Matheology § 022

[WM]

All rational numbers of the unit interval [0, 1] can be covered by
countably many intervals, such that the n-th rational is covered by an
interval of measure 1/10^n. There remain countably many complementary
intervals of measure 8/9 in total.
Does each of the complementary intervals contain only one
irrational number? Then there would be only countably many which could
be covered by another set of countably many intervals of measure 1/9.
My question: Can this contradiction be formalized in ZFC?
[user31686, StackExchange, 15. 4. 2012]
http://math.stackexchange.com/questions/132022/formalizing-an-idea/150674#150674

It has been mentioned already that the irrationals x of the set X of
the remaining part of measure 8/9 (or more), that is not covered by
your intervals, form a totally disconnected space, so called "Cantor
dust". Every particle x in X is separated from every other particle x'
in X by at least one rational q_n, and, as every q_n is covered by an
interval I_n, it is separated by at least one interval I_n. Since the
end points a_n and b_n of the I_n are rational numbers too, also
being covered by their own intervals, the particles of Cantor dust can
only be limits of infinite sequences of endpoints of overlapping
intervals I_n. (If they don't overlap, then the limits must come
earlier, but in any case infinitely many endpoints are required to
form a limit.) Such an infinite set of overlapping intervals is called
a cluster. In principle, given a fixed enumeration of the rationals,
we can calculate every cluster C_k and the limits of its union. Since
two clusters are disjoint (by their limits), there are only countably
many clusters (disjoint subsets of the countable set of intervals
I_n). Therefore, every irrational x can be put in bijection with the
cluster lying right of it, say, between x and its next right
neighbour. So, by this bijection we prove that the set of uncovered
irrational numbers x in X is countable.
[Stentor Schicklgruber, StackExchange (2012)]
http://math.stackexchange.com/users/32353/stentor-schicklgruber

Let all rational numbers q_n of the interval (0,oo) be covered by
intervals I_n =[s_n, t_n ] of measure |I_n| = 2^-n, such that q_n is
the center of I_n. Then there remain uncountably many irrational
numbers as uncovered "Cantor-dust". Every uncovered irrational x_a
must be separated from every uncovered irrational x_b by at least one
rational, hence by at least one interval I_n covering that rational.
But as the end points s_n and t_n of the I_n also are rational numbers
and also are covered by their own intervals, the irrationals x_a can
only be limits of infinite sequences (s_μ) or (t_ν) of endpoints of
overlapping intervals I_n. In principle we can calculate the limit x_a
of every such sequence (s_μ) or (t_ν) of endpoints of overlapping
intervals. Therefore, every irrational x_a can be put in bijection
with the infinite set of intervals lying right of it, say, between x_a
and its right neighbour x_b. There are countably many disjoint sets
like {t | t in (t_ν)} of elements of the sequences (t_ν) converging to
one of the x_a. By this bijection we get a countable set of not
covered irrational numbers x_a. Where are the other irrational numbers
that are not covered by intervals I_n? Nowhere. Uncountability is
contradicted.
[Quidquid pro quo, MathOverflow (2012)]

Regards, WM

[Uergil]

In article
<fa443090-fc4c-4fe5...@ec4g2000vbb.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> All rational numbers of the unit interval [0, 1] can be covered by
> countably many intervals, such that the n-th rational is covered by an
> interval of measure 1/10^n. There remain countably many complementary
> intervals of measure 8/9 in total.

That 8/9 would only be the case if those given intervals were pairwise
intersecting in at most one point, which given that your sequence are
originaly required to be midpoints does not happen.
--
"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

[WM]

On 29 Mai, 17:30, Uergil <Uer...@uer.net> wrote:
> In article
> <fa443090-fc4c-4fe5-9f0d-42aa80f9a...@ec4g2000vbb.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > All rational numbers of the unit interval [0, 1] can be covered by
> > countably many intervals, such that the n-th rational is covered by an
> > interval of measure 1/10^n. There remain countably many complementary
> > intervals of measure 8/9  in total.
>
> That 8/9 would only be the case if those given intervals were pairwise
> intersecting in at most one point, which given that your sequence are
> originaly required to be midpoints does not happen.

