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Matheology § 092
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WM
Jul 22, 2012, 5:51:59 AM7/22/12
to
Matheology § 092
My second best proof contradicting set theory
1)
Define a sequence of points p_n in the unit interval
p_n = 1/n.
These points define intervals
A_k = [1/n, 1/(n+1)] for odd n
and B_j = [1/n, 1/(n+1)] for even n.
The intervals of sort A ==== and B ---- are alternating. If the points
are denoted by n, we have something like the following configuration.
...7--6==5--4====3---------2===========1
Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
number of intervals remains the same.
...7--6==5--3====4---------2===========1
The intervals remain alternating. And in particular the number of
intervals cannot increase.
2)
Define a set of intervals I_n in the unit interval such that interval
I_n has length
|I_n| = 10^-n and covers the rational number q_n of a suitable
enumeration of all rational numbers of the unit interval.
Then the union of all I_n has measure < 1/9. The remaining part of
the unit interval has measure > 8/9 and is split into uncountably many
singletons. A sketch of the intervals I_n ~~~~~~~ is given here:
...a b~~~~~~~~~c~~~~~~~~~d~~~~~~e f~~~g~~~h i~~~j k~~~l
We cannot exclude intervals within intervals like c~~~d within b~~~e
or, alternatively, overlapping intervals like b~~~d and c~~~~e and
also adjacent intervals like f~~~g and g~~~h.
3)
Let the endpoints p_n of the configuration described in (1) move in an
arbitrary way, say powered by little ants or by the Gods of
matheology. Then it cannot be excluded that the p_n and the endpoints
of the I_n of (2) will conincide (no particular order is required).
...a b~~~~~~~~~c~~~~~~~~~d~~~~~~e f~~~g~~~h i~~~j k~~~l
...3=7--------------11=========5--------12=4---2===9-8==10-6==1
As our theorem shows, there will be not more than aleph_0 intervals in
the end position. This includes the set of I_n and the set of
intervals in the complement. In case that intervals fall into
intervals, the complete number can be reduced. In no case it can grow.
Therefore the assertion of uncountably many degenerate intervals (so
called singletons - but there cannot exist irrational singletons
without rational numbers separating them) in the complement has been
contradicted.
Regards, WM
My second best proof contradicting set theory
1)
Define a sequence of points p_n in the unit interval
p_n = 1/n.
These points define intervals
A_k = [1/n, 1/(n+1)] for odd n
and B_j = [1/n, 1/(n+1)] for even n.
The intervals of sort A ==== and B ---- are alternating. If the points
are denoted by n, we have something like the following configuration.
...7--6==5--4====3---------2===========1
Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
number of intervals remains the same.
...7--6==5--3====4---------2===========1
The intervals remain alternating. And in particular the number of
intervals cannot increase.
2)
Define a set of intervals I_n in the unit interval such that interval
I_n has length
|I_n| = 10^-n and covers the rational number q_n of a suitable
enumeration of all rational numbers of the unit interval.
Then the union of all I_n has measure < 1/9. The remaining part of
the unit interval has measure > 8/9 and is split into uncountably many
singletons. A sketch of the intervals I_n ~~~~~~~ is given here:
...a b~~~~~~~~~c~~~~~~~~~d~~~~~~e f~~~g~~~h i~~~j k~~~l
We cannot exclude intervals within intervals like c~~~d within b~~~e
or, alternatively, overlapping intervals like b~~~d and c~~~~e and
also adjacent intervals like f~~~g and g~~~h.
3)
Let the endpoints p_n of the configuration described in (1) move in an
arbitrary way, say powered by little ants or by the Gods of
matheology. Then it cannot be excluded that the p_n and the endpoints
of the I_n of (2) will conincide (no particular order is required).
...a b~~~~~~~~~c~~~~~~~~~d~~~~~~e f~~~g~~~h i~~~j k~~~l
...3=7--------------11=========5--------12=4---2===9-8==10-6==1
As our theorem shows, there will be not more than aleph_0 intervals in
the end position. This includes the set of I_n and the set of
intervals in the complement. In case that intervals fall into
intervals, the complete number can be reduced. In no case it can grow.
Therefore the assertion of uncountably many degenerate intervals (so
called singletons - but there cannot exist irrational singletons
without rational numbers separating them) in the complement has been
contradicted.
Regards, WM
Virgil
Jul 22, 2012, 4:12:21 PM7/22/12
to
In article
<00999ff7-a0d7-4e36...@n5g2000vbb.googlegroups.com>,
According to WM's argument, it would have been equally impossible in
the original interval for there to be one of only countably many
rationals between any two of uncountably many irrationals, but it is not
only possible but actual!.
WM just does not grok dense orderings.
There is no reason why with countably many open intervals of small
enough cumulative length there cannot be at least one such interval
between any two of the uncountably many irrationals not covered, those
intervals having the same sort of density property as the rationals
have, i.e., between any two such intervals there could be others, just
as between any two rationals there are others..
At least WM would have to provide proof that this cannot happen, which
proof, even if it existed, would be far beyond WMs capability to produce.
For example, consider the Cantor "middle thirds" set in which between
any two of the deleted open intervals there are other deleted open
intervals.
--
<00999ff7-a0d7-4e36...@n5g2000vbb.googlegroups.com>,
the original interval for there to be one of only countably many
rationals between any two of uncountably many irrationals, but it is not
only possible but actual!.
WM just does not grok dense orderings.
There is no reason why with countably many open intervals of small
enough cumulative length there cannot be at least one such interval
between any two of the uncountably many irrationals not covered, those
intervals having the same sort of density property as the rationals
have, i.e., between any two such intervals there could be others, just
as between any two rationals there are others..
At least WM would have to provide proof that this cannot happen, which
proof, even if it existed, would be far beyond WMs capability to produce.
For example, consider the Cantor "middle thirds" set in which between
any two of the deleted open intervals there are other deleted open
intervals.
--
WM
Jul 23, 2012, 5:27:15 AM7/23/12
to
On 22 Jul., 22:12, Virgil <vir...@ligriv.com> wrote:
> In article
> <00999ff7-a0d7-4e36-81f0-209d8c61a...@n5g2000vbb.googlegroups.com>,
Of course. But now i have proved it.
chance to increase their cardinality during movement. Otherwise you
would have to falsify my theorem.
longing for so long.
Regards, WM
> In article
> <00999ff7-a0d7-4e36-81f0-209d8c61a...@n5g2000vbb.googlegroups.com>,
>
> There is no reason why with countably many open intervals of small
> enough cumulative length there cannot be at least one such interval
> between any two of the uncountably many irrationals not covered, those
> intervals having the same sort of density property as the rationals
> have, i.e., between any two such intervals there could be others, just
> as between any two rationals there are others..
Alas, the intervals of (1) are countable alltogether. And there is no
> There is no reason why with countably many open intervals of small
> enough cumulative length there cannot be at least one such interval
> between any two of the uncountably many irrationals not covered, those
> intervals having the same sort of density property as the rationals
> have, i.e., between any two such intervals there could be others, just
> as between any two rationals there are others..
chance to increase their cardinality during movement. Otherwise you
would have to falsify my theorem.
>
> At least WM would have to provide proof that this cannot happen, which
> proof, even if it existed, would be far beyond WMs capability to produce.
Raise you eyes a bit. Then you can see this proof you have been
> At least WM would have to provide proof that this cannot happen, which
> proof, even if it existed, would be far beyond WMs capability to produce.
longing for so long.
Regards, WM
FredJeffries
Jul 23, 2012, 11:32:31 AM7/23/12
to
WM
Jul 23, 2012, 1:58:06 PM7/23/12
to
Uergil
Jul 23, 2012, 6:22:47 PM7/23/12
to
In article
<730b536e-151e-44fb...@a16g2000vby.googlegroups.com>,
Not outside of WMatheologys , you haven/t!
irrational points, so that between any two irrationals there is a
rational, why cannot countably many open clusters do the same?
Since the rational centered intervals cover every rational but have
cumulative lengths much less than that of [0,1], their open clusters
must be scattered and still infinite in number.
If one shrinks every open cluster to a single point, e.g., any one of
the rationals in it, by sliding uncovered points, the result will be
essentially a real interval of length greater than 8/9.
recognized as a valid proof of WM's claims.
Among other things, you still have to prove that the set of clusters
cannot be not dense-in-itself (in the sense that between any two
clusters there must be another cluster).
While it is fairly obvious that the set of clusters need not be
dense-in-itself, WM's argument fails unless he can prove that the set of
clusters CANNOT be dense-in-itself.
--
"Ignorance is preferable to error, and he is less
remote from the- truth who believes nothing than
he who believes what is wrong.
Thomas Jefferson
<730b536e-151e-44fb...@a16g2000vby.googlegroups.com>,
> >
> > There is no reason why with countably many open intervals of small
> > enough cumulative length there cannot be at least one such interval
> > between any two of the uncountably many irrationals not covered, those
> > intervals having the same sort of density property as the rationals
> > have, i.e., between any two such intervals there could be others, just
> > as between any two rationals there are others..
>
> Alas, the intervals of (1) are countable alltogether. And there is no
> chance to increase their cardinality during movement. Otherwise you
> would have to falsify my theorem.
If countably many rational points can "separate" uncountably many
> > There is no reason why with countably many open intervals of small
> > enough cumulative length there cannot be at least one such interval
> > between any two of the uncountably many irrationals not covered, those
> > intervals having the same sort of density property as the rationals
> > have, i.e., between any two such intervals there could be others, just
> > as between any two rationals there are others..
>
> Alas, the intervals of (1) are countable alltogether. And there is no
> chance to increase their cardinality during movement. Otherwise you
> would have to falsify my theorem.
irrational points, so that between any two irrationals there is a
rational, why cannot countably many open clusters do the same?
Since the rational centered intervals cover every rational but have
cumulative lengths much less than that of [0,1], their open clusters
must be scattered and still infinite in number.
If one shrinks every open cluster to a single point, e.g., any one of
the rationals in it, by sliding uncovered points, the result will be
essentially a real interval of length greater than 8/9.
> >
> > At least WM would have to provide proof that this cannot happen, which
> > proof, even if it existed, would be far beyond WMs capability to produce.
>
> Raise you eyes a bit. Then you can see this proof you have been
> longing for so long.
I can see a claim of proof but nothing that outside WMatheology would be
> > At least WM would have to provide proof that this cannot happen, which
> > proof, even if it existed, would be far beyond WMs capability to produce.
>
> Raise you eyes a bit. Then you can see this proof you have been
> longing for so long.
recognized as a valid proof of WM's claims.
Among other things, you still have to prove that the set of clusters
cannot be not dense-in-itself (in the sense that between any two
clusters there must be another cluster).
While it is fairly obvious that the set of clusters need not be
dense-in-itself, WM's argument fails unless he can prove that the set of
clusters CANNOT be dense-in-itself.
--
"Ignorance is preferable to error, and he is less
remote from the- truth who believes nothing than
he who believes what is wrong.
Thomas Jefferson
Virgil
Jul 23, 2012, 6:42:12 PM7/23/12
to
> On Jul 23, 2:27�am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Alas, the intervals of (1) are countable alltogether. And there is no
> > chance to increase their cardinality during movement. Otherwise you
> > would have to falsify my theorem.
WM has yet to present us with anything more than conjectures since his
>
> > Alas, the intervals of (1) are countable alltogether. And there is no
> > chance to increase their cardinality during movement. Otherwise you
> > would have to falsify my theorem.
alleged proofs of his conjectures are always flawed.
For example: In his covering of the rationals in (0,1) by a sequence of
intervals whose centers and endpoints are rationals and whose lengths
add up to some epsilon less than 1, this resolves into a countable set
of open intervals of even smaller total length.
But in order for WM to claim what he has claimed about such a
construction, he must prove that it is impossible to end up with "dense"
clusters in the sense of having another cluster between any two
clusters, in the same way that there is always another rational between
any two if them.
It seems quite possible to arrange a sequence of such intervals so that
no two clusters have a common endpoint, just as occurs with the Cantor
set. So a proof of its impossibility is required before one can, as WM
so blithely does, assume it.
--
WM
Jul 24, 2012, 5:26:17 AM7/24/12
to
On 24 Jul., 00:22, Uergil <Uer...@uer.net> wrote:
> In article
> <730b536e-151e-44fb-a865-754d2226a...@a16g2000vby.googlegroups.com>,
You must not believe that there is nothing outside other than the cell
you sit in.
> so that between any two irrationals there is a
> rational, why cannot countably many open clusters do the same?
It is claimed by some or most matheologians, that countably many
rationals could separate uncountably many irrationals. But in my
example I start with countably many intervals which during the proof
do not increase in number.
A_k and B_j.
Regards, WM
> In article
> <730b536e-151e-44fb-a865-754d2226a...@a16g2000vby.googlegroups.com>,
you sit in.
>
>
> > > There is no reason why with countably many open intervals of small
> > > enough cumulative length there cannot be at least one such interval
> > > between any two of the uncountably many irrationals not covered, those
> > > intervals having the same sort of density property as the rationals
> > > have, i.e., between any two such intervals there could be others, just
> > > as between any two rationals there are others..
>
> > Alas, the intervals of (1) are countable alltogether. And there is no
> > chance to increase their cardinality during movement. Otherwise you
> > would have to falsify my theorem.
>
> If countably many rational points can "separate" uncountably many
> irrational points,
Of course they cannot.
>
> > > There is no reason why with countably many open intervals of small
> > > enough cumulative length there cannot be at least one such interval
> > > between any two of the uncountably many irrationals not covered, those
> > > intervals having the same sort of density property as the rationals
> > > have, i.e., between any two such intervals there could be others, just
> > > as between any two rationals there are others..
>
> > Alas, the intervals of (1) are countable alltogether. And there is no
> > chance to increase their cardinality during movement. Otherwise you
> > would have to falsify my theorem.
>
> If countably many rational points can "separate" uncountably many
> irrational points,
> so that between any two irrationals there is a
> rational, why cannot countably many open clusters do the same?
rationals could separate uncountably many irrationals. But in my
example I start with countably many intervals which during the proof
do not increase in number.
>
> Since the rational centered intervals cover every rational but have
> cumulative lengths much less than that of [0,1], their open clusters
> must be scattered and still infinite in number.
But it can't. Therefore your "Since" is wrong.
> Since the rational centered intervals cover every rational but have
> cumulative lengths much less than that of [0,1], their open clusters
> must be scattered and still infinite in number.
>
> If one shrinks every open cluster to a single point, e.g., any one of
> the rationals in it, by sliding uncovered points, the result will be
> essentially a real interval of length greater than 8/9.
and of measure 0. That is a contradiction, isn't it?
> If one shrinks every open cluster to a single point, e.g., any one of
> the rationals in it, by sliding uncovered points, the result will be
> essentially a real interval of length greater than 8/9.
>
> Among other things, you still have to prove that the set of clusters
> cannot be not dense-in-itself (in the sense that between any two
> clusters there must be another cluster).
There are no clusters in my proof. Only a countable set of intervals
> Among other things, you still have to prove that the set of clusters
> cannot be not dense-in-itself (in the sense that between any two
> clusters there must be another cluster).
A_k and B_j.
>
> While it is fairly obvious that the set of clusters need not be
> dense-in-itself, WM's argument fails unless he can prove that the set of
> clusters CANNOT be dense-in-itself.
There are no clusters in my proof.
> While it is fairly obvious that the set of clusters need not be
> dense-in-itself, WM's argument fails unless he can prove that the set of
> clusters CANNOT be dense-in-itself.
Regards, WM
Virgil
Jul 24, 2012, 6:27:59 PM7/24/12
to
In article
<33f9929c-e95b-468e...@cu1g2000vbb.googlegroups.com>,
There is all of the standard mathematics that WM excludes from his
WMytheology, and WM's "proofs" are not valid anywhere but in his own
WMytheology.
uncountably many irrationals there is at lest one of countably many
rationals. At least in the standard model of the real field.
determining each interval, and end with countably many clusters of
intervals, why cannot the clusters do what the rationals do, namely be
"dense" in the sense that between any two clusters lies at least one
other cluster?
To CLAIM that this is impossible, as WM does, is NOT a proof that it
cannot happen.
clusters is finite and of total measure so much less than 1, then not
every rational can have been covered, so there must be infinitely many
of them, and only finitely many of them longer that any positive epsilon.
So what is to prevent there from being between any two clusters another
cluster?
1/10^n have collectively covered no more than 1/9 of [0,1] altogether so
there are 8/9 of it left to be covered.
There may well be uncountably many uncovered points which, together with
one point per cluster, forms an interval of strictly positive measure.
At least until is is PROVED, and not merely claimed, that it is
impossible to always have between any two clusters another cluster.
connected unions of them, which you have yourself called clusters, so
that there ARE clusters.
common, but still manage to cover every rationals in [0,1]?
Every such maximal connected union of the sets I_n is a cluster, and
since each I_n contains more thatn one rational point, such unions exist.
At least outside of WMYtheology.
--
<33f9929c-e95b-468e...@cu1g2000vbb.googlegroups.com>,
WMytheology, and WM's "proofs" are not valid anywhere but in his own
WMytheology.
> >
> >
> > > > There is no reason why with countably many open intervals of small
> > > > enough cumulative length there cannot be at least one such interval
> > > > between any two of the uncountably many irrationals not covered, those
> > > > intervals having the same sort of density property as the rationals
> > > > have, i.e., between any two such intervals there could be others, just
> > > > as between any two rationals there are others..
> >
> > > Alas, the intervals of (1) are countable alltogether. And there is no
> > > chance to increase their cardinality during movement. Otherwise you
> > > would have to falsify my theorem.
> >
> > If countably many rational points can "separate" uncountably many
> > irrational points,
>
> Of course they cannot.
They can, and do, in standard mathematics, i.e., between any two of
> >
> > > > There is no reason why with countably many open intervals of small
> > > > enough cumulative length there cannot be at least one such interval
> > > > between any two of the uncountably many irrationals not covered, those
> > > > intervals having the same sort of density property as the rationals
> > > > have, i.e., between any two such intervals there could be others, just
> > > > as between any two rationals there are others..
> >
> > > Alas, the intervals of (1) are countable alltogether. And there is no
> > > chance to increase their cardinality during movement. Otherwise you
> > > would have to falsify my theorem.
> >
> > If countably many rational points can "separate" uncountably many
> > irrational points,
>
> Of course they cannot.
uncountably many irrationals there is at lest one of countably many
rationals. At least in the standard model of the real field.
>
> > so that between any two irrationals there is a
> > rational, why cannot countably many open clusters do the same?
>
> It is claimed by some or most matheologians, that countably many
> rationals could separate uncountably many irrationals. But in my
> example I start with countably many intervals which during the proof
> do not increase in number.
Since you start with one interval for each rational and one rational
> > so that between any two irrationals there is a
> > rational, why cannot countably many open clusters do the same?
>
> It is claimed by some or most matheologians, that countably many
> rationals could separate uncountably many irrationals. But in my
> example I start with countably many intervals which during the proof
> do not increase in number.
determining each interval, and end with countably many clusters of
intervals, why cannot the clusters do what the rationals do, namely be
"dense" in the sense that between any two clusters lies at least one
other cluster?
To CLAIM that this is impossible, as WM does, is NOT a proof that it
cannot happen.
> >
> > Since the rational centered intervals cover every rational but have
> > cumulative lengths much less than that of [0,1], their open clusters
> > must be scattered and still infinite in number.
>
> But it can't. Therefore your "Since" is wrong.
WHY can't it be? Certainly if the number of those open interval
> > Since the rational centered intervals cover every rational but have
> > cumulative lengths much less than that of [0,1], their open clusters
> > must be scattered and still infinite in number.
>
> But it can't. Therefore your "Since" is wrong.
clusters is finite and of total measure so much less than 1, then not
every rational can have been covered, so there must be infinitely many
of them, and only finitely many of them longer that any positive epsilon.
So what is to prevent there from being between any two clusters another
cluster?
> >
> > If one shrinks every open cluster to a single point, e.g., any one of
> > the rationals in it, by sliding uncovered points, the result will be
> > essentially a real interval of length greater than 8/9.
>
> and of measure 0. That is a contradiction, isn't it?
Why would what is left be of measure 0? Your intervals, I_n, of length
> > If one shrinks every open cluster to a single point, e.g., any one of
> > the rationals in it, by sliding uncovered points, the result will be
> > essentially a real interval of length greater than 8/9.
>
> and of measure 0. That is a contradiction, isn't it?
1/10^n have collectively covered no more than 1/9 of [0,1] altogether so
there are 8/9 of it left to be covered.
There may well be uncountably many uncovered points which, together with
one point per cluster, forms an interval of strictly positive measure.
At least until is is PROVED, and not merely claimed, that it is
impossible to always have between any two clusters another cluster.
> >
> > Among other things, you still have to prove that the set of clusters
> > cannot be not dense-in-itself (in the sense that between any two
> > clusters there must be another cluster).
>
> There are no clusters in my proof. Only a countable set of intervals
> A_k and B_j.
If your intervals are allowed to overlap then there must be maximal
> > Among other things, you still have to prove that the set of clusters
> > cannot be not dense-in-itself (in the sense that between any two
> > clusters there must be another cluster).
>
> There are no clusters in my proof. Only a countable set of intervals
> A_k and B_j.
connected unions of them, which you have yourself called clusters, so
that there ARE clusters.
> >
> > While it is fairly obvious that the set of clusters need not be
> > dense-in-itself, WM's argument fails unless he can prove that the set of
> > clusters CANNOT be dense-in-itself.
>
> There are no clusters in my proof.
Are you claiming that no two of your intervals, I_n, have any points in
> > While it is fairly obvious that the set of clusters need not be
> > dense-in-itself, WM's argument fails unless he can prove that the set of
> > clusters CANNOT be dense-in-itself.
>
> There are no clusters in my proof.
common, but still manage to cover every rationals in [0,1]?
Every such maximal connected union of the sets I_n is a cluster, and
since each I_n contains more thatn one rational point, such unions exist.
At least outside of WMYtheology.
--
WM
Jul 25, 2012, 3:41:40 AM7/25/12
to
On 25 Jul., 00:27, Virgil <vir...@ligriv.com> wrote:
> > > If countably many rational points can "separate" uncountably many
> > > irrational points,
>
> > Of course they cannot.
>
> They can, and do, in standard mathematics, i.e., between any two of
> uncountably many irrationals there is at lest one of countably many
> rationals. At least in the standard model of the real field.
Have you any doubt in my theorem?
> > > If countably many rational points can "separate" uncountably many
> > > irrational points,
>
> > Of course they cannot.
>
> They can, and do, in standard mathematics, i.e., between any two of
> uncountably many irrationals there is at lest one of countably many
> rationals. At least in the standard model of the real field.
If not, then there is the proof that not more than countably many
intervals are possible.
>
>
>
> > > so that between any two irrationals there is a
> > > rational, why cannot countably many open clusters do the same?
>
> > It is claimed by some or most matheologians, that countably many
> > rationals could separate uncountably many irrationals. But in my
> > example I start with countably many intervals which during the proof
> > do not increase in number.
>
> Since you start with one interval for each rational and one rational
> determining each interval, and end with countably many clusters of
> intervals, why cannot the clusters do what the rationals do, namely be
> "dense" in the sense that between any two clusters lies at least one
> other cluster?
>
> To CLAIM that this is impossible, as WM does, is NOT a proof that it
> cannot happen.
I do not claim anything. I have proved that in the final state the
>
>
> > > so that between any two irrationals there is a
> > > rational, why cannot countably many open clusters do the same?
>
> > It is claimed by some or most matheologians, that countably many
> > rationals could separate uncountably many irrationals. But in my
> > example I start with countably many intervals which during the proof
> > do not increase in number.
>
> Since you start with one interval for each rational and one rational
> determining each interval, and end with countably many clusters of
> intervals, why cannot the clusters do what the rationals do, namely be
> "dense" in the sense that between any two clusters lies at least one
> other cluster?
