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WM
May 22, 2012, 4:18:32 AM5/22/12
to
Belsa Zarkin asked in Mathematics.StackExchange:
All rational numbers of the unit interval [0, 1] can be covered by
countably many intervals, such that the n-th rational is covered by an
interval I_n of measure 1/10^n. There remain countably many
complementary intervals of measure 8/9 in total.
Does each of the complementary intervals contain only one
irrational number? Then there would be only countably many which could
be covered by another set of countably many intervals of measure 1/9.
Is there at least one of the complementary intervals containing
more than one irrational number? Then there are at least two
irrational numbers without a rational between them. That is
mathematically impossible.
There is no interval left behind. Every interval contains at least one
rational number, so if you remove all rational numbers (let alone a
bunch of intervals containing all of them), there can be no interval
left over. – BRUNO
There is a mistake in your argument, which follows from the fact that
the complement of this union contains no interval. ...The irrational
numbers form a totally disconnected space, namely every connected
component is a singleton.- A KARAGILA
In a totally disconnected space there must be points or intervals
disconnecting it. The number of these points or intervals is countable
and in bijection with the remaining intervals. The bijection is the
same as that from N to Q. (For instance at every step n the
configuration of intervals could be determined in principle.) If you
deny the validity of this bijection for the limit, why don't you deny
the validity for the limit of the bijection from N to Q? – Belsa
Zarkin yesterday
The definition of a totally disconnected space is a space in which
every connected space is a singleton. E.g. a discrete space. Are you
saying that every discrete space is countable? – A KARAGILA
This discussion lasted for a while. Meanwhile it has been deleted.
Correct mathematics is this:
There are countably many intervals I_n of measure 10^-n such that
I_n covers the rational q_n. Then 1/9 (or less) of the unit interval
is covered. In the remaining 8/9 (or
more) there are uncountably many irrationals. But every two
irrationals have a rational between each other. That implies two
irrationals have at least one interval I_n between each other (because
there are no rationals outside of intervals I_n). That implies two
irrationals have at least one of aleph_0 endpoints I_n_1 or I_n_2 of
intervals I_n between each other. These endpoints can be considered
structuring and enumerating a Cantor-list. The only difference is that
the enumeration does not follow the natural order of N. But the number
of naturals does not change by reordering. So we have uncountably many
irrationals separated by countably many endpoints. That is a
contradiction similar to uncountably many entries in a Cantor-list or
uncountably many terms in a sequence.
Regards, WM
In future this series will appear in sci.logic solely.
All rational numbers of the unit interval [0, 1] can be covered by
countably many intervals, such that the n-th rational is covered by an
interval I_n of measure 1/10^n. There remain countably many
complementary intervals of measure 8/9 in total.
Does each of the complementary intervals contain only one
irrational number? Then there would be only countably many which could
be covered by another set of countably many intervals of measure 1/9.
Is there at least one of the complementary intervals containing
more than one irrational number? Then there are at least two
irrational numbers without a rational between them. That is
mathematically impossible.
There is no interval left behind. Every interval contains at least one
rational number, so if you remove all rational numbers (let alone a
bunch of intervals containing all of them), there can be no interval
left over. – BRUNO
There is a mistake in your argument, which follows from the fact that
the complement of this union contains no interval. ...The irrational
numbers form a totally disconnected space, namely every connected
component is a singleton.- A KARAGILA
In a totally disconnected space there must be points or intervals
disconnecting it. The number of these points or intervals is countable
and in bijection with the remaining intervals. The bijection is the
same as that from N to Q. (For instance at every step n the
configuration of intervals could be determined in principle.) If you
deny the validity of this bijection for the limit, why don't you deny
the validity for the limit of the bijection from N to Q? – Belsa
Zarkin yesterday
The definition of a totally disconnected space is a space in which
every connected space is a singleton. E.g. a discrete space. Are you
saying that every discrete space is countable? – A KARAGILA
This discussion lasted for a while. Meanwhile it has been deleted.
Correct mathematics is this:
There are countably many intervals I_n of measure 10^-n such that
I_n covers the rational q_n. Then 1/9 (or less) of the unit interval
is covered. In the remaining 8/9 (or
more) there are uncountably many irrationals. But every two
irrationals have a rational between each other. That implies two
irrationals have at least one interval I_n between each other (because
there are no rationals outside of intervals I_n). That implies two
irrationals have at least one of aleph_0 endpoints I_n_1 or I_n_2 of
intervals I_n between each other. These endpoints can be considered
structuring and enumerating a Cantor-list. The only difference is that
the enumeration does not follow the natural order of N. But the number
of naturals does not change by reordering. So we have uncountably many
irrationals separated by countably many endpoints. That is a
contradiction similar to uncountably many entries in a Cantor-list or
uncountably many terms in a sequence.
Regards, WM
In future this series will appear in sci.logic solely.
William Elliot
May 22, 2012, 4:53:30 AM5/22/12
to
On Tue, 22 May 2012, WM wrote:
> Belsa Zarkin asked in Mathematics.StackExchange:
>
> All rational numbers of the unit interval [0, 1] can be covered by
> countably many intervals, such that the n-th rational is covered by an
> interval I_n of measure 1/10^n. There remain countably many
> complementary intervals of measure 8/9 in total.
The complement of an interval usually isn't an interval.
> Belsa Zarkin asked in Mathematics.StackExchange:
>
> All rational numbers of the unit interval [0, 1] can be covered by
> countably many intervals, such that the n-th rational is covered by an
> interval I_n of measure 1/10^n. There remain countably many
> complementary intervals of measure 8/9 in total.
> Does each of the complementary intervals contain only one
> irrational number? Then there would be only countably many which could
> be covered by another set of countably many intervals of measure 1/9.
> Is there at least one of the complementary intervals containing
> more than one irrational number?
> Then there are at least two irrational numbers without a rational
> between them. That is mathematically impossible.
WM
May 22, 2012, 5:04:38 AM5/22/12
to
On 22 Mai, 10:53, William Elliot <ma...@panix.com> wrote:
> On Tue, 22 May 2012, WM wrote:
> > Belsa Zarkin asked in Mathematics.StackExchange:
>
> > All rational numbers of the unit interval [0, 1] can be covered by
> > countably many intervals, such that the n-th rational is covered by an
> > interval I_n of measure 1/10^n. There remain countably many
> > complementary intervals of measure 8/9 in total.
>
> The complement of an interval usually isn't an interval.
Interval is here but an abbreviation for Intervals and points. Even
> On Tue, 22 May 2012, WM wrote:
> > Belsa Zarkin asked in Mathematics.StackExchange:
>
> > All rational numbers of the unit interval [0, 1] can be covered by
> > countably many intervals, such that the n-th rational is covered by an
> > interval I_n of measure 1/10^n. There remain countably many
> > complementary intervals of measure 8/9 in total.
>
> The complement of an interval usually isn't an interval.
the empty interval is usually called an interval.
>
> > Does each of the complementary intervals contain only one
> > irrational number? Then there would be only countably many which could
> > be covered by another set of countably many intervals of measure 1/9.
>
> No, the complement of each interval contains uncountably many irrationals.
That is just under investigation.
> > Does each of the complementary intervals contain only one
> > irrational number? Then there would be only countably many which could
> > be covered by another set of countably many intervals of measure 1/9.
>
> No, the complement of each interval contains uncountably many irrationals.
>
> > Is there at least one of the complementary intervals containing
> > more than one irrational number?
>
> Yes.
irrationals have a rational between them. As every rational q_n is
covered by an interval I_n, we can conclude that every pair x and x'
of irrationals is separated by least one interval I_n. Hence x and x'
cannot belong to the same complementary interval.
>
> > Then there are at least two irrational numbers without a rational
> > between them. That is mathematically impossible.
>
> Prove it or cram it.
The proof has been given. I repeat: As every rational q_n is covered
> > Then there are at least two irrational numbers without a rational
> > between them. That is mathematically impossible.
>
> Prove it or cram it.
by an interval I_n, we can conclude that every pair x and x' of
irrationals is separated by least one interval I_n. Hence x and x'
cannot belong to the same complementary interval. This holds for every
pair of irrationals.
A very simple conclusion. What do you not understand?
Regards, WM
William Elliot
May 22, 2012, 6:45:59 AM5/22/12
to
On Tue, 22 May 2012, WM wrote:
> > > Belsa Zarkin asked in Mathematics.StackExchange:
> >
> > > All rational numbers of the unit interval [0, 1] can be covered by
> > > countably many intervals, such that the n-th rational is covered by an
> > > interval I_n of measure 1/10^n. There remain countably many
> > > complementary intervals of measure 8/9 in total.
> >
> > The complement of an interval usually isn't an interval.
>
> Interval is here but an abbreviation for Intervals and points. Even
> the empty interval is usually called an interval.
An interval is an order convex subset and usually the complement
> > > Belsa Zarkin asked in Mathematics.StackExchange:
> >
> > > All rational numbers of the unit interval [0, 1] can be covered by
> > > countably many intervals, such that the n-th rational is covered by an
> > > interval I_n of measure 1/10^n. There remain countably many
> > > complementary intervals of measure 8/9 in total.
> >
> > The complement of an interval usually isn't an interval.
>
> Interval is here but an abbreviation for Intervals and points. Even
> the empty interval is usually called an interval.
of an interval is not order convex.
> > > Does each of the complementary intervals contain only one
> > > irrational number? Then there would be only countably many which could
> > > be covered by another set of countably many intervals of measure 1/9.
> >
> > No, the complement of each interval contains uncountably many irrationals.
>
> That is just under investigation.
>
> > > Is there at least one of the complementary intervals containing
> > > more than one irrational number?
> >
> > Yes.
>
> Impossible in mathematics. In mathematics we prove that two different
> irrationals have a rational between them. As every rational q_n is
> covered by an interval I_n, we can conclude that every pair x and x'
> of irrationals is separated by least one interval I_n. Hence x and x'
> cannot belong to the same complementary interval.
> > > Then there are at least two irrational numbers without a rational
> > > between them. That is mathematically impossible.
> >
> > Prove it or cram it.
>
> The proof has been given. I repeat: As every rational q_n is covered
> by an interval I_n, we can conclude that every pair x and x' of
> irrationals is separated by least one interval I_n. Hence x and x'
> cannot belong to the same complementary interval. This holds for every
> pair of irrationals.
> A very simple conclusion. What do you not understand?
>
WM
May 22, 2012, 7:15:15 AM5/22/12
to
On 22 Mai, 12:45, William Elliot <ma...@panix.com> wrote:
> On Tue, 22 May 2012, WM wrote:
> > > > Belsa Zarkin asked in Mathematics.StackExchange:
>
> > > > All rational numbers of the unit interval [0, 1] can be covered by
> > > > countably many intervals, such that the n-th rational is covered by an
> > > > interval I_n of measure 1/10^n. There remain countably many
> > > > complementary intervals of measure 8/9 in total.
>
> > > The complement of an interval usually isn't an interval.
>
> > Interval is here but an abbreviation for Intervals and points. Even
> > the empty interval is usually called an interval.
>
> An interval is an order convex subset and usually the complement
> of an interval is not order convex.
>
That is true but undisputed and does not help to answer the problem.
> On Tue, 22 May 2012, WM wrote:
> > > > Belsa Zarkin asked in Mathematics.StackExchange:
>
> > > > All rational numbers of the unit interval [0, 1] can be covered by
> > > > countably many intervals, such that the n-th rational is covered by an
> > > > interval I_n of measure 1/10^n. There remain countably many
> > > > complementary intervals of measure 8/9 in total.
>
> > > The complement of an interval usually isn't an interval.
>
> > Interval is here but an abbreviation for Intervals and points. Even
> > the empty interval is usually called an interval.
>
> An interval is an order convex subset and usually the complement
> of an interval is not order convex.
>
You have to explain how uncountably many irrational points can exist
in the interval [0, 1] such that every point is separated from every
point by at least one interval I_n that is constructed around a
rational number q_n.
> > > > Does each of the complementary intervals contain only one
> > > > irrational number? Then there would be only countably many which could
> > > > be covered by another set of countably many intervals of measure 1/9.
>
> > > No, the complement of each interval contains uncountably many irrationals.
>
> > That is just under investigation.