In fact there remains more than 8/9. And with the approach of quidquid
pro quo there remains even an infinity. That's not a very deep
insight. Now go an and discover all those uncovered irrationals.

Regards, WM

[Uergil]

In article
<8beddac3-30dc-45e0...@fr28g2000vbb.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 29 Mai, 17:30, Uergil <Uer...@uer.net> wrote:
> > In article
> > <fa443090-fc4c-4fe5-9f0d-42aa80f9a...@ec4g2000vbb.googlegroups.com>,
> >
> >  WM <mueck...@rz.fh-augsburg.de> wrote:
> > > All rational numbers of the unit interval [0, 1] can be covered by
> > > countably many intervals, such that the n-th rational is covered by an
> > > interval of measure 1/10^n. There remain countably many complementary
> > > intervals of measure 8/9  in total.
> >
> > That 8/9 would only be the case if those given intervals were pairwise
> > intersecting in at most one point, which given that your sequence are
> > originaly required to be midpoints does not happen.
>
> In fact there remains more than 8/9.

I just said that!

[WM]

On 29 Mai, 19:31, Uergil <Uer...@uer.net> wrote:
> In article

> <8beddac3-30dc-45e0-8314-a37b4f907...@fr28g2000vbb.googlegroups.com>,

>
>
>
>
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 29 Mai, 17:30, Uergil <Uer...@uer.net> wrote:
> > > In article
> > > <fa443090-fc4c-4fe5-9f0d-42aa80f9a...@ec4g2000vbb.googlegroups.com>,
>
> > >  WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > All rational numbers of the unit interval [0, 1] can be covered by
> > > > countably many intervals, such that the n-th rational is covered by an
> > > > interval of measure 1/10^n. There remain countably many complementary
> > > > intervals of measure 8/9  in total.
>
> > > That 8/9 would only be the case if those given intervals were pairwise
> > > intersecting in at most one point, which given that your sequence are
> > > originaly required to be midpoints does not happen.
>
> > In fact there remains more than 8/9.
>
> I just said that!
> --

Fine. And now go an and discover all those uncovered irrationals.

Regards, WM

[Uergil]

In article
<992429af-23ca-4c60...@q2g2000vbv.googlegroups.com>,

Why? Did you lose them?

[WM]

On 29 Mai, 23:35, Uergil <Uer...@uer.net> wrote:

> > > > > That 8/9 would only be the case if those given intervals were pairwise
> > > > > intersecting in at most one point, which given that your sequence are
> > > > > originaly required to be midpoints does not happen.
>
> > > > In fact there remains more than 8/9.
>
> > > I just said that!
> > > --
>
> > Fine. And now go an and discover all those uncovered irrationals.
>
> Why? Did you lose them?

I never said there were more than countably many. But you said so. Now
go on and show where we can find them.

Regards, WM

[Uergil]

In article
<fcfc2e1e-59fa-4774...@ec4g2000vbb.googlegroups.com>,

Unless WM claims that his sequence of I_n's covers the whole of [0,1],
"most" of them are outside those intervals.

Each of then will be the limit of an increasing sequence of upper bounds
of your I_n's and a decreasing sequence of lower bounds of them.

Since your initial infinite sequence of all rationals has uncountably
many infinite subsequences, this is no problem.

[WM]

On 30 Mai, 21:15, Uergil <Uer...@uer.net> wrote:
> In article

> <fcfc2e1e-59fa-4774-8f68-d4c9ef9ad...@ec4g2000vbb.googlegroups.com>,

>
>
>
>
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 29 Mai, 23:35, Uergil <Uer...@uer.net> wrote:
>
> > > > > > > That 8/9 would only be the case if those given intervals were
> > > > > > > pairwise
> > > > > > > intersecting in at most one point, which given that your sequence
> > > > > > > are
> > > > > > > originaly required to be midpoints does not happen.
>
> > > > > > In fact there remains more than 8/9.
>
> > > > > I just said that!
> > > > > --
>
> > > > Fine. And now go an and discover all those uncovered irrationals.
>
> > > Why? Did you lose them?
>
> > I never said there were more than countably many. But you said so. Now
> > go on and show where we can find them.
>
> > Regards, WM
>
> Unless WM claims that his sequence of I_n's covers the whole of [0,1],
> "most" of them are outside those intervals.
>
> Each of then will be the limit of an increasing sequence of upper bounds
> of your I_n's and a decreasing sequence of lower bounds of them.
>
> Since your initial infinite sequence of all rationals has uncountably
> many infinite subsequences, this is no problem.

