing straight lines, without the intersection
> > point,
> >
> Hold it right there. Give me a *geometric* (meaning diagrammatic) construction of "the two crossing straight lines, without the intersection point".
>
> Not possible to diagrammatically remove one point from a geometric construction of a line segment. The problem is you are mixing algebra and geometry here. You have tacitly accepted the existence of open sets of real numbers in an algebraic sense and are inserting your belief into a geometric construction.
> >
> >are certainly what you are looking for.
> > IBM would agree, that the upper bound of the sequence
> >
> > (0,10),(0,15), (0,17 1/2), etc.. is 20. Even IBM
> > Switzerland would agree, and that 20 is not an upper
> > end of one of the intervalls in the sequence.
> >
> > Its the same as:
> >
> > 0.999... = 1
> >
> Again you are making the mistake of tacitly accepting the existence of the infinite sequence of intervals in question in an algebraic sense and inserting that belief into a geometric construction. I am saying that there is no geometric (which, for me, is diagrammatic) construction of such a sequence of intervals that excludes the stated upper bound 20 (which can be thought of as an interval of zero length).
>
{[0,10],[0,15],[0,17.5],.....}
Note that all the intervals in your original sequence should be closed for a diagrammatic construction. What you end up doing is to construct [0,20] geometrically and then suppose that this interval can be excluded from such a construction (which is really an algebraic assumption without any geometric equivalent).
Regards,RS
Jim Burns
Nov 21, 2019, 6:14:29 PM11/21/19
to
On 11/19/2019 12:10 PM, R. Srinivasan wrote:
> Actually Mueckenheim does have a paradox here, and
> it can be made clear by looking at the following
> version of Zeno's dichotomy paradox.
Actually, what Mueckenheim has is not a paradox.
However, it serves nicely as a warning against trusting too much
to intuition. Intuition might be manipulated by choosing how
one describes the situation at hand. That's WM's one weird
trick.
A different intuition can be called upon by using a different
description. For example:
-- A natural number k is in k endsegments.
-- There are more than k endsegments.
-- Therefore, k cannot be in all endsegments.
-- k is not in the set of those natural numbers which are
in all endsegments == the intersection of all endsegments.
-- Generalizing, the intersection of all endsegments
is empty.
[...]
infinite sets.
It can't be stated without natural numbers, though. And
the domain of _what we consider natural numbers_ is
infinite, for all reasonable interpretations of what that
means. I don't know if your requirements rule out
infinite _domains_ .
Being an element j of an endsegment E(k) = { k,k+1,k+2,... }
is equivalent to the statement j >= k.
Being an element j of each of a finite initial sequence
of endsegments E(1),E(2),...,E(k) is equivalent to a
finite conjunction of similar statements
j e E(1) & j e E(2) & ... & j e E(k)
which is equivalent to
j >= 1 & j >= 2 & ... & j >= k
which is equivalent to
Ai =< k: j >= i
The first half of the "paradox" is
Ak:( Intrsct{ E(1),E(2),...,E(k) } ~= {} )
Ak, Ej:( j e Intrsct{ E(1),E(2),...,E(k) } )
This becomes
Ak,Ej:( Ai =< k: j >= i )
The second half of the "paradox" is
Intrsct{ E(1),E(2), E(3),... } = {}
This is equivalent to the "infinite conjunction"
~Ek:( k >= 1 & k >= 2 & k >= 3 & ... )
This is better written
~Ek,Aj: k >= j
This is the set-free form of the two halves of the "paradox"
Ak,Ej:( Ai =< k: j >= i )
~Ek,Aj: k >= j
They don't look especially contradictory that way.
However, those are equivalent to these
Ak,Ej:( j >= k )
Ak,Ej:( j > k )
Even less contradictory-looking that way.
Recall what the "paradox" is:
Ak:( Intrsct{ E(1),E(2),...,E(k) } ~= {} )
Intrsct{ E(1),E(2), E(3),... } = {}
This is not a paradox.
It's only a couple formulas transformed to give them
a misleading superficial contradictoriness.
And the paradox they don't present can be expressed
without sets.
In my town, there's a State Fair every summer, and one
of the traditional exhibits is the Butter Cow -- a
sculpture of a cow (along with some humans we wish
to celebrate, for whatever reason) sculpted out of butter.
The Butter Cow looks (kinda) like a cow.
Nonetheless, the Butter Cow is not a cow.
There's a lesson here.
> Actually Mueckenheim does have a paradox here, and
> it can be made clear by looking at the following
> version of Zeno's dichotomy paradox.
Actually, what Mueckenheim has is not a paradox.
However, it serves nicely as a warning against trusting too much
to intuition. Intuition might be manipulated by choosing how
one describes the situation at hand. That's WM's one weird
trick.
A different intuition can be called upon by using a different
description. For example:
-- A natural number k is in k endsegments.
-- There are more than k endsegments.
-- Therefore, k cannot be in all endsegments.
-- k is not in the set of those natural numbers which are
in all endsegments == the intersection of all endsegments.
-- Generalizing, the intersection of all endsegments
is empty.
[...]
> I claim that these paradoxes arise essentially because
> of the existence of infinite sets and quantification over
> these. We need to restrict infinite classes to be proper
> of the existence of infinite sets and quantification over
> classes and ban quantification over proper classes in
> order to avoid these paradoxes.
The Mueckenheim "paradox" can be re-stated without
> order to avoid these paradoxes.
infinite sets.
It can't be stated without natural numbers, though. And
the domain of _what we consider natural numbers_ is
infinite, for all reasonable interpretations of what that
means. I don't know if your requirements rule out
infinite _domains_ .
Being an element j of an endsegment E(k) = { k,k+1,k+2,... }
is equivalent to the statement j >= k.
Being an element j of each of a finite initial sequence
of endsegments E(1),E(2),...,E(k) is equivalent to a
finite conjunction of similar statements
j e E(1) & j e E(2) & ... & j e E(k)
which is equivalent to
j >= 1 & j >= 2 & ... & j >= k
which is equivalent to
Ai =< k: j >= i
The first half of the "paradox" is
Ak:( Intrsct{ E(1),E(2),...,E(k) } ~= {} )
Ak, Ej:( j e Intrsct{ E(1),E(2),...,E(k) } )
This becomes
Ak,Ej:( Ai =< k: j >= i )
The second half of the "paradox" is
Intrsct{ E(1),E(2), E(3),... } = {}
This is equivalent to the "infinite conjunction"
~Ek:( k >= 1 & k >= 2 & k >= 3 & ... )
This is better written
~Ek,Aj: k >= j
This is the set-free form of the two halves of the "paradox"
Ak,Ej:( Ai =< k: j >= i )
~Ek,Aj: k >= j
They don't look especially contradictory that way.