>
> Only by WM
> > The proof has been given. I repeat: As every rational q_n is covered
> > by an interval I_n, we can conclude that every pair x and x' of
> > irrationals is separated by least one interval I_n. Hence x and x'
> > cannot belong to the same complementary interval. This holds for every
> > pair of irrationals.
> > A very simple conclusion. What do you not understand?
>
> Why dogmatic bullying is so important for you
your sight. As it is disproved by the above proof, I would like to
know which step you do not accept. Your remark that the remaining
irrationals do not form intervals (but Cantor dust) does not really
address the problem, does it? In my opinion every particle of Cantor
dust is separated by a rational q_n, and therefore by an interval I_n,
from every other particle of Cantor dust.
Regards, WM
Uirgil
May 22, 2012, 4:02:51 PM5/22/12
to
In article
<f8b390e3-cded-44ca...@v10g2000vbe.googlegroups.com>,
contradiction involved in having an uncountable linear ordering having
a countable linear sub-ordering dense within it.
It is, for example, trivial that between any two non-terminating
decimals (having infinitely many non-zero digits) there is a terminating
decimal (having only finitely many non-zero digits), and that the former
are uncountable and the latter countable.
<f8b390e3-cded-44ca...@v10g2000vbe.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> Correct mathematics is this:
> There are countably many intervals I_n of measure 10^-n such that
> I_n covers the rational q_n. Then 1/9 (or less) of the unit interval
> is covered. In the remaining 8/9 (or
> more) there are uncountably many irrationals. But every two
> irrationals have a rational between each other. That implies two
> irrationals have at least one interval I_n between each other (because
> there are no rationals outside of intervals I_n). That implies two
> irrationals have at least one of aleph_0 endpoints I_n_1 or I_n_2 of
> intervals I_n between each other. These endpoints can be considered
> structuring and enumerating a Cantor-list. The only difference is that
> the enumeration does not follow the natural order of N. But the number
> of naturals does not change by reordering. So we have uncountably many
> irrationals separated by countably many endpoints. That is a
> contradiction similar to uncountably many entries in a Cantor-list or
> uncountably many terms in a sequence.
That last sentence is NOT CORRECT, mathematically. There is no
> Correct mathematics is this:
> There are countably many intervals I_n of measure 10^-n such that
> I_n covers the rational q_n. Then 1/9 (or less) of the unit interval
> is covered. In the remaining 8/9 (or
> more) there are uncountably many irrationals. But every two
> irrationals have a rational between each other. That implies two
> irrationals have at least one interval I_n between each other (because
> there are no rationals outside of intervals I_n). That implies two
> irrationals have at least one of aleph_0 endpoints I_n_1 or I_n_2 of
> intervals I_n between each other. These endpoints can be considered
> structuring and enumerating a Cantor-list. The only difference is that
> the enumeration does not follow the natural order of N. But the number
> of naturals does not change by reordering. So we have uncountably many
> irrationals separated by countably many endpoints. That is a
> contradiction similar to uncountably many entries in a Cantor-list or
> uncountably many terms in a sequence.
contradiction involved in having an uncountable linear ordering having
a countable linear sub-ordering dense within it.
It is, for example, trivial that between any two non-terminating
decimals (having infinitely many non-zero digits) there is a terminating
decimal (having only finitely many non-zero digits), and that the former
are uncountable and the latter countable.
Uirgil
May 22, 2012, 4:19:53 PM5/22/12
to
In article
<dc2813f8-c73a-4563...@5g2000vbf.googlegroups.com>,
> > > Then there are at least two irrational numbers without a rational
> > > between them. That is mathematically impossible.
> >
> > Prove it or cram it.
>
> The proof has been given. I repeat: As every rational q_n is covered
> by an interval I_n, we can conclude that every pair x and x' of
> irrationals is separated by least one interval I_n. Hence x and x'
> cannot belong to the same complementary interval. This holds for every
> pair of irrationals.
>
> A very simple conclusion. What do you not understand?
Why you claiming what is so obviously false.
Unless EVERY pair of irrationals is separated by EVERY ONE of your
I_n's, which does not ever happen, for some pairs of irrationals there
will be LOTS of your "complimentary intervals' containing both.
And as every complimentary interval is necessarily of positive length,
being of form [0,q) or (q,1] for some rational q with 0 < q < 1, they
are each proved to contain infinitely many irrationals.
<dc2813f8-c73a-4563...@5g2000vbf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 22 Mai, 10:53, William Elliot <ma...@panix.com> wrote:
> > On Tue, 22 May 2012, WM wrote:
snip
> On 22 Mai, 10:53, William Elliot <ma...@panix.com> wrote:
> > On Tue, 22 May 2012, WM wrote:
> > > Then there are at least two irrational numbers without a rational
> > > between them. That is mathematically impossible.
> >
> > Prove it or cram it.
>
> The proof has been given. I repeat: As every rational q_n is covered
> by an interval I_n, we can conclude that every pair x and x' of
> irrationals is separated by least one interval I_n. Hence x and x'
> cannot belong to the same complementary interval. This holds for every
> pair of irrationals.
>
> A very simple conclusion. What do you not understand?
Unless EVERY pair of irrationals is separated by EVERY ONE of your
I_n's, which does not ever happen, for some pairs of irrationals there
will be LOTS of your "complimentary intervals' containing both.
And as every complimentary interval is necessarily of positive length,
being of form [0,q) or (q,1] for some rational q with 0 < q < 1, they
are each proved to contain infinitely many irrationals.
WM
May 22, 2012, 4:22:24 PM5/22/12
to
On 22 Mai, 22:02, Uirgil <uir...@uirgil.ur> wrote:
> In article
> <f8b390e3-cded-44ca-8761-6e16d7d21...@v10g2000vbe.googlegroups.com>,
> are uncountable and the latter countable.- Zitierten Text ausblenden -
No, that is not trivial but wrong. However it is widely believed to be
correct. Therefore I devised the example which should open your eyes.
1) Every rational q_n of the real axis (-oo, oo) is covered with an
interval I_n of length 10^-n.
2) The total measure is 1/9.
3) Irrational numbers of the remaining part of R cannot belong to
intervals but must exist as Cantor-dust in a totally disconnected
space.
4) The elements of this space must be disconnected such that every two
irrationals are separated by at least one interval I_n.
Do you see an error in these 4 steps?
Regards, WM
> In article
> <f8b390e3-cded-44ca-8761-6e16d7d21...@v10g2000vbe.googlegroups.com>,
No, that is not trivial but wrong. However it is widely believed to be
correct. Therefore I devised the example which should open your eyes.
1) Every rational q_n of the real axis (-oo, oo) is covered with an
interval I_n of length 10^-n.
2) The total measure is 1/9.
3) Irrational numbers of the remaining part of R cannot belong to
intervals but must exist as Cantor-dust in a totally disconnected
space.
4) The elements of this space must be disconnected such that every two
irrationals are separated by at least one interval I_n.
Do you see an error in these 4 steps?
Regards, WM
Uirgil
May 22, 2012, 4:35:53 PM5/22/12
to
In article
<f8775a5f-debf-4af3...@m10g2000vbn.googlegroups.com>,
Ennumerate the rationals in [0,1] as {q_n: n in N}
and let I_n = [ q_n - 1/(2*10^n), q_n + 1/(2*10^n)]
Given any pair of irrationals a and b with a < b , the interval
[(3*a + b)/4, (a + 3*b)/4] is the middle half of [a,b] containing
infinitely many q_n's, thus [a,b] contains infinitely many I-n's.
disproving.
I would like to
> know which step you do not accept. Your remark that the remaining
> irrationals do not form intervals (but Cantor dust) does not really
> address the problem, does it? In my opinion every particle of Cantor
> dust is separated by a rational q_n, and therefore by an interval I_n,
> from every other particle of Cantor dust.
Every irrational IS separated from every other irrational by infinitely
many rationals.
Perhaps tht Cantor dust has blown into WM's eyes, blinding him, as for
some reason or unreason, he seems to be blind on such matters.
<f8775a5f-debf-4af3...@m10g2000vbn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 22 Mai, 12:45, William Elliot <ma...@panix.com> wrote:
> > On Tue, 22 May 2012, WM wrote:
> > > > > Belsa Zarkin asked in Mathematics.StackExchange:
> >
> > > > > All rational numbers of the unit interval [0, 1] can be covered by
> > > > > countably many intervals, such that the n-th rational is covered by
> > > > > an
> > > > > interval I_n of measure 1/10^n. There remain countably many
> > > > > complementary intervals of measure 8/9 in total.
> >
> > > > The complement of an interval usually isn't an interval.
> >
> > > Interval is here but an abbreviation for Intervals and points. Even
> > > the empty interval is usually called an interval.
> >
> > An interval is an order convex subset and usually the complement
> > of an interval is not order convex.
> >
>
> That is true but undisputed and does not help to answer the problem.
> You have to explain how uncountably many irrational points can exist
> in the interval [0, 1] such that every point is separated from every
> point by at least one interval I_n that is constructed around a
> rational number q_n.
You did it yourself! have you forgot so soon?
> On 22 Mai, 12:45, William Elliot <ma...@panix.com> wrote:
> > On Tue, 22 May 2012, WM wrote:
> > > > > Belsa Zarkin asked in Mathematics.StackExchange:
> >
> > > > > All rational numbers of the unit interval [0, 1] can be covered by
> > > > > countably many intervals, such that the n-th rational is covered by
> > > > > an
> > > > > interval I_n of measure 1/10^n. There remain countably many
> > > > > complementary intervals of measure 8/9 in total.
> >
> > > > The complement of an interval usually isn't an interval.
> >
> > > Interval is here but an abbreviation for Intervals and points. Even
> > > the empty interval is usually called an interval.
> >
> > An interval is an order convex subset and usually the complement
> > of an interval is not order convex.
> >
>
> That is true but undisputed and does not help to answer the problem.
> You have to explain how uncountably many irrational points can exist
> in the interval [0, 1] such that every point is separated from every
> point by at least one interval I_n that is constructed around a
> rational number q_n.
Ennumerate the rationals in [0,1] as {q_n: n in N}
and let I_n = [ q_n - 1/(2*10^n), q_n + 1/(2*10^n)]
Given any pair of irrationals a and b with a < b , the interval
[(3*a + b)/4, (a + 3*b)/4] is the middle half of [a,b] containing
infinitely many q_n's, thus [a,b] contains infinitely many I-n's.
>
> > > > > Does each of the complementary intervals contain only one
> > > > > irrational number? Then there would be only countably many which
> > > > > could
> > > > > be covered by another set of countably many intervals of measure 1/9.
> >
> > > > No, the complement of each interval contains uncountably many
> > > > irrationals.
> >
> > > That is just under investigation.
> >
> > Only by WM
>
> If you do not want to investigate this problem, why do you post here?
>
> > > The proof has been given. I repeat: As every rational q_n is covered
> > > by an interval I_n, we can conclude that every pair x and x' of
> > > irrationals is separated by least one interval I_n. Hence x and x'
> > > cannot belong to the same complementary interval. This holds for every
> > > pair of irrationals.
> > > A very simple conclusion. What do you not understand?
> >
> > Why dogmatic bullying is so important for you
>
> Please note, I would really like to understand how you try to explain
> your sight. As it is disproved by the above proof
As the "above proof" has itself been disproved, it is incapable of
> > > > > Does each of the complementary intervals contain only one
> > > > > irrational number? Then there would be only countably many which
> > > > > could
> > > > > be covered by another set of countably many intervals of measure 1/9.
> >
> > > > No, the complement of each interval contains uncountably many
> > > > irrationals.
> >
> > > That is just under investigation.
> >
> > Only by WM
>
> If you do not want to investigate this problem, why do you post here?
>
> > > The proof has been given. I repeat: As every rational q_n is covered
> > > by an interval I_n, we can conclude that every pair x and x' of
> > > irrationals is separated by least one interval I_n. Hence x and x'
> > > cannot belong to the same complementary interval. This holds for every
> > > pair of irrationals.
> > > A very simple conclusion. What do you not understand?
> >
> > Why dogmatic bullying is so important for you
>
> Please note, I would really like to understand how you try to explain
> your sight. As it is disproved by the above proof
disproving.
I would like to
> know which step you do not accept. Your remark that the remaining
> irrationals do not form intervals (but Cantor dust) does not really
> address the problem, does it? In my opinion every particle of Cantor
> dust is separated by a rational q_n, and therefore by an interval I_n,
> from every other particle of Cantor dust.
many rationals.