>
> To CLAIM that this is impossible, as WM does, is NOT a proof that it
> cannot happen.
endpoints of (1) and (2) coincide and that there are only countably
many intervals.
Regards, WM
Virgil
Jul 25, 2012, 1:48:11 PM7/25/12
to
In article
<c532496a-30b6-4695...@x21g2000vbc.googlegroups.com>,
clusters there is another.
separate uncountably many irrationals" in the sense that between any two
irrationals there is a rational (in fact, countably many rationals).
there is another cluster, just like between any two rationals there is
another rational.
And if this IS the case, then your claims all fail.
--
<c532496a-30b6-4695...@x21g2000vbc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Jul., 00:27, Virgil <vir...@ligriv.com> wrote:
>
> > > > If countably many rational points can "separate" uncountably many
> > > > irrational points,
> >
> > > Of course they cannot.
> >
> > They can, and do, in standard mathematics, i.e., between any two of
> > uncountably many irrationals there is at lest one of countably many
> > rationals. At least in the standard model of the real field.
>
> Have you any doubt in my theorem?
> If not, then there is the proof that not more than countably many
> intervals are possible.
Countably many clusters of intervals are quite enough if between any two
> On 25 Jul., 00:27, Virgil <vir...@ligriv.com> wrote:
>
> > > > If countably many rational points can "separate" uncountably many
> > > > irrational points,
> >
> > > Of course they cannot.
> >
> > They can, and do, in standard mathematics, i.e., between any two of
> > uncountably many irrationals there is at lest one of countably many
> > rationals. At least in the standard model of the real field.
>
> Have you any doubt in my theorem?
> If not, then there is the proof that not more than countably many
> intervals are possible.
clusters there is another.
> >
> >
> >
> > > > so that between any two irrationals there is a
> > > > rational, why cannot countably many open clusters do the same?
> >
> > > It is claimed by some or most matheologians, that countably many
> > > rationals could separate uncountably many irrationals. But in my
> > > example I start with countably many intervals which during the proof
> > > do not increase in number.
Which in no way refutes that "that countably many rationals could
> >
> >
> > > > so that between any two irrationals there is a
> > > > rational, why cannot countably many open clusters do the same?
> >
> > > It is claimed by some or most matheologians, that countably many
> > > rationals could separate uncountably many irrationals. But in my
> > > example I start with countably many intervals which during the proof
> > > do not increase in number.
separate uncountably many irrationals" in the sense that between any two
irrationals there is a rational (in fact, countably many rationals).
> >
> > Since you start with one interval for each rational and one rational
> > determining each interval, and end with countably many clusters of
> > intervals, why cannot the clusters do what the rationals do, namely be
> > "dense" in the sense that between any two clusters lies at least one
> > other cluster?
> >
> > To CLAIM that this is impossible, as WM does, is NOT a proof that it
> > cannot happen.
>
> I do not claim anything. I have proved that in the final state the
> endpoints of (1) and (2) coincide and that there are only countably
> many intervals.
Which does not eliminate the possibility that between any two clusters
> > Since you start with one interval for each rational and one rational
> > determining each interval, and end with countably many clusters of
> > intervals, why cannot the clusters do what the rationals do, namely be
> > "dense" in the sense that between any two clusters lies at least one
> > other cluster?
> >
> > To CLAIM that this is impossible, as WM does, is NOT a proof that it
> > cannot happen.
>
> I do not claim anything. I have proved that in the final state the
> endpoints of (1) and (2) coincide and that there are only countably
> many intervals.
there is another cluster, just like between any two rationals there is
another rational.
And if this IS the case, then your claims all fail.
--
WM
Jul 25, 2012, 3:43:57 PM7/25/12
to
On 25 Jul., 19:48, Virgil <vir...@ligriv.com> wrote:
> > Have you any doubt in my theorem?
> > If not, then there is the proof that not more than countably many
> > intervals are possible.
>
> Countably many clusters of intervals are quite enough if between any two
> clusters there is another.
My theorem guarantees that a countable set of intervals A_k and B_j
> > Have you any doubt in my theorem?
> > If not, then there is the proof that not more than countably many
> > intervals are possible.
>
> Countably many clusters of intervals are quite enough if between any two
> clusters there is another.
covers the complete unit interval. After the ants have made all
endpoints coincide, some if these A_k and B_j are coveríng the I_n.
The remaining intervals cover the complement. By my proof these
intervals are limited in cardinality to a countable set.
>
>
> Which in no way refutes that "that countably many rationals could
> separate uncountably many irrationals" in the sense that between any two
> irrationals there is a rational (in fact, countably many rationals).
Not in my Gedankenexperiment. There the intervals A_k and B_j are and
>
> Which in no way refutes that "that countably many rationals could
> separate uncountably many irrationals" in the sense that between any two
> irrationals there is a rational (in fact, countably many rationals).
remain alternating.
>
>
>
> > I have proved that in the final state the
> > endpoints of (1) and (2) coincide and that there are only countably
> > many intervals.
>
> Which does not eliminate the possibility that between any two clusters
> there is another cluster, just like between any two rationals there is
> another rational.
There are no clusters.
>
>
> > I have proved that in the final state the
> > endpoints of (1) and (2) coincide and that there are only countably
> > many intervals.
>
> Which does not eliminate the possibility that between any two clusters
> there is another cluster, just like between any two rationals there is
> another rational.
>
> And if this IS the case, then your claims all fail.
But there are no clusters but only a countable set of alternating
> And if this IS the case, then your claims all fail.
intervals.
Regards, WM
Virgil
Jul 25, 2012, 4:28:11 PM7/25/12
to
In article
<d21b4e8f-4bd8-456d...@m7g2000vbc.googlegroups.com>,
[0,1] in an interval of positive length with no two of those intervals
having any points in common.?
If any two or more of them have a point in common then their union is
also an interval, and will union with other intervals having points in
common with one of them until no other interval has any points in common
with any interval in that connected union of intervals.
Such a connected union of intervals I shall call a cluster.
Every rational point must be a member of some cluster.
interval of strictly positive length without any two such intervals
having any elements in common?
The original intervals with a different rational as the midpoint of each
one of them, cannot be all disjoint because each contains infinitely
many rationals, thus clusters of those intervals must exist, being
maximal connected unions of such intervals, and being open intervals.
And in order for WM's claims about his of his covering of the rationals
to be valid, it is necessary, but probably not sufficient, for him to
prove that it is impossible that for every two such clusters there is
another between them.
--
<d21b4e8f-4bd8-456d...@m7g2000vbc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Jul., 19:48, Virgil <vir...@ligriv.com> wrote:
>
> > > Have you any doubt in my theorem?
> > > If not, then there is the proof that not more than countably many
> > > intervals are possible.
> >
> > Countably many clusters of intervals are quite enough if between any two
> > clusters there is another.
>
> My theorem guarantees that a countable set of intervals A_k and B_j
> covers the complete unit interval. After the ants have made all
> endpoints coincide, some if these A_k and B_j are cover�ng the I_n.
> On 25 Jul., 19:48, Virgil <vir...@ligriv.com> wrote:
>
> > > Have you any doubt in my theorem?
> > > If not, then there is the proof that not more than countably many
> > > intervals are possible.
> >
> > Countably many clusters of intervals are quite enough if between any two
> > clusters there is another.
>
> My theorem guarantees that a countable set of intervals A_k and B_j
> covers the complete unit interval. After the ants have made all
> The remaining intervals cover the complement. By my proof these
> intervals are limited in cardinality to a countable set.
Claimed, but not actually proven.
> intervals are limited in cardinality to a countable set.
> >
> >
> > Which in no way refutes that "that countably many rationals could
> > separate uncountably many irrationals" in the sense that between any two
> > irrationals there is a rational (in fact, countably many rationals).
>
> Not in my Gedankenexperiment. There the intervals A_k and B_j are and
> remain alternating.
Proof? I see none!
> >
> > Which in no way refutes that "that countably many rationals could
> > separate uncountably many irrationals" in the sense that between any two
> > irrationals there is a rational (in fact, countably many rationals).
>
> Not in my Gedankenexperiment. There the intervals A_k and B_j are and
> remain alternating.
> >
> >
> >
> > > I have proved that in the final state the
> > > endpoints of (1) and (2) coincide and that there are only countably
> > > many intervals.
> >
> > Which does not eliminate the possibility that between any two clusters
> > there is another cluster, just like between any two rationals there is
> > another rational.
>
> There are no clusters.
Then you are claiming that you have enclosed each and every rational in
> >
> >
> > > I have proved that in the final state the
> > > endpoints of (1) and (2) coincide and that there are only countably
> > > many intervals.
> >
> > Which does not eliminate the possibility that between any two clusters
> > there is another cluster, just like between any two rationals there is
> > another rational.
>
> There are no clusters.
[0,1] in an interval of positive length with no two of those intervals
having any points in common.?
If any two or more of them have a point in common then their union is
also an interval, and will union with other intervals having points in
common with one of them until no other interval has any points in common
with any interval in that connected union of intervals.
Such a connected union of intervals I shall call a cluster.
Every rational point must be a member of some cluster.
> >
> > And if this IS the case, then your claims all fail.
>
> But there are no clusters but only a countable set of alternating
> intervals.
Does WM claims that one can cover each rational in [0,1] with an
> > And if this IS the case, then your claims all fail.
>
> But there are no clusters but only a countable set of alternating
> intervals.
interval of strictly positive length without any two such intervals
having any elements in common?
The original intervals with a different rational as the midpoint of each
one of them, cannot be all disjoint because each contains infinitely
many rationals, thus clusters of those intervals must exist, being
maximal connected unions of such intervals, and being open intervals.
And in order for WM's claims about his of his covering of the rationals
to be valid, it is necessary, but probably not sufficient, for him to
prove that it is impossible that for every two such clusters there is
another between them.
--
WM
Jul 26, 2012, 3:54:17 AM7/26/12
to
> <d21b4e8f-4bd8-456d-b4a6-f91acb5bb...@m7g2000vbc.googlegroups.com>,
...===1___2=====3__4==5___________6...
Let the endpoint 4 move left:
...===1___2=====3__4==5___________6...
...===1___2=====3_4===5___________6...
...===1___2=====34====5___________6...
...===1___2===4_3=====5___________6...
...===1___2=4____3=====5___________6...
...===1_4=2______3=====5___________6...
The intervals remain alrenating and their number is constant.
> > There are no clusters.
>
> Then you are claiming that you have enclosed each and every rational in
> [0,1] in an interval of positive length with no two of those intervals
> having any points in common.?
No. Of course there are intervals in intervals or overlapping
intervals. But they have no meaning other then existing there. They
don't produce bew intervals or new numbers or "limits".
complemment cannot increase but on,y decrease. Because between two
overlapping intervals, there is no space for an uncovered irrational
number.
> Such a connected union of intervals I shall call a cluster.
> Every rational point must be a member of some cluster.
Maybe. No problem. Clusters do not increase the number of possible
places for uncovered irrationals.
Regards, WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 25 Jul., 19:48, Virgil <vir...@ligriv.com> wrote:
>
> > > > Have you any doubt in my theorem?
> > > > If not, then there is the proof that not more than countably many
> > > > intervals are possible.
>
> > > Countably many clusters of intervals are quite enough if between any two
> > > clusters there is another.
>
> > My theorem guarantees that a countable set of intervals A_k and B_j
> > covers the complete unit interval. After the ants have made all
> > endpoints coincide, some if these A_k and B_j are coveríng the I_n.
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 25 Jul., 19:48, Virgil <vir...@ligriv.com> wrote:
>
> > > > Have you any doubt in my theorem?
> > > > If not, then there is the proof that not more than countably many
> > > > intervals are possible.
>
> > > Countably many clusters of intervals are quite enough if between any two
> > > clusters there is another.
>
> > My theorem guarantees that a countable set of intervals A_k and B_j
> > covers the complete unit interval. After the ants have made all
> > The remaining intervals cover the complement. By my proof these
> > intervals are limited in cardinality to a countable set.
>
> Claimed, but not actually proven.
>
>
>
> > > Which in no way refutes that "that countably many rationals could
> > > separate uncountably many irrationals" in the sense that between any two
> > > irrationals there is a rational (in fact, countably many rationals).
>
> > Not in my Gedankenexperiment. There the intervals A_k and B_j are and
> > remain alternating.
>
> Proof? I see none!
Let exist intervals A ====, B ____ with endpoints n:
> > intervals are limited in cardinality to a countable set.
>
> Claimed, but not actually proven.
>
>
>
> > > Which in no way refutes that "that countably many rationals could
> > > separate uncountably many irrationals" in the sense that between any two
> > > irrationals there is a rational (in fact, countably many rationals).
>
> > Not in my Gedankenexperiment. There the intervals A_k and B_j are and
> > remain alternating.
>
> Proof? I see none!
...===1___2=====3__4==5___________6...
Let the endpoint 4 move left:
...===1___2=====3__4==5___________6...
...===1___2=====3_4===5___________6...
...===1___2=====34====5___________6...
...===1___2===4_3=====5___________6...
...===1___2=4____3=====5___________6...
...===1_4=2______3=====5___________6...
The intervals remain alrenating and their number is constant.
> > There are no clusters.
>
> Then you are claiming that you have enclosed each and every rational in
> [0,1] in an interval of positive length with no two of those intervals
> having any points in common.?
intervals. But they have no meaning other then existing there. They
don't produce bew intervals or new numbers or "limits".
>
> If any two or more of them have a point in common then their union is
> also an interval,
Of course. But by that proces the number of intervals of the
> If any two or more of them have a point in common then their union is
> also an interval,
complemment cannot increase but on,y decrease. Because between two
overlapping intervals, there is no space for an uncovered irrational
number.
> Such a connected union of intervals I shall call a cluster.
> Every rational point must be a member of some cluster.
places for uncovered irrationals.
Regards, WM
Virgil
Jul 26, 2012, 4:31:13 PM7/26/12
to
In article
<0aa280e2-5a1e-4a81...@z19g2000vbe.googlegroups.com>,
You mean that your overlapping intervals do not have unions which are
also intervals? What a strange world you live in.
unioning them when they overlap each other does not necessarily decrease
the 'number' of intervals, which can, and must, remain equally infinite.
And, unless WM claims to cover all of [0,1] with intervals of cumulative
length 1/9, such unions must leave at least 8/9 of the interval [0,1]
uncovered.
line, is dense, WM's claims all go down the tubes. We can then replace
each cluster by one of its irrational points and those, together with
the uncovered irrational points becomes an ordered set like a real
interval of length >= 8/9 (order-isomorphic to such a set).
--
<0aa280e2-5a1e-4a81...@z19g2000vbe.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 25 Jul., 22:28, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <d21b4e8f-4bd8-456d-b4a6-f91acb5bb...@m7g2000vbc.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 25 Jul., 19:48, Virgil <vir...@ligriv.com> wrote:
> >
> > > > > Have you any doubt in my theorem?
> > > > > If not, then there is the proof that not more than countably many
> > > > > intervals are possible.
> >
> > > > Countably many clusters of intervals are quite enough if between any two
> > > > clusters there is another.
> >
> > > My theorem guarantees that a countable set of intervals A_k and B_j
> > > covers the complete unit interval. After the ants have made all
> > > endpoints coincide, some if these A_k and B_j are cover�ng the I_n.
> On 25 Jul., 22:28, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <d21b4e8f-4bd8-456d-b4a6-f91acb5bb...@m7g2000vbc.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 25 Jul., 19:48, Virgil <vir...@ligriv.com> wrote:
> >
> > > > > Have you any doubt in my theorem?
> > > > > If not, then there is the proof that not more than countably many
> > > > > intervals are possible.
> >
> > > > Countably many clusters of intervals are quite enough if between any two
> > > > clusters there is another.
> >
> > > My theorem guarantees that a countable set of intervals A_k and B_j
> > > covers the complete unit interval. After the ants have made all
also intervals? What a strange world you live in.
> >
> > If any two �or more of them have a point in common then their union is
> > also an interval,
>
> Of course. But by that proces the number of intervals of the
> complemment cannot increase but on,y decrease. Because between two
> overlapping intervals, there is no space for an uncovered irrational
> number.
If the number of those intervals, I_n, is infinite, as it is, then
> > If any two �or more of them have a point in common then their union is
> > also an interval,
>
> Of course. But by that proces the number of intervals of the
> complemment cannot increase but on,y decrease. Because between two
> overlapping intervals, there is no space for an uncovered irrational
> number.
unioning them when they overlap each other does not necessarily decrease
the 'number' of intervals, which can, and must, remain equally infinite.
And, unless WM claims to cover all of [0,1] with intervals of cumulative
length 1/9, such unions must leave at least 8/9 of the interval [0,1]
uncovered.
>
> > Such a connected union of intervals I shall call a cluster.
> > Every rational point must be a member of some cluster.
>
> Maybe. No problem. Clusters do not increase the number of possible
> places for uncovered irrationals.
But if the set of such clusters, ordered by their positions on the real
> > Such a connected union of intervals I shall call a cluster.
> > Every rational point must be a member of some cluster.
>
> Maybe. No problem. Clusters do not increase the number of possible
> places for uncovered irrationals.
line, is dense, WM's claims all go down the tubes. We can then replace
each cluster by one of its irrational points and those, together with
the uncovered irrational points becomes an ordered set like a real
interval of length >= 8/9 (order-isomorphic to such a set).
--
WM
Aug 6, 2012, 7:19:33 AM8/6/12
to
> > > > My theorem guarantees that a countable set of intervals A_k and B_j
> > > > covers the complete unit interval. After the ants have made all
> > > > endpoints coincide, some if these A_k and B_j are cover ng the I_n.
> > > > The remaining intervals cover the complement. By my proof these
> > > > intervals are limited in cardinality to a countable set.
>
> > > Claimed, but not actually proven.
>
> > > > > Which in no way refutes that "that countably many rationals could
> > > > > separate uncountably many irrationals" in the sense that between any two
> > > > > irrationals there is a rational (in fact, countably many rationals).
>
> > > > Not in my Gedankenexperiment. There the intervals A_k and B_j are and
> > > > remain alternating.
>
> > > Proof? I see none!
>
> > Let exist intervals A ====, B ____ with endpoints n:
>
> > ...===1___2=====3__4==5___________6...
>
> > Let the endpoint 4 move left:
> > ...===1___2=====3__4==5___________6...
> > ...===1___2=====3_4===5___________6...
> > ...===1___2=====34====5___________6...
> > ...===1___2===4_3=====5___________6...
> > ...===1___2=4____3=====5___________6...
> > ...===1_4=2______3=====5___________6...
>
> > The intervals remain alternating and their number is constant.
> > > > covers the complete unit interval. After the ants have made all
> > > > endpoints coincide, some if these A_k and B_j are cover ng the I_n.
> > > > The remaining intervals cover the complement. By my proof these
> > > > intervals are limited in cardinality to a countable set.
>
> > > Claimed, but not actually proven.
>
> > > > > Which in no way refutes that "that countably many rationals could
> > > > > separate uncountably many irrationals" in the sense that between any two
> > > > > irrationals there is a rational (in fact, countably many rationals).
>
> > > > Not in my Gedankenexperiment. There the intervals A_k and B_j are and
> > > > remain alternating.
>
> > > Proof? I see none!
>
> > Let exist intervals A ====, B ____ with endpoints n:
>
> > ...===1___2=====3__4==5___________6...
>
> > Let the endpoint 4 move left:
> > ...===1___2=====3__4==5___________6...
> > ...===1___2=====3_4===5___________6...
> > ...===1___2=====34====5___________6...
> > ...===1___2===4_3=====5___________6...
> > ...===1___2=4____3=====5___________6...
> > ...===1_4=2______3=====5___________6...
>
Do you understand and accept that?
>
> > > > There are no clusters.
>
> > > Then you are claiming that you have enclosed each and every rational in
> > > [0,1] in an interval of positive length with no two of those intervals
> > > having any points in common.?
>
> > No. Of course there are intervals in intervals or overlapping
> > intervals. But they have no meaning other then existing there. They
> > don't produce new intervals or new numbers or "limits".
> > > > There are no clusters.
>
> > > Then you are claiming that you have enclosed each and every rational in
> > > [0,1] in an interval of positive length with no two of those intervals
> > > having any points in common.?
>
> > No. Of course there are intervals in intervals or overlapping
> > intervals. But they have no meaning other then existing there. They
>
You misunderstand. There may be many overlapping intervals forming a
unions. But if two or more intervals overlap, then the gaps (of the
complement) between them disappear. That means the number of intervals
in the complement cannot increase.
> What a strange world you live in.
>
>
>
> > > If any two or more of them have a point in common then their union is
> > > also an interval,
>
> > Of course. But by that process the number of intervals of the
>
>
> > > If any two or more of them have a point in common then their union is
> > > also an interval,
>
> > complement cannot increase but only decrease. Because between two
> > overlapping intervals, there is no space for an uncovered irrational
> > number.
>
> If the number of those intervals, I_n, is infinite, as it is, then
> unioning them when they overlap each other does not necessarily decrease
> the 'number' of intervals,
Who expects that? The number cannot increase. Only that is required
> > number.
>
> If the number of those intervals, I_n, is infinite, as it is, then
> unioning them when they overlap each other does not necessarily decrease
> the 'number' of intervals,
for my proof, because we start with two countable sets. Important is
that we never arrive at an uncountable subset.
> which can, and must, remain equally infinite.
> And, unless WM claims to cover all of [0,1] with intervals of cumulative
> length 1/9, such unions must leave at least 8/9 of the interval [0,1]
> uncovered.
sets of alternating intervals A_k and B_j. As my proof shows, this
situation does never change.
>
>
>
> > > Such a connected union of intervals I shall call a cluster.
> > > Every rational point must be a member of some cluster.
>
> > Maybe. No problem. Clusters do not increase the number of possible
> > places for uncovered irrationals.
>
> But if the set of such clusters, ordered by their positions on the real
> line, is dense, WM's claims all go down the tubes.
There is no if and when. Initially and finally the the complete
>
>
> > > Such a connected union of intervals I shall call a cluster.
> > > Every rational point must be a member of some cluster.
>
> > Maybe. No problem. Clusters do not increase the number of possible
> > places for uncovered irrationals.
>
> But if the set of such clusters, ordered by their positions on the real
> line, is dense, WM's claims all go down the tubes.
interval [0, 1] is covered by the countable sets of alternating
intervals A_k and B_j.
> We can then replace
> each cluster by one of its irrational points and those, together with
> the uncovered irrational points becomes an ordered set like a real
> interval of length >= 8/9 (order-isomorphic to such a set).
an error in my construction or accept that the notion of countability
is nonsense.
Or perhaps there is another lurker who can point out the "first error"
Virgil
Aug 6, 2012, 3:46:27 PM8/6/12
to
In article
<8008c8fa-b772-4cd9...@d8g2000yqj.googlegroups.com>,
If countably many rationals can have an uncountable relative compliment
in [0,1], the why cannot your countably many intervals also have an
uncountable relative compliment?
Since the Cantor set does it, you must PROVE rather than merely assume,
as you have been doing, that your construction cannot do the same.
>
> > What a strange world you live in.
>
> Strange is only that you do not understand such an easy matter.
When I see that something like the Cantor set can have countably many
open intervals in [0,1] and still have an uncountable relative
compliment, I do not trust your construction to act differently without
seeing a proof that it acts as differently.
So far WM has provided no such proof, so that I declare that his claimed
result is not yet proven,
countable sets from them without ever proving that what is not included
within his countable sets is also countable.
Since the Cantor set shows that one can remove countably many intervals
from [0,1] and still leave uncountably many points in place, WM is
obligated to prove that that cannot happen in his case.
Wm has yet to prove any such thing.