However, those are equivalent to these
Ak,Ej:( j >= k )
Ak,Ej:( j > k )
Even less contradictory-looking that way.
Recall what the "paradox" is:
Ak:( Intrsct{ E(1),E(2),...,E(k) } ~= {} )
Intrsct{ E(1),E(2), E(3),... } = {}
This is not a paradox.
It's only a couple formulas transformed to give them
a misleading superficial contradictoriness.
And the paradox they don't present can be expressed
without sets.
In my town, there's a State Fair every summer, and one
of the traditional exhibits is the Butter Cow -- a
sculpture of a cow (along with some humans we wish
to celebrate, for whatever reason) sculpted out of butter.
The Butter Cow looks (kinda) like a cow.
Nonetheless, the Butter Cow is not a cow.
There's a lesson here.
Transfinite Numbers
Nov 21, 2019, 7:56:02 PM11/21/19
to
Well you just remove the point. It is like in
AP brain fartos famous confusion. You can check
yourself, is this geometry sentence correct:
between two points lies a third new point
Actually the more correct formulation is:
between two distinct points lies a third new point
So in geometry we think that the points are
equipped with some equality, and we can both
decide for two points p,q equality p=q and
apartness p=/=q. So the diagram of a line intersection
without the line intersection point is simply the
diagram colored, whereby the two lines minus the
intersection point have the color green and the
intersection point has the color red.
AP brain fartos famous confusion. You can check
yourself, is this geometry sentence correct:
between two points lies a third new point
Actually the more correct formulation is:
between two distinct points lies a third new point
So in geometry we think that the points are
equipped with some equality, and we can both
decide for two points p,q equality p=q and
apartness p=/=q. So the diagram of a line intersection
without the line intersection point is simply the
diagram colored, whereby the two lines minus the
intersection point have the color green and the
intersection point has the color red.
Transfinite Numbers
Nov 21, 2019, 8:14:42 PM11/21/19
to
Dont tell me you cannot tell apart 10 from 20,
15 from 20, 17 1/2 from 20 and so on.
Thats what you guys do all the time. Tell apart
0.9 from 1, 0.99 from 1, 0.999 from 1, and
then use Mückenschuss to jump to the conclusion
0.999... =/= 1. Not very convincing.
Nothing todo with algebra vs geometry, both domains
have some apartness ingrained. You could of course
give up LEM for equality, not require this anymore:
x = y v x =/= y.
Better then use two different relations:
x = y x # y
But sooner or later doing this and sliding into
constructive analysis, you will need to accept
Turing’s argument for the undecidability of
the halting problem. Otherwise you will run
short of examples where x = y is not decidable.
Usually x # y is easier than x = y.
Here is a nice real number:
https://en.wikipedia.org/wiki/Specker_sequence
15 from 20, 17 1/2 from 20 and so on.
Thats what you guys do all the time. Tell apart
0.9 from 1, 0.99 from 1, 0.999 from 1, and
then use Mückenschuss to jump to the conclusion
0.999... =/= 1. Not very convincing.
Nothing todo with algebra vs geometry, both domains
have some apartness ingrained. You could of course
give up LEM for equality, not require this anymore:
x = y v x =/= y.
Better then use two different relations:
x = y x # y
But sooner or later doing this and sliding into
constructive analysis, you will need to accept
Turing’s argument for the undecidability of
the halting problem. Otherwise you will run
short of examples where x = y is not decidable.
Usually x # y is easier than x = y.
Here is a nice real number:
https://en.wikipedia.org/wiki/Specker_sequence
Ganzhinterseher
Nov 22, 2019, 3:02:48 AM11/22/19
to
Am Freitag, 22. November 2019 00:14:29 UTC+1 schrieb Jim Burns:
> On 11/19/2019 12:10 PM, R. Srinivasan wrote:
>
> > Actually Mueckenheim does have a paradox here, and
> > it can be made clear by looking at the following
> > version of Zeno's dichotomy paradox.
>
> Actually, what Mueckenheim has is not a paradox.
But it requires that the sequence of endsegments "can jump from infinite to empty without ever going through finite!" [Zelos Malum in "Three facts that appear incompatible without dark numbers", sci.math (20 Nov 2019)] My reply: It appears so because the endsegments producing the "jump" remain invisible. All definable natural numbers belong to FISONs. Therefore all definable endsegments are in (1) and are superfluous in (2).
> On 11/19/2019 12:10 PM, R. Srinivasan wrote:
>
> > Actually Mueckenheim does have a paradox here, and
> > it can be made clear by looking at the following
> > version of Zeno's dichotomy paradox.
>
> Actually, what Mueckenheim has is not a paradox.
>
> However, it serves nicely as a warning against trusting too much
> to intuition.
"Intuition" ike this:
> However, it serves nicely as a warning against trusting too much
> to intuition.
∩{E(1), E(2), ..., E(k)} = E(k) for all k (1)
But
∩{E(1), E(2), ...} = {} (2)
where all k appearing in (1) can be omitted. What remains if not intuition?
Regards, WM
Ganzhinterseher
Nov 22, 2019, 4:46:54 AM11/22/19
to
Am Donnerstag, 21. November 2019 17:11:35 UTC+1 schrieb Transfinite Numbers:
> If you look closely:
> https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
>
> The two crossing straight lines, without the intersection
> point, are certainly what you are looking for.
> IBM would agree, that the upper bound of the sequence
>
> (0,10),(0,15), (0,17 1/2), etc.. is 20. Even IBM
> Switzerland would agree, and that 20 is not an upper
> end of one of the intervalls in the sequence.
>
> Its the same as:
>
> 0.999... = 1
>
For all digits 9 the result is not 1. Don't blur correct mathematics by your infinity delusions.
> If you look closely:
> https://en.wikipedia.org/wiki/Zeno's_paradoxes#/media/File:Race_between_Achilles_and_the_tortoise.gif
>
> The two crossing straight lines, without the intersection
> point, are certainly what you are looking for.
> IBM would agree, that the upper bound of the sequence
>
> (0,10),(0,15), (0,17 1/2), etc.. is 20. Even IBM
> Switzerland would agree, and that 20 is not an upper
> end of one of the intervalls in the sequence.
>
> Its the same as:
>
> 0.999... = 1
>
Regards WM
Transfinite Numbers
Nov 22, 2019, 7:11:09 AM11/22/19
to
Don't blur correct mathematics by your infinity delusions,
aka Mückenschuss of Muckamatik.