Perhaps tht Cantor dust has blown into WM's eyes, blinding him, as for
some reason or unreason, he seems to be blind on such matters.
WM
May 22, 2012, 4:28:28 PM5/22/12
to
On 22 Mai, 22:19, Uirgil <uir...@uirgil.ur> wrote:
> In article
> <dc2813f8-c73a-4563-add5-48feb4d47...@5g2000vbf.googlegroups.com>,
> In article
You misunderstand. The complementary intervals do not contain any
rationals q_n. They cannot because all q_n are covered by intervals
I_n. Therefore the complementary intervals are merely single points.
(Limits of sequences of intervals I_n.)
Regards, WM
Uirgil
May 22, 2012, 6:42:36 PM5/22/12
to
In article
<9db0b271-e437-4e8a...@b26g2000vbt.googlegroups.com>,
Then your original definition was sufficiently ambiguous as not to make
that clear.
<9db0b271-e437-4e8a...@b26g2000vbt.googlegroups.com>,
that clear.
Uirgil
May 22, 2012, 6:52:49 PM5/22/12
to
In article
<23bcdffd-d442-496c...@p27g2000vbl.googlegroups.com>,
entries and order lexicographically, as usual, it is true, unless WM or
someone can provide two such sequences with no terminating sequence
between them.
Well can you, WM?
Actually I had intended my above statement to be about binary sequences
rather than decimals, and for binaries, it is trivially true.
<23bcdffd-d442-496c...@p27g2000vbl.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 22 Mai, 22:02, Uirgil <uir...@uirgil.ur> wrote:
> > In article
> > <f8b390e3-cded-44ca-8761-6e16d7d21...@v10g2000vbe.googlegroups.com>,
> >
> On 22 Mai, 22:02, Uirgil <uir...@uirgil.ur> wrote:
> > In article
> > <f8b390e3-cded-44ca-8761-6e16d7d21...@v10g2000vbe.googlegroups.com>,
> >
> > It is, for example, trivial that between any two non-terminating
> > decimals (having infinitely many non-zero digits) there is a terminating
> > decimal (having only finitely many non-zero digits), and that the former
> > are uncountable and the latter countable.- Zitierten Text ausblenden -
>
> No, that is not trivial but wrong. However it is widely believed to be
> correct. Therefore I devised the example which should open your eyes.
If we limit non-terminating to mean having infinitely many nonzero
> > decimals (having infinitely many non-zero digits) there is a terminating
> > decimal (having only finitely many non-zero digits), and that the former
> > are uncountable and the latter countable.- Zitierten Text ausblenden -
>
> No, that is not trivial but wrong. However it is widely believed to be
> correct. Therefore I devised the example which should open your eyes.
entries and order lexicographically, as usual, it is true, unless WM or
someone can provide two such sequences with no terminating sequence
between them.
Well can you, WM?
Actually I had intended my above statement to be about binary sequences
rather than decimals, and for binaries, it is trivially true.
WM
May 23, 2012, 1:20:28 AM5/23/12
to
On 23 Mai, 00:42, Uirgil <uir...@uirgil.ur> wrote:
> In article
> <9db0b271-e437-4e8a-83e2-e0708f325...@b26g2000vbt.googlegroups.com>,
> In article
I said: "every rational q_n is covered by an interval I_n". So, after
having understood now, what is your objection?
Regards, WM
Ralf Bader
May 23, 2012, 1:38:37 AM5/23/12
to
reals, contrary to your usual assertions about Zermelo's blunder which
really is yours. Why? Mind the wise words of your Chancellor Mrs. Merkel:
"Man muß die Dinge vom Ende her denken". (One should think through matters
from the end) So take the set I of the irrationals in the unit interval.
Order them, and then intersperse that ordering with separators in such a way
that between any two different irrationals x, y there always is at least one
separator. First, take the usual ordering of I according to absolute value.
Then with some 9 separators, cleverly interspersed, any two x, y are
separated if their absolute values differ by at least 1/10. With an
additional 90 separators, cleverly interspersed, any two x, y are separated
if their absolute values differ by at least 1/100. And so on. In each step,
a finite number of separators is interspersed, and the total set of
separators required is a countable union of finite sets which is countable.
And then, any two different x, y are separated by some (in fact, infinitely
many) separators because their absolute values differ by some inverse power
of 10.
Now take the well-ordering of I which you discovered. If you want to
intersperse this well-ordering with separators in such a way that there is a
separator between any two different irrationals x, y, then there must be a
separator between any x and the least y larger than x in that well-ordering.
And associating to any x that separator gives an injective mapping from I to
the set of separators required in this case, so "at least as many"
separators as there are elements of I are required in this case. And that
something like this would be necessary to separate any different x, y is
what you are always implicitly assuming, for example in the "above proof".
But this is wrong in the case of the extremely non-well ordering according
to absolute value, yet it is correct in the case of the well-ordering of I
you seem to have discovered.
WM
May 23, 2012, 1:56:36 AM5/23/12
to
On 23 Mai, 07:38, Ralf Bader <ba...@nefkom.net> wrote:
> But this is wrong in the case of the extremely non-well ordering according
> to absolute value, yet it is correct in the case of the well-ordering of I
> you seem to have discovered.- Zitierten Text ausblenden -
>
There are countably many finite intervals I_n, each one covering a
rational q_n. The cardinality of this set does not change by any
manipulation. Every particle of Cantor-dust is separated from every
other particle by at least one of such intervals. So we have in fact a
well-ordering of the Cantor dust. And in order to avoid this
contardiction we must believe that countably many finite intervals
become uncountable? Or that uncountably many particles of Cantor-dust
are "connected", i.e. not separated by intervals I_n? Of course, there
are people who can believe everything. But I prefer mathematics
without abiogenesis of intervals or dust particles.
Regards, WM
> WM wrote:
Why? Mind the wise words of your Chancellor Mrs. Merkel:
> "Man muß die Dinge vom Ende her denken". (One should think through matters
> from the end)
In the infinite that is impossible.
Why? Mind the wise words of your Chancellor Mrs. Merkel:
> "Man muß die Dinge vom Ende her denken". (One should think through matters
> from the end)
> But this is wrong in the case of the extremely non-well ordering according
> to absolute value, yet it is correct in the case of the well-ordering of I
>
There are countably many finite intervals I_n, each one covering a
rational q_n. The cardinality of this set does not change by any
manipulation. Every particle of Cantor-dust is separated from every
other particle by at least one of such intervals. So we have in fact a
well-ordering of the Cantor dust. And in order to avoid this
contardiction we must believe that countably many finite intervals
become uncountable? Or that uncountably many particles of Cantor-dust
are "connected", i.e. not separated by intervals I_n? Of course, there
are people who can believe everything. But I prefer mathematics
without abiogenesis of intervals or dust particles.
Regards, WM
WM
May 23, 2012, 8:19:19 AM5/23/12
to
On 23 Mai, 00:52, Uirgil <uir...@uirgil.ur> wrote:
> In article
> <23bcdffd-d442-496c-89ee-6ef715157...@p27g2000vbl.googlegroups.com>,
> In article
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 22 Mai, 22:02, Uirgil <uir...@uirgil.ur> wrote:
> > > In article
> > > <f8b390e3-cded-44ca-8761-6e16d7d21...@v10g2000vbe.googlegroups.com>,
>
> > > It is, for example, trivial that between any two non-terminating
> > > decimals (having infinitely many non-zero digits) there is a terminating
> > > decimal (having only finitely many non-zero digits), and that the former
> > > are uncountable and the latter countable.- Zitierten Text ausblenden -
>
> > No, that is not trivial but wrong. However it is widely believed to be
> > correct. Therefore I devised the example which should open your eyes.
>
> If we limit non-terminating to mean having infinitely many nonzero
> entries and order lexicographically, as usual, it is true, unless WM or
> someone can provide two such sequences with no terminating sequence
> between them.
>
> Well can you, WM?
You know that only countably many irrationals can be "provided". All
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 22 Mai, 22:02, Uirgil <uir...@uirgil.ur> wrote:
> > > In article
> > > <f8b390e3-cded-44ca-8761-6e16d7d21...@v10g2000vbe.googlegroups.com>,
>
> > > It is, for example, trivial that between any two non-terminating
> > > decimals (having infinitely many non-zero digits) there is a terminating
> > > decimal (having only finitely many non-zero digits), and that the former
> > > are uncountable and the latter countable.- Zitierten Text ausblenden -
>
> > No, that is not trivial but wrong. However it is widely believed to be
> > correct. Therefore I devised the example which should open your eyes.
>
> If we limit non-terminating to mean having infinitely many nonzero
> entries and order lexicographically, as usual, it is true, unless WM or
> someone can provide two such sequences with no terminating sequence
> between them.
>
> Well can you, WM?
the other irrationals do not have any magnitude (otherwise this could
be defined). So they do not exist on the real axis. So you have no
contradiction with regards to my proof.
>
You need not get anxious. Cantor's uncountably many numbers cannot be
found on the real axis (and cannot be found anywhere else). It is
simply a set of real numbers that are neither real nor numbers but
angels of matheology. That has the advantage that they cannot be
contradicted. Neither by well-orderings nor by extreme unwell
orderings. (Now we have not only extreme sports but also extreme
unwell-orders.)
> Actually I had intended my above statement to be about binary sequences
> rather than decimals, and for binaries, it is trivially true.
number. For that sake you need a finite definition like "0.111 and
then only zeros" or the like.
Regards, WM
Tonico
May 23, 2012, 8:55:09 AM5/23/12
to
On May 22, 12:04 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 22 Mai, 10:53, William Elliot <ma...@panix.com> wrote:
>
> > On Tue, 22 May 2012, WM wrote:
> > > Belsa Zarkin asked in Mathematics.StackExchange:
>
> > > All rational numbers of the unit interval [0, 1] can be covered by
> > > countably many intervals, such that the n-th rational is covered by an
> > > interval I_n of measure 1/10^n. There remain countably many
> > > complementary intervals of measure 8/9 in total.
>
> > The complement of an interval usually isn't an interval.
>
> Interval is here but an abbreviation for Intervals and points. Even
> the empty interval is usually called an interval.
**** What?!?!?!! An interval (the emtpy one) is called....and
> On 22 Mai, 10:53, William Elliot <ma...@panix.com> wrote:
>
> > On Tue, 22 May 2012, WM wrote:
> > > Belsa Zarkin asked in Mathematics.StackExchange:
>
> > > All rational numbers of the unit interval [0, 1] can be covered by
> > > countably many intervals, such that the n-th rational is covered by an
> > > interval I_n of measure 1/10^n. There remain countably many
> > > complementary intervals of measure 8/9 in total.
>
> > The complement of an interval usually isn't an interval.
>
> Interval is here but an abbreviation for Intervals and points. Even
> the empty interval is usually called an interval.
interval?!? Ot, these evil, mischevious and awful mathematicians...!
Of course, that "interval" is an abbreviation for "interval and
points" is as logical and sound as saying that "WM is abb. for Wise
Man from the Crankhood of Trollland"...Really. :) ****
>
>
> > > Does each of the complementary intervals contain only one
> > > irrational number? Then there would be only countably many which could
> > > be covered by another set of countably many intervals of measure 1/9.
>
> > No, the complement of each interval contains uncountably many irrationals.
>
> That is just under investigation.
on fire these days, WM! ****
>
> > > Is there at least one of the complementary intervals containing
> > > more than one irrational number?
>
> > Yes.
>
> Impossible in mathematics. In mathematics we prove that two different
> irrationals have a rational between them. As every rational q_n is
> covered by an interval I_n, we can conclude that every pair x and x'
> of irrationals is separated by least one interval I_n. Hence x and x'
> cannot belong to the same complementary interval.
not usually an interval yet you, as usual, try to confund your readers
with the same name...and, as usual, you're dead wrong, and your
logical skills have reached depths of abbyss that won't be easy to
reach for anyone else. ****
>
> > > Then there are at least two irrational numbers without a rational
> > > between them. That is mathematically impossible.
>
> > Prove it or cram it.
>
> The proof has been given. I repeat: As every rational q_n is covered
> by an interval I_n, we can conclude that every pair x and x' of
> irrationals is separated by least one interval I_n. Hence x and x'
> cannot belong to the same complementary interval. This holds for every
> pair of irrationals.