Therefore his claims are NOT established as having to occuranywhere
outsinde of his WMytheology.
rational. Where did those A_k and B_j come from?
original I_n's as "clusters", then there is nothing to show that they
are not densely ordered, that between any two of them there could be
another.
In which case, just like in the Cantor set, the compliment may be
uncountable.
for which the Cantor set is a counter-example, Thus your claims have not
yet been established
consecutively rather than densely ordered, i.e., that for every cluster,
except possible end clusters, there was a next larger and a next smaller
cluster, but this is obviously nonsense as it would limit things to
finitely many clusters.
What you never have been able to disprove is the possibility that the
set of clusters is densely ordered, so that between any two clusters
there are other clusters, which, like with the densely ordered rationals
would leave space for uncountably many uncovered irrationals.
Once again WM makes claims but fails to provide satisfactory proofs of
them. At least not satisfactory outside of his WMytheology!
--
<8008c8fa-b772-4cd9...@d8g2000yqj.googlegroups.com>,
in [0,1], the why cannot your countably many intervals also have an
uncountable relative compliment?
Since the Cantor set does it, you must PROVE rather than merely assume,
as you have been doing, that your construction cannot do the same.
>
> > What a strange world you live in.
>
> Strange is only that you do not understand such an easy matter.
open intervals in [0,1] and still have an uncountable relative
compliment, I do not trust your construction to act differently without
seeing a proof that it acts as differently.
So far WM has provided no such proof, so that I declare that his claimed
result is not yet proven,
> >
> >
> >
> > > > If any two or more of them have a point in common then their union is
> > > > also an interval,
> >
> > > Of course. But by that process the number of intervals of the
> > > complement cannot increase but only decrease. Because between two
> > > overlapping intervals, there is no space for an uncovered irrational
> > > number.
> >
> > If the number of those intervals, I_n, �is infinite, as it is, then
> > unioning them when they overlap each other does not necessarily decrease
> > the 'number' of intervals,
>
> Who expects that? The number cannot increase. Only that is required
> for my proof, because we start with two countable sets. Important is
> that we never arrive at an uncountable subset.
WM starts with one countable set, the rationals, and creates other
> >
> >
> > > > If any two or more of them have a point in common then their union is
> > > > also an interval,
> >
> > > Of course. But by that process the number of intervals of the
> > > complement cannot increase but only decrease. Because between two
> > > overlapping intervals, there is no space for an uncovered irrational
> > > number.
> >
> > If the number of those intervals, I_n, �is infinite, as it is, then
> > unioning them when they overlap each other does not necessarily decrease
> > the 'number' of intervals,
>
> Who expects that? The number cannot increase. Only that is required
> for my proof, because we start with two countable sets. Important is
> that we never arrive at an uncountable subset.
countable sets from them without ever proving that what is not included
within his countable sets is also countable.
Since the Cantor set shows that one can remove countably many intervals
from [0,1] and still leave uncountably many points in place, WM is
obligated to prove that that cannot happen in his case.
Wm has yet to prove any such thing.
Therefore his claims are NOT established as having to occuranywhere
outsinde of his WMytheology.
>
>
> > which can, and must, remain equally infinite.
> > And, unless WM claims to cover all of [0,1] with intervals of cumulative
> > length 1/9, such unions must leave at least 8/9 of the interval [0,1]
> > uncovered.
>
> Initially the complete interval [0, 1] is covered by the countable
> sets of alternating intervals A_k and B_j. As my proof shows, this
> situation does never change.
Originally WM only had intervals I_n, of length 10^(-n), one for each
>
> > which can, and must, remain equally infinite.
> > And, unless WM claims to cover all of [0,1] with intervals of cumulative
> > length 1/9, such unions must leave at least 8/9 of the interval [0,1]
> > uncovered.
>
> Initially the complete interval [0, 1] is covered by the countable
> sets of alternating intervals A_k and B_j. As my proof shows, this
> situation does never change.
rational. Where did those A_k and B_j come from?
> >
> >
> >
> > > > Such a connected union of intervals I shall call a cluster.
> > > > Every rational point must be a member of some cluster.
> >
> > > Maybe. No problem. Clusters do not increase the number of possible
> > > places for uncovered irrationals.
> >
> > But if the set of such clusters, ordered by their positions on the real
> > line, is dense, WM's claims all go down the tubes.
>
> There is no if and when. Initially and finally the the complete
> interval [0, 1] is covered by the countable sets of alternating
> intervals A_k and B_j.
Where do these come from? If one set, say the A_k's is derived from the
> >
> >
> > > > Such a connected union of intervals I shall call a cluster.
> > > > Every rational point must be a member of some cluster.
> >
> > > Maybe. No problem. Clusters do not increase the number of possible
> > > places for uncovered irrationals.
> >
> > But if the set of such clusters, ordered by their positions on the real
> > line, is dense, WM's claims all go down the tubes.
>
> There is no if and when. Initially and finally the the complete
> interval [0, 1] is covered by the countable sets of alternating
> intervals A_k and B_j.
original I_n's as "clusters", then there is nothing to show that they
are not densely ordered, that between any two of them there could be
another.
In which case, just like in the Cantor set, the compliment may be
uncountable.
>
> > We can then replace
> > each cluster by one of its irrational points and those, together with
> > the uncovered irrational points becomes an ordered set like a real
> > interval of length >= 8/9 (order-isomorphic to such a set).
>
> If there is a contradictory result, you either should be able to show
> an error in my construction or accept that the notion of countability
> is nonsense.
I show that you have not proved you claims and that you have made claims
> > We can then replace
> > each cluster by one of its irrational points and those, together with
> > the uncovered irrational points becomes an ordered set like a real
> > interval of length >= 8/9 (order-isomorphic to such a set).
>
> If there is a contradictory result, you either should be able to show
> an error in my construction or accept that the notion of countability
> is nonsense.
for which the Cantor set is a counter-example, Thus your claims have not
yet been established
>
> Or perhaps there is another lurker who can point out the "first error"
> in my proof.
In your original "proof", you claimed that your set of clusters was
> Or perhaps there is another lurker who can point out the "first error"
> in my proof.
consecutively rather than densely ordered, i.e., that for every cluster,
except possible end clusters, there was a next larger and a next smaller
cluster, but this is obviously nonsense as it would limit things to
finitely many clusters.
What you never have been able to disprove is the possibility that the
set of clusters is densely ordered, so that between any two clusters
there are other clusters, which, like with the densely ordered rationals
would leave space for uncountably many uncovered irrationals.
Once again WM makes claims but fails to provide satisfactory proofs of
them. At least not satisfactory outside of his WMytheology!
--
WM
Aug 7, 2012, 4:06:36 PM8/7/12
to
On 6 Aug., 21:46, Virgil <vir...@ligriv.com> wrote:
> > > > > Proof? I see none!
>
> > > > Let exist intervals A ====, B ____ with endpoints n:
>
> > > > ...===1___2=====3__4==5___________6...
>
> > > > Let the endpoint 4 move left:
> > > > ...===1___2=====3__4==5___________6...
> > > > ...===1___2=====3_4===5___________6...
> > > > ...===1___2=====34====5___________6...
> > > > ...===1___2===4_3=====5___________6...
> > > > ...===1___2=4____3=====5___________6...
> > > > ...===1_4=2______3=====5___________6...
>
> > > > The intervals remain alternating and their number is constant.
>
> > Do you understand and accept that?
Do you understand and accept that?
> > > > > Proof? I see none!
>
> > > > Let exist intervals A ====, B ____ with endpoints n:
>
> > > > ...===1___2=====3__4==5___________6...
>
> > > > Let the endpoint 4 move left:
> > > > ...===1___2=====3__4==5___________6...
> > > > ...===1___2=====3_4===5___________6...
> > > > ...===1___2=====34====5___________6...
> > > > ...===1___2===4_3=====5___________6...
> > > > ...===1___2=4____3=====5___________6...
> > > > ...===1_4=2______3=====5___________6...
>
> > > > The intervals remain alternating and their number is constant.
>
> > Do you understand and accept that?
Do you understand and accept that?
> If countably many rationals can have an uncountable relative compliment
> in [0,1], the why cannot your countably many intervals also have an
> uncountable relative compliment?
>
> Since the Cantor set does it, you must PROVE rather than merely assume,
> as you have been doing, that your construction cannot do the same.
Do you understand that the number of intervals A_k and B_j cannot
> in [0,1], the why cannot your countably many intervals also have an
> uncountable relative compliment?
>
> Since the Cantor set does it, you must PROVE rather than merely assume,
> as you have been doing, that your construction cannot do the same.
increase during the process of moving endpoints?
Do you understand that the endpoints can move such that they coincide
with the endpoints of the intervals I_n?
> > Initially the complete interval [0, 1] is covered by the countable
> > sets of alternating intervals A_k and B_j. As my proof shows, this
> > situation does never change.
>
> Originally WM only had intervals I_n, of length 10^(-n), one for each
> rational. Where did those A_k and B_j come from?
Define a sequence of points p_n in the unit interval: p_n = 1/n.
These points define intervals
A_k = [1/n, 1/(n+1)] for odd n
and
B_j = [1/n, 1/(n+1)] for even n.
Regards, WM
These points define intervals
A_k = [1/n, 1/(n+1)] for odd n
and
B_j = [1/n, 1/(n+1)] for even n.
SPQR
Aug 7, 2012, 6:27:54 PM8/7/12
to
In article
<c1a382ae-e4a5-4f5f...@f2g2000vbm.googlegroups.com>,
intervals alone or the b===c intervals alone will cover all rationals?
If so, you are wrong, at least unless in one set the sum lengths is 0
and the other the sum of lengths is 1.
> >
> > > Do you understand and accept that?
>
> Do you understand and accept that?
Since you are then changing the lengths of your interval, they may well
end up having cumulative lengths equal to or greater than 1.
accumulation other than 0 and 1, they cannot in any way be relevant to
having an interval centered at each rational.
It is not at all clear to me what those ...===1___2=====3__4==5____6
intervals are supposed to demonstrate, other than WM's incompetence.
> Do you understand that the endpoints can move such that they coincide
> with the endpoints of the intervals I_n?
No! At least not without reversing the original order of infinitely many
consecutive pairs of endpoints.
their endpoints are moved, as long as their original order is preserved.
Do you really mean something like
A_k = [1/k, 1/(k+1)] for odd k
and
B_j = [1/j, 1/(j+1)] for even
And if so the sum of lengths of all A_j and B_k intervals will be 1, as
long as their union is [0,1[.
Thus WM again shows us the foolishness that living in WMytheolgy induces.
<c1a382ae-e4a5-4f5f...@f2g2000vbm.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 6 Aug., 21:46, Virgil <vir...@ligriv.com> wrote:
>
> > > > > > Proof? I see none!
> >
> > > > > Let exist intervals A ====, B ____ with endpoints n:
> >
> > > > > ...===1___2=====3__4==5___________6...
> >
> > > > > Let the endpoint 4 move left:
> > > > > ...===1___2=====3__4==5___________6...
> > > > > ...===1___2=====3_4===5___________6...
> > > > > ...===1___2=====34====5___________6...
> > > > > ...===1___2===4_3=====5___________6...
> > > > > ...===1___2=4____3=====5___________6...
> > > > > ...===1_4=2______3=====5___________6...
> >
> > > > > The intervals remain alternating and their number is constant.
Do you claim that with such alternating intervals either the a___b
> On 6 Aug., 21:46, Virgil <vir...@ligriv.com> wrote:
>
> > > > > > Proof? I see none!
> >
> > > > > Let exist intervals A ====, B ____ with endpoints n:
> >
> > > > > ...===1___2=====3__4==5___________6...
> >
> > > > > Let the endpoint 4 move left:
> > > > > ...===1___2=====3__4==5___________6...
> > > > > ...===1___2=====3_4===5___________6...
> > > > > ...===1___2=====34====5___________6...
> > > > > ...===1___2===4_3=====5___________6...
> > > > > ...===1___2=4____3=====5___________6...
> > > > > ...===1_4=2______3=====5___________6...
> >
> > > > > The intervals remain alternating and their number is constant.
intervals alone or the b===c intervals alone will cover all rationals?
If so, you are wrong, at least unless in one set the sum lengths is 0
and the other the sum of lengths is 1.
> >
> > > Do you understand and accept that?
>
> Do you understand and accept that?
end up having cumulative lengths equal to or greater than 1.
>
> > If countably many rationals can have an uncountable relative compliment
> > in [0,1], the why cannot your countably many intervals also have an
> > uncountable relative compliment?
> >
> > Since the Cantor set does it, you must PROVE rather than merely assume,
> > as you have been doing, that your construction cannot do the same.
>
> Do you understand that the number of intervals A_k and B_j cannot
> increase during the process of moving endpoints?
Since the A_k and B_j endpoints collectively can have no points of
> > If countably many rationals can have an uncountable relative compliment
> > in [0,1], the why cannot your countably many intervals also have an
> > uncountable relative compliment?
> >
> > Since the Cantor set does it, you must PROVE rather than merely assume,
> > as you have been doing, that your construction cannot do the same.
>
> Do you understand that the number of intervals A_k and B_j cannot
> increase during the process of moving endpoints?
accumulation other than 0 and 1, they cannot in any way be relevant to
having an interval centered at each rational.
It is not at all clear to me what those ...===1___2=====3__4==5____6
intervals are supposed to demonstrate, other than WM's incompetence.
> Do you understand that the endpoints can move such that they coincide
> with the endpoints of the intervals I_n?
consecutive pairs of endpoints.
>
>
> > > Initially the complete interval [0, 1] is covered by the countable
> > > sets of alternating intervals A_k and B_j. As my proof shows, this
> > > situation does never change.
Then the sum of lengths of those intervals will equal 1 no matter how
>
> > > Initially the complete interval [0, 1] is covered by the countable
> > > sets of alternating intervals A_k and B_j. As my proof shows, this
> > > situation does never change.
their endpoints are moved, as long as their original order is preserved.
> >
> > Originally WM only had intervals I_n, of length 10^(-n), one for each
> > rational. Where did those A_k and B_j come from?
>
> They were generated independently:
> Define a sequence of points p_n in the unit interval: p_n = 1/n.
> These points define intervals
> A_k = [1/n, 1/(n+1)] for odd n
> and
> B_j = [1/n, 1/(n+1)] for even n.
How are j and k supposed to be related to n?
> > Originally WM only had intervals I_n, of length 10^(-n), one for each
> > rational. Where did those A_k and B_j come from?
>
> They were generated independently:
> Define a sequence of points p_n in the unit interval: p_n = 1/n.
> These points define intervals
> A_k = [1/n, 1/(n+1)] for odd n
> and
> B_j = [1/n, 1/(n+1)] for even n.
Do you really mean something like
A_k = [1/k, 1/(k+1)] for odd k
and
B_j = [1/j, 1/(j+1)] for even
And if so the sum of lengths of all A_j and B_k intervals will be 1, as
long as their union is [0,1[.
Thus WM again shows us the foolishness that living in WMytheolgy induces.
WM
Aug 8, 2012, 9:05:46 AM8/8/12
to
On 8 Aug., 00:27, SPQR <S...@SPQR.org> wrote:
> > > > > > Let exist intervals A ====, B ____ with endpoints n:
>
> > > > > > ...===1___2=====3__4==5___________6...
>
> > > > > > Let the endpoint 4 move left:
> > > > > > ...===1___2=====3__4==5___________6...
> > > > > > ...===1___2=====3_4===5___________6...
> > > > > > ...===1___2=====34====5___________6...
> > > > > > ...===1___2===4_3=====5___________6...
> > > > > > ...===1___2=4____3=====5___________6...
> > > > > > ...===1_4=2______3=====5___________6...
>
> > > > > > The intervals remain alternating and their number is constant.
>
> Do you claim that with such alternating intervals either the a___b
> intervals alone or the b===c intervals alone will cover all rationals?
No. A rational number will be covered by an interval A_k === or by an
> > > > > > Let exist intervals A ====, B ____ with endpoints n:
>
> > > > > > ...===1___2=====3__4==5___________6...
>
> > > > > > Let the endpoint 4 move left:
> > > > > > ...===1___2=====3__4==5___________6...
> > > > > > ...===1___2=====3_4===5___________6...
> > > > > > ...===1___2=====34====5___________6...
> > > > > > ...===1___2===4_3=====5___________6...
> > > > > > ...===1___2=4____3=====5___________6...
> > > > > > ...===1_4=2______3=====5___________6...
>
> > > > > > The intervals remain alternating and their number is constant.
>
> Do you claim that with such alternating intervals either the a___b
> intervals alone or the b===c intervals alone will cover all rationals?
interval B_j or, if the rational number is just an endpoint, by both
adjacent intervals, because all intervals are closed and an endpoint
between two adjacent intervals belongs to both.
>
> If so, you are wrong, at least unless in one set the sum lengths is 0
> and the other the sum of lengths is 1.
indicated in the upper line by ~~~, the intervals A_k by ==== and B_j
by ___
...a b~~~~~~~~~c~~~~~~~~~d~~~~~~e f~~~g~~~h i~~~j k~~~l
> Since you are then changing the lengths of your interval, they may well
> end up having cumulative lengths equal to or greater than 1.
>
neighbour grows, The intervals remain alternating, their number
remains the same and the measure of their union remains the same,
namely length 1.
> > Do you understand that the endpoints can move such that they coincide
> > with the endpoints of the intervals I_n?
>
> No! At least not without reversing the original order of infinitely many
> consecutive pairs of endpoints.
points when moving in an arbitrary way will conicide with aleph_0
fixed points.
>
>
>
> > > > Initially the complete interval [0, 1] is covered by the countable
> > > > sets of alternating intervals A_k and B_j. As my proof shows, this
> > > > situation does never change.
>
> Then the sum of lengths of those intervals will equal 1 no matter how
> their endpoints are moved, as long as their original order is preserved.
Of course.
>
>
> > > > Initially the complete interval [0, 1] is covered by the countable
> > > > sets of alternating intervals A_k and B_j. As my proof shows, this
> > > > situation does never change.
>
> Then the sum of lengths of those intervals will equal 1 no matter how
> their endpoints are moved, as long as their original order is preserved.
> > > Originally WM only had intervals I_n, of length 10^(-n), one for each
> > > rational. Where did those A_k and B_j come from?
>
> > They were generated independently:
> > Define a sequence of points p_n in the unit interval: p_n = 1/n.
> > These points define intervals
> > A_k = [1/n, 1/(n+1)] for odd n
> > and
> > B_j = [1/n, 1/(n+1)] for even n.
>
> How are j and k supposed to be related to n?
>
j = n/2
A_1 = [1/1, 1/2]
B_1 = [1/2, 1/3]
A_2 = [1/3, 1/4]
B_2 = [1/4, 1/5]
...
Regards, WM
Virgil
Aug 8, 2012, 2:31:28 PM8/8/12
to
In article
<55d149a5-d688-4cfd...@a9g2000vbn.googlegroups.com>,
I your do not see that your new set of intervals in any way resolves the
problems with your original I_n: that the set of clusters may be such
that between any two clusters there are other clusters, and thus the set
of clusters regarded as rational points along with the set of uncovered
irrational points is ordered like a real interval of length of at least
8/9 that of [0,1].
In which case, your construction has proven nothing that you claimed.
--
<55d149a5-d688-4cfd...@a9g2000vbn.googlegroups.com>,
problems with your original I_n: that the set of clusters may be such
that between any two clusters there are other clusters, and thus the set
of clusters regarded as rational points along with the set of uncovered
irrational points is ordered like a real interval of length of at least
8/9 that of [0,1].
In which case, your construction has proven nothing that you claimed.
--
WM
Aug 8, 2012, 4:59:18 PM8/8/12
to
On 8 Aug., 20:31, Virgil <vir...@ligriv.com> wrote:
> I your do not see that your new set of intervals in any way resolves the
> problems with your original I_n: that the set of clusters may be such
> that between any two clusters there are other clusters, and thus the set
> of clusters regarded as rational points along with the set of uncovered
> irrational points is ordered like a real interval of length of at least
> 8/9 that of [0,1].
>
> In which case, your construction has proven nothing that you claimed.
My construction proves that if the endpoints of the A-B-intervals
> I your do not see that your new set of intervals in any way resolves the
> problems with your original I_n: that the set of clusters may be such
> that between any two clusters there are other clusters, and thus the set
> of clusters regarded as rational points along with the set of uncovered
> irrational points is ordered like a real interval of length of at least
> 8/9 that of [0,1].
>
> In which case, your construction has proven nothing that you claimed.
coincide with the endpoints of the I-intervals, then there we have
countably many intervals of I_n and complementary intervals together.
No clusters. Nothing else. the question is only: Is there a mechanism
that prevents the coincidence of endpoints?
Regards, WM
Virgil
Aug 9, 2012, 12:40:48 AM8/9/12
to
In article
<fe61474c-a697-4745...@y1g2000vbx.googlegroups.com>,
clusters regardless. How you move around your A-B-intervals, will not
change that.
> Nothing else. the question is only: Is there a mechanism
> that prevents the coincidence of endpoints?
If you allow the axiom of choice, the endpoints of the A-B-intervals can
be rearranged almost any way one likes, but that does not destroy the
order properties of the original I_n construction, it merely creates a
new but unrelated ordering.
What you would have to prove is that essentially all of the relevant
order properties of the original I_n structure would have to be
preserved by your massive reordering, which you have not done, and
cannot do.
--
<fe61474c-a697-4745...@y1g2000vbx.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 8 Aug., 20:31, Virgil <vir...@ligriv.com> wrote:
>
> > I your do not see that your new set of intervals in any way resolves the
> > problems with your original I_n: that the set of clusters may be such
> > that between any two clusters there are other clusters, and thus the set
> > of clusters regarded as rational points along with the set of uncovered
> > irrational points is ordered like a real interval of length of at least
> > 8/9 that of [0,1].
> >
> > In which case, your construction has proven nothing that you claimed.
>
> My construction proves that if the endpoints of the A-B-intervals
> coincide with the endpoints of the I-intervals, then there we have
> countably many intervals of I_n and complementary intervals together.
> No clusters.
The clusters of the original I_n intervals will continue to exist as
> On 8 Aug., 20:31, Virgil <vir...@ligriv.com> wrote:
>
> > I your do not see that your new set of intervals in any way resolves the
> > problems with your original I_n: that the set of clusters may be such
> > that between any two clusters there are other clusters, and thus the set
> > of clusters regarded as rational points along with the set of uncovered
> > irrational points is ordered like a real interval of length of at least
> > 8/9 that of [0,1].
> >
> > In which case, your construction has proven nothing that you claimed.
>
> My construction proves that if the endpoints of the A-B-intervals
> coincide with the endpoints of the I-intervals, then there we have
> countably many intervals of I_n and complementary intervals together.
> No clusters.
clusters regardless. How you move around your A-B-intervals, will not
change that.
> Nothing else. the question is only: Is there a mechanism
> that prevents the coincidence of endpoints?
be rearranged almost any way one likes, but that does not destroy the
order properties of the original I_n construction, it merely creates a
new but unrelated ordering.
What you would have to prove is that essentially all of the relevant
order properties of the original I_n structure would have to be
preserved by your massive reordering, which you have not done, and
cannot do.
--
WM
Aug 9, 2012, 6:34:16 AM8/9/12
to
On 9 Aug., 06:40, Virgil <vir...@ligriv.com> wrote:
> The clusters of the original I_n intervals will continue to exist as
> clusters regardless. How you move around your A-B-intervals, will not
> change that.
There are no clusters. There are points on the real axis. This proof
> The clusters of the original I_n intervals will continue to exist as
> clusters regardless. How you move around your A-B-intervals, will not
> change that.
concerns geometry - nothing else.
>
> > Nothing else. the question is only: Is there a mechanism
> > that prevents the coincidence of endpoints?