This is a finite intersection:
∩{E(1), E(2), ..., E(k)} = E(k) for all k (1)
And this is an infinite intersection,
(1) doesn't say (2) is wrong:
∩{E(1), E(2), ...} = {} (2)
Its the same with, again (1) doesn't say (2) is wrong:
For all digits 9 the result is not 1.
0.999...999 =/= 1 for all k (1)
\-- k --/
And this:
0.999... = 1 (2)
Whats wrong with you? Too much booze in the morning.
aka Mückenschuss of Muckamatik.
This is a finite intersection:
∩{E(1), E(2), ..., E(k)} = E(k) for all k (1)
(1) doesn't say (2) is wrong:
∩{E(1), E(2), ...} = {} (2)
For all digits 9 the result is not 1.
\-- k --/
And this:
0.999... = 1 (2)
Whats wrong with you? Too much booze in the morning.
Jim Burns
Nov 22, 2019, 1:52:43 PM11/22/19
to
(@WM -- I mistakenly emailed this to your private
account. I intended this discussion to be public.
I'm sorry for any trouble to you,)
On 11/22/2019 3:02 AM, Ganzhinterseher wrote:
> Am Freitag, 22. November 2019 00:14:29 UTC+1
> schrieb Jim Burns:
>> On 11/19/2019 12:10 PM, R. Srinivasan wrote:
>>> Actually Mueckenheim does have a paradox here, and
>>> it can be made clear by looking at the following
>>> version of Zeno's dichotomy paradox.
>>
>> Actually, what Mueckenheim has is not a paradox.
>
> But it requires that the sequence of endsegments
> "can jump from infinite to empty without ever going
> through finite!" [Zelos Malum in "Three facts that
> appear incompatible without dark numbers", sci.math
> (20 Nov 2019)]
A better metaphor for the change is, instead of some
infinite leap, a change in topic. We _were_ talking about
a collection of end segments _with_ a last end segment.
_Now_ we are talking about a collection of end segments
_without_ a last end segment.
> My reply: It appears so because the endsegments producing
> the "jump" remain invisible. All definable natural numbers
> belong to FISONs. Therefore all definable endsegments are
> in (1) and are superfluous in (2).
Your reply introduces "natural numbers" which are not
natural numbers. But only some of the time?
You seem to be "resolving" your inconsistencies by
retconning what you mean by "natural number". That's fine
for comic books, but this is mathematics.
----
The following claims are generally part of what is
claimed about "natural numbers".
If you won't agree to these, you are refusing to talk about
natural numbers. You are instead only talking to yourself,
about whatever you have dreamed up. And whatever problems
you see -- even if they are really problems -- are only
problems with what you have dreamed up.
0 is a natural.
(Or, call it something else. The name doesn't matter.)
For each natural k, there is a unique successor Sk,
which is also a natural, such that
-- Sk is not a successor to any other natural
other than k.
For each natural k, there is a sequence seq[k]
of naturals such that
-- seq[k] has finitely many entries
-- seq[k] begins with 0 and ends with k
-- each entry j (other than k) is immediately followed
in the sequence by Sj
-- Sk is not equal to any entry of seq[k].
For example, for 1000000, there is seq[1000000].
0, 1, 2, ..., 999998, 999999, 1000000
seq[1000000] is finite.
It begins with 0 and ends with 1000000.
0 is immediately followed by 1, 1 by 2, etc.
1000001 is not equal to any entry.
----
I. Each natural k is finitely many steps from 0.
II. There are infinitely many naturals (because there is
a 'new' Sk after each k).
I and II are just a matter of _what we mean_ by "natural".
In particular, there is no place for your dark numbers.
You're vague about what it means to be "definable",
but it seems clear that "k is definable" if and only if
"there is a seq[k] as described". Then, _according to_
_what we mean_ , there are no dark numbers.
>> However, it serves nicely as a warning against trusting
>> too much to intuition.
>
> "Intuition" ike this:
No, I'm referring to your _intuition_ that there is some
problem with (1) and (2). Maybe you've read a couple of the
many _arguments_ that have been posted for you explaining
why that's not a problem?
> ∩{E(1), E(2), ..., E(k)} = E(k) for all k (1)
This is equivalent to
Aj,Ak:(( Ai =< k: j >= i ) <-> ( j >= k))
I think all we need to prove this is transitivity
( j>=k ) -> ( k>=i -> j>=i )
> But
> ∩{E(1), E(2), ...} = {} (2)
This is equivalent to
~Ej,Ak: j >= k
Its proof is just a matter of there not being
a largest natural.
But you can't equivocate on what a _natural_ is.
> where all k appearing in (1) can be omitted.
> What remains if not intuition?
Consider the end segments
{ 1,2,3,4,... }
{ 2,3,4,5,... }
{ 3,4,5,6,... }
{ 4,5,6,7,... }
..
1 is not in all end segments.
2 is not in all end segments.
3 is not in all end segments.
4 is not in all end segments.
..
This is a matter of what we mean by "natural":
There are no natural other than the ones with finite
sequences, such as 0,1,2,...,998,999,1000.
All of those naturals (with finite sequences) are not in
the intersection of all end segments. And all of those
naturals are _all of the naturals_
Therefore, the intersection of all end segments is empty.
account. I intended this discussion to be public.
I'm sorry for any trouble to you,)
On 11/22/2019 3:02 AM, Ganzhinterseher wrote:
> Am Freitag, 22. November 2019 00:14:29 UTC+1
> schrieb Jim Burns:
>> On 11/19/2019 12:10 PM, R. Srinivasan wrote:
>>> Actually Mueckenheim does have a paradox here, and
>>> it can be made clear by looking at the following
>>> version of Zeno's dichotomy paradox.
>>
>> Actually, what Mueckenheim has is not a paradox.
>
> But it requires that the sequence of endsegments
> "can jump from infinite to empty without ever going
> through finite!" [Zelos Malum in "Three facts that
> appear incompatible without dark numbers", sci.math
> (20 Nov 2019)]
infinite leap, a change in topic. We _were_ talking about
a collection of end segments _with_ a last end segment.
_Now_ we are talking about a collection of end segments
_without_ a last end segment.
> My reply: It appears so because the endsegments producing
> the "jump" remain invisible. All definable natural numbers
> belong to FISONs. Therefore all definable endsegments are
> in (1) and are superfluous in (2).
natural numbers. But only some of the time?