>
> A very simple conclusion. What do you not understand?
feel deeply ashamed of making an ass of yourself over and ober and
over and...and over and ober, and how is it possible for one single
human being to be so utterly impermeable to knowledge, logic and
common mathematical sense.
Tonio
Pd. Chau!
>
> Regards, WM
Uirgil
May 24, 2012, 2:25:56 AM5/24/12
to
In article
<5f866fe0-a9ae-448a...@w10g2000vbc.googlegroups.com>,
countably many of them is met!
>
> > Actually I had intended my above statement to be about binary sequences
> > rather than decimals, and for binaries, it is trivially true.
>
> For binary sequences it is trivially true that none of them defines a
> number. For that sake you need a finite definition like "0.111 and
> then only zeros" or the like.
Since at least countably many infinite binary sequences can be finitely
defined, there are infinitely many counterexamples to your claim that
<5f866fe0-a9ae-448a...@w10g2000vbc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 23 Mai, 00:52, Uirgil <uir...@uirgil.ur> wrote:
> > In article
> > <23bcdffd-d442-496c-89ee-6ef715157...@p27g2000vbl.googlegroups.com>,
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 22 Mai, 22:02, Uirgil <uir...@uirgil.ur> wrote:
> > > > In article
> > > > <f8b390e3-cded-44ca-8761-6e16d7d21...@v10g2000vbe.googlegroups.com>,
> >
> > > > It is, for example, trivial that between any two non-terminating
> > > > decimals (having infinitely many non-zero digits) there is a terminating
> > > > decimal (having only finitely many non-zero digits), and that the former
> > > > are uncountable and the latter countable.- Zitierten Text ausblenden -
> >
> > > No, that is not trivial but wrong. However it is widely believed to be
> > > correct. Therefore I devised the example which should open your eyes.
> >
> > If we limit non-terminating to mean having infinitely many nonzero
> > entries and order lexicographically, as usual, it is true, unless WM or
> > someone can provide two such sequences with no terminating sequence
> > between them.
> >
> > Well can you, WM?
>
> You know that only countably many irrationals can be "provided".
Since I am only speaking about two irrationals, the restriction to
> On 23 Mai, 00:52, Uirgil <uir...@uirgil.ur> wrote:
> > In article
> > <23bcdffd-d442-496c-89ee-6ef715157...@p27g2000vbl.googlegroups.com>,
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 22 Mai, 22:02, Uirgil <uir...@uirgil.ur> wrote:
> > > > In article
> > > > <f8b390e3-cded-44ca-8761-6e16d7d21...@v10g2000vbe.googlegroups.com>,
> >
> > > > It is, for example, trivial that between any two non-terminating
> > > > decimals (having infinitely many non-zero digits) there is a terminating
> > > > decimal (having only finitely many non-zero digits), and that the former
> > > > are uncountable and the latter countable.- Zitierten Text ausblenden -
> >
> > > No, that is not trivial but wrong. However it is widely believed to be
> > > correct. Therefore I devised the example which should open your eyes.
> >
> > If we limit non-terminating to mean having infinitely many nonzero
> > entries and order lexicographically, as usual, it is true, unless WM or
> > someone can provide two such sequences with no terminating sequence
> > between them.
> >
> > Well can you, WM?
>
> You know that only countably many irrationals can be "provided".
countably many of them is met!
>
> > Actually I had intended my above statement to be about binary sequences
> > rather than decimals, and for binaries, it is trivially true.
>
> For binary sequences it is trivially true that none of them defines a
> number. For that sake you need a finite definition like "0.111 and
> then only zeros" or the like.
defined, there are infinitely many counterexamples to your claim that
Uirgil
May 24, 2012, 2:27:37 AM5/24/12
to
In article
<481d5d13-5cd9-4bba...@w13g2000vbc.googlegroups.com>,
<481d5d13-5cd9-4bba...@w13g2000vbc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 23 Mai, 07:38, Ralf Bader <ba...@nefkom.net> wrote:
> > WM wrote:
> Why? Mind the wise words of your Chancellor Mrs. Merkel:
> > "Man muß die Dinge vom Ende her denken". (One should think through matters
> > from the end)
>
>
> In the infinite that is impossible.
Not at all. Many infinities have at least one end
> On 23 Mai, 07:38, Ralf Bader <ba...@nefkom.net> wrote:
> > WM wrote:
> Why? Mind the wise words of your Chancellor Mrs. Merkel:
> > "Man muß die Dinge vom Ende her denken". (One should think through matters
> > from the end)
>
>
> In the infinite that is impossible.
Uirgil
May 24, 2012, 2:31:02 AM5/24/12
to
In article
<65dc68ad-7594-4268...@5g2000vbf.googlegroups.com>,
You did not make clear that your complimentation was with respect to the
union of all those rational intervals, rather than one compliment for
each interval.
<65dc68ad-7594-4268...@5g2000vbf.googlegroups.com>,
union of all those rational intervals, rather than one compliment for
each interval.
Message has been deleted
WM
May 24, 2012, 2:53:06 AM5/24/12
to
On 24 Mai, 08:27, Uirgil <uir...@uirgil.ur> wrote:
> In article
> <481d5d13-5cd9-4bba-ace5-48fc5ef8b...@w13g2000vbc.googlegroups.com>,
> In article
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 23 Mai, 07:38, Ralf Bader <ba...@nefkom.net> wrote:
> > > WM wrote:
> > Why? Mind the wise words of your Chancellor Mrs. Merkel:
> > > "Man muß die Dinge vom Ende her denken". (One should think through matters
> > > from the end)
>
> > In the infinite that is impossible.
>
> Not at all. Many infinities have at least one end
And the sausage has two.
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 23 Mai, 07:38, Ralf Bader <ba...@nefkom.net> wrote:
> > > WM wrote:
> > Why? Mind the wise words of your Chancellor Mrs. Merkel:
> > > "Man muß die Dinge vom Ende her denken". (One should think through matters
> > > from the end)
>
> > In the infinite that is impossible.
>
> Not at all. Many infinities have at least one end
But one of them is the beginning. Bader obviously meant the second one
that is not present in endless sets. - except for matheologians.
Regards, WM
WM
May 24, 2012, 3:03:20 AM5/24/12
to
On 24 Mai, 08:31, Uirgil <uir...@uirgil.ur> wrote:
> In article
> <65dc68ad-7594-4268-be98-7095e3369...@5g2000vbf.googlegroups.com>,
> In article
Now it should be clear that at most 1/9 of the unit interval (or even
of the complete real line) is covered by intervals I_n. That means at
least 8/9 (infinity) is not covered by intervals I_n. This remaining
part does not contain any rational numbers. Therefore either two
irrationals of that remaining part must be connected or disconnected
by one of the intervals I_n. The former is not possible in mathematics
(perhaps it is possible in matheology - but that is not of interest
for me). The second means that we have either only countably many
irrationals or we are forced to believe that there are more
irrationals than rationals and nevertheless every pair of irrationals
is separated by an I_n.
Of course under such circumstances there will never appear a
contradiction in ZFC. Why should 3 < 4 and 3 > 4 not hold
simultaneously, if only infinitely many such relations are concerned.
But, as I already said, ZFC and its high priests, matheologians and
mythologicans are irrelevant for my position.
(This series is not designed to convince them but to show to those who
are not yet infected by this intellectual disease how finished
infinity can lead astray.)
Regards, WM
WM
May 24, 2012, 3:47:57 AM5/24/12
to
On 24 Mai, 08:25, Uirgil <uir...@uirgil.ur> wrote:
> > > If we limit non-terminating to mean having infinitely many nonzero
> > > entries and order lexicographically, as usual, it is true, unless WM or
> > > someone can provide two such sequences with no terminating sequence
> > > between them.
> > > Well can you, WM?
> > You know that only countably many irrationals can be "provided".
> Since I am only speaking about two irrationals, the restriction to
> countably many of them is met!
Of course there is a rational between every two irrationals that can
> > > If we limit non-terminating to mean having infinitely many nonzero
> > > entries and order lexicographically, as usual, it is true, unless WM or
> > > someone can provide two such sequences with no terminating sequence
> > > between them.
> > > Well can you, WM?
> > You know that only countably many irrationals can be "provided".
> Since I am only speaking about two irrationals, the restriction to
> countably many of them is met!
be identified.
This shows, among other arguments, that there are only countably many
irrationals that can be identified.
> > > Actually I had intended my above statement to be about binary sequences
> > > rather than decimals, and for binaries, it is trivially true.
> > For binary sequences it is trivially true that none of them defines a
> > number. For that sake you need a finite definition like "0.111 and
> > then only zeros" or the like.
> Since at least countably many infinite binary sequences can be finitely
> defined,
> there are infinitely many counterexamples to your claim that
Wrong. The infinite sequence does not define a number. The number
defines an infinite sequence in the sense that every digit d_n is
followed by another digit d_m. But there are never "all digits".
Your error is very simple. From A ==> B you conclude B ==> A.
A number defines an infinite string of digits.
But an infinite string of digits does not define a number (because it
is never finished - and definitions define things only when the
definitions have been completely spelled out.
Think over this simpler example: "Everybody who lives in New York
lives in USA." But you reverse this to "everybody who lives in USA
lives in New York". Is that correct?
Regards, WM
Alan Smaill
May 24, 2012, 10:15:01 AM5/24/12
to
WM <muec...@rz.fh-augsburg.de> writes:
> You know that only countably many irrationals can be "provided". All
> the other irrationals do not have any magnitude (otherwise this could
> be defined). So they do not exist on the real axis. So you have no
> contradiction with regards to my proof.
Is this meant to be some sort of logical argument?
> You know that only countably many irrationals can be "provided". All
> the other irrationals do not have any magnitude (otherwise this could
> be defined). So they do not exist on the real axis. So you have no
> contradiction with regards to my proof.
From what assumptions?
By denying the need for axioms except when matheology comes into
play, you leave yourself with your own trail of dust and
non-mathematics.
There are plenty of proposals out there that do genuine maths, (eg
Weihrauch's text on computable analysis) and avoid some of the
counter-intuitive consquences of the usual position, where at least we
can see what the accepted principles are.
> Regards, WM
>
--
Alan Smaill
WM
May 24, 2012, 10:28:16 AM5/24/12
to
On 24 Mai, 16:15, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
distiguished. Numbers that cannot be distinguished (by >, =, <) cannot
be disproven to exist.
Have you meanwhile accepted that countably many finite intervals
cannot separate uncountably many separated points?
Or is this bold conclusion not covered by axioms and logic?
Regards, WM
> WM <mueck...@rz.fh-augsburg.de> writes:
> > You know that only countably many irrationals can be "provided". All
> > the other irrationals do not have any magnitude (otherwise this could
> > be defined). So they do not exist on the real axis. So you have no
> > contradiction with regards to my proof.
>
> Is this meant to be some sort of logical argument?
> From what assumptions?
No assumptions. It is clear that only countably many numbers can be
> > You know that only countably many irrationals can be "provided". All
> > the other irrationals do not have any magnitude (otherwise this could
> > be defined). So they do not exist on the real axis. So you have no
> > contradiction with regards to my proof.
>
> Is this meant to be some sort of logical argument?
> From what assumptions?
distiguished. Numbers that cannot be distinguished (by >, =, <) cannot
be disproven to exist.
Have you meanwhile accepted that countably many finite intervals
cannot separate uncountably many separated points?
Or is this bold conclusion not covered by axioms and logic?
Regards, WM
Ralf Bader
May 24, 2012, 12:39:21 PM5/24/12
to
WM wrote:
> On 23 Mai, 07:38, Ralf Bader <ba...@nefkom.net> wrote:
>> WM wrote:
> Why? Mind the wise words of your Chancellor Mrs. Merkel:
>> "Man muß die Dinge vom Ende her denken". (One should think through
>> matters from the end)
>
>
> In the infinite that is impossible.
You need not prove every day that you are too stupid for these matters.
> On 23 Mai, 07:38, Ralf Bader <ba...@nefkom.net> wrote:
>> WM wrote:
> Why? Mind the wise words of your Chancellor Mrs. Merkel:
>> "Man muß die Dinge vom Ende her denken". (One should think through
>> matters from the end)
>
>
> In the infinite that is impossible.