The limit cannot be an inner point of a cluster, because I consider
only irrationals that are not covered by an interval I_n.

The limit cannot be in some distance of every cluster, because then it
would have a neighbour (an endpoint of a cluster) without a rational
between both. That is not mathematics.

The limits can only be endpoints of clusters. That restricts them to a
countable set.

Can it be formulated more lucid?

Regards, WM

[Uergil]

In article
<70f6cff4-fa8c-4d81...@e20g2000vbm.googlegroups.com>,

All those irrationals that are contained in some I_n will be inner
points of some cluster.

And there is nothing to prevent an irrational not in any I_n from being
the limit point of an increasing sequence of upper endpoints of I_n's or
of a decreasing sequence of lower endpoints of I_n's.

>
> The limit cannot be in some distance of every cluster, because then it
> would have a neighbour (an endpoint of a cluster) without a rational
> between both. That is not mathematics.

It is your objection that is not mathematics.

Your only "cluster" is apparently the union of all your I_n intervals
containing all rational, so that every irrational must have infinitely
many I_n's on each side of it, even if it is itself in one or more of
the I_n's.

>
> The limits can only be endpoints of clusters. That restricts them to a
> countable set.

Claimed. but unproven. At least by my understanding of what you mean by
"cluster". If by "cluster" you mean a connected subset of the union of
all the I_n, by what argument do you claim countability?

>
> Can it be formulated more lucid?

If your arguments were framed correctly, they could be far more lucid.

[WM]

On 31 Mai, 01:33, Uergil <Uer...@uer.net> wrote:

> > The limit cannot be an inner point of a cluster, because I consider
> > only irrationals that are not covered by an interval I_n.
>
> All those irrationals that are contained in some I_n will be inner
> points of some cluster.

Yes.

>
> And there is nothing to prevent an irrational not in any I_n from being
> the limit point of an increasing sequence of upper endpoints of I_n's or
> of a decreasing sequence of lower endpoints of I_n's.
>

Yes.

> > The limits can only be endpoints of clusters. That restricts them to a
> > countable set.
>
> Claimed. but unproven. At least by my understanding of what you mean by
> "cluster". If by "cluster" you mean a connected subset of the union of
> all the I_n, by what argument do you claim countability?

The set of disjoint subsets of a countable set is countable.

Regards, WM

[Mike Terry]

"WM" <muec...@rz.fh-augsburg.de> wrote in message
news:fa443090-fc4c-4fe5...@ec4g2000vbb.googlegroups.com...

> All rational numbers of the unit interval [0, 1] can be covered by
> countably many intervals, such that the n-th rational is covered by an
> interval of measure 1/10^n. There remain countably many complementary
> intervals of measure 8/9 in total.

I'm not sure if this post represents your latest thoughts after feedback
from others, but it seems to match what I found in a PDF file on your web
site. If there's a more up to date version please let me know. Anyway to
comment on this post...

Not countably many - just like even though a tree has countably many nodes
there are uncountably many paths through the tree...

> Does each of the complementary intervals contain only one
> irrational number?

Yes. If it contained two points there would be a rational inbetween,
contradiction.

> Then there would be only countably many which could
> be covered by another set of countably many intervals of measure 1/9.

No, that's not right as there needn't be only countably many complementary
intervals.

> My question: Can this contradiction be formalized in ZFC?
> [user31686, StackExchange, 15. 4. 2012]
>
http://math.stackexchange.com/questions/132022/formalizing-an-idea/150674#15
0674
>
> It has been mentioned already that the irrationals x of the set X of
> the remaining part of measure 8/9 (or more), that is not covered by
> your intervals, form a totally disconnected space, so called "Cantor
> dust". Every particle x in X is separated from every other particle x'
> in X by at least one rational q_n, and, as every q_n is covered by an
> interval I_n, it is separated by at least one interval I_n. Since the
> end points a_n and b_n of the I_n are rational numbers too, also
> being covered by their own intervals, the particles of Cantor dust can
> only be limits of infinite sequences of endpoints of overlapping
> intervals I_n.