>
> If you allow the axiom of choice, the endpoints of the A-B-intervals can
> be rearranged almost any way one likes, but that does not destroy the
> order properties of the original I_n construction, it merely creates a
> new but unrelated ordering.
It creates all endpoints of the I_n construction.
> > Nothing else. the question is only: Is there a mechanism
> > that prevents the coincidence of endpoints?
>
> If you allow the axiom of choice, the endpoints of the A-B-intervals can
> be rearranged almost any way one likes, but that does not destroy the
> order properties of the original I_n construction, it merely creates a
> new but unrelated ordering.
>
> What you would have to prove is that essentially all of the relevant
> order properties of the original I_n structure would have to be
> preserved by your massive reordering, which you have not done, and
> cannot do.
All endpoints are there. If there is anything else in the original
> What you would have to prove is that essentially all of the relevant
> order properties of the original I_n structure would have to be
> preserved by your massive reordering, which you have not done, and
> cannot do.
arrangement of the I_n, then it is not explicitly stated and does not
belong to mathematics.
Regards, WM
Virgil
Aug 9, 2012, 3:18:43 PM8/9/12
to
In article
<af5f7e63-6d12-44e1...@z11g2000yqa.googlegroups.com>,
to be pairwise disjoint, which can only happen if no interval of
positive length contains more than one rational number.
> There are points on the real axis. This proof
> concerns geometry - nothing else.
Your argument is far short of being a proof
order to create the I_n construction. There need not be any two of the
I_n intervals having a common endpoint but seclusive intervals in your
A_k-B_j sequence all do, there is no way to make one into the other
without destroying the first's structure.
moved discontinuously is not enough.
> If there is anything else in the original
> arrangement of the I_n, then it is not explicitly stated and does not
> belong to mathematics.
If it is deducible from the original arrangements by mathematicians,
even if not by WM, then it does belong to mathematics, and is binding,
even on non-mathematicians like WM.
The conversion from WM's A_K-B_j arrangement to his I_n arrangement
has not been show capable of being done by only continuous deformations,
so that none of what WM claims for it has been shown to hold.
--
<af5f7e63-6d12-44e1...@z11g2000yqa.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 9 Aug., 06:40, Virgil <vir...@ligriv.com> wrote:
>
> > The clusters of the original I_n intervals will continue to exist as
> > clusters regardless. How you move around your A-B-intervals, will not
> > change that.
>
> There are no clusters.
In order for there not to be any clusters, your I_n intervals would have
> On 9 Aug., 06:40, Virgil <vir...@ligriv.com> wrote:
>
> > The clusters of the original I_n intervals will continue to exist as
> > clusters regardless. How you move around your A-B-intervals, will not
> > change that.
>
> There are no clusters.
to be pairwise disjoint, which can only happen if no interval of
positive length contains more than one rational number.
> There are points on the real axis. This proof
> concerns geometry - nothing else.
> >
> > > Nothing else. the question is only: Is there a mechanism
> > > that prevents the coincidence of endpoints?
> >
> > If you allow the axiom of choice, the endpoints of the A-B-intervals can
> > be rearranged almost any way one likes, but that does not destroy the
> > order properties of the original I_n construction, it merely creates a
> > new but unrelated ordering.
>
> It creates all endpoints of the I_n construction.
But in the process, destroys all properties of the A_k-B_j ordering in
> > > Nothing else. the question is only: Is there a mechanism
> > > that prevents the coincidence of endpoints?
> >
> > If you allow the axiom of choice, the endpoints of the A-B-intervals can
> > be rearranged almost any way one likes, but that does not destroy the
> > order properties of the original I_n construction, it merely creates a
> > new but unrelated ordering.
>
> It creates all endpoints of the I_n construction.
order to create the I_n construction. There need not be any two of the
I_n intervals having a common endpoint but seclusive intervals in your
A_k-B_j sequence all do, there is no way to make one into the other
without destroying the first's structure.
> >
> > What you would have to prove is that essentially �all of the relevant
> > order properties of the original I_n structure would have to be
> > preserved by your massive reordering, which you have not done, and
> > cannot do.
>
> All endpoints are there.
Which is hardly enough. Even all points being there, by having been
> > What you would have to prove is that essentially �all of the relevant
> > order properties of the original I_n structure would have to be
> > preserved by your massive reordering, which you have not done, and
> > cannot do.
>
> All endpoints are there.
moved discontinuously is not enough.
> If there is anything else in the original
> arrangement of the I_n, then it is not explicitly stated and does not
> belong to mathematics.
even if not by WM, then it does belong to mathematics, and is binding,
even on non-mathematicians like WM.
The conversion from WM's A_K-B_j arrangement to his I_n arrangement
has not been show capable of being done by only continuous deformations,
so that none of what WM claims for it has been shown to hold.
--
WM
Aug 9, 2012, 4:53:06 PM8/9/12
to
> <af5f7e63-6d12-44e1-9348-f70051e32...@z11g2000yqa.googlegroups.com>,
sequences of disjoint intervals. If you like to call joint intervals
clusters, we can do so. But that does not change anything about my
proof.
not. I do not see any obstacle. That results in a configuration having
all endpoints and only countably many intervals with and without
rationals.
completely arbitrary way. Nothing with the A-B-intervals changes. Here
you can see it again. Observe point 5:
...
7-6=5-----4=====3--------------------2================================1
...
7-6===5-4=====3--------------------2================================1
...
7-6=====4---5==3--------------------2================================1
...
7-6=====4---------3==5--------------2================================1
...
7-6=====4---------3===========2---------5==========================1
> > All endpoints are there.
>
> Which is hardly enough. Even all points being there, by having been
> moved discontinuously is not enough.
Look at the drawing above.
because I composed it.
Regards, WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 9 Aug., 06:40, Virgil <vir...@ligriv.com> wrote:
>
> > > The clusters of the original I_n intervals will continue to exist as
> > > clusters regardless. How you move around your A-B-intervals, will not
> > > change that.
>
> > There are no clusters.
>
> In order for there not to be any clusters, your I_n intervals would have
> to be pairwise disjoint, which can only happen if no interval of
> positive length contains more than one rational number.
Joint intervals form an interval. Under cluster we understood
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 9 Aug., 06:40, Virgil <vir...@ligriv.com> wrote:
>
> > > The clusters of the original I_n intervals will continue to exist as
> > > clusters regardless. How you move around your A-B-intervals, will not
> > > change that.
>
> > There are no clusters.
>
> In order for there not to be any clusters, your I_n intervals would have
> to be pairwise disjoint, which can only happen if no interval of
> positive length contains more than one rational number.
sequences of disjoint intervals. If you like to call joint intervals
clusters, we can do so. But that does not change anything about my
proof.
>
> > There are points on the real axis. This proof
> > concerns geometry - nothing else.
>
> Your argument is far short of being a proof
The question is only whether a coincidence of endpoints is possible or
> > There are points on the real axis. This proof
> > concerns geometry - nothing else.
>
> Your argument is far short of being a proof
not. I do not see any obstacle. That results in a configuration having
all endpoints and only countably many intervals with and without
rationals.
>
>
> > It creates all endpoints of the I_n construction.
>
> But in the process, destroys all properties of the A_k-B_j ordering in
> order to create the I_n construction. There need not be any two of the
> I_n intervals having a common endpoint but seclusive intervals in your
> A_k-B_j sequence all do, there is no way to make one into the other
> without destroying the first's structure.
You are in error. I showed it. Look it up. The endpoints can move in a
>
> > It creates all endpoints of the I_n construction.
>
> But in the process, destroys all properties of the A_k-B_j ordering in
> order to create the I_n construction. There need not be any two of the
> I_n intervals having a common endpoint but seclusive intervals in your
> A_k-B_j sequence all do, there is no way to make one into the other
> without destroying the first's structure.
completely arbitrary way. Nothing with the A-B-intervals changes. Here
you can see it again. Observe point 5:
...
7-6=5-----4=====3--------------------2================================1
...
7-6===5-4=====3--------------------2================================1
...
7-6=====4---5==3--------------------2================================1
...
7-6=====4---------3==5--------------2================================1
...
7-6=====4---------3===========2---------5==========================1
> > All endpoints are there.
>
> Which is hardly enough. Even all points being there, by having been
> moved discontinuously is not enough.
>
> > If there is anything else in the original
> > arrangement of the I_n, then it is not explicitly stated and does not
> > belong to mathematics.
>
> If it is deducible from the original arrangements by mathematicians,
> even if not by WM, then it does belong to mathematics, and is binding,
> even on non-mathematicians like WM.
There is nothing deductible from the original arrangement. I know it
> > If there is anything else in the original
> > arrangement of the I_n, then it is not explicitly stated and does not
> > belong to mathematics.
>
> If it is deducible from the original arrangements by mathematicians,
> even if not by WM, then it does belong to mathematics, and is binding,
> even on non-mathematicians like WM.
because I composed it.
>
> The conversion from WM's A_K-B_j arrangement to his I_n arrangement
> has not been show capable of being done by only continuous deformations,
> so that none of what WM claims for it has been shown to hold.
If you have not yet understood, then look again at the drawing above.
> The conversion from WM's A_K-B_j arrangement to his I_n arrangement
> has not been show capable of being done by only continuous deformations,
> so that none of what WM claims for it has been shown to hold.
Regards, WM
William Hughes
Aug 9, 2012, 5:19:44 PM8/9/12
to
On Jul 22, 6:51 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> Matheology § 092
>
> My second best proof contradicting set theory
>
> 1)
> Matheology § 092
>
> My second best proof contradicting set theory
>
> 1)
> Define a sequence of points p_n in the unit interval
> p_n = 1/n.
> These points define intervals
> A_k = [1/n, 1/(n+1)] for odd n
> and B_j = [1/n, 1/(n+1)] for even n.
> p_n = 1/n.
> These points define intervals
> A_k = [1/n, 1/(n+1)] for odd n
> and B_j = [1/n, 1/(n+1)] for even n.
> The intervals of sort A ==== and B ---- are alternating. If the points
> are denoted by n, we have something like the following configuration.
> ...7--6==5--4====3---------2===========1
>
> Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
> number of intervals remains the same.
> ...7--6==5--3====4---------2===========1
> The intervals remain alternating. And in particular the number of
> intervals cannot increase.
From this Theorem we immediately get:
> are denoted by n, we have something like the following configuration.
> ...7--6==5--4====3---------2===========1
>
> Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
> number of intervals remains the same.
> ...7--6==5--3====4---------2===========1
> The intervals remain alternating. And in particular the number of
> intervals cannot increase.
If any finite set of the p_n are moved then the
number of intervals does not increase.
Look! Over There! A Pink Elephant
If any set of the p_n are moved then the number
of intervals does not increase.
Virgil
Aug 9, 2012, 6:55:28 PM8/9/12
to
In article
<a3bf5fa7-a582-46ed...@n13g2000vby.googlegroups.com>,
themselves.
While it is trivially possible to ennumerate the endpoints of the I_n
intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the
properties of those intervals, so that WM's enumeration also does prove
change anything.
> I do not see any obstacle.
That WM does not see an obstacle is hardly proof that there is none.
> That results in a configuration having
> all endpoints and only countably many intervals with and without
> rationals.
That you move your A_k to P_(2*k-1) and your B_J to P_(2*j) proves
nothing.
> I showed it.
Also WRONG!
> Look it up. The endpoints can move in a
> completely arbitrary way. Nothing with the A-B-intervals changes.
Everything of the A_B_ intervals changes or everything of the I_n
intervals changes, or both.
In any case you have not, and cannot prove that "nothing changes", which
proof is necessary before you can claim that nothing changes
Here
> you can see it again. Observe point 5:
>
> ...
> 7-6=5-----4=====3--------------------2================================1
> ...
> 7-6===5-4=====3--------------------2================================1
> ...
> 7-6=====4---5==3--------------------2================================1
> ...
> 7-6=====4---------3==5--------------2================================1
> ...
> 7-6=====4---------3===========2---------5==========================1
>
> > > All endpoints are there.
> >
> > Which is hardly enough. Even all points being there, by having been
> > moved discontinuously is not enough.
>
> Look at the drawing above.
> >
> > > If there is anything else in the original
> > > arrangement of the I_n, then it is not explicitly stated and does not
> > > belong to mathematics.
> >
> > If it is deducible from the original arrangements by mathematicians,
> > even if not by WM, then it does belong to mathematics, and is binding,
> > even on non-mathematicians like WM.
>
> There is nothing deductible from the original arrangement. I know it
> because I composed it.
Your claim of knowledge about things you claim to have constructed is
not evidence and is not proof.
Consideration of WM's misrepresentation of, say, complete infinite
binary trees, justifies rejection of WM's claim to be able to claim
things without actually proving them.
All I see is that WM is able to swap the positions of two adjacent
labels.
Since this, however often done. does not affect the order type of the
A-B set of points (all of which orderings will be well-ordered, and not
any other ordering) and the standard indexing of endpoints of the I_n
are certainly not a well-ordering of their positions in [0,1].
Thus WM again claims what he would be incapable of proving even if it
were true.
--
<a3bf5fa7-a582-46ed...@n13g2000vby.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 9 Aug., 21:18, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <af5f7e63-6d12-44e1-9348-f70051e32...@z11g2000yqa.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 9 Aug., 06:40, Virgil <vir...@ligriv.com> wrote:
> >
> > > > The clusters of the original I_n intervals will continue to exist as
> > > > clusters regardless. How you move around your A-B-intervals, will not
> > > > change that.
> >
> > > There are no clusters.
> >
> > In order for there not to be any clusters, your I_n intervals would have
> > to be pairwise disjoint, which can only happen if no interval of
> > positive length contains more than one rational number.
>
> Joint intervals form an interval. Under cluster we understood
> sequences of disjoint intervals. If you like to call joint intervals
> clusters, we can do so. But that does not change anything about my
> proof.
> >
> > > There are points on the real axis. This proof
> > > concerns geometry - nothing else.
> >
> > Your argument is far short of being a proof
>
> The question is only whether a coincidence of endpoints is possible or
> not.
In which case you can say nothing about anything but those points
> On 9 Aug., 21:18, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <af5f7e63-6d12-44e1-9348-f70051e32...@z11g2000yqa.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 9 Aug., 06:40, Virgil <vir...@ligriv.com> wrote:
> >
> > > > The clusters of the original I_n intervals will continue to exist as
> > > > clusters regardless. How you move around your A-B-intervals, will not
> > > > change that.
> >
> > > There are no clusters.
> >
> > In order for there not to be any clusters, your I_n intervals would have
> > to be pairwise disjoint, which can only happen if no interval of
> > positive length contains more than one rational number.
>
> Joint intervals form an interval. Under cluster we understood
> sequences of disjoint intervals. If you like to call joint intervals
> clusters, we can do so. But that does not change anything about my
> proof.
> >
> > > There are points on the real axis. This proof
> > > concerns geometry - nothing else.
> >
> > Your argument is far short of being a proof
>
> The question is only whether a coincidence of endpoints is possible or
> not.
themselves.
While it is trivially possible to ennumerate the endpoints of the I_n
intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the
properties of those intervals, so that WM's enumeration also does prove
change anything.
> I do not see any obstacle.
> That results in a configuration having
> all endpoints and only countably many intervals with and without
> rationals.
nothing.
> >
> >
> > > It creates all endpoints of the I_n construction.
> >
> > But in the process, destroys all properties of the A_k-B_j ordering in
> > order to create the I_n construction. There need not be any two of the
> > I_n intervals having a common endpoint but cosecutive intervals in your
> >
> > > It creates all endpoints of the I_n construction.
> >
> > But in the process, destroys all properties of the A_k-B_j ordering in
> > order to create the I_n construction. There need not be any two of the
> > A_k-B_j sequence all do, there is no way to make one into the other
> > without destroying the first's structure.
>
> You are in error.
WRONG!
> > without destroying the first's structure.
>
> You are in error.
> I showed it.
Also WRONG!
> Look it up. The endpoints can move in a
> completely arbitrary way. Nothing with the A-B-intervals changes.
intervals changes, or both.
In any case you have not, and cannot prove that "nothing changes", which
proof is necessary before you can claim that nothing changes
Here
> you can see it again. Observe point 5:
>
> ...
> 7-6=5-----4=====3--------------------2================================1
> ...
> 7-6===5-4=====3--------------------2================================1
> ...
> 7-6=====4---5==3--------------------2================================1
> ...
> 7-6=====4---------3==5--------------2================================1
> ...
> 7-6=====4---------3===========2---------5==========================1
>
> > > All endpoints are there.
> >
> > Which is hardly enough. Even all points being there, by having been
> > moved discontinuously is not enough.
>
> Look at the drawing above.
> >
> > > If there is anything else in the original
> > > arrangement of the I_n, then it is not explicitly stated and does not
> > > belong to mathematics.
> >
> > If it is deducible from the original arrangements by mathematicians,
> > even if not by WM, then it does belong to mathematics, and is binding,
> > even on non-mathematicians like WM.
>
> There is nothing deductible from the original arrangement. I know it
> because I composed it.
not evidence and is not proof.
Consideration of WM's misrepresentation of, say, complete infinite
binary trees, justifies rejection of WM's claim to be able to claim
things without actually proving them.
> >
> > The conversion from WM's A_K-B_j arrangement to his I_n arrangement
> > has not been show capable of being done by only continuous deformations,
> > so that none of what WM claims for it has been shown to hold.
>
> If you have not yet understood, then look again at the drawing above.
I see it, but it proves nothing except that WM is a fairly sloppy typist.
> > The conversion from WM's A_K-B_j arrangement to his I_n arrangement
> > has not been show capable of being done by only continuous deformations,
> > so that none of what WM claims for it has been shown to hold.
>
> If you have not yet understood, then look again at the drawing above.
All I see is that WM is able to swap the positions of two adjacent
labels.
Since this, however often done. does not affect the order type of the
A-B set of points (all of which orderings will be well-ordered, and not
any other ordering) and the standard indexing of endpoints of the I_n
are certainly not a well-ordering of their positions in [0,1].
Thus WM again claims what he would be incapable of proving even if it
were true.
--
WM
Aug 10, 2012, 11:52:42 AM8/10/12
to
On 9 Aug., 23:19, William Hughes <wpihug...@gmail.com> wrote:
> On Jul 22, 6:51 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
> > Matheology § 092
>
> > My second best proof contradicting set theory
>
> > 1)
> > Define a sequence of points p_n in the unit interval
> > p_n = 1/n.
> > These points define intervals
> > A_k = [1/n, 1/(n+1)] for odd n
> > and B_j = [1/n, 1/(n+1)] for even n.
> > The intervals of sort A ==== and B ---- are alternating. If the points
> > are denoted by n, we have something like the following configuration.
> > ...7--6==5--4====3---------2===========1
>
> > Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
> > number of intervals remains the same.
> > ...7--6==5--3====4---------2===========1
> > The intervals remain alternating. And in particular the number of
> > intervals cannot increase.
>
> From this Theorem we immediately get:
>
> If any finite set of the p_n are moved then the
> number of intervals does not increase.
That means the increase of intervals can be put into a sequence:
> On Jul 22, 6:51 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
> > Matheology § 092
>
> > My second best proof contradicting set theory
>
> > 1)
> > Define a sequence of points p_n in the unit interval
> > p_n = 1/n.
> > These points define intervals
> > A_k = [1/n, 1/(n+1)] for odd n
> > and B_j = [1/n, 1/(n+1)] for even n.
> > The intervals of sort A ==== and B ---- are alternating. If the points
> > are denoted by n, we have something like the following configuration.
> > ...7--6==5--4====3---------2===========1
>
> > Theorem: If two neighbouring points p_n and p_(n+1) are exchanged, the
> > number of intervals remains the same.
> > ...7--6==5--3====4---------2===========1
> > The intervals remain alternating. And in particular the number of
> > intervals cannot increase.
>
> From this Theorem we immediately get:
>
> If any finite set of the p_n are moved then the
> number of intervals does not increase.
0, 0, 0, ....
that is 0 for every finite n.
>
> Look! Over There! A Pink Elephant
The limit of above sequence is uncountable.
> Look! Over There! A Pink Elephant
No, I do not look over there but I use mathematics, in particular
analysis. The limit of a sequenece of zeros is zero.
By the way, why do you believe in the equinumerousity of N and Q?
For every finite initial segment we have equinumerousity. Look! Over
there!
Or why do you believe in uncountability?
For every finite initial segment of the list we can exclude the
presence of the diagonal number. Look! Over There!
Do you recognize that your argument disqualifies all of Cantor's?
Regards, WM
WM
Aug 10, 2012, 11:57:27 AM8/10/12
to
On 10 Aug., 00:55, Virgil <vir...@ligriv.com> wrote:
> > The question is only whether a coincidence of endpoints is possible or
> > not.
>
> In which case you can say nothing about anything but those points
> themselves.
The A and B intervals show that the space between the points is filled
> > The question is only whether a coincidence of endpoints is possible or
> > not.
>
> In which case you can say nothing about anything but those points
> themselves.
with countably many intervals.
>
> While it is trivially possible to ennumerate the endpoints of the I_n
> intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the
> properties of those intervals, so that WM's enumeration also does prove
> change anything.
But there are the A-B-Intervals fitting just the same situation.
> While it is trivially possible to ennumerate the endpoints of the I_n
> intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the
> properties of those intervals, so that WM's enumeration also does prove
> change anything.
>
> That you move your A_k to P_(2*k-1) and your B_J to P_(2*j) proves
> nothing.
You forget that for every move of a point the number of intervals
> That you move your A_k to P_(2*k-1) and your B_J to P_(2*j) proves
> nothing.
remain the same.
> > > A_k-B_j sequence all do, there is no way to make one into the other
> > > without destroying the first's structure.
>
> > You are in error.
>
> WRONG!
...
7-6=5-----4=====3--------------------2================================1
...
7-6===5-4=====3--------------------2================================1
...
7-6=====4---5==3--------------------2================================1
...
7-6=====4---------3==5--------------2================================1
...
7-6=====4---------3===========2---------5==========================1
Induction proves this for every finite number of points. The increase
7-6=5-----4=====3--------------------2================================1
...
7-6===5-4=====3--------------------2================================1
...
7-6=====4---5==3--------------------2================================1
...
7-6=====4---------3==5--------------2================================1
...
7-6=====4---------3===========2---------5==========================1
of intervals is always 0.
Analysis shows that the sequence of interval-increases
0, 0, 0, ... has the limit 0.
Regards, WM
Virgil
Aug 10, 2012, 4:17:48 PM8/10/12
to
In article
<4bdd0ac5-137e-4a1f...@m13g2000vbd.googlegroups.com>,
filled with only finitely many intervals, but the space between any two
points is filled with infinitely many I_n intervals, and your moving one
point at a time will never change that.
intervals in an interval whereas no A-B-interval overlaps any other
A-B-interval.
Further, almost all I_n intervals have infinitely many predecessors (I_n
intervals starting closer to 0, so no such interval can ever become one
of your A-B-intervals by your one at a time endpoint swaps.
of them. You one at a time processing never exhausts an infinite task.
After each of your "steps", as much remains to be done as before that
step, so the job can never be diminished that way.
job remaining cannot even be diminished, much less completed.
--
<4bdd0ac5-137e-4a1f...@m13g2000vbd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 10 Aug., 00:55, Virgil <vir...@ligriv.com> wrote:
>
> > > The question is only whether a coincidence of endpoints is possible or
> > > not.
> >
> > In which case you can say nothing about anything but those points
> > themselves.
>
> The A and B intervals show that the space between the points is filled
> with countably many intervals.
Your A and B intervals would require that every space between points be
> On 10 Aug., 00:55, Virgil <vir...@ligriv.com> wrote:
>
> > > The question is only whether a coincidence of endpoints is possible or
> > > not.
> >
> > In which case you can say nothing about anything but those points
> > themselves.
>
> The A and B intervals show that the space between the points is filled
> with countably many intervals.
filled with only finitely many intervals, but the space between any two
points is filled with infinitely many I_n intervals, and your moving one
point at a time will never change that.