You seem to be "resolving" your inconsistencies by
retconning what you mean by "natural number". That's fine
for comic books, but this is mathematics.
----
The following claims are generally part of what is
claimed about "natural numbers".
If you won't agree to these, you are refusing to talk about
natural numbers. You are instead only talking to yourself,
about whatever you have dreamed up. And whatever problems
you see -- even if they are really problems -- are only
problems with what you have dreamed up.
0 is a natural.
(Or, call it something else. The name doesn't matter.)
For each natural k, there is a unique successor Sk,
which is also a natural, such that
-- Sk is not a successor to any other natural
other than k.
For each natural k, there is a sequence seq[k]
of naturals such that
-- seq[k] has finitely many entries
-- seq[k] begins with 0 and ends with k
-- each entry j (other than k) is immediately followed
in the sequence by Sj
-- Sk is not equal to any entry of seq[k].
For example, for 1000000, there is seq[1000000].
0, 1, 2, ..., 999998, 999999, 1000000
seq[1000000] is finite.
It begins with 0 and ends with 1000000.
0 is immediately followed by 1, 1 by 2, etc.
1000001 is not equal to any entry.
----
I. Each natural k is finitely many steps from 0.
II. There are infinitely many naturals (because there is
a 'new' Sk after each k).
I and II are just a matter of _what we mean_ by "natural".
In particular, there is no place for your dark numbers.
You're vague about what it means to be "definable",
but it seems clear that "k is definable" if and only if
"there is a seq[k] as described". Then, _according to_
_what we mean_ , there are no dark numbers.
>> However, it serves nicely as a warning against trusting
>> too much to intuition.
>
> "Intuition" ike this:
problem with (1) and (2). Maybe you've read a couple of the
many _arguments_ that have been posted for you explaining
why that's not a problem?
> ∩{E(1), E(2), ..., E(k)} = E(k) for all k (1)
Aj,Ak:(( Ai =< k: j >= i ) <-> ( j >= k))
I think all we need to prove this is transitivity
( j>=k ) -> ( k>=i -> j>=i )
> But
> ∩{E(1), E(2), ...} = {} (2)
~Ej,Ak: j >= k
Its proof is just a matter of there not being
a largest natural.
But you can't equivocate on what a _natural_ is.
> where all k appearing in (1) can be omitted.
> What remains if not intuition?
{ 1,2,3,4,... }
{ 2,3,4,5,... }
{ 3,4,5,6,... }
{ 4,5,6,7,... }
..
1 is not in all end segments.
2 is not in all end segments.
3 is not in all end segments.
4 is not in all end segments.
..
This is a matter of what we mean by "natural":
There are no natural other than the ones with finite
sequences, such as 0,1,2,...,998,999,1000.
All of those naturals (with finite sequences) are not in
the intersection of all end segments. And all of those
naturals are _all of the naturals_
Therefore, the intersection of all end segments is empty.
Ganzhinterseher
Nov 23, 2019, 7:03:53 AM11/23/19
to
Am Freitag, 22. November 2019 13:11:09 UTC+1 schrieb Transfinite Numbers:
> This is a finite intersection:
>
> ∩{E(1), E(2), ..., E(k)} = E(k) for all k (1)
>
> And this is an infinite intersection,
> (1) doesn't say (2) is wrong:
>
> ∩{E(1), E(2), ...} = {} (2)
But (1) says that you cannot define any endsegment that appears in (2) because all definable endsegments are in (1).
> This is a finite intersection:
>
> ∩{E(1), E(2), ..., E(k)} = E(k) for all k (1)
>
> And this is an infinite intersection,
> (1) doesn't say (2) is wrong:
>
> ∩{E(1), E(2), ...} = {} (2)
>
> Its the same with, again (1) doesn't say (2) is wrong:
Don't use handwaving claims- Define any endsegment required in (2).
> Its the same with, again (1) doesn't say (2) is wrong:
>
> For all digits 9 the result is not 1.
>
> 0.999...999 =/= 1 for all k (1)
> \-- k --/
>
> And this:
>
> 0.999... = 1 (2)
Every digit at a definable position fails. Dp you believe in dark positions?
> For all digits 9 the result is not 1.
>
> 0.999...999 =/= 1 for all k (1)
> \-- k --/
>
> And this:
>
> 0.999... = 1 (2)
Regards, WM
Ganzhinterseher
Nov 23, 2019, 7:24:57 AM11/23/19
to
Am Freitag, 22. November 2019 19:52:43 UTC+1 schrieb Jim Burns:
> (@WM -- I mistakenly emailed this to your private
> account. I intended this discussion to be public.
> I'm sorry for any trouble to you,)
No matter. I never look into that account. That's why I switched to Hinterseher.
> (@WM -- I mistakenly emailed this to your private
> account. I intended this discussion to be public.
> I'm sorry for any trouble to you,)
> > But it requires that the sequence of endsegments
> > "can jump from infinite to empty without ever going
> > through finite!" [Zelos Malum in "Three facts that
> > appear incompatible without dark numbers", sci.math
> > (20 Nov 2019)]
>
> A better metaphor for the change is, instead of some
> infinite leap, a change in topic. We _were_ talking about
> a collection of end segments _with_ a last end segment.
> _Now_ we are talking about a collection of end segments
> _without_ a last end segment.
>
> > My reply: It appears so because the endsegments producing
> > the "jump" remain invisible. All definable natural numbers
> > belong to FISONs. Therefore all definable endsegments are
> > in (1) and are superfluous in (2).
>
> Your reply introduces "natural numbers" which are not
> natural numbers. But only some of the time?
My answer refers to infinity which is not infinite. The set of all definable natnumbers does not make infinity complete.
> > My reply: It appears so because the endsegments producing
> > the "jump" remain invisible. All definable natural numbers
> > belong to FISONs. Therefore all definable endsegments are
> > in (1) and are superfluous in (2).
>
> Your reply introduces "natural numbers" which are not
> natural numbers. But only some of the time?
>
> You seem to be "resolving" your inconsistencies by
> retconning what you mean by "natural number".
The inconsistency is at your side:
> You seem to be "resolving" your inconsistencies by
> retconning what you mean by "natural number".
This infinite sequence
{1} = {1}
{1} U {1, 2} = {1, 2}
{1} U {1, 2} U {1, 2, 3} = {1, 2, 3}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} = {1, 2, 3, 4,}
{1} U {1, 2} U {1, 2, 3} U {1, 2, 3, 4} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5}
...
contains all definabe FISONs but no infinite union. If you claim that the infinite union existst and is ℕ with |ℕ| = ℵo, larger than every n, then each of the above FISONs can be omitted. What remains to make up an infinite union? No definable FISON available.