So [0,1] has been partitioned into isolated irrationals (the "complementary
intervals"), and a countable set of disjoint "clusters".

Between any two clusters we will have at least one isolated irrational, and
between any two isolated irrationals we will have at least one cluster.

The clusters are open intervals, whose endpoints must be isolated
irrationals, but it needn't be the case that every isolated irrational is
the end point of a cluster.

For example, for an isolated irrational x, we could have two infinite
sequences of clusters C_1, C_2, C3,... and D_1, D_2, D3,... with
C_1 < C_2 < C_3 < ... < x < ... < D_3 < D_2 < D_1
and x is the least upper bound of Union(C_n) and the greatest lower bound of
Union(D_n)

(I've introduced the obvious ordering relationship on the clusters and
isolated irrationals.)

Note that x is not the end point of any cluster.

> (If they don't overlap, then the limits must come
> earlier, but in any case infinitely many endpoints are required to
> form a limit.) Such an infinite set of overlapping intervals is called
> a cluster. In principle, given a fixed enumeration of the rationals,
> we can calculate every cluster C_k and the limits of its union. Since
> two clusters are disjoint (by their limits), there are only countably
> many clusters (disjoint subsets of the countable set of intervals
> I_n).

Yes.

> Therefore, every irrational x can be put in bijection with the
> cluster lying right of it, say, between x and its next right
> neighbour.

No, there needn't be a cluster lying to the right of x. Or rather, there
needn't be any cluster directly adjacent to x. As with my example above,
there is no cluster *immediately* to the right of x. Of course, all the D_n
are "to the right of x", but not *adjacent*, and the key point is you can't
choose one of them to be uniquely associated with x in a bijection...

> So, by this bijection we prove that the set of uncovered
> irrational numbers x in X is countable.

Wrong.

> [Stentor Schicklgruber, StackExchange (2012)]
> http://math.stackexchange.com/users/32353/stentor-schicklgruber
>
> Let all rational numbers q_n of the interval (0,oo) be covered by
> intervals I_n =[s_n, t_n ] of measure |I_n| = 2^-n, such that q_n is
> the center of I_n. Then there remain uncountably many irrational
> numbers as uncovered "Cantor-dust". Every uncovered irrational x_a
> must be separated from every uncovered irrational x_b by at least one
> rational, hence by at least one interval I_n covering that rational.
> But as the end points s_n and t_n of the I_n also are rational numbers
> and also are covered by their own intervals, the irrationals x_a can

> only be limits of infinite sequences (s_�ｿｽ) or (t_�ｿｽ) of endpoints of

> overlapping intervals I_n. In principle we can calculate the limit x_a

> of every such sequence (s_�ｿｽ) or (t_�ｿｽ) of endpoints of overlapping

> intervals. Therefore, every irrational x_a can be put in bijection
> with the infinite set of intervals lying right of it,

This seems just a repeat of above?

To say correctly what the above is perhaps trying to say: Strictly, the
bijection is between the isolated irrationals, and the "cuts" in the set of
*clusters*. This is precisely similar to Dedekind's construction of the
reals as "cuts" in the rationals. Note that with Dedekind's construction of
the reals there are only countably many rationals, but somehow the cuts
create uncountably many reals!

> say, between x_a
> and its right neighbour x_b.

No, x_a does not have a "right neighbour" x_b.

> There are countably many disjoint sets

> like {t | t in (t_�ｿｽ)} of elements of the sequences (t_�ｿｽ) converging to

> one of the x_a. By this bijection we get a countable set of not
> covered irrational numbers x_a.

This is wrong. Just as with Dedekind's construction of R as cuts in Q, the
fact that the set of Clusters is countable does not imply that the "cuts" of
this set form a countable set.

> Where are the other irrational numbers
> that are not covered by intervals I_n? Nowhere. Uncountability is
> contradicted.