> >
> > While it is trivially possible to ennumerate the endpoints of the I_n
> > intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the
> > properties of those intervals, so that WM's enumeration also does prove
> > change anything.
>
> But there are the A-B-Intervals fitting just the same situation.
Note that every I_n interval overlaps infinitely many other I_n
> > While it is trivially possible to ennumerate the endpoints of the I_n
> > intervals as, say, P_(2*n-1) and P_(2*n), that in no way changes the
> > properties of those intervals, so that WM's enumeration also does prove
> > change anything.
>
> But there are the A-B-Intervals fitting just the same situation.
intervals in an interval whereas no A-B-interval overlaps any other
A-B-interval.
Further, almost all I_n intervals have infinitely many predecessors (I_n
intervals starting closer to 0, so no such interval can ever become one
of your A-B-intervals by your one at a time endpoint swaps.
> >
> > That you move your A_k to P_(2*k-1) and your B_J to P_(2*j) proves
> > nothing.
>
> You forget that for every move of a point the number of intervals
> remain the same.
And the number of I_n's unaccounted for remain the same, infinitely many
> > That you move your A_k to P_(2*k-1) and your B_J to P_(2*j) proves
> > nothing.
>
> You forget that for every move of a point the number of intervals
> remain the same.
of them. You one at a time processing never exhausts an infinite task.
After each of your "steps", as much remains to be done as before that
step, so the job can never be diminished that way.
>
> > > > A_k-B_j sequence all do, there is no way to make one into the other
> > > > without destroying the first's structure.
> >
> > > You are in error.
> >
> > WRONG!
>
> Proof for one point:
> ...
> 7-6=5-----4=====3--------------------2================================1
> ...
> 7-6===5-4=====3--------------------2================================1
> ...
> 7-6=====4---5==3--------------------2================================1
> ...
> 7-6=====4---------3==5--------------2================================1
> ...
> 7-6=====4---------3===========2---------5==========================1
>
> Induction proves this for every finite number of points. The increase
> of intervals is always 0.
But after each step there are still infinitely many steps to go, so the
> > > > A_k-B_j sequence all do, there is no way to make one into the other
> > > > without destroying the first's structure.
> >
> > > You are in error.
> >
> > WRONG!
>
> Proof for one point:
> ...
> 7-6=5-----4=====3--------------------2================================1
> ...
> 7-6===5-4=====3--------------------2================================1
> ...
> 7-6=====4---5==3--------------------2================================1
> ...
> 7-6=====4---------3==5--------------2================================1
> ...
> 7-6=====4---------3===========2---------5==========================1
>
> Induction proves this for every finite number of points. The increase
> of intervals is always 0.
job remaining cannot even be diminished, much less completed.
--
Virgil
Aug 10, 2012, 4:36:10 PM8/10/12
to
In article
<f1ed786e-54c8-4591...@n13g2000vby.googlegroups.com>,
If uncountable , then not zero after all!
mathematicians anywhere outside of his own little world of WMytheology.
done for N and Q, then they satisfy the standard mathematical definition
of equinumerousity.
Equivalently if one can construct injections from each to the other, as
has often been done for N and Q, then they also satisfy the standard
mathematical definition of equinumerousity.
neither a surjection of N onto it nor an injection from it to N.
Which is the standard test for a set being uncountable.
At least outside of WMytheology.
> For every finite initial segment of the list we can exclude the
> presence of the diagonal number. Look! Over There!
Since your "over there" looks remarkably like WMytheology, I prefer to
continue looking only at standard mathematics.
true, if at all, only in WM' WMytheology.
And WM has yet to show any part of Cantor's results to be false in
standard mathematics, or, indeed, anywhere outside of his own
WMytheology.
--
<f1ed786e-54c8-4591...@n13g2000vby.googlegroups.com>,
>
> No, I do not look over there but I use mathematics, in particular
> analysis. The limit of a sequenece of zeros is zero.
What WM claims is mathematics is not recognized as such by actual
> No, I do not look over there but I use mathematics, in particular
> analysis. The limit of a sequenece of zeros is zero.
mathematicians anywhere outside of his own little world of WMytheology.
>
> By the way, why do you believe in the equinumerousity of N and Q?
> For every finite initial segment we have equinumerousity. Look! Over
> there!
If one can construct a bijection between two sets, as has often been
> By the way, why do you believe in the equinumerousity of N and Q?
> For every finite initial segment we have equinumerousity. Look! Over
> there!
done for N and Q, then they satisfy the standard mathematical definition
of equinumerousity.
Equivalently if one can construct injections from each to the other, as
has often been done for N and Q, then they also satisfy the standard
mathematical definition of equinumerousity.
>
> Or why do you believe in uncountability?
Because we can find a set which allows injection of N into it but
> Or why do you believe in uncountability?
neither a surjection of N onto it nor an injection from it to N.
Which is the standard test for a set being uncountable.
At least outside of WMytheology.
> For every finite initial segment of the list we can exclude the
> presence of the diagonal number. Look! Over There!
continue looking only at standard mathematics.
>
> Do you recognize that your argument disqualifies all of Cantor's?
I do not recognize as true in standard mathematics those things that are
> Do you recognize that your argument disqualifies all of Cantor's?
true, if at all, only in WM' WMytheology.
And WM has yet to show any part of Cantor's results to be false in
standard mathematics, or, indeed, anywhere outside of his own
WMytheology.
--
William Hughes
Aug 10, 2012, 10:44:06 PM8/10/12
to
Look! Over There! A Pink Elephant
WM
Aug 11, 2012, 3:06:37 AM8/11/12
to
There is the number of changes of the interval number for 1, 2,
3, ..., n, ... moving points.
And there is the limit of changes of the interval number, namely for
infinitely many moving points.
This sequence can be understood mathematically by so-called analysis:
The terms of the sequence consist of numbers and the sequence has a
number as its limit
0, 0, 0, ... --> 0
The limit of these numbers 0 is 0 and that is the number of the limit.
No escape possible.
Regards, WM
WM
Aug 11, 2012, 5:28:51 AM8/11/12
to
On 10 Aug., 22:17, Virgil <vir...@ligriv.com> wrote:
> > The A and B intervals show that the space between the points is filled
> > with countably many intervals.
>
> Your A and B intervals would require that every space between points be
> filled with only finitely many intervals,
No. Only in the initial configuration the space between eps and 1 is
> > The A and B intervals show that the space between the points is filled
> > with countably many intervals.
>
> Your A and B intervals would require that every space between points be
> filled with only finitely many intervals,
filled with finitely many A-B-intervals. But every space between two
points will, in the final configuration, contain infinitely many
endpoints and thus infinitely many intervals.
Compare Cantor's enumeration of the rationals: Up to every natural
number n, the number of rationals q_1 to q_n is finite.
> but the space between any two
> points is filled with infinitely many I_n intervals, and your moving one
> point at a time will never change that.
ants.
Why should moving one point at a time never change that? Compare
Tristram Shandy. Compare the increasing skill of my ants, such that
the n-th move requires 1/2^n years only.
You should recognize that these arguments of yours are invalid. Try
again.
> Note that every I_n interval overlaps infinitely many other I_n
> intervals in an interval whereas no A-B-interval overlaps any other
> A-B-interval.
each other, then in this union of intervals, there are infinitely many
endpoints of the A-B-intervals - but not a single irrational number of
the complement.
>
> Further, almost all I_n intervals have infinitely many predecessors (I_n
> intervals starting closer to 0, so no such interval can ever become one
> of your A-B-intervals by your one at a time endpoint swaps.
turns out insufficient, then we have infinitely many years. Note: Only
the *possibility* that my ants will accomplish the required
configuration of endpoints invalidates the basis of set theory.
>
> After each of your "steps", as much remains to be done as before that
> step, so the job can never be diminished that way.
That is a very good argument. Why not apply it to Cantors counting of
> After each of your "steps", as much remains to be done as before that
> step, so the job can never be diminished that way.
rationals? Or to his "proof" of uncountability? But no simultaneity is
required. Therefore I have hired aleph_0 little ants. Each one is
instructed to put, on command, its point on an endpoint of the
intervals I_n. (You must know: According to Cantor set theory also
covers biology: Der dritte Theil bringt die Anwendungen der
Mengenlehre auf die Naturwissenschaften: Physik, Chemie, Mineralogie,
Botanik, Zoologie, Anthropologie, Biologie, Physiologie, Medizin etc.
Ist also das, was die Engländer „Natural philosophy" nennen. 1912,
Sept. 20, Cantor to Hilbert)
>
> But after each step there are still infinitely many steps to go, so the
> job remaining cannot even be diminished, much less completed.
(1, 2, 3, ..., n)
(q_1, q_2, q_3, ... q_n)
and what many conclude from it.
Regards, WM
WM
Aug 11, 2012, 5:24:55 AM8/11/12
to
On 10 Aug., 22:36, Virgil <vir...@ligriv.com> wrote:
> > By the way, why do you believe in the equinumerousity of N and Q?
> > For every finite initial segment we have equinumerousity. Look! Over
> > there!
>
> If one can construct a bijection between two sets, as has often been
> done for N and Q, then they satisfy the standard mathematical definition
> of equinumerousity
If equinumerousity exists by definition, then that is not enough to
> > By the way, why do you believe in the equinumerousity of N and Q?
> > For every finite initial segment we have equinumerousity. Look! Over
> > there!
>
> If one can construct a bijection between two sets, as has often been
> done for N and Q, then they satisfy the standard mathematical definition
> of equinumerousity
imply bijectability?
If I define that there are aleph_0 endpoints of the I_n and aleph_0
endpoints of the A-B-intervals then that is not sufficient to prove
bijectability, i.e., that they can be paired?
So you insist that a bijection implies equinumerousity, but that
equinumerousity does not imply bijectability?
>
> I prefer to
> continue looking only at standard mathematics.
In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
> I prefer to
> continue looking only at standard mathematics.
Regards, WM
Virgil
Aug 11, 2012, 1:25:56 PM8/11/12
to
In article
<a657ca8a-4459-47bf...@m13g2000vbd.googlegroups.com>,
between those sets has been proved to exist.
> If I define that there are aleph_0 endpoints of the I_n and aleph_0
> endpoints of the A-B-intervals then that is not sufficient to prove
> bijectability, i.e., that they can be paired?
Proof of equinumerousity does not prove anything else.
It does not prove anything about the order properties of the sets in
question. And it does not prove that your alleged method of reordering
of the endpoints of the I_n is possible.
There is no way to rearrange the endpoints of the I_n to you A-B pattern
without destroying what the I-n were designed to do.
Among other things:
(1) the cumulative length of the I_n intervals is less than 1/9 where
the cumulative length of the A-B intervals is 1.
(2) the midpoints, and very likely the endpoints, of the I_n intervals
are dense in [0,1] but the A_B interval endpoints have only one
accumulation point.
Yet WM claims that, by a sequence of operations each moving one point at
a time or swapping the position of two points, he can convert one to the
other. The problem being that such an infinite process always has
infinitely farther to go before it gets any closer to being finished.
For example, in removing the rationals, one at a time, from [0,1], what
is left after any finite number of steps remains dense in [0,1].
Thus the proper "limit", if any, would have to be a dense in [0,1].
At lest in WMytheology.
>
> Regards, WM
--
<a657ca8a-4459-47bf...@m13g2000vbd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 10 Aug., 22:36, Virgil <vir...@ligriv.com> wrote:
>
> > > By the way, why do you believe in the equinumerousity of N and Q?
> > > For every finite initial segment we have equinumerousity. Look! Over
> > > there!
> >
> > If one can construct a bijection between two sets, as has often been
> > done for N and Q, then they satisfy the standard mathematical definition
> > of equinumerousity
>
> If equinumerousity exists by definition, then that is not enough to
> imply bijectability?
Equinumerousity between two sets is only known to exist when a bijection
> On 10 Aug., 22:36, Virgil <vir...@ligriv.com> wrote:
>
> > > By the way, why do you believe in the equinumerousity of N and Q?
> > > For every finite initial segment we have equinumerousity. Look! Over
> > > there!
> >
> > If one can construct a bijection between two sets, as has often been
> > done for N and Q, then they satisfy the standard mathematical definition
> > of equinumerousity
>
> If equinumerousity exists by definition, then that is not enough to
> imply bijectability?
between those sets has been proved to exist.
> If I define that there are aleph_0 endpoints of the I_n and aleph_0
> endpoints of the A-B-intervals then that is not sufficient to prove
> bijectability, i.e., that they can be paired?
It does not prove anything about the order properties of the sets in
question. And it does not prove that your alleged method of reordering
of the endpoints of the I_n is possible.
>
> So you insist that a bijection implies equinumerousity, but that
> equinumerousity does not imply bijectability?
No I do not!
> So you insist that a bijection implies equinumerousity, but that
> equinumerousity does not imply bijectability?
> >
> > I prefer to
> > continue looking only at standard mathematics.
>
> In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
Which is totally irrelevant to WM's problem, of proving his false claims.
> > I prefer to
> > continue looking only at standard mathematics.
>
> In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
There is no way to rearrange the endpoints of the I_n to you A-B pattern
without destroying what the I-n were designed to do.
Among other things:
(1) the cumulative length of the I_n intervals is less than 1/9 where
the cumulative length of the A-B intervals is 1.
(2) the midpoints, and very likely the endpoints, of the I_n intervals
are dense in [0,1] but the A_B interval endpoints have only one
accumulation point.
Yet WM claims that, by a sequence of operations each moving one point at
a time or swapping the position of two points, he can convert one to the
other. The problem being that such an infinite process always has
infinitely farther to go before it gets any closer to being finished.
For example, in removing the rationals, one at a time, from [0,1], what
is left after any finite number of steps remains dense in [0,1].
Thus the proper "limit", if any, would have to be a dense in [0,1].
At lest in WMytheology.
>
> Regards, WM
--
Virgil
Aug 11, 2012, 1:32:15 PM8/11/12
to
In article
<1d0f5a53-c69a-4241...@h5g2000vbl.googlegroups.com>,
>
> > Note that every I_n interval overlaps infinitely many other I_n
> > intervals in an interval whereas no A-B-interval overlaps any other
> > A-B-interval.
>
> Why do you mention that? If infinitely many I_n overlap or include
> each other, then in this union of intervals, there are infinitely many
> endpoints of the A-B-intervals - but not a single irrational number of
> the complement.
Which sows you cannot do what you claim to do.
--
<1d0f5a53-c69a-4241...@h5g2000vbl.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 10 Aug., 22:17, Virgil <vir...@ligriv.com> wrote:
>
> > > The A and B intervals show that the space between the points is filled
> > > with countably many intervals.
> >
> > Your A and B intervals would require that every space between points be
> > filled with only finitely many intervals,
>
> No. Only in the initial configuration the space between eps and 1 is
> filled with finitely many A-B-intervals. But every space between two
> points will, in the final configuration, contain infinitely many
> endpoints and thus infinitely many intervals.
Claimed but not proven possible.
> On 10 Aug., 22:17, Virgil <vir...@ligriv.com> wrote:
>
> > > The A and B intervals show that the space between the points is filled
> > > with countably many intervals.
> >
> > Your A and B intervals would require that every space between points be
> > filled with only finitely many intervals,
>
> No. Only in the initial configuration the space between eps and 1 is
> filled with finitely many A-B-intervals. But every space between two
> points will, in the final configuration, contain infinitely many
> endpoints and thus infinitely many intervals.
>
> Compare Cantor's enumeration of the rationals: Up to every natural
> number n, the number of rationals q_1 to q_n is finite.
>
> > but the space between any two
> > points is filled with infinitely many I_n intervals, and your moving one
> > point at a time will never change that.
>
> Why should I move one point at a time? I may employ infinitely many
> ants.
That is what you described doing!
> Compare Cantor's enumeration of the rationals: Up to every natural
> number n, the number of rationals q_1 to q_n is finite.
>
> > but the space between any two
> > points is filled with infinitely many I_n intervals, and your moving one
> > point at a time will never change that.
>
> Why should I move one point at a time? I may employ infinitely many
> ants.
>
> Why should moving one point at a time never change that? Compare
> Tristram Shandy. Compare the increasing skill of my ants, such that
> the n-th move requires 1/2^n years only.
>
> You should recognize that these arguments of yours are invalid. Try
> again.
No more invalid than your arguments.
> Why should moving one point at a time never change that? Compare
> Tristram Shandy. Compare the increasing skill of my ants, such that
> the n-th move requires 1/2^n years only.
>
> You should recognize that these arguments of yours are invalid. Try
> again.
>
> > Note that every I_n interval overlaps infinitely many other I_n
> > intervals in an interval whereas no A-B-interval overlaps any other
> > A-B-interval.
>
> Why do you mention that? If infinitely many I_n overlap or include
> each other, then in this union of intervals, there are infinitely many
> endpoints of the A-B-intervals - but not a single irrational number of
> the complement.
Virgil
Aug 11, 2012, 1:39:11 PM8/11/12
to
In article
<56a5f418-d5ff-4505...@m13g2000vbd.googlegroups.com>,
known whether all rationals are still covered, at which point your
argument fails.
with consecutive endpoints and remain so forever, merely being renamed
does not change that structure.
> The limit of these numbers 0 is 0 and that is the number of the limit.
> No escape possible.
There is no need to escape from a place one has never been and can never
be.
--
<56a5f418-d5ff-4505...@m13g2000vbd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > That means the increase of intervals can be put into a sequence:
> >
> > > 0, 0, 0, ....
> >
> > > that is 0 for every finite n.
> >
> > Thus the limit of the number of changes is 0
> >
> > Look! Over There! A Pink Elephant
> >
> > Thus the number of changes of the limit is 0
>
> There is no "change of limit".
>
> There is the number of changes of the interval number for 1, 2,
> 3, ..., n, ... moving points.
>
> And there is the limit of changes of the interval number, namely for
> infinitely many moving points.
As soon as any one endpoint of the I_n has been moved, it is no longer
> On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > That means the increase of intervals can be put into a sequence:
> >
> > > 0, 0, 0, ....
> >
> > > that is 0 for every finite n.
> >
> > Thus the limit of the number of changes is 0
> >
> > Look! Over There! A Pink Elephant
> >
> > Thus the number of changes of the limit is 0
>
> There is no "change of limit".
>
> There is the number of changes of the interval number for 1, 2,
> 3, ..., n, ... moving points.
>
> And there is the limit of changes of the interval number, namely for
> infinitely many moving points.
known whether all rationals are still covered, at which point your
argument fails.
>
> This sequence can be understood mathematically by so-called analysis:
> The terms of the sequence consist of numbers and the sequence has a
> number as its limit
> 0, 0, 0, ... --> 0
Actually the real terms of your A-B sequence are a sequence of intervals
> This sequence can be understood mathematically by so-called analysis:
> The terms of the sequence consist of numbers and the sequence has a
> number as its limit
> 0, 0, 0, ... --> 0
with consecutive endpoints and remain so forever, merely being renamed
does not change that structure.
> The limit of these numbers 0 is 0 and that is the number of the limit.
> No escape possible.
be.
--
William Hughes
Aug 11, 2012, 1:46:43 PM8/11/12
to
On Aug 11, 4:06 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
>
> > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > That means the increase of intervals can be put into a sequence:
>
> > > 0, 0, 0, ....
>
> > > that is 0 for every finite n.
>
> > Thus the limit of the number of changes is 0
>
> > Look! Over There! A Pink Elephant
>
> > Thus the number of changes of the limit is 0
>
> There is no "change of limit".
>
> There is the number of changes of the interval number for 1, 2,
> 3, ..., n, ... moving points.
The phrase
> On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
>
> > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > That means the increase of intervals can be put into a sequence:
>
> > > 0, 0, 0, ....
>
> > > that is 0 for every finite n.
>
> > Thus the limit of the number of changes is 0
>
> > Look! Over There! A Pink Elephant
>
> > Thus the number of changes of the limit is 0
>
> There is no "change of limit".
>
> There is the number of changes of the interval number for 1, 2,
> 3, ..., n, ... moving points.
"number of changes of the interval number" is nonsense.
There are a number of possible things you might mean by
this ambiguous phrase. Since a popular MO with you is to
use two different interpretations and claim a contradiction
I will not guess.
WM
Aug 12, 2012, 12:39:27 PM8/12/12
to
On 11 Aug., 19:46, William Hughes <wpihug...@gmail.com> wrote:
> On Aug 11, 4:06 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
> > On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
>
> > > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > That means the increase of intervals can be put into a sequence:
>
> > > > 0, 0, 0, ....
>
> > > > that is 0 for every finite n.
>
> > > Thus the limit of the number of changes is 0
>
> > > Look! Over There! A Pink Elephant
>
> > > Thus the number of changes of the limit is 0
>
> > There is no "change of limit".
>
> > There is the number of changes of the interval number for 1, 2,
> > 3, ..., n, ... moving points.
>
> The phrase
> "number of changes of the interval number" is nonsense.
My construction has aleph_0 endpoints and aleph_0 intervals (of type A
> On Aug 11, 4:06 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
> > On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
>
> > > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > That means the increase of intervals can be put into a sequence:
>
> > > > 0, 0, 0, ....
>
> > > > that is 0 for every finite n.
>
> > > Thus the limit of the number of changes is 0
>
> > > Look! Over There! A Pink Elephant
>
> > > Thus the number of changes of the limit is 0
>
> > There is no "change of limit".
>
> > There is the number of changes of the interval number for 1, 2,
> > 3, ..., n, ... moving points.
>
> The phrase
> "number of changes of the interval number" is nonsense.
and B) between them.
The number of changes of the number of intervals is zero for every
move of every endpoint in every direction.
> There are a number of possible things you might mean by
> this ambiguous phrase.
But your "Look over there is" is nonsense. If a sequences of sets
could increase in another manner than its cardinality, then we would
possible have for the set of initial segments (1, 2, ..., n) of the
natural numbers:
The limit oo of the sequence of cardinalities need not be identical to
the cardinality of the limit N of the sequence. There was no
justification at all that aleph_0 was the first transfinite
cardinality. It could as well be 2^2^2^aleph_0 or -10 or something
else..
> I will not guess.
You will stop with your silly "Look over there" until you think that
the serious objection against countability implied by your arguing
will have been forgotten, and then you will start again with that
nonsense.
Regards, WM
WM
Aug 12, 2012, 1:24:30 PM8/12/12
to
On 11 Aug., 19:25, Virgil <vir...@ligriv.com> wrote:
> > So you insist that a bijection implies equinumerousity, but that
> > equinumerousity does not imply bijectability?
>
> No I do not!
>
>
>
> > > I prefer to
> > > continue looking only at standard mathematics.
>
> > In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
>
> Which is totally irrelevant to WM's problem, of proving his false claims.
>
> There is no way to rearrange the endpoints of the I_n to you A-B pattern
I need only equinumerousity to prove that a bijection is possible,
> > So you insist that a bijection implies equinumerousity, but that
> > equinumerousity does not imply bijectability?
>
> No I do not!
>
>
>
> > > I prefer to
> > > continue looking only at standard mathematics.
>
> > In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
>
> Which is totally irrelevant to WM's problem, of proving his false claims.
>
> There is no way to rearrange the endpoints of the I_n to you A-B pattern
because (few lines above) you answered the question:
> So you insist that a bijection implies equinumerousity, but that
> equinumerousity does not imply bijectability?
by
> equinumerousity does not imply bijectability?
No I do not!
> without destroying what the I-n were designed to do.
nothing destroyed.
Regards, WM
William Hughes
Aug 12, 2012, 2:47:01 PM8/12/12
to
On Aug 12, 1:39 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
The number of changes of the number of intervals is zero for every
move of every finite set of endpoints in every direction
The number of changes of the number of intervals is zero for every
Look! Over There! A Pink Elephant!