LudovicoVan
May 24, 2012, 1:07:28 PM5/24/12
to
"Ralf Bader" wrote in message news:jplns3$ea5$1...@news.m-online.net...
Quite stupid was your remark: your Chancellor's wise words happen to be the
perfect slogan for those who will never change anything at all.
-LV
perfect slogan for those who will never change anything at all.
-LV
WM
May 24, 2012, 12:47:48 PM5/24/12
to
infinite ends. Not everybody is capable of doing so. But I am not
jealous.
With your recently published rule, however, "this is wrong in the case
of the extremely non-well ordering according
to absolute value" you seem to be doomed of failure. Have you received
any support? Are you aware of the fact that the intervals of my proof
have been counted and as such cannot get un-well ordrerd, neither a
bit nor extremely? And that includes their borders too. All of them.
And further space is not available for the "uncountably many"
irrationals.
Regards, WM
Virgil
May 24, 2012, 5:46:35 PM5/24/12
to
In article
<79abf159-1bdf-4026...@w24g2000vby.googlegroups.com>,
WM has yet to produce a proof on his own of any claim of his own that is
not flawed, and usually fatally flawed.
--
<79abf159-1bdf-4026...@w24g2000vby.googlegroups.com>,
not flawed, and usually fatally flawed.
--
Uirgil
May 24, 2012, 5:49:02 PM5/24/12
to
In article
<ab157e7b-f27e-4d06...@w13g2000vbc.googlegroups.com>,
> Numbers that cannot be distinguished (by >, =, <) cannot
> be disproven to exist.
>
> Have you meanwhile accepted that countably many finite intervals
> cannot separate uncountably many separated points?
A claim that is sufficiently ambiguous as to be meaningless.
<ab157e7b-f27e-4d06...@w13g2000vbc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mai, 16:15, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > You know that only countably many irrationals can be "provided". All
> > > the other irrationals do not have any magnitude (otherwise this could
> > > be defined). So they do not exist on the real axis. So you have no
> > > contradiction with regards to my proof.
> >
> > Is this meant to be some sort of logical argument?
> > From what assumptions?
>
> No assumptions. It is clear that only countably many numbers can be
> distiguished.
What "is clear" to WM is an assumption by WM.
> On 24 Mai, 16:15, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > You know that only countably many irrationals can be "provided". All
> > > the other irrationals do not have any magnitude (otherwise this could
> > > be defined). So they do not exist on the real axis. So you have no
> > > contradiction with regards to my proof.
> >
> > Is this meant to be some sort of logical argument?
> > From what assumptions?
>
> No assumptions. It is clear that only countably many numbers can be
> distiguished.
> Numbers that cannot be distinguished (by >, =, <) cannot
> be disproven to exist.
>
> Have you meanwhile accepted that countably many finite intervals
> cannot separate uncountably many separated points?
Uirgil
May 24, 2012, 5:52:21 PM5/24/12
to
In article
<604f9d11-a8e2-4ae0...@eh4g2000vbb.googlegroups.com>,
Wm now claims mind-reading among his talents!
<604f9d11-a8e2-4ae0...@eh4g2000vbb.googlegroups.com>,
Uirgil
May 24, 2012, 5:59:39 PM5/24/12
to
In article
<15fa1f67-3401-46a7...@v24g2000vbx.googlegroups.com>,
can only be between a countable set of smaller irrationals and a
cuntable set of larger irrationals, and that does not follow merely from
the density of the rationals .
those infinite sequences in turn define numbers.,
<15fa1f67-3401-46a7...@v24g2000vbx.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mai, 08:25, Uirgil <uir...@uirgil.ur> wrote:
>
>
> > > > If we limit non-terminating to mean having infinitely many nonzero
> > > > entries and order lexicographically, as usual, it is true, unless WM or
> > > > someone can provide two such sequences with no terminating sequence
> > > > between them.
>
> > > > Well can you, WM?
>
> > > You know that only countably many irrationals can be "provided".
>
> > Since I am only speaking about two irrationals, the restriction to
> > countably many of them is met!
>
> Of course there is a rational between every two irrationals that can
> be identified.
> This shows, among other arguments, that there are only countably many
> irrationals that can be identified.
No it does not. What you are claiming would require that one rational
> On 24 Mai, 08:25, Uirgil <uir...@uirgil.ur> wrote:
>
>
> > > > If we limit non-terminating to mean having infinitely many nonzero
> > > > entries and order lexicographically, as usual, it is true, unless WM or
> > > > someone can provide two such sequences with no terminating sequence
> > > > between them.
>
> > > > Well can you, WM?
>
> > > You know that only countably many irrationals can be "provided".
>
> > Since I am only speaking about two irrationals, the restriction to
> > countably many of them is met!
>
> Of course there is a rational between every two irrationals that can
> be identified.
> This shows, among other arguments, that there are only countably many
> irrationals that can be identified.
can only be between a countable set of smaller irrationals and a
cuntable set of larger irrationals, and that does not follow merely from
the density of the rationals .
>
> > > > Actually I had intended my above statement to be about binary sequences
> > > > rather than decimals, and for binaries, it is trivially true.
>
> > > For binary sequences it is trivially true that none of them defines a
> > > number. For that sake you need a finite definition like "0.111 and
> > > then only zeros" or the like.
>
> > Since at least countably many infinite binary sequences can be finitely
> > defined,
>
> Not only "at least" but also "at most".
Claimed but not proven.
> > > > Actually I had intended my above statement to be about binary sequences
> > > > rather than decimals, and for binaries, it is trivially true.
>
> > > For binary sequences it is trivially true that none of them defines a
> > > number. For that sake you need a finite definition like "0.111 and
> > > then only zeros" or the like.
>
> > Since at least countably many infinite binary sequences can be finitely
> > defined,
>
> Not only "at least" but also "at most".
>
> > there are infinitely many counterexamples to your claim that
> > none of them defines a number.-
>
> Wrong. The infinite sequence does not define a number. The number
> defines an infinite sequence in the sense that every digit d_n is
> followed by another digit d_m. But there are never "all digits".
>
> Your error is very simple. From A ==> B you conclude B ==> A.
> A number defines an infinite string of digits.
> But an infinite string of digits does not define a number (because it
> is never finished - and definitions define things only when the
> definitions have been completely spelled out.
Since one can finitely define certain infinite sequences of digits,
> > there are infinitely many counterexamples to your claim that
> > none of them defines a number.-
>
> Wrong. The infinite sequence does not define a number. The number
> defines an infinite sequence in the sense that every digit d_n is
> followed by another digit d_m. But there are never "all digits".
>
> Your error is very simple. From A ==> B you conclude B ==> A.
> A number defines an infinite string of digits.
> But an infinite string of digits does not define a number (because it
> is never finished - and definitions define things only when the
> definitions have been completely spelled out.
those infinite sequences in turn define numbers.,
Uirgil
May 24, 2012, 6:04:32 PM5/24/12
to
In article
<4ab62dd8-e558-4411...@5g2000vbf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mai, 08:25, Uirgil <uir...@uirgil.ur> wrote:
> > In article
> > <5f866fe0-a9ae-448a-852e-404fbb3d5...@w10g2000vbc.googlegroups.com>,
> >
> >
> >
> > > > Actually I had intended my above statement to be about binary sequences
> > > > rather than decimals, and for binaries, it is trivially true.
> >
> > > For binary sequences it is trivially true that none of them defines a
> > > number. For that sake you need a finite definition like "0.111 and
> > > then only zeros" or the like.
Then you concede that the infinite binary sequence that you just defined
in tern defines a number?
> > there are infinitely many counterexamples to your claim that
mathematics.
<4ab62dd8-e558-4411...@5g2000vbf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mai, 08:25, Uirgil <uir...@uirgil.ur> wrote:
> > In article
> >
> >
> >
> >
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 23 Mai, 00:52, Uirgil <uir...@uirgil.ur> wrote:
.....
> >
> >
> >
> >
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 23 Mai, 00:52, Uirgil <uir...@uirgil.ur> wrote:
> >
> >
> >
> > > > Actually I had intended my above statement to be about binary sequences
> > > > rather than decimals, and for binaries, it is trivially true.
> >
> > > For binary sequences it is trivially true that none of them defines a
> > > number. For that sake you need a finite definition like "0.111 and
> > > then only zeros" or the like.
in tern defines a number?
> > there are infinitely many counterexamples to your claim that
> > none of them defines a number.-
>
> Wrong. The infinite sequence does not define a number. The number
> defines an infinite sequence in the sence that any digit d_n is
>
> Wrong. The infinite sequence does not define a number. The number
> followed by another digit d_m. But there are never "all digits".
Maybe not in WM's world, but his world is much to small to hold all of
mathematics.
Ralf Bader
May 24, 2012, 6:41:08 PM5/24/12
to
--
Neueste Forschungsergebnisse aus deutschen Spitzenhochschulen. Heute von
Prof. Dr. Wolfgang Mückenheim, Mathematikkoryphäe der FH Augsburg, aus
seiner Postille "Physical constraints of numbers": "Even some single
numbers smaller than 2^10^100 ... do not exist."
WM
May 25, 2012, 8:20:29 AM5/25/12
to
On 25 Mai, 00:41, Ralf Bader <ba...@nefkom.net> wrote:
>
> You are babbling incredibly stupid nonsense.
>
Wouldn't you say the same if your intelligence was below the threshold
>
> You are babbling incredibly stupid nonsense.
>
of any comprehension? Can you at least follow my logic here?
Nevertheless, what don't you understand in my simple argument? Perhaps
I can help you?
Regards, WM
WM
May 25, 2012, 8:08:38 AM5/25/12
to
On 24 Mai, 23:59, Uirgil <uir...@uirgil.ur> wrote:
>
> > > Since at least countably many infinite binary sequences can be finitely
> > > defined,
>
> > Not only "at least" but also "at most".
>
> Claimed but not proven.
Proven - but not comprehensible in matheology.
>
> > > Since at least countably many infinite binary sequences can be finitely
> > > defined,
>
> > Not only "at least" but also "at most".
>
> Claimed but not proven.
>
> > Your error is very simple. From A ==> B you conclude B ==> A.
> > A number defines an infinite string of digits.
> > But an infinite string of digits does not define a number (because it
> > is never finished - and definitions define things only when the
> > definitions have been completely spelled out.
>
> Since one can finitely define certain infinite sequences of digits,
> those infinite sequences in turn define numbers
No. The definition has to be finite. An infinite sequence does never
> > Your error is very simple. From A ==> B you conclude B ==> A.
> > A number defines an infinite string of digits.
> > But an infinite string of digits does not define a number (because it
> > is never finished - and definitions define things only when the
> > definitions have been completely spelled out.
>
> Since one can finitely define certain infinite sequences of digits,
> those infinite sequences in turn define numbers
define anything, because a definition is a message, at least to
oneself, and a message does not convey information before the end
signal has received.
Regards, WM
Ralf Bader
May 25, 2012, 3:17:30 PM5/25/12
to
"Die Natur hat schon häufig natürliche Zahlen zerlegt, zum Beispiel...die
acht Beine einer Spinne in die vier Himmelsrichtungen." Prof. Dr. W.
Mückenheim, Mathematikkoryphäe der "Hochschule Augsburg", am 01.10.09 in
de.sci.mathematik
WM
May 25, 2012, 4:24:09 PM5/25/12
to
On 25 Mai, 21:17, Ralf Bader <ba...@nefkom.net> wrote:
> WM wrote:
> > On 25 Mai, 00:41, Ralf Bader <ba...@nefkom.net> wrote:
>
> >> You are babbling incredibly stupid nonsense.
>
> > Wouldn't you say the same if your intelligence was below the threshold
> > of any comprehension? Can you at least follow my logic here?
>
> > Nevertheless, what don't you understand in my simple argument? Perhaps
> > I can help you?
>
> You are babbling incredibly stupid nonsense.
Would you say the same if your intelligence was not below the
> WM wrote:
> > On 25 Mai, 00:41, Ralf Bader <ba...@nefkom.net> wrote:
>
> >> You are babbling incredibly stupid nonsense.
>
> > Wouldn't you say the same if your intelligence was below the threshold
> > of any comprehension? Can you at least follow my logic here?
>
> > Nevertheless, what don't you understand in my simple argument? Perhaps
> > I can help you?