Not after correcting a couple of simple misconceptions!

Regards,
Mike.

[WM]

On 5 Jun., 01:24, "Mike Terry"
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message

>
> news:fa443090-fc4c-4fe5...@ec4g2000vbb.googlegroups.com...
>
> > All rational numbers of the unit interval [0, 1] can be covered by
> > countably many intervals, such that the n-th rational is covered by an
> > interval of measure 1/10^n. There remain countably many complementary
> > intervals of measure 8/9  in total.
>
> I'm not sure if this post represents your latest thoughts after feedback
> from others, but it seems to match what I found in a PDF file on your web
> site.  If there's a more up to date version please let me know.  Anyway to
> comment on this post...
>
> Not countably many - just like even though a tree has countably many nodes
> there are uncountably many paths through the tree...

Wrong. If the real axis contains countably many intervals, then there
are at most countably many intervals in the complement.

The Binary Tree has not uncountably many paths. Proof: Construct the
complete Binary Tree using only all countably many finite paths each
of being appended by an infinite tail of zeros (or, alternatively ones
or some other fixed choice). From the result you cannot determine what
kind of paths has been used and what paths are missing. Therefore you
assertion is not anything that can be proved by mathematics of digit
sequences.

>
> >    Does each of the complementary intervals contain only one
> > irrational number?
>
> Yes.  If it contained two points there would be a rational inbetween,
> contradiction.

So each one is a singleton. Why is it a singleton? What makes it a
singleton? Intervals around rationals.

>
> For example, for an isolated irrational x, we could have two infinite
> sequences of clusters C_1, C_2, C3,... and D_1, D_2, D3,... with
>     C_1 < C_2 < C_3 < ... < x < ... < D_3 < D_2 < D_1
> and x is the least upper bound of Union(C_n) and the greatest lower bound of
> Union(D_n)
>
> (I've introduced the obvious ordering relationship on the clusters and
> isolated irrationals.)
>
> Note that x is not the end point of any cluster.

Since x is a singleton, there must be something that makes it a
singleton. And there is no other means than an interval or cluster.

>
> > (If they don't overlap, then the limits must come
> > earlier, but in any case infinitely many endpoints are required to
> > form a limit.) Such an infinite set of overlapping intervals is called
> > a cluster. In principle, given a fixed enumeration of the rationals,
> > we can calculate every cluster C_k and the limits of its union. Since
> > two clusters are disjoint (by their limits), there are only countably
> > many clusters (disjoint subsets of the countable set of intervals
> > I_n).
>
> Yes.
>
> > Therefore, every irrational x can be put in bijection with the
> > cluster lying right of it, say, between x and its next right
> > neighbour.
>
> No, there needn't be a cluster lying to the right of x.  Or rather, there
> needn't be any cluster directly adjacent to x.

Then x is not a singleton.

 As with my example above,
> there is no cluster *immediately* to the right of x.  Of course, all the D_n
> are "to the right of x", but not *adjacent*, and the key point is you can't
> choose one of them to be uniquely associated with x in a bijection...

It is not necessary to choose an adjacent one. It is important that
there *is* an adjacent one, because otherwise x is not a singleton.
And every interval is extended, i.e., it is not only a point.
Therefore x is even isolated.

> > say, between x_a
> > and its right neighbour x_b.
>
> No, x_a does not have a "right neighbour" x_b.

Yes, because x_a = x_b.

> > There are countably many disjoint sets

> > like {t | t in (t_ )} of elements of the sequences (t_ ) converging to

> > one of the x_a. By this bijection we get a countable set of not
> > covered irrational numbers x_a.
>
> This is wrong.  Just as with Dedekind's construction of R as cuts in Q, the
> fact that the set of Clusters is countable does not imply that the "cuts" of
> this set form a countable set.

Of course the cuts form a countable set. Every cut is defined by a
finite word. The set of finite words is countable.

Regards, WM

[WM]

On 5 Jun., 01:24, "Mike Terry"
<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message

> This is wrong.  Just as with Dedekind's construction of R as cuts in Q, the
> fact that the set of Clusters is countable does not imply that the "cuts" of
> this set form a countable set.