Virgil
Aug 12, 2012, 6:38:54 PM8/12/12
to
In article
<1941f5af-31dc-4ac1...@t7g2000vbe.googlegroups.com>,
Then also nothing newe is proved by WM's arguments.. The
"equinummreoisity" or "bijectability" of the sets of points in question
has effectively nothing to so with the order properties of those sets,
since we already know, e.g., that the naturals and the rationals are
equinumerous but are quite different as ordered sets
Note that each of the set of rationals in [0,1] and the set of
irrationals in [0, 1] is dense in the other even though there is no
bijection between them and thy are not equinumerous.
--
<1941f5af-31dc-4ac1...@t7g2000vbe.googlegroups.com>,
"equinummreoisity" or "bijectability" of the sets of points in question
has effectively nothing to so with the order properties of those sets,
since we already know, e.g., that the naturals and the rationals are
equinumerous but are quite different as ordered sets
Note that each of the set of rationals in [0,1] and the set of
irrationals in [0, 1] is dense in the other even though there is no
bijection between them and thy are not equinumerous.
--
Virgil
Aug 12, 2012, 7:01:43 PM8/12/12
to
In article
<233b3daa-ee40-4b6d...@h5g2000vbl.googlegroups.com>,
intervals must be an upper endpoint of exactly one interval and the
lower endpoint of another such interval, which is nonsense. or that
their order properties are irrelevant, which mis even ore nonsensical.
intervals A-B intervals, but that order properties of those two sets of
intervals differ and that difference is not destroyed by your
'bijection'.
> The limit oo of the sequence of cardinalities
There are only two equal cardinalities involved, but their equality is
irrelevant to other properties of the set of I_n intervals.
>
> You will stop with your silly "Look over there"
Wy should he stop when you are the one claiming that your argument
proves something else quite different and irrelevant to it.
> until you think that
> the serious objection against countability implied by your arguing
> will have been forgotten, and then you will start again with that
> nonsense.
Countability is NOT the issue. The isue is whether countability proves
anything else, and in this cans it does NOT prove that the set of I_n
intervals behaves as WM claims it behaves.
WM wishes to claim that after removing the countably many I_n intervals,
with I_n of length 1/10^n, from [0,1], one will have only countably many
points remaining.
But his maunderings prove no such thing.
--
<233b3daa-ee40-4b6d...@h5g2000vbl.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 11 Aug., 19:46, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 11, 4:06 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > That means the increase of intervals can be put into a sequence:
> >
> > > > > 0, 0, 0, ....
> >
> > > > > that is 0 for every finite n.
> >
> > > > Thus the limit of the number of changes is 0
> >
> > > > Look! Over There! A Pink Elephant
> >
> > > > Thus the number of changes of the limit is 0
> >
> > > There is no "change of limit".
> >
> > > There is the number of changes of the interval number for 1, 2,
> > > 3, ..., n, ... moving points.
> >
> > The phrase
> > "number of changes of the interval number" is nonsense.
>
> My construction has aleph_0 endpoints and aleph_0 intervals (of type A
> and B) between them.
>
> The number of changes of the number of intervals is zero for every
> move of every endpoint in every direction.
You are effectively arguing that all but one endpoint of one of the I_n
> On 11 Aug., 19:46, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 11, 4:06 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 11 Aug., 04:44, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Aug 10, 12:52 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > That means the increase of intervals can be put into a sequence:
> >
> > > > > 0, 0, 0, ....
> >
> > > > > that is 0 for every finite n.
> >
> > > > Thus the limit of the number of changes is 0
> >
> > > > Look! Over There! A Pink Elephant
> >
> > > > Thus the number of changes of the limit is 0
> >
> > > There is no "change of limit".
> >
> > > There is the number of changes of the interval number for 1, 2,
> > > 3, ..., n, ... moving points.
> >
> > The phrase
> > "number of changes of the interval number" is nonsense.
>
> My construction has aleph_0 endpoints and aleph_0 intervals (of type A
> and B) between them.
>
> The number of changes of the number of intervals is zero for every
> move of every endpoint in every direction.
intervals must be an upper endpoint of exactly one interval and the
lower endpoint of another such interval, which is nonsense. or that
their order properties are irrelevant, which mis even ore nonsensical.
>
> > There are a number of possible things you might mean by
> > this ambiguous phrase.
>
> There is no number of things because every possible change is zero.
The no changes have occurred, and your argument is nothing.
> > There are a number of possible things you might mean by
> > this ambiguous phrase.
>
> There is no number of things because every possible change is zero.
>
> But your "Look over there is" is nonsense. If a sequences of sets
> could increase in another manner than its cardinality,
It is not that the NUMBER of I_n intervals differs from the NUMBER of
> But your "Look over there is" is nonsense. If a sequences of sets
> could increase in another manner than its cardinality,
intervals A-B intervals, but that order properties of those two sets of
intervals differ and that difference is not destroyed by your
'bijection'.
> The limit oo of the sequence of cardinalities
irrelevant to other properties of the set of I_n intervals.
>
> You will stop with your silly "Look over there"
proves something else quite different and irrelevant to it.
> until you think that
> the serious objection against countability implied by your arguing
> will have been forgotten, and then you will start again with that
> nonsense.
anything else, and in this cans it does NOT prove that the set of I_n
intervals behaves as WM claims it behaves.
WM wishes to claim that after removing the countably many I_n intervals,
with I_n of length 1/10^n, from [0,1], one will have only countably many
points remaining.
But his maunderings prove no such thing.
--
WM
Aug 13, 2012, 10:29:07 AM8/13/12
to
On 12 Aug., 20:47, William Hughes <wpihug...@gmail.com> wrote:
> On Aug 12, 1:39 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> The number of changes of the number of intervals is zero for every
> move of every finite set of endpoints in every direction
>
> Look! Over There! A Pink Elephant!
If infinitely many points move (or if one point moves infinitely
> On Aug 12, 1:39 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> The number of changes of the number of intervals is zero for every
> move of every finite set of endpoints in every direction
>
> Look! Over There! A Pink Elephant!
often?), the number of intervals-changes is not only infinite but
uncountable.
There are some rules of matheology that have to be learned peu a peu.
Regards, WM
WM
Aug 13, 2012, 10:32:19 AM8/13/12
to
> <1941f5af-31dc-4ac1-a417-94f30684c...@t7g2000vbe.googlegroups.com>,
> The
> "equinummreoisity" or "bijectability" of the sets of points in question
> has effectively nothing to so with the order properties of those sets,
Therefore it is clear that the number of intervals does not increase
when the endpoints are moved.
> since we already know, e.g., that the naturals and the rationals are
> equinumerous but are quite different as ordered sets
We know that there are as many intervals of sort A and B as there are
endpoints. The moving of endpoints does not change this
equinumerousity, does it?
Regards, WM
>
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 11 Aug., 19:25, Virgil <vir...@ligriv.com> wrote:
>
> > > > So you insist that a bijection implies equinumerousity, but that
> > > > equinumerousity does not imply bijectability?
>
> > > No I do not!
>
> > > > > I prefer to
> > > > > continue looking only at standard mathematics.
>
> > > > In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
>
> > > Which is totally irrelevant to WM's problem, of proving his false claims.
>
> > > There is no way to rearrange the endpoints of the I_n to you A-B pattern
>
> > I need only equinumerousity to prove that a bijection is possible,
> > because (few lines above) you answered the question:
> > > So you insist that a bijection implies equinumerousity, but that
> > > equinumerousity does not imply bijectability?
> > by
> > No I do not!
>
> > > without destroying what the I-n were designed to do.
>
> > The I_n remain what they are. There is nothing changed and, therefore,
> > nothing destroyed.
>
> Then also nothing newe is proved by WM's arguments.
Nothing is new, but unknown to many.
>
>
>
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 11 Aug., 19:25, Virgil <vir...@ligriv.com> wrote:
>
> > > > So you insist that a bijection implies equinumerousity, but that
> > > > equinumerousity does not imply bijectability?
>
> > > No I do not!
>
> > > > > I prefer to
> > > > > continue looking only at standard mathematics.
>
> > > > In standard mathematics the limit of the sequence 0, 0, 0, ... is 0.
>
> > > Which is totally irrelevant to WM's problem, of proving his false claims.
>
> > > There is no way to rearrange the endpoints of the I_n to you A-B pattern
>
> > I need only equinumerousity to prove that a bijection is possible,
> > because (few lines above) you answered the question:
> > > So you insist that a bijection implies equinumerousity, but that
> > > equinumerousity does not imply bijectability?
> > by
> > No I do not!
>
> > > without destroying what the I-n were designed to do.
>
> > The I_n remain what they are. There is nothing changed and, therefore,
> > nothing destroyed.
>
> Then also nothing newe is proved by WM's arguments.
> The
> "equinummreoisity" or "bijectability" of the sets of points in question
> has effectively nothing to so with the order properties of those sets,
when the endpoints are moved.
> since we already know, e.g., that the naturals and the rationals are
> equinumerous but are quite different as ordered sets
endpoints. The moving of endpoints does not change this
equinumerousity, does it?
>
> Note that each of the set of rationals in [0,1] and the set of
> irrationals in [0, 1] is dense in the other even though there is no
> bijection between them and thy are not equinumerous.
Since my proof shows the contrary, your claim is invalid.
> Note that each of the set of rationals in [0,1] and the set of
> irrationals in [0, 1] is dense in the other even though there is no
> bijection between them and thy are not equinumerous.
Regards, WM
Virgil
Aug 13, 2012, 2:49:03 PM8/13/12
to
In article
<3b329b5d-8254-460a...@a9g2000vbn.googlegroups.com>,
But how do you move endpoints in the A-B intervals so as to leave
uncovered all the points not covered by the I_n intervals? Unless you
can show how to do that, your arguments fail.
leave uncovered precisely those points left uncovered by the I_n
intervals, your arguments fail.
And you certainly have not done so yet!
A_B endpoints so as to leave uncovered every point left uncovered by the
I_n intervals
--
<3b329b5d-8254-460a...@a9g2000vbn.googlegroups.com>,
uncovered all the points not covered by the I_n intervals? Unless you
can show how to do that, your arguments fail.
>
> > since we already know, e.g., that the naturals and the rationals are
> > equinumerous but are quite different as ordered sets
>
> We know that there are as many intervals of sort A and B as there are
> endpoints. The moving of endpoints does not change this
> equinumerousity, does it?
Irrelevant! Unless you can show how to move your A_B endpoints so as to
> > since we already know, e.g., that the naturals and the rationals are
> > equinumerous but are quite different as ordered sets
>
> We know that there are as many intervals of sort A and B as there are
> endpoints. The moving of endpoints does not change this
> equinumerousity, does it?
leave uncovered precisely those points left uncovered by the I_n
intervals, your arguments fail.
And you certainly have not done so yet!
> >
> > Note that each of the set of rationals in [0,1] and the set of
> > irrationals in [0, 1] is dense in the other even though there is no
> > bijection between them and thy are not equinumerous.
>
> Since my proof shows the contrary, your claim is invalid.
You "proof" is not a proof until you can at least show how to move your
> > Note that each of the set of rationals in [0,1] and the set of
> > irrationals in [0, 1] is dense in the other even though there is no
> > bijection between them and thy are not equinumerous.
>
> Since my proof shows the contrary, your claim is invalid.
A_B endpoints so as to leave uncovered every point left uncovered by the
I_n intervals
--
Virgil
Aug 13, 2012, 3:01:37 PM8/13/12
to
In article
<ff5249cb-54cf-4274...@n13g2000vby.googlegroups.com>,
WW is the only one who can ever 'learn' any of the rules of his
matheology, as he makes them up as he goes along.
Whereas re the rules of standard mathematics, WM does not even learn
them peu a peu.
Note that WM's A-B intervals cover all but one point of [0,1], whereas
his I_n intervals leave at least 8/9 of [0,1] uncovered, which even in
WMs world works out to be at least countably many points uncovered..
And WM has not yet explained how his method of moving the endpoints of
the A-B intervals can ever leave more than just one point of [0,1]
uncovered, much less countably many, which is itself much less than 8/9
of the points in [0,1].
--
<ff5249cb-54cf-4274...@n13g2000vby.googlegroups.com>,
matheology, as he makes them up as he goes along.
Whereas re the rules of standard mathematics, WM does not even learn
them peu a peu.
Note that WM's A-B intervals cover all but one point of [0,1], whereas
his I_n intervals leave at least 8/9 of [0,1] uncovered, which even in
WMs world works out to be at least countably many points uncovered..
And WM has not yet explained how his method of moving the endpoints of
the A-B intervals can ever leave more than just one point of [0,1]
uncovered, much less countably many, which is itself much less than 8/9
of the points in [0,1].
--
William Hughes
Aug 13, 2012, 10:49:49 PM8/13/12
to
On Aug 13, 11:29 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 12 Aug., 20:47, William Hughes <wpihug...@gmail.com> wrote:
>
> > On Aug 12, 1:39 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > The number of changes of the number of intervals is zero for every
> > move of every finite set of endpoints in every direction
>
> > Look! Over There! A Pink Elephant!
> > > The number of changes of the number of intervals is zero for every
> > > move of every endpoint in every direction.
> On 12 Aug., 20:47, William Hughes <wpihug...@gmail.com> wrote:
>
> > On Aug 12, 1:39 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > The number of changes of the number of intervals is zero for every
> > move of every finite set of endpoints in every direction
>
> > Look! Over There! A Pink Elephant!
> > > The number of changes of the number of intervals is zero for every
>
> If infinitely many points move (or if one point moves infinitely
> often?), the number of intervals-changes is not only infinite but
> uncountable.
to your final state requires the movement
of an infinite number of points. A theorem that only deals with
the movement of finite numbers of points
(even if it deals with every finite number of points)
is not applicable.
WM
Aug 14, 2012, 3:40:19 AM8/14/12
to
On 14 Aug., 04:49, William Hughes <wpihug...@gmail.com> wrote:
>
> Your problem is that a change from your initial state
> to your final state requires the movement
> of an infinite number of points.
Your objection is completely unfounded, wishful thinking. Since
>
> Your problem is that a change from your initial state
> to your final state requires the movement
> of an infinite number of points.
translation is a continuous process (even for the movement of an
uncountable number of points of some body) your objection is
unfounded. In no way is it possible to create additional intervals by
moving endpoints - how many ever.
And remember: My proof holds for every finite n, that means for all
infinitely many n in N.
> A theorem that only deals with
> the movement of finite numbers of points
> (even if it deals with every finite number of points)
> is not applicable.
will fail too. The latter holds for every line (and its predecessors),
i.e., for every finite number of lines.
Finally: In the real line we have countably many rationals, i.e.,
countably many right interval ends and countably many left interval
ends. Although we have no information about the intervals, it is clear
to everybody not yet blinded by Cantor's trick, that there cannot be
more intervals than endpoints.
Regards, WM
WM
Aug 14, 2012, 3:44:34 AM8/14/12
to
On 13 Aug., 20:49, Virgil <vir...@ligriv.com> wrote:
> You "proof" is not a proof until you can at least show how to move your
> A_B endpoints so as to leave uncovered every point left uncovered by the
> I_n intervals
It is sufficient to have the possibility that all endpoints of A-B-
> You "proof" is not a proof until you can at least show how to move your
> A_B endpoints so as to leave uncovered every point left uncovered by the
> I_n intervals
intervals will get in coincidence with all endpoints of the I-
intervals. Further translation is continuous, even for the uncountably
many points of your body. Otherwise it might happen that you pass a
door and uncountably many Virgils will enter the next room. By this
example you can Cantor's matheology in its full perversity.
Regards, WM
William Hughes
Aug 14, 2012, 8:58:13 AM8/14/12
to
On Aug 14, 4:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> My proof holds for every finite n, that means for all
> infinitely many n in N.
> My proof holds for every finite n, that means for all
> infinitely many n in N.
Look! Over There! A Pink Elephant!
That means for the set N.
William Hughes
Aug 14, 2012, 7:51:28 AM8/14/12
to
> My proof holds for every finite n, that means for all
> infinitely many n in N.
> infinitely many n in N.
Look! Over There! A Pink Elephant!
That mean for the set N.
Virgil
Aug 14, 2012, 1:49:25 PM8/14/12
to
In article
<c0bba27e-7b8f-4340...@f17g2000vbz.googlegroups.com>,
In standard mathematics it proves nothing.
--
<c0bba27e-7b8f-4340...@f17g2000vbz.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 13 Aug., 20:49, Virgil <vir...@ligriv.com> wrote:
>
> > You "proof" is not a proof until you can at least show how to move your
> > A_B endpoints so as to leave uncovered every point left uncovered by the
> > I_n intervals
>
> It is sufficient to have the possibility that all endpoints of A-B-
> intervals will get in coincidence with all endpoints of the I-
> intervals.
It is only sufficient in WMytheology.
> On 13 Aug., 20:49, Virgil <vir...@ligriv.com> wrote:
>
> > You "proof" is not a proof until you can at least show how to move your
> > A_B endpoints so as to leave uncovered every point left uncovered by the
> > I_n intervals
>
> It is sufficient to have the possibility that all endpoints of A-B-
> intervals will get in coincidence with all endpoints of the I-
> intervals.
In standard mathematics it proves nothing.
--
Virgil
Aug 14, 2012, 2:05:15 PM8/14/12
to
In article
<6c350faf-6bcf-48b4...@b12g2000yqp.googlegroups.com>,
intervals need be moved in order to bring all infinitely many of those
endpoints into matching up with the I_n intervals?
Which infinitely many A-N endpoints does WM claim are already in their
correct I_n positions?
> Since
> translation is a continuous process (even for the movement of an
> uncountable number of points of some body) your objection is
> unfounded.
His objection is WELL founded, and your opbjection to it is the only
thing unfounded here.
>In no way is it possible to create additional intervals by
> moving endpoints - how many ever.
It is also impossible to eliminate intervals, as would be necessary to
do what WM claims to be able to do, change from a sequence of always
consecutive A-B intervals covering all of [0,1[ to a sequence of
non-consecutive I_n intervals covering less than 1/9 of [0,1].
>
> > �A theorem that only deals with
>
> Finally: In the real line we have countably many rationals, i.e.,
> countably many right interval ends and countably many left interval
> ends. Although we have no information about the intervals, it is clear
> to everybody not yet blinded by Cantor's trick, that there cannot be
> more intervals than endpoints.
If WM is claiming that the number of real points in the real interval
[0,1] is countable, let us see him actually count them by constructing
either a surjection from N to [O,1] or an injection from [0,1] to N.
>
Until he has done one or the other, he is merely bowing hot air from hi
WMythematical underworld to our Standard mathematical overworld.
--
<6c350faf-6bcf-48b4...@b12g2000yqp.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 14 Aug., 04:49, William Hughes <wpihug...@gmail.com> wrote:
>
> >
> > Your problem is that a change from your initial state
> > to your final state requires the movement
> > of an infinite number of points.
>
> Your objection is completely unfounded, wishful thinking.
Does WM claim that only finitely many of the endpoints of the A-B
> On 14 Aug., 04:49, William Hughes <wpihug...@gmail.com> wrote:
>
> >
> > Your problem is that a change from your initial state
> > to your final state requires the movement
> > of an infinite number of points.
>
> Your objection is completely unfounded, wishful thinking.
intervals need be moved in order to bring all infinitely many of those
endpoints into matching up with the I_n intervals?
Which infinitely many A-N endpoints does WM claim are already in their
correct I_n positions?
> Since
> translation is a continuous process (even for the movement of an
> uncountable number of points of some body) your objection is
> unfounded.
thing unfounded here.
>In no way is it possible to create additional intervals by
> moving endpoints - how many ever.
do what WM claims to be able to do, change from a sequence of always
consecutive A-B intervals covering all of [0,1[ to a sequence of
non-consecutive I_n intervals covering less than 1/9 of [0,1].
>
> And remember: My proof holds for every finite n, that means for all
> infinitely many n in N.
But your "proof" does not prove what it claims to prove.
> And remember: My proof holds for every finite n, that means for all
> infinitely many n in N.
>
> > �A theorem that only deals with
> > the movement of finite numbers of points
> > (even if it deals with every finite number of points)
> > is not applicable.
>
> First if it is not applicable and valid, then Cantors diagonal proof
> will fail too.
Not to those who understand it.
> > (even if it deals with every finite number of points)
> > is not applicable.
>
> First if it is not applicable and valid, then Cantors diagonal proof
> will fail too.
>
> Finally: In the real line we have countably many rationals, i.e.,
> countably many right interval ends and countably many left interval
> ends. Although we have no information about the intervals, it is clear
> to everybody not yet blinded by Cantor's trick, that there cannot be
> more intervals than endpoints.
[0,1] is countable, let us see him actually count them by constructing
either a surjection from N to [O,1] or an injection from [0,1] to N.
>
Until he has done one or the other, he is merely bowing hot air from hi
WMythematical underworld to our Standard mathematical overworld.
--
WM
Aug 14, 2012, 4:45:23 PM8/14/12
to
Compare Cantor's diagonal argument. It holds for every line L_n that
has a natural index n.
Regards, WM
WM
Aug 14, 2012, 4:53:00 PM8/14/12
to
On 14 Aug., 20:05, Virgil <vir...@ligriv.com> wrote:
> >In no way is it possible to create additional intervals by
> > moving endpoints - how many ever.
>
> It is also impossible to eliminate intervals,
It is possible to eliminate intervals, namely if some I_n fall inside
> >In no way is it possible to create additional intervals by
> > moving endpoints - how many ever.
>
> It is also impossible to eliminate intervals,
of each other.
> > And remember: My proof holds for every finite n, that means for all
> > infinitely many n in N.
>
> But your "proof" does not prove what it claims to prove.
remain countably many A-B-intervals and their endpoints cover every
endpoint of the I-intervals.
> > First if it is not applicable and valid, then Cantors diagonal proof
> > will fail too.
>
> Not to those who understand it.
contradiction with my A-B-intervals. You only parrot what you have
been taught.
>
>
>
> > Finally: In the real line we have countably many rationals, i.e.,
> > countably many right interval ends and countably many left interval
> > ends. Although we have no information about the intervals, it is clear
> > to everybody not yet blinded by Cantor's trick, that there cannot be
> > more intervals than endpoints.
>
> If WM is claiming that the number of real points in the real interval
> [0,1] is countable,
There are obviously not more points.
>
>
> > Finally: In the real line we have countably many rationals, i.e.,
> > countably many right interval ends and countably many left interval
> > ends. Although we have no information about the intervals, it is clear
> > to everybody not yet blinded by Cantor's trick, that there cannot be
> > more intervals than endpoints.
>
> If WM is claiming that the number of real points in the real interval
> [0,1] is countable,
> let us see him actually count them by constructing
> either a surjection from N to [O,1] or an injection from [0,1] to N.
have been constructed since 1890) it is not possible to construct a
bijection using "all" real numbers.
Regards, WM
Sam Sung
Aug 14, 2012, 5:07:45 PM8/14/12
to
WM schrieb:
LOL, we know WM believes that only some (few) numbers "really" "exist".
William Hughes
Aug 14, 2012, 5:42:54 PM8/14/12
to
On Aug 14, 5:45 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 14 Aug., 14:58, William Hughes <wpihug...@gmail.com> wrote:
>
> > On Aug 14, 4:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > My proof holds for every finite n, that means for all
> > > infinitely many n in N.
>
> > Look! Over There! A Pink Elephant!
>
> > That means for the set N.
>
> That means for all points
Look! Over There! A Pink Elephant.
> On 14 Aug., 14:58, William Hughes <wpihug...@gmail.com> wrote:
>
> > On Aug 14, 4:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > My proof holds for every finite n, that means for all
> > > infinitely many n in N.
>
> > Look! Over There! A Pink Elephant!
>
> > That means for the set N.