>
> You are babbling incredibly stupid nonsense.
threshold of any comprehension?
Sorry it's unfair to ask. Of course you cannot answer that question.
Nevertheless it's funny that I just heard nearly the same objection
with respect to your "extremely non-well ordering according to
absolute value". Of course the borders of well-ordered intervals are
well-ordered too, aren't they?
Regards, WM
Uirgil
May 25, 2012, 4:25:47 PM5/25/12
to
In article
<88ce96c3-63e4-47e8...@z19g2000vbe.googlegroups.com>,
nowhere nearly the same.
<88ce96c3-63e4-47e8...@z19g2000vbe.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mai, 23:59, Uirgil <uir...@uirgil.ur> wrote:
> >
> > > > Since at least countably many infinite binary sequences can be finitely
> > > > defined,
> >
> > > Not only "at least" but also "at most".
> >
> > Claimed but not proven.
>
> Proven - but not comprehensible in math
What WM claims to have proven and what WM has actually proven are
> On 24 Mai, 23:59, Uirgil <uir...@uirgil.ur> wrote:
> >
> > > > Since at least countably many infinite binary sequences can be finitely
> > > > defined,
> >
> > > Not only "at least" but also "at most".
> >
> > Claimed but not proven.
>
> Proven - but not comprehensible in math
nowhere nearly the same.
> >
> > > Your error is very simple. From A ==> B you conclude B ==> A.
> > > A number defines an infinite string of digits.
> > > But an infinite string of digits does not define a number (because it
> > > is never finished - and definitions define things only when the
> > > definitions have been completely spelled out.
> >
> > Since one can finitely define certain infinite sequences of digits,
> > those infinite sequences in turn define numbers
>
> No.
Yes!
> > > Your error is very simple. From A ==> B you conclude B ==> A.
> > > A number defines an infinite string of digits.
> > > But an infinite string of digits does not define a number (because it
> > > is never finished - and definitions define things only when the
> > > definitions have been completely spelled out.
> >
> > Since one can finitely define certain infinite sequences of digits,
> > those infinite sequences in turn define numbers
>
> No.
Ralf Bader
May 25, 2012, 7:12:08 PM5/25/12
to
Virgil
May 25, 2012, 7:21:42 PM5/25/12
to
In article
<edc6a694-2461-4147...@b26g2000vbt.googlegroups.com>,
posting anything, so WM's insult falsifies itself.
--
<edc6a694-2461-4147...@b26g2000vbt.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Mai, 21:17, Ralf Bader <ba...@nefkom.net> wrote:
> > WM wrote:
> > > On 25 Mai, 00:41, Ralf Bader <ba...@nefkom.net> wrote:
> >
> > >> You are babbling incredibly stupid nonsense.
> >
> > > Wouldn't you say the same if your intelligence was below the threshold
> > > of any comprehension? Can you at least follow my logic here?
> >
> > > Nevertheless, what don't you understand in my simple argument? Perhaps
> > > I can help you?
> >
> > You are babbling incredibly stupid nonsense.
>
> Would you say the same if your intelligence was not below the
> threshold of any comprehension?
If that were truly the case, then he would not have been capable of
> On 25 Mai, 21:17, Ralf Bader <ba...@nefkom.net> wrote:
> > WM wrote:
> > > On 25 Mai, 00:41, Ralf Bader <ba...@nefkom.net> wrote:
> >
> > >> You are babbling incredibly stupid nonsense.
> >
> > > Wouldn't you say the same if your intelligence was below the threshold
> > > of any comprehension? Can you at least follow my logic here?
> >
> > > Nevertheless, what don't you understand in my simple argument? Perhaps
> > > I can help you?
> >
> > You are babbling incredibly stupid nonsense.
>
> Would you say the same if your intelligence was not below the
> threshold of any comprehension?
posting anything, so WM's insult falsifies itself.
--
WM
May 26, 2012, 8:58:20 AM5/26/12
to
On 26 Mai, 01:21, Virgil <vir...@ligriv.com> wrote:
> In article
> <edc6a694-2461-4147-9264-ddb57fbbb...@b26g2000vbt.googlegroups.com>,
Are you really sure?
Regards, WM
> In article
> <edc6a694-2461-4147-9264-ddb57fbbb...@b26g2000vbt.googlegroups.com>,
Regards, WM
WM
May 26, 2012, 3:52:59 PM5/26/12
to
Why is my proof better than the argument that countably many rational
numbers cannot separate uncountably many reals?
The usual counter argument is: A (necessarily countable) sequence can
have uncountably many limiting points. Although this sounds lime
incredibly stupid nonsense, set theorists must claim it, because of
the unit interval and the enumeration of the rationals therein.
And it is not easy to disprove.
But my argument is much stronger. The covering of all rationals q_n
with closed intervals I_n of measure 10^-n leaves uncountably many
irrationals x_alpha uncovered. They cannot be connected but must be
disconnected by at least one rational, hence by at least one interval
I_n.
But they cannot exist as endpoints of intervals since the endpoints
are rationals. Hence the x_alpha can only exist as limits of sequences
of endpoints. However, they cannot exist _between_ two endpoints that
belong to the same sequence because the intervals belonging to one and
the same sequence, with limit x_alpha say, must overlap. Otherwise
they do not form one and the same sequence. Therefore infinitely many
endpoints of a sequence alpha are necessary to make up a place for the
irrational x_alpha. Therefore there must be more rationals than
irrationals.
Regards, WM
numbers cannot separate uncountably many reals?
The usual counter argument is: A (necessarily countable) sequence can
have uncountably many limiting points. Although this sounds lime
incredibly stupid nonsense, set theorists must claim it, because of
the unit interval and the enumeration of the rationals therein.
And it is not easy to disprove.
But my argument is much stronger. The covering of all rationals q_n
with closed intervals I_n of measure 10^-n leaves uncountably many
irrationals x_alpha uncovered. They cannot be connected but must be
disconnected by at least one rational, hence by at least one interval
I_n.
But they cannot exist as endpoints of intervals since the endpoints
are rationals. Hence the x_alpha can only exist as limits of sequences
of endpoints. However, they cannot exist _between_ two endpoints that
belong to the same sequence because the intervals belonging to one and
the same sequence, with limit x_alpha say, must overlap. Otherwise
they do not form one and the same sequence. Therefore infinitely many
endpoints of a sequence alpha are necessary to make up a place for the
irrational x_alpha. Therefore there must be more rationals than
irrationals.
Regards, WM
Virgil
May 26, 2012, 4:37:55 PM5/26/12
to
In article
<48eb70db-7c99-4023...@b1g2000vbb.googlegroups.com>,
known that between any two reals there are infinitely many rationals.
interior real of the unit interval, is the limit of a sequence of
rational upper ends of your intervals below it and simultaneoulsly is
the limit of a sequence of rational lower ends of your intervals above
it.
The thing is that any sequence of rationals which is dense in [0,1]
will have strictly increasing subsequences converging to each value in
(0,1] and strictly decreasing subsequences converging to each value in
[0,1).
Note that the set of subsequences of an infinite sequence obviously has
the same cardinality as the set of infinite subsets of N.
Thus any contradictions exist only in WMs head, and not in the example
he cited.
--
<48eb70db-7c99-4023...@b1g2000vbb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> Why is my proof better than the argument that countably many rational
> numbers cannot separate uncountably many reals?
>
> The usual counter argument is: A (necessarily countable) sequence can
> have uncountably many limiting points. Although this sounds lime
> incredibly stupid nonsense, set theorists must claim it, because of
> the unit interval and the enumeration of the rationals therein.
>
> And it is not easy to disprove.
>
> But my argument is much stronger. The covering of all rationals q_n
> with closed intervals I_n of measure 10^-n leaves uncountably many
> irrationals x_alpha uncovered. They cannot be connected but must be
> disconnected by at least one rational, hence by at least one interval
> I_n.
Since the rationals are known to be dense in the reals, it is already
> Why is my proof better than the argument that countably many rational
> numbers cannot separate uncountably many reals?
>
> The usual counter argument is: A (necessarily countable) sequence can
> have uncountably many limiting points. Although this sounds lime
> incredibly stupid nonsense, set theorists must claim it, because of
> the unit interval and the enumeration of the rationals therein.
>
> And it is not easy to disprove.
>
> But my argument is much stronger. The covering of all rationals q_n
> with closed intervals I_n of measure 10^-n leaves uncountably many
> irrationals x_alpha uncovered. They cannot be connected but must be
> disconnected by at least one rational, hence by at least one interval
> I_n.
known that between any two reals there are infinitely many rationals.
>
> But they cannot exist as endpoints of intervals since the endpoints
> are rationals. Hence the x_alpha can only exist as limits of sequences
> of endpoints.
Granted! But it is easily seen that each irrational, and indeed each
> But they cannot exist as endpoints of intervals since the endpoints
> are rationals. Hence the x_alpha can only exist as limits of sequences
> of endpoints.
interior real of the unit interval, is the limit of a sequence of
rational upper ends of your intervals below it and simultaneoulsly is
the limit of a sequence of rational lower ends of your intervals above
it.
The thing is that any sequence of rationals which is dense in [0,1]
will have strictly increasing subsequences converging to each value in
(0,1] and strictly decreasing subsequences converging to each value in
[0,1).
Note that the set of subsequences of an infinite sequence obviously has
the same cardinality as the set of infinite subsets of N.
Thus any contradictions exist only in WMs head, and not in the example
he cited.
--
Virgil
May 26, 2012, 4:39:13 PM5/26/12
to
In article
<17f854df-9bbe-4cc8...@s9g2000vbg.googlegroups.com>,
As so often is the case, WM's claim disproves itself.
--
<17f854df-9bbe-4cc8...@s9g2000vbg.googlegroups.com>,
--
Jürgen R.
May 26, 2012, 8:17:09 PM5/26/12
to
"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag
news:48eb70db-7c99-4023...@b1g2000vbb.googlegroups.com...
Did gyre and gimble in the wabe;
All mimsy were the borogoves,
And the mome raths outgrabe.
The Jabberwocky
>
> Regards, WM
Ralf Bader
May 26, 2012, 11:09:21 PM5/26/12
to
WM wrote:
> Why is my proof better than the argument that countably many rational
> numbers cannot separate uncountably many reals?
>
> The usual counter argument is: A (necessarily countable) sequence can
> have uncountably many limiting points. Although this sounds lime
> incredibly stupid nonsense, set theorists must claim it, because of
> the unit interval and the enumeration of the rationals therein.
>
> And it is not easy to disprove.
>
> But my argument is much stronger. The covering of all rationals q_n
> with closed intervals I_n of measure 10^-n leaves uncountably many
> irrationals x_alpha uncovered. They cannot be connected but must be
> disconnected by at least one rational, hence by at least one interval
> I_n.
>
> But they cannot exist as endpoints of intervals since the endpoints
> are rationals. Hence the x_alpha can only exist as limits of sequences
> of endpoints. However, they cannot exist _between_ two endpoints that
> belong to the same sequence because the intervals belonging to one and
> the same sequence, with limit x_alpha say, must overlap.
It is easy, applying usual procedures, to see that any x_alpha is a limit of
> Why is my proof better than the argument that countably many rational
> numbers cannot separate uncountably many reals?
>
> The usual counter argument is: A (necessarily countable) sequence can
> have uncountably many limiting points. Although this sounds lime
> incredibly stupid nonsense, set theorists must claim it, because of
> the unit interval and the enumeration of the rationals therein.
>
> And it is not easy to disprove.
>
> But my argument is much stronger. The covering of all rationals q_n
> with closed intervals I_n of measure 10^-n leaves uncountably many
> irrationals x_alpha uncovered. They cannot be connected but must be
> disconnected by at least one rational, hence by at least one interval
> I_n.
>
> But they cannot exist as endpoints of intervals since the endpoints
> are rationals. Hence the x_alpha can only exist as limits of sequences
> of endpoints. However, they cannot exist _between_ two endpoints that
> belong to the same sequence because the intervals belonging to one and
> the same sequence, with limit x_alpha say, must overlap.
endpoints of a subsequence of (I_n) whose members are pairwise non-
overlapping.