Consider the unit interval bent to a circle such that 0 is the same
point as 1.

Now define intervals, each by two points. Are you able to define
countably many intervals while the complement contains uncountably
many? How would you distinguish between intervals and intervals of the
complement?

Note, you need not to union overlapping intervals. Leave each one as
it stand. Is there anything uncountable? But if, after all,
overlapping intervals are unioned, can the intervals of the
compelement grow in number to uncountability? Reducing one number
increases the other number.

Regards, WM

[Mike Terry]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:6cb84549-1e6b-4efe...@j9g2000vbk.googlegroups.com...

> On 5 Jun., 01:24, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
> >
news:fa443090-fc4c-4fe5...@ec4g2000vbb.googlegroups.com...
> >
> > > All rational numbers of the unit interval [0, 1] can be covered by
> > > countably many intervals, such that the n-th rational is covered by an
> > > interval of measure 1/10^n. There remain countably many complementary
> > > intervals of measure 8/9 in total.
> >
> > I'm not sure if this post represents your latest thoughts after feedback
> > from others, but it seems to match what I found in a PDF file on your
web
> > site. If there's a more up to date version please let me know. Anyway to
> > comment on this post...
> >
> > Not countably many - just like even though a tree has countably many
nodes
> > there are uncountably many paths through the tree...
>
> Wrong. If the real axis contains countably many intervals, then there
> are at most countably many intervals in the complement.

No, you're just failing to visualise what is possible with intervals.
Because you can't visualise this, you just keep asserting it's impossible.

Anyway, for this thread it suffices for me to point out that your claim that
there are only countably many singletons in the complement requires proof,
and that you've not yet provided any proof, only assertions.

>
> The Binary Tree has not uncountably many paths. Proof: Construct the
> complete Binary Tree using only all countably many finite paths each
> of being appended by an infinite tail of zeros (or, alternatively ones
> or some other fixed choice). From the result you cannot determine what
> kind of paths has been used and what paths are missing. Therefore you
> assertion is not anything that can be proved by mathematics of digit
> sequences.
> >
> > > Does each of the complementary intervals contain only one
> > > irrational number?
> >
> > Yes. If it contained two points there would be a rational inbetween,
> > contradiction.
>
> So each one is a singleton. Why is it a singleton? What makes it a
> singleton?

The definition of a singleton! There is no neighbourhood of x which is
contained in the complement.

Or put another way, every neighbourhood of x contains a point belonging to a
cluster.

I think I should explain more carefully what this means. It means that if
you give me a neighbourhood N of x, THEN I can find a cluster C that
intersects N. Note that my choice of C is made AFTER you give me N, so I'm
free to choose a different C for each different N you give me. It does not
mean that there is a single cluster C such that C intersects every neighbour
hood of x. The order of qualifiers is important!

> Intervals around rationals.

You could say this of course, but it does not mean the singletons are
"directly adjacent" to any clusters - there can be infinite sequences of
clusters approaching closer and closer to x. [as suggested in my C_n, D_n
example below]

>
> >
> > For example, for an isolated irrational x, we could have two infinite
> > sequences of clusters C_1, C_2, C3,... and D_1, D_2, D3,... with
> > C_1 < C_2 < C_3 < ... < x < ... < D_3 < D_2 < D_1
> > and x is the least upper bound of Union(C_n) and the greatest lower
bound of
> > Union(D_n)
> >
> > (I've introduced the obvious ordering relationship on the clusters and
> > isolated irrationals.)
> >
> > Note that x is not the end point of any cluster.
>
> Since x is a singleton, there must be something that makes it a
> singleton. And there is no other means than an interval or cluster.

No, that's wrong. It can be an infinite sequence of clusters as above.