>
> That means for all points
That mean for the set of all points
Virgil
Aug 14, 2012, 7:51:58 PM8/14/12
to
In article
<4f797ef1-6115-4f52...@g2g2000vba.googlegroups.com>,
I_m's, WM cannot possibly do what he claims to do, and if each I_n has
both endpoints in common with endpoints of other I_m's, they cannot
collectively cover every rational.
standard mathematics.
has existed, or ever will exist, WM is just flat out lying again.
There are, in fact, a number of proofs, valid everywhere outside
WMytheology, that no such mappings can exist.
--
<4f797ef1-6115-4f52...@g2g2000vba.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 14 Aug., 20:05, Virgil <vir...@ligriv.com> wrote:
>
> > >In no way is it possible to create additional intervals by
> > > moving endpoints - how many ever.
> >
> > It is also impossible to eliminate intervals,
>
> It is possible to eliminate intervals, namely if some I_n fall inside
> of each other.
>
> > > And remember: My proof holds for every finite n, that means for all
> > > infinitely many n in N.
> >
> > But your "proof" does not prove what it claims to prove.
>
> Of course it proves that coutably many alternating A-B-intervals
> remain countably many A-B-intervals and their endpoints cover every
> endpoint of the I-intervals.
>
> > > First if it is not applicable and valid, then Cantors diagonal proof
> > > will fail too.
> >
> > Not to those who understand it.
>
> You do not belong to that set. Otherwise you would understand the
> contradiction with my A-B-intervals. You only parrot what you have
> been taught.
Unless each I_n has both endpoints in common with endpoints of other
> On 14 Aug., 20:05, Virgil <vir...@ligriv.com> wrote:
>
> > >In no way is it possible to create additional intervals by
> > > moving endpoints - how many ever.
> >
> > It is also impossible to eliminate intervals,
>
> It is possible to eliminate intervals, namely if some I_n fall inside
> of each other.
>
> > > And remember: My proof holds for every finite n, that means for all
> > > infinitely many n in N.
> >
> > But your "proof" does not prove what it claims to prove.
>
> Of course it proves that coutably many alternating A-B-intervals
> remain countably many A-B-intervals and their endpoints cover every
> endpoint of the I-intervals.
>
> > > First if it is not applicable and valid, then Cantors diagonal proof
> > > will fail too.
> >
> > Not to those who understand it.
>
> You do not belong to that set. Otherwise you would understand the
> contradiction with my A-B-intervals. You only parrot what you have
> been taught.
I_m's, WM cannot possibly do what he claims to do, and if each I_n has
both endpoints in common with endpoints of other I_m's, they cannot
collectively cover every rational.
> >
> >
> >
> > > Finally: In the real line we have countably many rationals, i.e.,
> > > countably many right interval ends and countably many left interval
> > > ends. Although we have no information about the intervals, it is clear
> > > to everybody not yet blinded by Cantor's trick, that there cannot be
> > > more intervals than endpoints.
> >
> > If WM is claiming that the number of real points in the real interval
> > [0,1] is countable,
>
> There are obviously not more points.
What is "obvious" in WMytheology is often false, like that claim, in
> >
> >
> > > Finally: In the real line we have countably many rationals, i.e.,
> > > countably many right interval ends and countably many left interval
> > > ends. Although we have no information about the intervals, it is clear
> > > to everybody not yet blinded by Cantor's trick, that there cannot be
> > > more intervals than endpoints.
> >
> > If WM is claiming that the number of real points in the real interval
> > [0,1] is countable,
>
> There are obviously not more points.
standard mathematics.
>
> > let us see him actually count them by constructing
> > either a surjection from N to [O,1] or an injection from [0,1] to N.
>
> Since only very few exist (about as many as Cantor diagonals actually
> have been constructed since 1890) it is not possible to construct a
> bijection using "all" real numbers.
Since no surjection from N to [O,1] or injection from [0,1] to N ever
> > let us see him actually count them by constructing
> > either a surjection from N to [O,1] or an injection from [0,1] to N.
>
> Since only very few exist (about as many as Cantor diagonals actually
> have been constructed since 1890) it is not possible to construct a
> bijection using "all" real numbers.
has existed, or ever will exist, WM is just flat out lying again.
There are, in fact, a number of proofs, valid everywhere outside
WMytheology, that no such mappings can exist.
--
Virgil
Aug 14, 2012, 7:59:27 PM8/14/12
to
In article
<b6449843-73a4-4a66...@q4g2000vbb.googlegroups.com>,
(1) for every n in N there is an m in N with m > n
means exactly the same as
(2) there is an m in in N and for every n in N, m > n
Which is just another example of his quantifier dyslexia, just one of
the many quirks of WMs mythematics.
--
<b6449843-73a4-4a66...@q4g2000vbb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> > > My proof holds for every finite n, that means for all
> > > infinitely many n in N.
WM is essentially claiming that
> > > My proof holds for every finite n, that means for all
> > > infinitely many n in N.
(1) for every n in N there is an m in N with m > n
means exactly the same as
(2) there is an m in in N and for every n in N, m > n
Which is just another example of his quantifier dyslexia, just one of
the many quirks of WMs mythematics.
--
WM
Aug 15, 2012, 7:29:58 AM8/15/12
to
If something holds for n+1 if it holds for n, and if it holds for 1,
then it holds for all natural numbers. Same is valid for my points p_n
and the indices, respectively. No blathering about sets please. That
can only contribute to your further confusion.
Regards, WM
WM
Aug 15, 2012, 7:32:39 AM8/15/12
to
> <b6449843-73a4-4a66-824f-49b697b10...@q4g2000vbb.googlegroups.com>,
I mean: If my proof holds for p_1 and if its validity for p_n implies
its validity for p_(n+1), as it does, then it holds for all p_n.
Nothing more, nothing less.
Regrads, WM
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > My proof holds for every finite n, that means for all
> > > > infinitely many n in N.
>
> WM is essentially claiming that
>
> (1) for every n in N there is an m in N with m > n
>
> means exactly the same as
>
> (2) there is an m in in N and for every n in N, m > n
>
You are in error. That is Cantor's approach.
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > My proof holds for every finite n, that means for all
> > > > infinitely many n in N.
>
> WM is essentially claiming that
>
> (1) for every n in N there is an m in N with m > n
>
> means exactly the same as
>
> (2) there is an m in in N and for every n in N, m > n
>
I mean: If my proof holds for p_1 and if its validity for p_n implies
its validity for p_(n+1), as it does, then it holds for all p_n.
Nothing more, nothing less.
Regrads, WM
William Hughes
Aug 15, 2012, 9:27:07 AM8/15/12
to
On Aug 15, 8:29 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 14 Aug., 23:42, William Hughes <wpihug...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>
> > On Aug 14, 5:45 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 14 Aug., 14:58, William Hughes <wpihug...@gmail.com> wrote:
>
> > > > On Aug 14, 4:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > > My proof holds for every finite n, that means for all
> > > > > infinitely many n in N.
>
> > > > Look! Over There! A Pink Elephant!
>
> > > > That means for the set N.
>
> > > That means for all points
>
> > Look! Over There! A Pink Elephant.
>
> > That mean for the set of all points
>
> Oh no. Please stop your talking of sets.
> If something holds for n+1 if it holds for n, and if it holds for 1,
> then it holds for all natural numbers. Same is valid for my points p_n
> and the indices, respectively.
P is true for every finite subset of N.
> On 14 Aug., 23:42, William Hughes <wpihug...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>
> > On Aug 14, 5:45 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 14 Aug., 14:58, William Hughes <wpihug...@gmail.com> wrote:
>
> > > > On Aug 14, 4:40 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > > My proof holds for every finite n, that means for all
> > > > > infinitely many n in N.
>
> > > > Look! Over There! A Pink Elephant!
>
> > > > That means for the set N.
>
> > > That means for all points
>
> > Look! Over There! A Pink Elephant.
>
> > That mean for the set of all points
>
> Oh no. Please stop your talking of sets.
> If something holds for n+1 if it holds for n, and if it holds for 1,
> then it holds for all natural numbers. Same is valid for my points p_n
> and the indices, respectively.
Look! Over There! A Pink Elephant
Virgil
Aug 15, 2012, 2:41:14 PM8/15/12
to
In article
<d40ca131-8178-41e7...@13g2000vbf.googlegroups.com>,
WM does not.
WM's claims keep jumping from individuals to sets of individuals without
justification.
--
<d40ca131-8178-41e7...@13g2000vbf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 15 Aug., 01:59, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <b6449843-73a4-4a66-824f-49b697b10...@q4g2000vbb.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > My proof holds for every finite n, that means for all
> > > > > infinitely many n in N.
> >
> > WM is essentially claiming that
> >
> > (1) for every n in N there is an m in N with m > n
> >
> > means exactly the same as
> >
> > (2) there is an m in in N and for every n in N, m > n
> >
>
> You are in error. That is Cantor's approach.
Not Hardly! Cantor, and real mathematicians all know the difference, but
> On 15 Aug., 01:59, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <b6449843-73a4-4a66-824f-49b697b10...@q4g2000vbb.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > My proof holds for every finite n, that means for all
> > > > > infinitely many n in N.
> >
> > WM is essentially claiming that
> >
> > (1) for every n in N there is an m in N with m > n
> >
> > means exactly the same as
> >
> > (2) there is an m in in N and for every n in N, m > n
> >
>
> You are in error. That is Cantor's approach.
WM does not.
>
> I mean: If my proof holds for p_1 and if its validity for p_n implies
> its validity for p_(n+1), as it does, then it holds for all p_n.
> Nothing more, nothing less.
But it does not necessarily hold for any set of p_n's.
> I mean: If my proof holds for p_1 and if its validity for p_n implies
> its validity for p_(n+1), as it does, then it holds for all p_n.
> Nothing more, nothing less.
WM's claims keep jumping from individuals to sets of individuals without
justification.
--
WM
Aug 15, 2012, 2:43:02 PM8/15/12
to
You mean Cantor's diagonal argument:
The antidiagonal is not in any finite initial segment of the list?
Or do you allude to his enumeration of the rationals?
For every pair of equipotent finite initial segment of Q and N wie
have equinumerousity?
Or do you even allude to the enumeration of the algebraic numbers?
Many possibilities.
But I for my part refrain from sets but I use induction:
p_1 is true.
p_n is true ==> p_(n+1) is true
Then p(n) is true for every natural number. No pink William required
to understand that.
Regards, WM
Virgil
Aug 15, 2012, 2:45:21 PM8/15/12
to
In article
<eca14eb9-fb41-4ff5...@q35g2000yqm.googlegroups.com>,
Same is valid for my points p_n
> and the indices, respectively. No blathering about sets please. That
> can only contribute to your further confusion.
Matbe to your confusion,, WM, since you do not seem to have any clear
understanding of sets or how they differ from their members.
By WMytheological standards, since every natural is finite, the set of
all naturals must also be finite.
--
<eca14eb9-fb41-4ff5...@q35g2000yqm.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 14 Aug., 23:42, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 14, 5:45�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 14 Aug., 14:58, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Aug 14, 4:40�am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > My proof holds for every finite n, that means for all
> > > > > infinitely many n in N.
> >
> > > > Look! Over There! A Pink Elephant!
> >
> > > > That means for the set N.
> >
> > > That means for all points
> >
> > Look! Over There! A Pink Elephant.
> >
> > That mean for the set of all points
>
> Oh no. Please stop your talking of sets.
> If something holds for n+1 if it holds for n, and if it holds for 1,
> then it holds for all natural numbers.
But not necessarily for any set of natural numbers.
> On 14 Aug., 23:42, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 14, 5:45�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 14 Aug., 14:58, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Aug 14, 4:40�am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > My proof holds for every finite n, that means for all
> > > > > infinitely many n in N.
> >
> > > > Look! Over There! A Pink Elephant!
> >
> > > > That means for the set N.
> >
> > > That means for all points
> >
> > Look! Over There! A Pink Elephant.
> >
> > That mean for the set of all points
>
> Oh no. Please stop your talking of sets.
> If something holds for n+1 if it holds for n, and if it holds for 1,
> then it holds for all natural numbers.
Same is valid for my points p_n
> and the indices, respectively. No blathering about sets please. That
> can only contribute to your further confusion.
understanding of sets or how they differ from their members.
By WMytheological standards, since every natural is finite, the set of
all naturals must also be finite.
--
Virgil
Aug 15, 2012, 5:02:19 PM8/15/12
to
In article
<585572ce-fb04-4a94...@l17g2000yqg.googlegroups.com>,
but since all positions in the list are finite positions, ...
finite subsets of an infinite set and all subsets of an infinite set.
>
> Regards, WM
--
<585572ce-fb04-4a94...@l17g2000yqg.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 15 Aug., 15:27, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 15, 8:29�am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 14 Aug., 23:42, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Aug 14, 5:45�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > On 14 Aug., 14:58, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > > > On Aug 14, 4:40�am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > > > My proof holds for every finite n, that means for all
> > > > > > > infinitely many n in N.
> >
> > > > > > Look! Over There! A Pink Elephant!
> >
> > > > > > That means for the set N.
> >
> > > > > That means for all points
> >
> > > > Look! Over There! A Pink Elephant.
> >
> > > > That mean for the set of all points
> >
> > > Oh no. Please stop your talking of sets.
> > > If something holds for n+1 if it holds for n, and if it holds for 1,
> > > then it holds for all natural numbers. Same is valid for my points p_n
> > > and the indices, respectively.
> >
> > P is true for every finite subset of N.
> >
> > Look! Over There! A Pink Elephant
> >
> > P is true for every subset of N.
>
> You mean Cantor's diagonal argument:
> The antidiagonal is not in any finite initial segment of the list?
Which means that it is not in any finite position in the list,
> On 15 Aug., 15:27, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 15, 8:29�am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 14 Aug., 23:42, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Aug 14, 5:45�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > On 14 Aug., 14:58, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > > > On Aug 14, 4:40�am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > > > > My proof holds for every finite n, that means for all
> > > > > > > infinitely many n in N.
> >
> > > > > > Look! Over There! A Pink Elephant!
> >
> > > > > > That means for the set N.
> >
> > > > > That means for all points
> >
> > > > Look! Over There! A Pink Elephant.
> >
> > > > That mean for the set of all points
> >
> > > Oh no. Please stop your talking of sets.
> > > If something holds for n+1 if it holds for n, and if it holds for 1,
> > > then it holds for all natural numbers. Same is valid for my points p_n
> > > and the indices, respectively.
> >
> > P is true for every finite subset of N.
> >
> > Look! Over There! A Pink Elephant
> >
> > P is true for every subset of N.
>
> You mean Cantor's diagonal argument:
> The antidiagonal is not in any finite initial segment of the list?
but since all positions in the list are finite positions, ...
>
> Or do you allude to his enumeration of the rationals?
> For every pair of equipotent finite initial segment of Q and N wie
> have equinumerousity?
Or at least equipotency.
> Or do you allude to his enumeration of the rationals?
> For every pair of equipotent finite initial segment of Q and N wie
> have equinumerousity?
>
> Or do you even allude to the enumeration of the algebraic numbers?
He alludes to a distinction which WM seems unable to make between all
> Or do you even allude to the enumeration of the algebraic numbers?
finite subsets of an infinite set and all subsets of an infinite set.
>
> Many possibilities.
>
> But I for my part refrain from sets but I use induction:
>
> p_1 is true.
> p_n is true ==> p_(n+1) is true
> Then p(n) is true for every natural number. No pink William required
> to understand that.
You then get p(n) for each n in N, but not necessarily p(N).
> Many possibilities.
>
> But I for my part refrain from sets but I use induction:
>
> p_1 is true.
> p_n is true ==> p_(n+1) is true
> Then p(n) is true for every natural number. No pink William required
> to understand that.
>
> Regards, WM
--
William Hughes
Aug 15, 2012, 8:19:22 PM8/15/12
to
On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> But I for my part refrain from sets but I use induction:
And thus you only prove things about singletona.
> But I for my part refrain from sets but I use induction:
You can go from singletons to finite sets.
Look! Over There! A Pink Elephant!
WM
Aug 16, 2012, 3:09:33 AM8/16/12
to
On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
> On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > But I for my part refrain from sets but I use induction:
>
> And thus you only prove things about singletona.
No, I prove things for every finite sequence of points (p_1, p_2, ...,
> On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > But I for my part refrain from sets but I use induction:
>
> And thus you only prove things about singletona.
p_n). That menas, there is no finite sequence that would oppose my
result.
But the sequence of all finite sequences is the infinite sequence.
And everybody who denies this proof should show a number of points
that does no follow my argument:
>
> You can go from singletons to finite sets.
>
> Look! Over There! A Pink Elephant!
>
> You can go from singletons to infinite sets.
A proof by induction is valid for all natural numbers n. As there are
> You can go from singletons to finite sets.
>
> Look! Over There! A Pink Elephant!
>
> You can go from singletons to infinite sets.
no points p_n with unnatural indices n, my proof holds for all points.
Regards, WM
WM
Aug 16, 2012, 4:23:20 AM8/16/12
to
On 15 Aug., 23:02, Virgil <vir...@ligriv.com> wrote:
>
> > You mean Cantor's diagonal argument:
> > The antidiagonal is not in any finite initial segment of the list?
>
> Which means that it is not in any finite position in the list,
> but since all positions in the list are finite positions, ..
Just as in my proof: Since all points p_n have a finite index n, the
>
> > You mean Cantor's diagonal argument:
> > The antidiagonal is not in any finite initial segment of the list?
>
> Which means that it is not in any finite position in the list,
> but since all positions in the list are finite positions, ..
invariability of intervals holds for every number of points with
finite index n..
>
>
> > Or do you even allude to the enumeration of the algebraic numbers?
>
> He alludes to a distinction which WM seems unable to make between all
> finite subsets of an infinite set and all subsets of an infinite set.
Cantor shows equipotency for all finite subsets - not more.
>
> > Or do you even allude to the enumeration of the algebraic numbers?
>
> He alludes to a distinction which WM seems unable to make between all
> finite subsets of an infinite set and all subsets of an infinite set.
>
>
>
> > Many possibilities.
>
> > But I for my part refrain from sets but I use induction:
>
> > p_1 is true.
> > p_n is true ==> p_(n+1) is true
> > Then p(n) is true for every natural number. No pink William required
> > to understand that.
>
> You then get p(n) for each n in N, but not necessarily p(N).
I do not need p(N). p is a point and has as index a natural number n.
>
>
> > Many possibilities.
>
> > But I for my part refrain from sets but I use induction:
>
> > p_1 is true.
> > p_n is true ==> p_(n+1) is true
> > Then p(n) is true for every natural number. No pink William required
> > to understand that.
>
> You then get p(n) for each n in N, but not necessarily p(N).
For all points p_1 to p_n my proof shows that the number of intervals
remains constant during movement.
Regards, WM
William Hughes
Aug 16, 2012, 8:01:13 AM8/16/12
to
On Aug 16, 4:09 am, WM <mueck...@rz.fh-augsburg.de> wrote:
<snip>
> ... I prove things for every finite sequence of points (p_1, p_2, ...,
sequence.
Look! Over There! A Pink Elephant!
Thus I prove something is true for the infinite sequence
<snip>
> ... I prove things for every finite sequence of points (p_1, p_2, ...,
> p_n). That menas, there is no finite sequence that would oppose my
> result.
> But the sequence of all finite sequences is the infinite sequence.
>
Thus I prove something is true for every element of the infinite
> result.
> But the sequence of all finite sequences is the infinite sequence.
>
sequence.
Look! Over There! A Pink Elephant!
Alan Smaill
Aug 16, 2012, 8:25:59 AM8/16/12
to
WM <muec...@rz.fh-augsburg.de> writes:
> On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
>> On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>>
>> > But I for my part refrain from sets but I use induction:
>>
>> And thus you only prove things about singletona.
>
> No, I prove things for every finite sequence of points (p_1, p_2, ...,
> p_n). That menas, there is no finite sequence that would oppose my
> result.
> But the sequence of all finite sequences is the infinite sequence.
>
> And everybody who denies this proof should show a number of points
> that does no follow my argument:
The sum of the first n numbers is finite (by induction).
> On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
>> On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>>
>> > But I for my part refrain from sets but I use induction:
>>
>> And thus you only prove things about singletona.
>
> No, I prove things for every finite sequence of points (p_1, p_2, ...,
> p_n). That menas, there is no finite sequence that would oppose my
> result.
> But the sequence of all finite sequences is the infinite sequence.
>
> And everybody who denies this proof should show a number of points
> that does no follow my argument:
Therefore, according to WM, the sum of all the natural numbers is finite.
With proofs like this, you can find inconsistencies wherever you look.
--
Alan Smaill
Virgil
Aug 16, 2012, 2:03:42 PM8/16/12
to
In article
<85dcae41-933a-46f0...@x16g2000yqh.googlegroups.com>,
a countable set of intervals, just as the set of A_n is countable, but
the Cantor set itself is uncountable, so your argument fall flat on its
silly face, the compliment of the union of all I_n may still be
uncountable despite all your hysterical claims to the contrary.
is not.
into conformity with endpoints of the the I_N intervals unless you can
show that each endpoint of an I_n interval is the endpoint of one and
only one other I_n interval in one long connected chain, like your A-B
intervals always are, but the I_n intervals cannot be..
You cannot disconnect A-B intervals from one another, as the I_n's are
disconnected, as long as they are requires to share endpoints the way
you require them to do.
--
<85dcae41-933a-46f0...@x16g2000yqh.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 15 Aug., 23:02, Virgil <vir...@ligriv.com> wrote:
>
> >
> > > You mean Cantor's diagonal argument:
> > > The antidiagonal is not in any finite initial segment of the list?
> >
> > Which means that it is not in any finite position in the list,
> > but since all positions in the list are finite positions, ..
>
> Just as in my proof: Since all points p_n have a finite index n, the
> invariability of intervals holds for every number of points with
> finite index n..
Your what? Note the compliment of the Cantor set in the unit interval is
> On 15 Aug., 23:02, Virgil <vir...@ligriv.com> wrote:
>
> >
> > > You mean Cantor's diagonal argument:
> > > The antidiagonal is not in any finite initial segment of the list?
> >
> > Which means that it is not in any finite position in the list,
> > but since all positions in the list are finite positions, ..
>
> Just as in my proof: Since all points p_n have a finite index n, the
> invariability of intervals holds for every number of points with
> finite index n..
a countable set of intervals, just as the set of A_n is countable, but
the Cantor set itself is uncountable, so your argument fall flat on its
silly face, the compliment of the union of all I_n may still be
uncountable despite all your hysterical claims to the contrary.
> >
> >
> > > Or do you even allude to the enumeration of the algebraic numbers?
> >
> > He alludes to a distinction which WM seems unable to make between all
> > finite subsets of an infinite set and all subsets of an infinite set.
>
> Cantor shows equipotency for all finite subsets - not more.
The set of finite subsets on N is countable, the set of all subsets of N
> >
> > > Or do you even allude to the enumeration of the algebraic numbers?
> >
> > He alludes to a distinction which WM seems unable to make between all
> > finite subsets of an infinite set and all subsets of an infinite set.
>
> Cantor shows equipotency for all finite subsets - not more.
is not.
> >
> >
> >
> > > Many possibilities.
> >
> > > But I for my part refrain from sets but I use induction:
> >
> > > p_1 is true.
> > > p_n is true ==> p_(n+1) is true
> > > Then p(n) is true for every natural number. No pink William required
> > > to understand that.
> >
> > You then get p(n) for each n in N, but not necessarily p(N).
>
> I do not need p(N). p is a point and has as index a natural number n.
> For all points p_1 to p_n my proof shows that the number of intervals
> remains constant during movement.
But your "movements" can never bring the endpoints of your A-B intervals
> >
> >
> > > Many possibilities.
> >
> > > But I for my part refrain from sets but I use induction:
> >
> > > p_1 is true.