(Let x_alpha = x. Let (q_i) be a sequence of rationals converging to x. Let
I_j_i be the interval covering q_i. By taking a subsequence of (q_i), it may
be arranged that j_(i+1) > j_i, for all i. Call that subsequence again
(q_i). Then any sequence (p_i) such that p_i is an endpoint of I_j_i also
converges to x. Take such a sequence, call it (p_i) and let J_i be the
interval one of whose endpoints p_i is. Then x is in the complement of the
union of the J_i's. Choose inductively a subsequence (p_s_i) of (p_i):
Assume p_s_k is already chosen. x has a positive distance d_k from the
compact set J = J_s_1 u ... u J_s_k. Then choose s_(k+1) large enough so
that J_s_(k+1) has length < d/10, say; and simultaneously large enough so
that p_s_(k+1) has distance < d/10, say, from x. Then J_s_(k+1) is in the
interval [x-d/5,x+d/5] which is disjoint from J. So the sequence (p_s_i) has
the required property.)
> Otherwise
> they do not form one and the same sequence. Therefore infinitely many
> endpoints of a sequence alpha are necessary to make up a place for the
> irrational x_alpha. Therefore there must be more rationals than
> irrationals.
mathematical argument. Maybe it is even correct. But your stronger argument
is surely a perfectly correct proof of the fact that:
WM
May 27, 2012, 4:21:38 AM5/27/12
to
> It is easy, applying usual procedures, to see that any x_alpha is a limit of
> endpoints of a subsequence of (I_n) whose members are pairwise non-
> overlapping.
That is easy in fact, but not doubted and not relevant for my
> endpoints of a subsequence of (I_n) whose members are pairwise non-
> overlapping.
argument.
Therefore I delete the following text.
>
> > Otherwise
> > they do not form one and the same sequence. Therefore infinitely many
> > endpoints of a sequence alpha are necessary to make up a place for the
> > irrational x_alpha. Therefore there must be more rationals than
> > irrationals.
>
> Oh yes. What I wrote above is an example of what I understand by a
> mathematical argument.
missed the topic).
My argument shows this: Every pair of irrationals x_a and x_b, that
are not covered by intervals I_n, is the limit x_a of a decreasing
sequence of left endpoints of overlapping (joint) intervals and the
limit x_b of an increasing sequence of endpoints of overlapping (also
joint) intervals, respectively. Therefore every pair of irrationals
that are not covered by intervals I_n has infinitely many intervals in
between. This is an upper estimation. You tried a lower estimation.
Look, your arguing is similar to this: I show that pi > 3, and you
try to falsify my result by proving pi < 1000.
Nevertheless, nice to see that there is a chance you could understand.
Regards, WM
WM
May 27, 2012, 9:04:44 AM5/27/12
to
On 27 Mai, 05:09, Ralf Bader <ba...@nefkom.net> wrote:
> WM wrote:
> > Why is my proof better than the argument that countably many rational
> > numbers cannot separate uncountably many reals?
>
Here I emphasize the important point: There are two non-empty sets, X
> > Why is my proof better than the argument that countably many rational
> > numbers cannot separate uncountably many reals?
>
and Y, of irrationals: those, y in Y, covered by the intervals I_n,
and those, x in X, not covered by such intervals. All irrationals are
limits of Cauchy-sequences of rationals.
But: The x in X, in addition, *must* be limits of sequences of
interval-ends such that all interval ends that lie between two
neighbouring x (they exist) do not belong to any other sequences of
interval-ends definig other x. The set of all interval ends is split
into disjunct subsets by the sequences, the limits of which are x in
X.
There are not more than countably many disjunct infinite subsets of a
countable set.
Regards, WM
Ralf Bader
May 27, 2012, 11:26:21 AM5/27/12
to
Ralf Bader
May 27, 2012, 11:26:55 AM5/27/12
to
Virgil
May 27, 2012, 2:11:55 PM5/27/12
to
In article
<82e180e8-d227-4389...@s9g2000vbg.googlegroups.com>,
There are not more than countably many pairwise disjoint finite sets
either.
But subsequences of a sequence, regarded as subsets of the set of terms
of the sequence, need not be disjoint, nor merely countable.
The set of infinite subsets of N, for example, is trivially not
countable.
--
<82e180e8-d227-4389...@s9g2000vbg.googlegroups.com>,
either.
But subsequences of a sequence, regarded as subsets of the set of terms
of the sequence, need not be disjoint, nor merely countable.
The set of infinite subsets of N, for example, is trivially not
countable.
--
Virgil
May 27, 2012, 2:23:16 PM5/27/12
to
In article
<5c2047eb-3db9-4df3...@cu1g2000vbb.googlegroups.com>,
be so in refutation of your argument, so your deletion is self serving
and logically unjustified.
prove that between any such x_a and x_b, there are no other irrationals
not covered be any intervals, wish I hold to be false. And even then, yu
have provided no adequate justification for your claim.
> Therefore every pair of irrationals
> that are not covered by intervals I_n has infinitely many intervals in
> between. This is an upper estimation. You tried a lower estimation.
>
> Look, your arguing is similar to this: I show that pi > 3, and you
> try to falsify my result by proving pi < 1000.
Actually you seem to be trying to show that pi < 3.
--
<5c2047eb-3db9-4df3...@cu1g2000vbb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Mai, 05:09, Ralf Bader <ba...@nefkom.net> wrote:
> > WM wrote:
> >
> > It is easy, applying usual procedures, to see that any x_alpha is a limit of
> > endpoints of a subsequence of (I_n) whose members are pairwise non-
> > overlapping.
>
> That is easy in fact, but not doubted and not relevant for my
> argument.
> Therefore I delete the following text.
While it may not be relevant in support of your argument, it might well
> On 27 Mai, 05:09, Ralf Bader <ba...@nefkom.net> wrote:
> > WM wrote:
> >
> > It is easy, applying usual procedures, to see that any x_alpha is a limit of
> > endpoints of a subsequence of (I_n) whose members are pairwise non-
> > overlapping.
>
> That is easy in fact, but not doubted and not relevant for my
> argument.
> Therefore I delete the following text.
be so in refutation of your argument, so your deletion is self serving
and logically unjustified.
>
> >
> > > Otherwise
> > > they do not form one and the same sequence. Therefore infinitely many
> > > endpoints of a sequence alpha are necessary to make up a place for the
> > > irrational x_alpha. Therefore there must be more rationals than
> > > irrationals.
> >
> > Oh yes. What I wrote above is an example of what I understand by a
> > mathematical argument.
>
> No question, but that has not been asked for. Thema verfehlt (you
> missed the topic).
>
> My argument shows this: Every pair of irrationals x_a and x_b, that
> are not covered by intervals I_n, is the limit x_a of a decreasing
> sequence of left endpoints of overlapping (joint) intervals and the
> limit x_b of an increasing sequence of endpoints of overlapping (also
> joint) intervals, respectively.
In order for your claim above to be true, it would first be necessary to
> >
> > > Otherwise
> > > they do not form one and the same sequence. Therefore infinitely many
> > > endpoints of a sequence alpha are necessary to make up a place for the
> > > irrational x_alpha. Therefore there must be more rationals than
> > > irrationals.
> >
> > Oh yes. What I wrote above is an example of what I understand by a
> > mathematical argument.
>
> No question, but that has not been asked for. Thema verfehlt (you
> missed the topic).
>
> My argument shows this: Every pair of irrationals x_a and x_b, that
> are not covered by intervals I_n, is the limit x_a of a decreasing
> sequence of left endpoints of overlapping (joint) intervals and the
> limit x_b of an increasing sequence of endpoints of overlapping (also
> joint) intervals, respectively.
prove that between any such x_a and x_b, there are no other irrationals
not covered be any intervals, wish I hold to be false. And even then, yu
have provided no adequate justification for your claim.
> Therefore every pair of irrationals
> that are not covered by intervals I_n has infinitely many intervals in
> between. This is an upper estimation. You tried a lower estimation.
>
> Look, your arguing is similar to this: I show that pi > 3, and you
> try to falsify my result by proving pi < 1000.
--
WM
May 27, 2012, 3:09:25 PM5/27/12
to
On 27 Mai, 20:23, Virgil <vir...@ligriv.com> wrote:
>
> In order for your claim above to be true, it would first be necessary to
> prove that between any such x_a and x_b, there are no other irrationals
> not covered be any intervals, wish I hold to be false.
As two irrationals must be separated by a rational and all rationals
>
> In order for your claim above to be true, it would first be necessary to
> prove that between any such x_a and x_b, there are no other irrationals
> not covered be any intervals, wish I hold to be false.
are centers of intervals, x_a and x_b must be separated by at least
one interval I_n. Take the nearest neighbours of this kind. They can
be determined in principle. Then, by definition, there are no further
x_c between them. As x_a and x_b cannot be smallest irrationals larger
than a rational interval end (such irrationals do not exist) they must
be limits of infinite sequences of interval ends.
> And even then, you
> have provided no adequate justification for your claim.
First look whether you can agree to the above argument.
Regards, WM
WM
May 27, 2012, 4:42:46 PM5/27/12
to
who understand the contradiction of uncountability. But is appears as
if you would belong to the last ones. Don't blame your misfortune on
me.
Regards, WM
Virgil
May 27, 2012, 6:19:21 PM5/27/12
to
In article
<6723dfc6-dd11-44c4...@s9g2000vbg.googlegroups.com>,
assumed, but no mathematician can ever prove beyond all doubt to
everyone the validity of all his or her assumptions.
Which claim of proving beyond all dounbt all of his own assumptions is
precisely what WM claims he can do.
THough he has never done it, and will, no doubt, continue his emulation
of Sysiphus.
--
<6723dfc6-dd11-44c4...@s9g2000vbg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
>
> What a pity! You had the chance to belong to the first mathematicians
> who understand the contradiction of uncountability. But is appears as
> if you would belong to the last ones. Don't blame your misfortune on
> me.
WM does not understand that all mathematics is conditional on what is
>
> What a pity! You had the chance to belong to the first mathematicians
> who understand the contradiction of uncountability. But is appears as
> if you would belong to the last ones. Don't blame your misfortune on
> me.
assumed, but no mathematician can ever prove beyond all doubt to
everyone the validity of all his or her assumptions.
Which claim of proving beyond all dounbt all of his own assumptions is
precisely what WM claims he can do.
THough he has never done it, and will, no doubt, continue his emulation
of Sysiphus.
--
Virgil
May 27, 2012, 6:30:01 PM5/27/12
to
In article
<6b01757b-19ab-4056...@q2g2000vbv.googlegroups.com>,
kind?
If your sequence of intervals [q_n - 1/10^n, q_n + 1/10^n] is to have
each rational as a midpoint of one interval, then every such interval
will have infinitely many others as subintervals.
So, similarly, between any two irrationals there will be infinitely many
such intervals, so for any x_a there will be an increasing sequence of
x_b's converging to it and for each X-b a decreasing sequence of x_a's
converging to it.
> They can
> be determined in principle.
Not with the properties that you ascribe to them.
> Then, by definition, there are no further
> x_c between them.
WRONG!
> As x_a and x_b cannot be smallest irrationals larger
> than a rational interval end (such irrationals do not exist) they must
> be limits of infinite sequences of interval ends.
And of interrval midpoints.
--
<6b01757b-19ab-4056...@q2g2000vbv.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 27 Mai, 20:23, Virgil <vir...@ligriv.com> wrote:
> >
> > In order for your claim above to be true, it would first be necessary to
> > prove that between any such x_a and x_b, there are no other irrationals
> > not covered be any intervals, wish I hold to be false.
>
> As two irrationals must be separated by a rational and all rationals
> are centers of intervals, x_a and x_b must be separated by at least
> one interval I_n. Take the nearest neighbours of this kind.
What makes you think that there can be any "nearest neighbors" of this
> On 27 Mai, 20:23, Virgil <vir...@ligriv.com> wrote:
> >
> > In order for your claim above to be true, it would first be necessary to
> > prove that between any such x_a and x_b, there are no other irrationals
> > not covered be any intervals, wish I hold to be false.
>
> As two irrationals must be separated by a rational and all rationals
> are centers of intervals, x_a and x_b must be separated by at least
> one interval I_n. Take the nearest neighbours of this kind.
kind?
If your sequence of intervals [q_n - 1/10^n, q_n + 1/10^n] is to have
each rational as a midpoint of one interval, then every such interval
will have infinitely many others as subintervals.