> In particular, there is no place for your dark numbers.
>
> You're vague about what it means to be "definable",
A natnumer is definable if we can reach consent what natnumber is meant. The simplest way is to have the decimal representation.
> You're vague about what it means to be "definable",
> but it seems clear that "k is definable" if and only if
> "there is a seq[k] as described". Then, _according to_
> _what we mean_ , there are no dark numbers.
> No, I'm referring to your _intuition_ that there is some
> problem with (1) and (2). Maybe you've read a couple of the
> many _arguments_ that have been posted for you explaining
> why that's not a problem?
>
> > ∩{E(1), E(2), ..., E(k)} = E(k) for all k (1)
>
> This is equivalent to
> Aj,Ak:(( Ai =< k: j >= i ) <-> ( j >= k))
>
> Therefore, the intersection of all end segments is empty.
Please find an endsegment with a definable natnumber as index that is required there. You can't. Required are there only undefinable endsegments. The usual answer by set theorists is this: "There must be infinitely many, but we cannot specify a concrete one." That's essentially what I claim. They do not understand (or pretend so) that all definable endsegments can be omitted, because they cannot believe that unspecifiable endsegments are unspecifiable.
> Therefore, the intersection of all end segments is empty.
Regards, WM
Transfinite Numbers
Nov 23, 2019, 7:47:06 AM11/23/19
to
Don't blur correct mathematics by your infinity delusions
R. Srinivasan
Nov 23, 2019, 9:25:42 AM11/23/19
to
Let us start from the sets
N = {1,2,3,...), E(n) = {n, n+1, n+2,...}, n=1,2,...
Let us define the operation RnX as "Remove the number n from the set X"
Consider the infinite sequence of operations
R1E(1)_R2E(2)_....RnE(n)_Rn+1E(n+1)....
Will this sequence of operations terminate? Note that if we start with N=E(1) and perform the operation R1E(1), we get E(2). We then perform R2E(2) to get E(3). Thus from the original set N, we have removed, one by one, each of the infinitely many natural numbers and so what you should end up with is the empty set.
Let us modify the above sequence of operations as follows. Define the operation
SX as "Subtract 1 from each element of X". Clearly SR1E(1)=E(1)=N. Now repeat this operation infinitely many times, i.e.,
SR1E(1)_SR(1)E(1)_.....
This infinite sequence of operations also removes, from the original set N=E(1), each of the natural numbers 1,2,3,.... To see this note that when the operations SR1E(1) is done twice, we have removed the numbers 1 and 2 from the original set N that we started with, etc.
But this sequence of operations cannot terminate and what remains is always the original set N. I think this intuition is what Mueckenheim is trying to capture.
It is paradoxical to claim that you can complete an infinite sequence of operations one at a time, that is, that an infinite counting process can be completed. I have stated the impossibility of completing such a process as the "nonclassical induction principle" in my eprint on Zeno's dichotomy paradox:
http://philsci-archive.pitt.edu/15799/
The only counter to Mueckenheim's argument is that the infinite intersection on {E(1),E(2),...} is not a counting process, i.e., it is not done "one at a time". But our intuition that intersection is empty comes from precisely such a counting process. I.e.,
1 is not in the infinite intersection.
2 is not in the infinite intersection.
If n is not in the infinite intersection, n+1 is not in the infinite intersection.
Therefore the infinite intersection is empty.
But there are alternative intuitions that the infinite intersection should be nonempty as I have noted. As far as I am concerned, quantification over infinite sets is the reason that such paradoxes arise.
Regards, RS
Transfinite Numbers
Nov 23, 2019, 12:56:17 PM11/23/19
to
This would be a Mükenschuss, to jump from, lets
write R^n(a) for { x | R^n(a,x) }, then:
forall n e N P(R^n(a))
To this here:
P(∩_n e N R^n(a))
forall and ∩ don't commute like this. There is no
such inference rule in math.
write R^n(a) for { x | R^n(a,x) }, then:
forall n e N P(R^n(a))
To this here:
P(∩_n e N R^n(a))
forall and ∩ don't commute like this. There is no
such inference rule in math.
Transfinite Numbers
Nov 23, 2019, 12:58:56 PM11/23/19
to
Mückenschuss is a fallacy.
FredJeffries
Nov 23, 2019, 3:34:05 PM11/23/19
to
On Saturday, November 23, 2019 at 6:25:42 AM UTC-8, R. Srinivasan wrote:
> When we perform infinitely many operations on an infinite set,
No human has ever yet 'performed infinitely many operations'
> When we perform infinitely many operations on an infinite set,
> we are always setting ourselves up for a paradox. Let me try to explain the paradox that Mueckenheim is trying to capture.
>
> Let us start from the sets
>
> N = {1,2,3,...), E(n) = {n, n+1, n+2,...}, n=1,2,...
>
> Let us define the operation RnX as "Remove the number n from the set X"
>
> Consider the infinite sequence of operations
>
> R1E(1)_R2E(2)_....RnE(n)_Rn+1E(n+1)....
>
> Will this sequence of operations terminate?
> Note that if we start with N=E(1) and perform the operation R1E(1), we get E(2). We then perform R2E(2) to get E(3). Thus from the original set N, we have removed, one by one, each of the infinitely many natural numbers
> and so what you should end up with is the empty set.
> Let us modify the above sequence of operations as follows. Define the operation
> SX as "Subtract 1 from each element of X". Clearly SR1E(1)=E(1)=N. Now repeat this operation infinitely many times, i.e.,
>
> SR1E(1)_SR(1)E(1)_.....
> This infinite sequence of operations also removes
, from the original set N=E(1), each of the natural numbers 1,2,3,.... To see this note that when the operations SR1E(1) is done twice, we have removed the numbers 1 and 2 from the original set N that we started with, etc.
>
> But this sequence of operations cannot terminate
> and what remains is always the original set N.
> I think this intuition is what Mueckenheim is trying to capture.
>
> It is paradoxical to claim that you can complete an infinite sequence of operations one at a time
Transfinite Numbers
Nov 23, 2019, 3:44:36 PM11/23/19
to
I agree with "gibberish" somehow. Many paradoxes are
not based on faulty reasoning. Rather on uncareful
assumptions that unexpectedly lead to some unwanted
results. For example the Russel Set paradox, that
basically shows that the universal class cannot be
a set, at least for example in ZFC, doesn't use any
faulty reasoning. So its very difficult to follow
these new Mückenheim Paradoxes when they use Mückenschuss
and similar. I don't think faulty reasonig gives
paradoxes. It only gives fault results.
not based on faulty reasoning. Rather on uncareful
assumptions that unexpectedly lead to some unwanted
results. For example the Russel Set paradox, that
basically shows that the universal class cannot be
a set, at least for example in ZFC, doesn't use any
faulty reasoning. So its very difficult to follow
these new Mückenheim Paradoxes when they use Mückenschuss
and similar. I don't think faulty reasonig gives
paradoxes. It only gives fault results.