If you don't think such a situation is possible, and that every x is the
endpoint of a cluster then by all means provide a proof of this - but an
assertion is not the same as a proof. And repeatedly stating "there must be
something that makes it a singleton" is not a proof! :)

> >
> > > (If they don't overlap, then the limits must come
> > > earlier, but in any case infinitely many endpoints are required to
> > > form a limit.) Such an infinite set of overlapping intervals is called
> > > a cluster. In principle, given a fixed enumeration of the rationals,
> > > we can calculate every cluster C_k and the limits of its union. Since
> > > two clusters are disjoint (by their limits), there are only countably
> > > many clusters (disjoint subsets of the countable set of intervals
> > > I_n).
> >
> > Yes.
> >
> > > Therefore, every irrational x can be put in bijection with the
> > > cluster lying right of it, say, between x and its next right
> > > neighbour.
> >
> > No, there needn't be a cluster lying to the right of x. Or rather, there
> > needn't be any cluster directly adjacent to x.
>
> Then x is not a singleton.

Why not? This is exactly what you need to prove rather than just assert.

>
>
> As with my example above,
> > there is no cluster *immediately* to the right of x. Of course, all the
D_n
> > are "to the right of x", but not *adjacent*, and the key point is you
can't
> > choose one of them to be uniquely associated with x in a bijection...
>
> It is not necessary to choose an adjacent one. It is important that
> there *is* an adjacent one, because otherwise x is not a singleton.

No, this is your same mistake again: assuming that each singleton must be
the endpoint of a cluster. [I see that several others have said the same
thing to you in other threads, so this seems to be the key sticking point.]

> And every interval is extended, i.e., it is not only a point.

Well, most people count the closed interval [x,x] as an interval consisting
of a single point. Obviously such intervals aren't extended. However, all
the intervals removed from the line (centred on the rationals) were by
definition extended. What remains could be said to be singleton intervals
(which aren't extended), or don't describe the {x} as intervals if you don't
want [x,x] to be counted as an interval...

> Therefore x is even isolated.
>
> > > say, between x_a
> > > and its right neighbour x_b.
> >
> > No, x_a does not have a "right neighbour" x_b.
>
> Yes, because x_a = x_b.

That doesn't make sense. You said

"Therefore, every irrational x_a can be put in bijection
with the infinite set of intervals lying right of it, say,

between x_a and its right neighbour x_b."

I said "No, x_a does not have a "right neighbour" x_b". [An irrational
(belonging to the complement) need not have a unique right neighbour as
there could be an infinite sequence of such, with no unique right
neighbour.] You said

"Yes, because x_a = x_b".

So you're saying look at the infinite set of intervals between x_a and x_b,
where x_a = x_b.

>
>
> > > There are countably many disjoint sets
> > > like {t | t in (t_ )} of elements of the sequences (t_ ) converging to
> > > one of the x_a. By this bijection we get a countable set of not
> > > covered irrational numbers x_a.
> >
> > This is wrong. Just as with Dedekind's construction of R as cuts in Q,
the
> > fact that the set of Clusters is countable does not imply that the
"cuts" of
> > this set form a countable set.
>
> Of course the cuts form a countable set. Every cut is defined by a
> finite word. The set of finite words is countable.

It's not enough that you just assert things like this. You must prove such
claims! E.g. you could start with saying how each cut is defined by a
finite word.


Regards,
Mike.

[Mike Terry]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:457adda6-1640-4078...@b26g2000vbt.googlegroups.com...

> On 5 Jun., 01:24, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>
> > This is wrong. Just as with Dedekind's construction of R as cuts in Q,
the
> > fact that the set of Clusters is countable does not imply that the
"cuts" of
> > this set form a countable set.
>
>
> Consider the unit interval bent to a circle such that 0 is the same
> point as 1.
>
> Now define intervals, each by two points. Are you able to define
> countably many intervals while the complement contains uncountably
> many? How would you distinguish between intervals and intervals of the
> complement?

Yes of course - the well known Cantor set is an example:

http://en.wikipedia.org/wiki/Cantor_set

I don't understand what you mean about "distinguish between intervals and
intervals of the complement". Obviously they are distinguished by not
belonging to any of the intervals that were removed. The intervals of the
complement are singleton intervals consisting of isolated single points {x}.

Ah, maybe I see what you're asking. Perhaps you're pointing out that a
bunch of intervals were removed, and another bunch remain at the end. The
intervals removed were countable, and an uncountable set remains! Well,
note that any set of disjoint intervals of non-zero extent will always be
countable, because each contains a rational. However, this isn't the case
with intervals of zero length (i.e. singleton sets {x}). Obviously we have
R = Union {{x}: x element of R}, and we have partitioned R into uncountably
many disjoint "intervals" {x}. [Or perhaps you don't believe R is
uncountable, but that would be your problem, not ours...]