So, similarly, between any two irrationals there will be infinitely many
such intervals, so for any x_a there will be an increasing sequence of
x_b's converging to it and for each X-b a decreasing sequence of x_a's
converging to it.
> They can
> be determined in principle.
> Then, by definition, there are no further
> x_c between them.
> As x_a and x_b cannot be smallest irrationals larger
> than a rational interval end (such irrationals do not exist) they must
> be limits of infinite sequences of interval ends.
>
>
> > And even then, you
> > have provided no adequate justification for your claim.
>
> First look whether you can agree to the above argument.
Your argument fails in many respects, some of which I have noted!
>
> > And even then, you
> > have provided no adequate justification for your claim.
>
> First look whether you can agree to the above argument.
--
YBM
May 27, 2012, 8:31:30 PM5/27/12
to
You'll be fired as a professor and you deserved it, you are a piece of
shit.
WM
May 28, 2012, 9:05:31 AM5/28/12
to
> <6b01757b-19ab-4056-b8b0-74a61018c...@q2g2000vbv.googlegroups.com>,
have an irrational limit to the left hand side and an irrational limit
to the right hand side of the real axis. These two limits are the
closest irrational numbers that are not covered by intervals and that
have C between each other.
> If your sequence of intervals [q_n - 1/10^n, q_n + 1/10^n] is to have
> each rational as a midpoint of one interval, then every such interval
> will have infinitely many others as subintervals.
Why not?
> > They can
> > be determined in principle.
>
> Not with the properties that you ascribe to them.
Perhaps you have problems to understand?
Regards, WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Mai, 20:23, Virgil <vir...@ligriv.com> wrote:
>
> > > In order for your claim above to be true, it would first be necessary to
> > > prove that between any such x_a and x_b, there are no other irrationals
> > > not covered be any intervals, wish I hold to be false.
>
> > As two irrationals must be separated by a rational and all rationals
> > are centers of intervals, x_a and x_b must be separated by at least
> > one interval I_n. Take the nearest neighbours of this kind.
>
> What makes you think that there can be any "nearest neighbors" of this
> kind?
A cluster C of overlapping intervals I_n with rational end points can
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Mai, 20:23, Virgil <vir...@ligriv.com> wrote:
>
> > > In order for your claim above to be true, it would first be necessary to
> > > prove that between any such x_a and x_b, there are no other irrationals
> > > not covered be any intervals, wish I hold to be false.
>
> > As two irrationals must be separated by a rational and all rationals
> > are centers of intervals, x_a and x_b must be separated by at least
> > one interval I_n. Take the nearest neighbours of this kind.
>
> What makes you think that there can be any "nearest neighbors" of this
> kind?
have an irrational limit to the left hand side and an irrational limit
to the right hand side of the real axis. These two limits are the
closest irrational numbers that are not covered by intervals and that
have C between each other.
> If your sequence of intervals [q_n - 1/10^n, q_n + 1/10^n] is to have
> each rational as a midpoint of one interval, then every such interval
> will have infinitely many others as subintervals.
> > They can
> > be determined in principle.
>
> Not with the properties that you ascribe to them.
>
> > Then, by definition, there are no further
> > x_c between them.
>
> WRONG!
Right by definition of closest neighbours.
> > Then, by definition, there are no further
> > x_c between them.
>
> WRONG!
Regards, WM
Jan Andres
May 29, 2012, 9:42:47 PM5/29/12
to
On 2012-05-24, WM <muec...@rz.fh-augsburg.de> wrote:
> On 24 Mai, 08:31, Uirgil <uir...@uirgil.ur> wrote:
>> In article
>> <65dc68ad-7594-4268-be98-7095e3369...@5g2000vbf.googlegroups.com>,
>>
>>
>>
>>
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:
>> > On 23 Mai, 00:42, Uirgil <uir...@uirgil.ur> wrote:
>> > > In article
>> > > <9db0b271-e437-4e8a-83e2-e0708f325...@b26g2000vbt.googlegroups.com>,
>>
>> > > WM <mueck...@rz.fh-augsburg.de> wrote:
>> > > > On 22 Mai, 22:19, Uirgil <uir...@uirgil.ur> wrote:
>> > > > > In article
>> > > > > <dc2813f8-c73a-4563-add5-48feb4d47...@5g2000vbf.googlegroups.com>,
>>
>> > > > > WM <mueck...@rz.fh-augsburg.de> wrote:
>> > > > > > On 22 Mai, 10:53, William Elliot <ma...@panix.com> wrote:
>> > > > > > > On Tue, 22 May 2012, WM wrote:
>>
>> > > > > snip
>>
>> > > > > > > > Then there are at least two irrational numbers without a rational
>> > > > > > > > between them. That is mathematically impossible.
>>
>> > > > > > > Prove it or cram it.
>>
>> > > > > > The proof has been given. I repeat: As every rational q_n is covered
>> > > > > > by an interval I_n, we can conclude that every pair x and x' of
>> > > > > > irrationals is separated by least one interval I_n. Hence x and x'
>> > > > > > cannot belong to the same complementary interval. This holds for every
>> > > > > > pair of irrationals.
>>
>> > > > > > A very simple conclusion. What do you not understand?
>>
>> > > > > Why you claiming what is so obviously false.
>>
>> > > > > Unless EVERY pair of irrationals is separated by EVERY ONE of your
>> > > > > I_n's, which does not ever happen, for some pairs of irrationals there
>> > > > > will be LOTS of your "complimentary intervals' containing both.
>>
>> > > > > And as every complimentary interval is necessarily of positive length,
>> > > > > being of form [0,q) or (q,1] for some rational q with 0 < q < 1, they
>> > > > > are each proved to contain infinitely many irrationals.
>>
>> > > > Without rationals between them?
>>
>> > > > You misunderstand. The complementary intervals do not contain any
>> > > > rationals q_n. They cannot because all q_n are covered by intervals
>> > > > I_n. Therefore the complementary intervals are merely single points.
>> > > > (Limits of sequences of intervals I_n.)
>>
>> > > Then your original definition was sufficiently ambiguous as not to make
>> > > that clear.-
>>
>> > I said: "every rational q_n is covered by an interval I_n". So, after
>> > having understood now, what is your objection?
>>
>> You did not make clear that your complimentation was with respect to the
>> union of all those rational intervals, rather than one compliment for
>> each interval.-
>
> Now it should be clear that at most 1/9 of the unit interval (or even
> of the complete real line) is covered by intervals I_n. That means at
> least 8/9 (infinity) is not covered by intervals I_n. This remaining
> part does not contain any rational numbers. Therefore either two
> irrationals of that remaining part must be connected or disconnected
> by one of the intervals I_n. The former is not possible in mathematics
> (perhaps it is possible in matheology - but that is not of interest
> for me). The second means that we have either only countably many
> irrationals or we are forced to believe that there are more
> irrationals than rationals and nevertheless every pair of irrationals
> is separated by an I_n.
What the heck makes you think they must be separated by exactly one of
your intervals? Inbetween any two distinct reals there are INFINITELY
MANY rationals and thus infinitely many of your I_n intervals. So what
we're talking about is an injection from |R^2 to the power set of |N.
That such an injection exists is a well-known result and is _of course_
consistent with the existence of uncountable infinity.
> On 24 Mai, 08:31, Uirgil <uir...@uirgil.ur> wrote:
>> In article
>> <65dc68ad-7594-4268-be98-7095e3369...@5g2000vbf.googlegroups.com>,
>>
>>
>>
>>
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:
>> > On 23 Mai, 00:42, Uirgil <uir...@uirgil.ur> wrote:
>> > > In article
>> > > <9db0b271-e437-4e8a-83e2-e0708f325...@b26g2000vbt.googlegroups.com>,
>>
>> > > WM <mueck...@rz.fh-augsburg.de> wrote:
>> > > > On 22 Mai, 22:19, Uirgil <uir...@uirgil.ur> wrote:
>> > > > > In article
>> > > > > <dc2813f8-c73a-4563-add5-48feb4d47...@5g2000vbf.googlegroups.com>,
>>
>> > > > > WM <mueck...@rz.fh-augsburg.de> wrote:
>> > > > > > On 22 Mai, 10:53, William Elliot <ma...@panix.com> wrote:
>> > > > > > > On Tue, 22 May 2012, WM wrote:
>>
>> > > > > snip
>>
>> > > > > > > > Then there are at least two irrational numbers without a rational
>> > > > > > > > between them. That is mathematically impossible.
>>
>> > > > > > > Prove it or cram it.
>>
>> > > > > > The proof has been given. I repeat: As every rational q_n is covered
>> > > > > > by an interval I_n, we can conclude that every pair x and x' of
>> > > > > > irrationals is separated by least one interval I_n. Hence x and x'
>> > > > > > cannot belong to the same complementary interval. This holds for every
>> > > > > > pair of irrationals.
>>
>> > > > > > A very simple conclusion. What do you not understand?
>>
>> > > > > Why you claiming what is so obviously false.
>>
>> > > > > Unless EVERY pair of irrationals is separated by EVERY ONE of your
>> > > > > I_n's, which does not ever happen, for some pairs of irrationals there
>> > > > > will be LOTS of your "complimentary intervals' containing both.
>>
>> > > > > And as every complimentary interval is necessarily of positive length,
>> > > > > being of form [0,q) or (q,1] for some rational q with 0 < q < 1, they
>> > > > > are each proved to contain infinitely many irrationals.
>>
>> > > > Without rationals between them?
>>
>> > > > You misunderstand. The complementary intervals do not contain any
>> > > > rationals q_n. They cannot because all q_n are covered by intervals
>> > > > I_n. Therefore the complementary intervals are merely single points.
>> > > > (Limits of sequences of intervals I_n.)
>>
>> > > Then your original definition was sufficiently ambiguous as not to make
>> > > that clear.-
>>
>> > I said: "every rational q_n is covered by an interval I_n". So, after
>> > having understood now, what is your objection?
>>
>> You did not make clear that your complimentation was with respect to the
>> union of all those rational intervals, rather than one compliment for
>> each interval.-
>
> Now it should be clear that at most 1/9 of the unit interval (or even
> of the complete real line) is covered by intervals I_n. That means at
> least 8/9 (infinity) is not covered by intervals I_n. This remaining
> part does not contain any rational numbers. Therefore either two
> irrationals of that remaining part must be connected or disconnected
> by one of the intervals I_n. The former is not possible in mathematics
> (perhaps it is possible in matheology - but that is not of interest
> for me). The second means that we have either only countably many
> irrationals or we are forced to believe that there are more
> irrationals than rationals and nevertheless every pair of irrationals
> is separated by an I_n.
What the heck makes you think they must be separated by exactly one of
your intervals? Inbetween any two distinct reals there are INFINITELY
MANY rationals and thus infinitely many of your I_n intervals. So what
we're talking about is an injection from |R^2 to the power set of |N.
That such an injection exists is a well-known result and is _of course_
consistent with the existence of uncountable infinity.
WM
May 30, 2012, 5:38:41 AM5/30/12
to
On 30 Mai, 03:42, Jan Andres <J...@nAndr.es> wrote:
rational is in an interval).
Inbetween any two distinct reals there are INFINITELY
> MANY rationals and thus infinitely many of your I_n intervals. So what
> we're talking about is an injection from |R^2 to the power set of |N.
Here we consider only those irrationals which are outside of all
intervals I_n. See for a concise formulation: Matheology § 022.
Regards, WM
> On 2012-05-24, WM <mueck...@rz.fh-augsburg.de> wrote:
>
The second means that we have either only countably many
> > irrationals or we are forced to believe that there are more
> > irrationals than rationals and nevertheless every pair of irrationals
> > is separated by an I_n.
>
> What the heck makes you think they must be separated by exactly one of
> your intervals?
They must be separated by at least one of the intervals (because every
>
The second means that we have either only countably many
> > irrationals or we are forced to believe that there are more
> > irrationals than rationals and nevertheless every pair of irrationals
> > is separated by an I_n.
>
> What the heck makes you think they must be separated by exactly one of
> your intervals?
rational is in an interval).
Inbetween any two distinct reals there are INFINITELY
> MANY rationals and thus infinitely many of your I_n intervals. So what
> we're talking about is an injection from |R^2 to the power set of |N.
intervals I_n. See for a concise formulation: Matheology § 022.
Regards, WM
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