Khong Dong
Nov 23, 2019, 3:58:33 PM11/23/19
to
On Saturday, 23 November 2019 13:34:05 UTC-7, FredJeffries wrote:
> On Saturday, November 23, 2019 at 6:25:42 AM UTC-8, R. Srinivasan wrote:
> > When we perform infinitely many operations on an infinite set,
>
> No human has ever yet 'performed infinitely many operations'
As a logical consequence, Goedel could not possibly have known what the order of
> On Saturday, November 23, 2019 at 6:25:42 AM UTC-8, R. Srinivasan wrote:
> > When we perform infinitely many operations on an infinite set,
>
> No human has ever yet 'performed infinitely many operations'
prime-magnitudes be: to mortal knowledge that's just an illusion, pacifying
the poor mind - opium of the mind, so to speak (though no politics intended).
R. Srinivasan
Nov 23, 2019, 4:50:20 PM11/23/19
to
On Sunday, November 24, 2019 at 2:04:05 AM UTC+5:30, FredJeffries wrote:
> On Saturday, November 23, 2019 at 6:25:42 AM UTC-8, R. Srinivasan wrote:
> > When we perform infinitely many operations on an infinite set,
>
> No human has ever yet 'performed infinitely many operations'
>
> > we are always setting ourselves up for a paradox. Let me try to explain the paradox that Mueckenheim is trying to capture.
> >
> > Let us start from the sets
> >
> > N = {1,2,3,...), E(n) = {n, n+1, n+2,...}, n=1,2,...
> >
> > Let us define the operation RnX as "Remove the number n from the set X"
> >
> > Consider the infinite sequence of operations
> >
> > R1E(1)_R2E(2)_....RnE(n)_Rn+1E(n+1)....
> >
> > Will this sequence of operations terminate?
>
> Of course not. YOU have claimed that it is 'infinite sequence of operations'. Infinite sequences do not terminate.
>
OK. So when Achilles chases a tortoise, does he or does he not cross an infinite sequence of points? I am asking can this infinite sequence of operations be completed? Does he, or does he not complete an infinite counting process when he crosses an infinite sequence of points? Say yes or no, and then we can proceed.
> On Saturday, November 23, 2019 at 6:25:42 AM UTC-8, R. Srinivasan wrote:
> > When we perform infinitely many operations on an infinite set,
>
> No human has ever yet 'performed infinitely many operations'
>
> > we are always setting ourselves up for a paradox. Let me try to explain the paradox that Mueckenheim is trying to capture.
> >
> > Let us start from the sets
> >
> > N = {1,2,3,...), E(n) = {n, n+1, n+2,...}, n=1,2,...
> >
> > Let us define the operation RnX as "Remove the number n from the set X"
> >
> > Consider the infinite sequence of operations
> >
> > R1E(1)_R2E(2)_....RnE(n)_Rn+1E(n+1)....
> >
> > Will this sequence of operations terminate?
>
> Of course not. YOU have claimed that it is 'infinite sequence of operations'. Infinite sequences do not terminate.
>
Regards,
RS
FredJeffries
Nov 23, 2019, 5:30:31 PM11/23/19
to
On Saturday, November 23, 2019 at 1:50:20 PM UTC-8, R. Srinivasan wrote:
> OK. So when Achilles chases a tortoise, does he or does he not cross an infinite sequence of points? I am asking can this infinite sequence of operations be completed? Does he, or does he not complete an infinite counting process when he crosses an infinite sequence of points? Say yes or no, and then we can proceed.
Your question is, at best, a false dichotomy; at worst, complete gibberish. Only a fool would accede to your demand to 'Say yes or no[sic]'
> OK. So when Achilles chases a tortoise, does he or does he not cross an infinite sequence of points? I am asking can this infinite sequence of operations be completed? Does he, or does he not complete an infinite counting process when he crosses an infinite sequence of points? Say yes or no, and then we can proceed.
I already gave you my response several days ago: give us precise definitions for your vague, ambiguous, misleading terms such as 'Achilles', 'cross', 'chase'.
And tell us why, since you concede that the interval [0, 1/2] CAN be 'traversed', you cannot use the same technique to 'traverse' the interval [1/2, 1] and, thus, the whole interval [0, 1]
If your Achilles is a human, is he able to 'cross' the interval from his back end to his front?
https://groups.google.com/forum/?hl=en#!original/sci.logic/tbNEmXQIfsY/ZsFw6NsECgAJ
FredJeffries
Nov 23, 2019, 5:48:24 PM11/23/19
to
On Wednesday, November 20, 2019 at 11:25:09 AM UTC-8, Ganzhinterseher wrote:
> Am Mittwoch, 20. November 2019 15:20:01 UTC+1 schrieb FredJeffries:
>
> No, it is only wrong in dirty mathematics. We will debate, experiment, prove and conjecture until some picture emerges that satisfies this wonderful taskmaster that is our intuition.
https://en.wikipedia.org/wiki/Hubris
> Am Mittwoch, 20. November 2019 15:20:01 UTC+1 schrieb FredJeffries:
> > On Wednesday, November 20, 2019 at 6:05:40 AM UTC-8, Transfinite Numbers wrote:
> >
> > > Does not solve the problem that your intuition is wrong.
> >
> > Intuition is almost always wrong.
> >
> > > Does not solve the problem that your intuition is wrong.
> >
>
> No, it is only wrong in dirty mathematics. We will debate, experiment, prove and conjecture until some picture emerges that satisfies this wonderful taskmaster that is our intuition.
https://en.wikipedia.org/wiki/Hubris
R. Srinivasan
Nov 23, 2019, 6:15:19 PM11/23/19
to
https://www.iep.utm.edu/zeno-par/#H2
\begin{quote}
More specifically, in the case of the paradoxes of motion such as the Achilles and the Dichotomy, Zeno's mistake was not his assuming there is a completed infinity of places for the runner to go, which was what Aristotle said was Zeno's mistake. Instead, Zeno's and Aristotle's mistake was in assuming that this is too many places (for the runner to go to in a finite time).