> > > p_n is true ==> p_(n+1) is true
> > > Then p(n) is true for every natural number. No pink William required
> > > to understand that.
> >
> > You then get p(n) for each n in N, but not necessarily p(N).
>
> I do not need p(N). p is a point and has as index a natural number n.
> For all points p_1 to p_n my proof shows that the number of intervals
> remains constant during movement.
into conformity with endpoints of the the I_N intervals unless you can
show that each endpoint of an I_n interval is the endpoint of one and
only one other I_n interval in one long connected chain, like your A-B
intervals always are, but the I_n intervals cannot be..
You cannot disconnect A-B intervals from one another, as the I_n's are
disconnected, as long as they are requires to share endpoints the way
you require them to do.
--
Virgil
Aug 16, 2012, 2:13:49 PM8/16/12
to
In article
<fde0eba1-9af6-4eb4...@r9g2000yqi.googlegroups.com>,
intervals conform. There are uncovered irrationals throughout [0,1] with
the I_n Intervals but no uncovered irrationals less that the upper bound
WM cannot seem to distinguish sets from their members.
> As there are
> no points p_n with unnatural indices n, my proof holds for all points.
But not for intervals, and since it is about intervals, your "proof" is
not a proof of what you allege it to prove.
There is no point to your reiterating your irrelevancies, as they do not
impress anyone outside of your WMytheological Wonderland.
--
<fde0eba1-9af6-4eb4...@r9g2000yqi.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > But I for my part refrain from sets but I use induction:
> >
> > And thus you only prove things about singletona.
>
> No, I prove things for every finite sequence of points (p_1, p_2, ...,
> p_n). That menas, there is no finite sequence that would oppose my
> result.
> But the sequence of all finite sequences is the infinite sequence.
But even if you make all the endpoints conform, you cannot make the
> On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > But I for my part refrain from sets but I use induction:
> >
> > And thus you only prove things about singletona.
>
> No, I prove things for every finite sequence of points (p_1, p_2, ...,
> p_n). That menas, there is no finite sequence that would oppose my
> result.
> But the sequence of all finite sequences is the infinite sequence.
intervals conform. There are uncovered irrationals throughout [0,1] with
the I_n Intervals but no uncovered irrationals less that the upper bound
of the endpoints of the A-B intervals
>
>
> And everybody who denies this proof should show a number of points
> that does no follow my argument:
We have shown that your argument is irrelevant.
> that does no follow my argument:
> >
> > You can go from singletons to finite sets.
> >
> > Look! Over There! A Pink Elephant!
> >
> > You can go from singletons to infinite sets.
>
> A proof by induction is valid for all natural numbers n.
But not necessarily for any set of natural numbers as set.
> > You can go from singletons to finite sets.
> >
> > Look! Over There! A Pink Elephant!
> >
> > You can go from singletons to infinite sets.
>
> A proof by induction is valid for all natural numbers n.
WM cannot seem to distinguish sets from their members.
> As there are
> no points p_n with unnatural indices n, my proof holds for all points.
not a proof of what you allege it to prove.
There is no point to your reiterating your irrelevancies, as they do not
impress anyone outside of your WMytheological Wonderland.
--
WM
Aug 17, 2012, 2:57:08 AM8/17/12
to
On 16 Aug., 14:01, William Hughes <wpihug...@gmail.com> wrote:
> On Aug 16, 4:09 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> <snip>
>
> > ... I prove things for every finite sequence of points (p_1, p_2, ...,
> > p_n). That menas, there is no finite sequence that would oppose my
> > result.
> > But the sequence of all finite sequences is the infinite sequence.
>
> Thus I prove something is true for every element of the infinite
> sequence.
For every initial segment of the infinite sequence!
> On Aug 16, 4:09 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> <snip>
>
> > ... I prove things for every finite sequence of points (p_1, p_2, ...,
> > p_n). That menas, there is no finite sequence that would oppose my
> > result.
> > But the sequence of all finite sequences is the infinite sequence.
>
> Thus I prove something is true for every element of the infinite
> sequence.
>
> Look! Over There! A Pink Elephant!
>
> Thus I prove something is true for the infinite sequence
There is no pink William required to conclude that - in mathematics.
> Look! Over There! A Pink Elephant!
>
> Thus I prove something is true for the infinite sequence
Regards, WM
WM
Aug 17, 2012, 2:57:43 AM8/17/12
to
On 16 Aug., 14:25, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
number is the first to give an infinite sum?
There is none. Therefore for all natural numbers n that you can define
individually, the sum 1 to n is defined too.
> Therefore, according to WM, the sum of all the natural numbers is finite.
There is no "all" with respect to natural numbers that you cannot
define.
But my proof does not use this distinction. Like "for all natural
numbers" the sum 1 to n can be calculated, "for all natural numbers"
my proof holds.
Regards, WM
> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
> >> On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> >> > But I for my part refrain from sets but I use induction:
>
> >> And thus you only prove things about singletona.
>
> > No, I prove things for every finite sequence of points (p_1, p_2, ...,
> > p_n). That menas, there is no finite sequence that would oppose my
> > result.
> > But the sequence of all finite sequences is the infinite sequence.
>
> > And everybody who denies this proof should show a number of points
> > that does no follow my argument:
>
> The sum of the first n numbers is finite (by induction).
What natural number is the last to give a finite sum? What natural
> > On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
> >> On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> >> > But I for my part refrain from sets but I use induction:
>
> >> And thus you only prove things about singletona.
>
> > No, I prove things for every finite sequence of points (p_1, p_2, ...,
> > p_n). That menas, there is no finite sequence that would oppose my
> > result.
> > But the sequence of all finite sequences is the infinite sequence.
>
> > And everybody who denies this proof should show a number of points
> > that does no follow my argument:
>
> The sum of the first n numbers is finite (by induction).
number is the first to give an infinite sum?
There is none. Therefore for all natural numbers n that you can define
individually, the sum 1 to n is defined too.
> Therefore, according to WM, the sum of all the natural numbers is finite.
define.
But my proof does not use this distinction. Like "for all natural
numbers" the sum 1 to n can be calculated, "for all natural numbers"
my proof holds.
Regards, WM
WM
Aug 17, 2012, 2:58:46 AM8/17/12
to
On 16 Aug., 20:13, Virgil <vir...@ligriv.com> wrote:
> But even if you make all the endpoints conform, you cannot make the
> intervals conform.
So intervals have existence independent of endpoints?
> But even if you make all the endpoints conform, you cannot make the
> intervals conform.
You give an excellent example of Matheology!
Regards, WM
Virgil
Aug 17, 2012, 4:20:17 AM8/17/12
to
In article
<a9eaf5a0-325e-4c84...@k3g2000vby.googlegroups.com>,
is always an endpoint of two intervals and each endpoint of any I_n is
almost certainly the endpoint of only one I_n interval, how do you claim
to turn all those infinitely many double endpoints into single
endpoints just by moving them around?
> You give an excellent example of Matheology!
Nowhere as good as any of yours, since I am always grounded in standard
mathematics, and you rarely are.
--
<a9eaf5a0-325e-4c84...@k3g2000vby.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 16 Aug., 20:13, Virgil <vir...@ligriv.com> wrote:
>
> > But even if you make all the endpoints conform, you cannot make the
> > intervals conform.
>
> So intervals have existence independent of endpoints?
Not at all. but since all but the first endpoint of your A_B intervals
> On 16 Aug., 20:13, Virgil <vir...@ligriv.com> wrote:
>
> > But even if you make all the endpoints conform, you cannot make the
> > intervals conform.
>
> So intervals have existence independent of endpoints?
is always an endpoint of two intervals and each endpoint of any I_n is
almost certainly the endpoint of only one I_n interval, how do you claim
to turn all those infinitely many double endpoints into single
endpoints just by moving them around?
> You give an excellent example of Matheology!
mathematics, and you rarely are.
--
Virgil
Aug 17, 2012, 4:28:52 AM8/17/12
to
In article
<656465d6-9603-4815...@r9g2000vby.googlegroups.com>,
prove exists.
two intervals.
For each I_n interval, its endpoints will usually be endpoints of only
one interval.
How does your moving A-B endpoint around, but always leaving each
endpoint but the first as an endpoint to two intervals, end up leaving
lots of those points as endpoints of only one interval?
Until you can explain that, your "proof" is poof!
--
<656465d6-9603-4815...@r9g2000vby.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 16 Aug., 14:25, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
> > >> On Aug 15, 3:43�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > >> > But I for my part refrain from sets but I use induction:
> >
> > >> And thus you only prove things about singletona.
> >
> > > No, I prove things for every finite sequence of points (p_1, p_2, ...,
> > > p_n). That menas, there is no finite sequence that would oppose my
> > > result.
> > > But the sequence of all finite sequences is the infinite sequence.
> >
> > > And everybody who denies this proof should show a number of points
> > > that does no follow my argument:
> >
> > The sum of the first n numbers is finite (by induction).
>
> What natural number is the last to give a finite sum? What natural
> number is the first to give an infinite sum?
> There is none. Therefore for all natural numbers n that you can define
> individually, the sum 1 to n is defined too.
>
> > Therefore, according to WM, the sum of all the natural numbers is finite.
>
> There is no "all" with respect to natural numbers that you cannot
> define.
Only WM would base an argument on the existence of that which he cannot
> On 16 Aug., 14:25, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
> > >> On Aug 15, 3:43�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > >> > But I for my part refrain from sets but I use induction:
> >
> > >> And thus you only prove things about singletona.
> >
> > > No, I prove things for every finite sequence of points (p_1, p_2, ...,
> > > p_n). That menas, there is no finite sequence that would oppose my
> > > result.
> > > But the sequence of all finite sequences is the infinite sequence.
> >
> > > And everybody who denies this proof should show a number of points
> > > that does no follow my argument:
> >
> > The sum of the first n numbers is finite (by induction).
>
> What natural number is the last to give a finite sum? What natural
> number is the first to give an infinite sum?
> There is none. Therefore for all natural numbers n that you can define
> individually, the sum 1 to n is defined too.
>
> > Therefore, according to WM, the sum of all the natural numbers is finite.
>
> There is no "all" with respect to natural numbers that you cannot
> define.
prove exists.
>
> But my proof does not use this distinction. Like "for all natural
> numbers" the sum 1 to n can be calculated, "for all natural numbers"
> my proof holds.
For your A-B intervals, each endpoint but the first is an endpoint of
> But my proof does not use this distinction. Like "for all natural
> numbers" the sum 1 to n can be calculated, "for all natural numbers"
> my proof holds.
two intervals.
For each I_n interval, its endpoints will usually be endpoints of only
one interval.
How does your moving A-B endpoint around, but always leaving each
endpoint but the first as an endpoint to two intervals, end up leaving
lots of those points as endpoints of only one interval?
Until you can explain that, your "proof" is poof!
--
Virgil
Aug 17, 2012, 4:32:57 AM8/17/12
to
In article
<a88f2171-0a58-449e...@f17g2000vbz.googlegroups.com>,
For your A-B intervals, each endpoint but the first is always an
endpoint of two consecutive intervals.
For each I_n interval, its endpoints will usually be endpoints of only
one interval, though may occasionally be of more than two intervals.
How does your moving A-B endpoint around, but always leaving each
endpoint but the first as an endpoint of exactly two intervals, end up
leaving lots of those points as endpoints of only one I_n interval, and
possible some as endpoints of three or more I_n intervals ?
<a88f2171-0a58-449e...@f17g2000vbz.googlegroups.com>,
endpoint of two consecutive intervals.
For each I_n interval, its endpoints will usually be endpoints of only
How does your moving A-B endpoint around, but always leaving each
leaving lots of those points as endpoints of only one I_n interval, and
possible some as endpoints of three or more I_n intervals ?
William Hughes
Aug 17, 2012, 7:01:40 AM8/17/12
to
> For every initial segment of the infinite sequence!
Look! Over There! A Pink Elephant!
For the infinite sequence
Look! Over There! A Pink Elephant!
WM
Aug 17, 2012, 7:46:08 AM8/17/12
to
finite initial segments. Every finite initial segment is the union of
all of its predecessors. If it is impossible to find an end of the
uniuon, then the sequence is unending or infinite (in Latin).
For further explanation of your mistake see the thread:
The finite infinite and the infinite infinite
Regards, WM
Alan Smaill
Aug 17, 2012, 11:46:07 AM8/17/12
to
WM <muec...@rz.fh-augsburg.de> writes:
> On 16 Aug., 14:25, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>> > On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
>> >> On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>>
>> >> > But I for my part refrain from sets but I use induction:
>>
>> >> And thus you only prove things about singletona.
>>
>> > No, I prove things for every finite sequence of points (p_1, p_2, ...,
>> > p_n). That menas, there is no finite sequence that would oppose my
>> > result.
>> > But the sequence of all finite sequences is the infinite sequence.
>>
>> > And everybody who denies this proof should show a number of points
>> > that does no follow my argument:
>>
>> The sum of the first n numbers is finite (by induction).
>
> What natural number is the last to give a finite sum? What natural
> number is the first to give an infinite sum?
> There is none. Therefore for all natural numbers n that you can define
> individually, the sum 1 to n is defined too.
Gosh, you got something right.
> On 16 Aug., 14:25, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>> > On 16 Aug., 02:19, William Hughes <wpihug...@gmail.com> wrote:
>> >> On Aug 15, 3:43 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>>
>> >> > But I for my part refrain from sets but I use induction:
>>
>> >> And thus you only prove things about singletona.
>>
>> > No, I prove things for every finite sequence of points (p_1, p_2, ...,
>> > p_n). That menas, there is no finite sequence that would oppose my
>> > result.
>> > But the sequence of all finite sequences is the infinite sequence.
>>
>> > And everybody who denies this proof should show a number of points
>> > that does no follow my argument:
>>
>> The sum of the first n numbers is finite (by induction).
>
> What natural number is the last to give a finite sum? What natural
> number is the first to give an infinite sum?
> There is none. Therefore for all natural numbers n that you can define
> individually, the sum 1 to n is defined too.
>> Therefore, according to WM, the sum of all the natural numbers is finite.
>
> There is no "all" with respect to natural numbers that you cannot
> define.
finite.
With arguments like this, WM finds contradictions everywhere.
>
> Regards, WM
>
--
Alan Smaill
William Hughes
Aug 17, 2012, 12:18:50 PM8/17/12
to
On Aug 17, 8:46 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 17 Aug., 13:01, William Hughes <wpihug...@gmail.com> wrote:
>
> > On Aug 17, 3:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > <snip>
>
> > > For every initial segment of the infinite sequence!
>
> > Look! Over There! A Pink Elephant!
>
> > For the infinite sequence
>
> Of course. The infinite sequence is not more than the union of all its
> finite initial segments.
Thus since every finite initial segment has a fixed maximum
> On 17 Aug., 13:01, William Hughes <wpihug...@gmail.com> wrote:
>
> > On Aug 17, 3:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > <snip>
>
> > > For every initial segment of the infinite sequence!
>
> > Look! Over There! A Pink Elephant!
>
> > For the infinite sequence
>
> Of course. The infinite sequence is not more than the union of all its
> finite initial segments.
element, the infinite sequence has a fixed maximum element.
WM
Aug 17, 2012, 4:13:33 PM8/17/12
to
But for those who believe in that property, the new thread about
infinities shows the contradictory consequences.
This thread has been finished as far as my contributions are
concerned.
Regards, WM
WM
Aug 17, 2012, 4:11:01 PM8/17/12
to
rubbish.
You simply are not able to recognize that every set of numbers that
you can identify, is finite. But we need not discuss your "finished
infinitities" in order to see that valid properties like the sum of 1
to n can be calculated for every element n of the complete set |N.
Likewise the possibility to move every finite set of points implies
the possibility to move all points - as long as you cannot specify a
point p_k that would spoil my proof.
I will no longer follow this thread. Your error (in case you really
believe in the nonsense that you claim here) is identified in the new
thread about infinities.
Regards, WM
Virgil
Aug 17, 2012, 4:40:45 PM8/17/12
to
In article
<86f5a589-eef0-47e4...@z6g2000vbc.googlegroups.com>,
It is remarkable how many of the things that exist in standard
mathematics do not exist in WM's mytheological world and how many of the
things that WM claims exist in his mytheological world do not exist in
standard mathematics.
--
<86f5a589-eef0-47e4...@z6g2000vbc.googlegroups.com>,
mathematics do not exist in WM's mytheological world and how many of the
things that WM claims exist in his mytheological world do not exist in
standard mathematics.
--
Virgil
Aug 17, 2012, 4:43:17 PM8/17/12
to
In article
<74b08473-75d8-459a...@q17g2000yqh.googlegroups.com>,
Hughes is not making any mistakes of his own, he is merely making fun of
WM's mistakes.
--
<74b08473-75d8-459a...@q17g2000yqh.googlegroups.com>,
WM's mistakes.
--
Virgil
Aug 17, 2012, 4:44:05 PM8/17/12
to
William Hughes
Aug 17, 2012, 4:42:34 PM8/17/12
to
Brave WM, he bravely ran away.
When danger reared its ugly head
he bravely turned his tail and fled
When danger reared its ugly head
he bravely turned his tail and fled
Alan Smaill
Aug 17, 2012, 5:38:16 PM8/17/12
to
> But for those who believe in that property, the new thread about
> infinities shows the contradictory consequences.
> This thread has been finished as far as my contributions are
> concerned.
Virgil
Aug 17, 2012, 6:08:08 PM8/17/12
to
In article
<be19e8b8-3061-45cb...@z6g2000vbc.googlegroups.com>,
> You simply are not able to recognize that every set of numbers that
> you can identify, is finite. But we need not discuss your "finished
> infinitities" in order to see that valid properties like the sum of 1
> to n can be calculated for every element n of the complete set |N.
> Likewise the possibility to move every finite set of points implies
> the possibility to move all points - as long as you cannot specify a
> point p_k that would spoil my proof.
> Your error (in case you really
> believe in the nonsense that you claim here) is identified in the new
> thread about infinities.
WM's error is that he just does not understand what is going on here.
WM does not have the power to make things work his way when standard
mathematics insists that they work some other way.
Which is most of the time.
Thus WM's world and the world of standard mathematics rarely coincide.
--
<be19e8b8-3061-45cb...@z6g2000vbc.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:
> On 17 Aug., 18:18, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 17, 8:46 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 17 Aug., 13:01, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Aug 17, 3:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > <snip>
> >
> > > > > For every initial segment of the infinite sequence!
> >
> > > > Look! Over There! A Pink Elephant!
> >
> > > > For the infinite sequence
> >
> > > Of course. The infinite sequence is not more than the union of all its
> > > finite initial segments.
> >
> > Thus since every finite initial segment has a fixed maximum
> > element, the infinite sequence has a fixed maximum element.
>
> No, as long as you believe in real infinity that comparison is
> rubbish.
But WM does not so believe, so that must hold in WMytheology.
> On 17 Aug., 18:18, William Hughes <wpihug...@gmail.com> wrote:
> > On Aug 17, 8:46 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 17 Aug., 13:01, William Hughes <wpihug...@gmail.com> wrote:
> >
> > > > On Aug 17, 3:57 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > > <snip>
> >
> > > > > For every initial segment of the infinite sequence!
> >
> > > > Look! Over There! A Pink Elephant!
> >
> > > > For the infinite sequence
> >
> > > Of course. The infinite sequence is not more than the union of all its
> > > finite initial segments.
> >
> > Thus since every finite initial segment has a fixed maximum
> > element, the infinite sequence has a fixed maximum element.
>
> No, as long as you believe in real infinity that comparison is
> rubbish.
> You simply are not able to recognize that every set of numbers that
> you can identify, is finite. But we need not discuss your "finished
> infinitities" in order to see that valid properties like the sum of 1
> to n can be calculated for every element n of the complete set |N.
> Likewise the possibility to move every finite set of points implies
> the possibility to move all points - as long as you cannot specify a
> point p_k that would spoil my proof.
For your A-B intervals, each endpoint but the first is always an
endpoint of two consecutive intervals.
For each I_n interval, its endpoints will usually be endpoints of only
one interval, though may also occasionally be of two or even more
endpoint of two consecutive intervals.
For each I_n interval, its endpoints will usually be endpoints of only
intervals.
How does your moving A-B endpoint around, but always leaving each
endpoint but the first as an endpoint of exactly two intervals, end up
leaving lots of those points as endpoints of only one I_n interval, and
possible some as endpoints of three or more I_n intervals ?
Until you can explain that, your "proof" is poof!
>
How does your moving A-B endpoint around, but always leaving each
endpoint but the first as an endpoint of exactly two intervals, end up
leaving lots of those points as endpoints of only one I_n interval, and
possible some as endpoints of three or more I_n intervals ?
Until you can explain that, your "proof" is poof!
>
> I will no longer follow this thread.
You have not really been "following it" so far!
> Your error (in case you really
> believe in the nonsense that you claim here) is identified in the new
> thread about infinities.
WM does not have the power to make things work his way when standard
mathematics insists that they work some other way.
Which is most of the time.
Thus WM's world and the world of standard mathematics rarely coincide.
--
WM
Aug 18, 2012, 2:23:07 AM8/18/12
to
On 17 Aug., 10:32, Virgil <vir...@ligriv.com> wrote:
> For your A-B intervals, each endpoint but the first is always an
> endpoint of two consecutive intervals.
Right. I call this the concertina principle.
> For your A-B intervals, each endpoint but the first is always an
> endpoint of two consecutive intervals.
>
> For each I_n interval, its endpoints will usually be endpoints of only
> one interval,
No. An endpoint of interval I_n is also endpoint of the complement of
> For each I_n interval, its endpoints will usually be endpoints of only
> one interval,
that I_n, which is either part of an interval I_m or of the complement
of all intervals UI_m.
> though may occasionally be of more than two intervals.
>
> How does your moving A-B endpoint around, but always leaving each
> endpoint but the first as an endpoint of exactly two intervals, end up
> leaving lots of those points as endpoints of only one I_n interval, and
> possible some as endpoints of three or more I_n intervals ?
other begins. My proof shows that the complement of UI_n consists of
not more than countably many intervals.
In case that three or more intervals I_n have the same endpoint, the
due intervals of the complement between these I_n do not exist.
Therefore they will not increase the number of intervals in the
complement.
But you need not bother at all. You stated that intervals are not
defined by endpoints. That amount to mathematics to be free of logic.
Regards, WM
WM
Aug 18, 2012, 2:25:59 AM8/18/12
to
On 17 Aug., 23:38, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> WM <mueck...@rz.fh-augsburg.de> writes:
> > There is no "all definable numbers".
>
> There are no non-definable natural numbers.
Every number that has a Kolomogorov complexity of 10^100 or larger is
> WM <mueck...@rz.fh-augsburg.de> writes:
> > There is no "all definable numbers".
>
> There are no non-definable natural numbers.
undefinable. Of course you are right: There are no such numbers. Where
should they exist and in which form?
Regards, WM
William Hughes
Aug 18, 2012, 8:01:28 AM8/18/12
to
On Aug 18, 3:25 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> Every number that has a Kolomogorov complexity of 10^100 or larger is
> undefinable. Of course you are right: There are no such numbers. Where
> should they exist and in which form?
Thus in Wolkenmuekenheim the infinite sequence is bounded
> Every number that has a Kolomogorov complexity of 10^100 or larger is
> undefinable. Of course you are right: There are no such numbers. Where
> should they exist and in which form?
above.. This explains a lot.
WM
Aug 18, 2012, 9:38:40 AM8/18/12
to
Do you really lack the cleverness to surpass every given threshold
with 10^100 bits? This explains a lot.
Regards, WM
WM
Aug 18, 2012, 9:37:49 AM8/18/12
to
Regards, WM
William Hughes
Aug 18, 2012, 9:56:22 AM8/18/12
to
is bounded above (this follows from the obvious fact that there
are less than 10^100 numbers with Kolmogorov complexity of less than
10^100)
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