\end{quote}
Is this author (Bradley Dowden, Professor of Philosophy at California State U., Sacramento) also talking gibberish? Do you agree with Prof. Dowden that the runner "goes to" a completed infinity of places in finite time? Or are you going to quibble about what "runner", "go to", etc., mean?
Also take a look at this paper:
http://www.pitt.edu/~jearman/EarmanNorton1996a.pdf
\begin{quote}
Black’s fallacy lies in confusion of two senses of “incompletable” and
its allure lies in the ease with which we can slide between the two senses.-
An infinite sequence of acts is incompletable in the sense that wc can
nominate no last act, the act that completes it. An infinite sequence of
acts may also be incompletable in the sense that we cannot carry out the
totality of all its acts, even though each act individually may be execut¬
able. This may become the case, for example, in the runner’s journey, if
the runner is required to spend equal time in each of the infinitely many
intervals. An infinite sequence of acts cannot be completed in the first
sense, but that certainly does not entail that it cannot be completed in the
second sense.
\end{quote}
Are Earman and Norton (professors of philosophy at U. Pittsburgh) also talking gibberish when they say that an "infinite sequence of acts" can be completed "in the second sense" noted above? Are you going to quibble with them about what "acts" mean, etc.?
There are umpteen other examples of claims amongst mathematicians, logicians, philosophers that infinite tasks (supertasks) are completeable. Are they all talking gibberish in your opinion?
Regards, RS
R. Srinivasan
Nov 23, 2019, 6:29:20 PM11/23/19
to
On Sunday, November 24, 2019 at 4:00:31 AM UTC+5:30, FredJeffries wrote:
> On Saturday, November 23, 2019 at 1:50:20 PM UTC-8, R. Srinivasan wrote:
>
> > OK. So when Achilles chases a tortoise, does he or does he not cross an infinite sequence of points? I am asking can this infinite sequence of operations be completed? Does he, or does he not complete an infinite counting process when he crosses an infinite sequence of points? Say yes or no, and then we can proceed.
>
> Your question is, at best, a false dichotomy; at worst, complete gibberish. Only a fool would accede to your demand to 'Say yes or no[sic]'
>
> I already gave you my response several days ago: give us precise definitions for your vague, ambiguous, misleading terms such as 'Achilles', 'cross', 'chase'.
>
> And tell us why, since you concede that the interval [0, 1/2] CAN be 'traversed', you cannot use the same technique to 'traverse' the interval [1/2, 1] and, thus, the whole interval [0, 1]
>
You do not understand what I am talking about here. The point is that *we assume* that the runner is able to cross [0,1/2]. Then *assume* he is able to cross [1/2,3/4], etc. Then try to derive a contradiction from the fact that the runner completes an infinite task as described. It is called "proof by contradiction". You have to see what *my* assumptions are, and what *my* conclusions are, and show me where I erred. This is not about using "the same technique" to cross the interval [1/2,1].
> On Saturday, November 23, 2019 at 1:50:20 PM UTC-8, R. Srinivasan wrote:
>
> > OK. So when Achilles chases a tortoise, does he or does he not cross an infinite sequence of points? I am asking can this infinite sequence of operations be completed? Does he, or does he not complete an infinite counting process when he crosses an infinite sequence of points? Say yes or no, and then we can proceed.
>
> Your question is, at best, a false dichotomy; at worst, complete gibberish. Only a fool would accede to your demand to 'Say yes or no[sic]'
>
> I already gave you my response several days ago: give us precise definitions for your vague, ambiguous, misleading terms such as 'Achilles', 'cross', 'chase'.
>
> And tell us why, since you concede that the interval [0, 1/2] CAN be 'traversed', you cannot use the same technique to 'traverse' the interval [1/2, 1] and, thus, the whole interval [0, 1]
>
Regards, RS
Jim Burns
Nov 23, 2019, 10:20:48 PM11/23/19
to
On 11/23/2019 9:25 AM, R. Srinivasan wrote:
> On Friday, November 22, 2019 at 4:44:29 AM UTC+5:30,
> Jim Burns wrote:
>> However, those are equivalent to these
>>
>> Ak,Ej:( j >= k )
>>
>> Ak,Ej:( j > k )
>>
>> Even less contradictory-looking that way.
>> Recall what the "paradox" is:
>>
>> Ak:( Intrsct{ E(1),E(2),...,E(k) } ~= {} )
>>
>> Intrsct{ E(1),E(2), E(3),... } = {}
>>
>> This is not a paradox.
> On Friday, November 22, 2019 at 4:44:29 AM UTC+5:30,
> Jim Burns wrote:
>> However, those are equivalent to these
>>
>> Ak,Ej:( j >= k )
>>
>> Ak,Ej:( j > k )
>>
>> Even less contradictory-looking that way.
>> Recall what the "paradox" is:
>>
>> Ak:( Intrsct{ E(1),E(2),...,E(k) } ~= {} )
>>
>> Intrsct{ E(1),E(2), E(3),... } = {}
>>
>> This is not a paradox.
> When we perform infinitely many operations on an infinite
> set,
We do not perform infinitely many operations.
> set,
We also do not make infinitely long claims.
Something we can do is make finite-length claims about
infinitely many individuals. We can do this with a variable
that ranges over an infinite domain.
~(Sx = 0) is equivalent to the claim ~(S0 = 0) and ~(SS0 = 0)
and ~(SSSSSSSSSSSS0 = 0) and infinitely more claims.
But ~(Sx = 0) is finite in length. There is no too-long-to-say
kind of paradox hidden in saying ~(Sx = 0).
Making the same claim for each individual in an infinite
domain is a limited sort of super-power. But there are some
useful things we can say with it.
A notable example is induction.
----
Define N to be the intersection of all pre-inductive
collections,
where "X is pre-inductive" means
-- 0 is in X, and
-- for all y, if y is in X, then Sy is in X.
----
We can prove
-- If X is pre-inductive, then N *IS* a subset of X.
-- If X is an incomplete (unequal-to-N) subset of N,
then N *IS NOT* a subset of X.
X cannot be both pre-inductive and an incomplete subset
of N, since N cannot both be a subset and not be a subset
of X.
Suppose that X is a pre-inductive subset of N.
We've shown that X cannot be a pre-inductive _incomplete_
subset of N, so X is the _complete_ subset of N, which is
N itself.
--- If a subset X of N is pre-inductive (that is to say,
if 0 is in X and, for all y, if y is in X, then Sy is in X),
then X = N.
----
We can show separately that N is pre-inductive.
----
Suppose we know
~(Sx = 0)
~(x = y) -> ~(Sx = Sy)
Then, the subset Y of N defined as the elements of N
which are not their own successor, ~(Sx = x), is pre-inductive.
~(S0 = 0)
0 is in Y
~(Sy = y) -> ~(SSy = Sy)
for all y, if y is in Y, then Sy is in Y
Y is pre-inductive.
The only subset of N which Y can be is N. Y = N.
Y, which is defined as the elements of N which are not
theor own successor, is N.
for all y in N, ~(Sy = y)
Induction.
----
Note that induction does not require infinitely many
operations to prove.
Transfinite Numbers
Nov 24, 2019, 4:05:54 AM11/24/19
to
Zeno was a time traveller. He saw the confusion in
sci.logic, and travelled back ca. -2500 years to
italy to compose a few paradoxes to even confuse more.
LMAO!
sci.logic, and travelled back ca. -2500 years to
italy to compose a few paradoxes to even confuse more.
LMAO!
Transfinite Numbers
Nov 24, 2019, 4:45:11 AM11/24/19
to
To create the paradox, you dont need a continuum.
Not with the numbers involved in a linear speed
mechanics. All you need is a variant of:
between two distinct points lies a third new point
This is also true for Q (the rational numbers),
not only for R (the real numbers):
forall p,q (p<q => exists r (p<r<q))
https://en.wikipedia.org/wiki/Dense_order
Achilles and the tortoise create such witnesses r.
So Zeno is not an optimal example and then to say
the "Standard Solution for the runners in the
Achilles Paradox and the Dichotomy Paradox, the
runner's path is a physical continuum that is
completed by using a positive, finite speed."
So this "Standard Solution" is sufficient but
not necessary, and hence not convincing.
Not with the numbers involved in a linear speed
mechanics. All you need is a variant of:
between two distinct points lies a third new point
not only for R (the real numbers):
forall p,q (p<q => exists r (p<r<q))
https://en.wikipedia.org/wiki/Dense_order
Achilles and the tortoise create such witnesses r.
So Zeno is not an optimal example and then to say
the "Standard Solution for the runners in the
Achilles Paradox and the Dichotomy Paradox, the
runner's path is a physical continuum that is
completed by using a positive, finite speed."
So this "Standard Solution" is sufficient but
not necessary, and hence not convincing.
Transfinite Numbers
Nov 24, 2019, 4:51:45 AM11/24/19
to
You only get the continuum if you look at all
possible achilles and the tortoise, and where
achilles and the tortoise vary their speed
in certain bounds. Just one linear speed setup
doesn't do the job. Otherwise you are stuck with
some pre continuua paradoxes, like:
2. The Paradoxes of Plurality
2.1 The Argument from Denseness
https://plato.stanford.edu/entries/paradox-zeno
possible achilles and the tortoise, and where
achilles and the tortoise vary their speed
in certain bounds. Just one linear speed setup
doesn't do the job. Otherwise you are stuck with
some pre continuua paradoxes, like:
2. The Paradoxes of Plurality
2.1 The Argument from Denseness
https://plato.stanford.edu/entries/paradox-zeno
Ganzhinterseher
Nov 24, 2019, 5:02:48 AM11/24/19
to
Regards, WM
R. Srinivasan
Nov 24, 2019, 9:16:40 AM11/24/19
to
On Sunday, November 24, 2019 at 8:50:48 AM UTC+5:30, Jim Burns wrote:
> On 11/23/2019 9:25 AM, R. Srinivasan wrote:
> > On Friday, November 22, 2019 at 4:44:29 AM UTC+5:30,
> > Jim Burns wrote:
>
> >> However, those are equivalent to these
> >>
> >> Ak,Ej:( j >= k )
> >>
> >> Ak,Ej:( j > k )
> >>
> >> Even less contradictory-looking that way.
> >> Recall what the "paradox" is:
> >>
> >> Ak:( Intrsct{ E(1),E(2),...,E(k) } ~= {} )
> >>
> >> Intrsct{ E(1),E(2), E(3),... } = {}
> >>
> >> This is not a paradox.
>
> > When we perform infinitely many operations on an infinite
> > set,
>
> We do not perform infinitely many operations.
>
> We also do not make infinitely long claims.
>
The claim that supertasks can be completed, to which I posted a couple of respectable links, for example, the claim that Achilles crosses a "completed infinity" of points one at a time, is equivalent to the claim that an infinite counting process can be completed, i.e., that the set of natural numbers N can be exhausted by removing its elements one at a time. Do you agree?
> On 11/23/2019 9:25 AM, R. Srinivasan wrote:
> > On Friday, November 22, 2019 at 4:44:29 AM UTC+5:30,
> > Jim Burns wrote:
>
> >> However, those are equivalent to these
> >>
> >> Ak,Ej:( j >= k )
> >>
> >> Ak,Ej:( j > k )
> >>
> >> Even less contradictory-looking that way.
> >> Recall what the "paradox" is:
> >>
> >> Ak:( Intrsct{ E(1),E(2),...,E(k) } ~= {} )
> >>
> >> Intrsct{ E(1),E(2), E(3),... } = {}
> >>
> >> This is not a paradox.
>
> > When we perform infinitely many operations on an infinite
> > set,
>
> We do not perform infinitely many operations.
>
> We also do not make infinitely long claims.
>
Regards, RS
Transfinite Numbers
Nov 24, 2019, 10:26:23 AM11/24/19
to
Achilles will not jump from some n to omega.
Inside this Zeno order, and order that only exists in
the mind of the observer,
we never have a position n, such that n+1 is the
finish line. But Achilles nevertheless reaches
the finish line.
How is this possible? Well the order of the Zeno
thought experiment is only in the mind of the observer,
possibly every ordinal which is not too large
can be put on a line, even more complex ordinals
than only omega+1 which is used in the Zeno paradox.
What is the largest ordinal that can be
put on a Q line, on a R line? Cantor discovered
transfinite set theory through such a problem in
his paper about sin/cos series,
his intuition guided him to the theory.
Inside this Zeno order, and order that only exists in
the mind of the observer,
we never have a position n, such that n+1 is the
finish line. But Achilles nevertheless reaches
the finish line.
How is this possible? Well the order of the Zeno
thought experiment is only in the mind of the observer,
possibly every ordinal which is not too large
can be put on a line, even more complex ordinals
than only omega+1 which is used in the Zeno paradox.
What is the largest ordinal that can be
put on a Q line, on a R line? Cantor discovered
transfinite set theory through such a problem in
his paper about sin/cos series,
his intuition guided him to the theory.
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