So perhaps the answer you're seeking is to point out the asymmetry between
the intervals removed (which are of non-zero extent) and the intervals that
remain (which are singleton sets)?

>
> Note, you need not to union overlapping intervals. Leave each one as
> it stand. Is there anything uncountable?

Yes. So this hasn't supported your case at all...

More to the point: if you believe only countably many intervals can remain
then since it is YOU making claims here, it is YOU who needs to prove this.
(Others don't need to prove anything or provide counter examples or what
not - the onus is on you to prove what you claim.)

Regards,
Mike.

[WM]

On 5 Jun., 20:11, "Mike Terry"

<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>
> news:457adda6-1640-4078...@b26g2000vbt.googlegroups.com...
>
>
>
>
>
> > On 5 Jun., 01:24, "Mike Terry"
> > <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>
> > > This is wrong. Just as with Dedekind's construction of R as cuts in Q,
> the
> > > fact that the set of Clusters is countable does not imply that the
> "cuts" of
> > > this set form a countable set.
>
> > Consider the unit interval bent to a circle such that 0 is the same
> > point as 1.
>
> > Now define intervals, each by two points. Are you able to define
> > countably many intervals while the complement contains uncountably
> > many? How would you distinguish between intervals and intervals of the
> > complement?
>
> Yes of course - the well known Cantor set is an example:

No. There you have only uncountably many *points* in the remaining
set. As there are also rationals like 1/4 and 3/10 remaining, that is
not an obvious contradiction.

But if in my example you happen, accidentally, to put all intervals of
length 10^-n around all rational numbers q_n (that is very improbable
but not impossible), then you have exactly countably many intervals
with rationals and uncountably many intervals including singletons
without rationals.

Now let me hide the scale. How would you distinguish between the
intervals?

>
>
> I don't understand what you mean about "distinguish between intervals and
> intervals of the complement".  Obviously they are distinguished by not
> belonging to any of the intervals that were removed.

There is nothing removed. All intervals remain where they are.

> The intervals of the
> complement are singleton intervals consisting of isolated single points {x}.

What isolates these points?

>
> Ah, maybe I see what you're asking.  Perhaps you're pointing out that a
> bunch of intervals were removed, and another bunch remain at the end.  The
> intervals removed were countable, and an uncountable set remains!  Well,
> note that any set of disjoint intervals of non-zero extent will always be
> countable, because each contains a rational.  However, this isn't the case
> with intervals of zero length (i.e. singleton sets {x}).

What makes them having zero length?

> Obviously we have
> R = Union {{x}: x element of R}, and we have partitioned R into uncountably
> many disjoint "intervals" {x}.  [Or perhaps you don't believe R is
> uncountable, but that would be your problem, not ours...]

I am open to every explanation. For instance if you distribute the
same amount of intervals in another configuration, will there also be
one kind countable and the other kind will not?

>
> So perhaps the answer you're seeking is to point out the asymmetry between
> the intervals removed (which are of non-zero extent) and the intervals that
> remain (which are singleton sets)?

As I said, nothing is removed. In addition you don't know the scale
but put the intervals by accident.

>
>
> More to the point: if you believe only countably many intervals can remain
> then since it is YOU making claims here, it is YOU who needs to prove this.
> (Others don't need to prove anything or provide counter examples or what
> not - the onus is on you to prove what you claim.)

Consider a circle. Mark two pints: then you have two intervals. Mark
another pair of points, this will again give you two intervals (not
counting the interrelation with the first marked points). Mark another
pair of points. That will give you again two intervals. Mark aleph_0
pairs of points. That will give you, according to your belief, aleph_0
intervals of the first kind and 2^aleph_0 intervals of the second
kind. But which is the first and which is the second kind? If you
can't answer this question, then I believe I have proved my claim.

Regards, WM

[WM]

On 5 Jun., 19:48, "Mike Terry"

<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>

> Or put another way, every neighbourhood of x contains a point belonging to a
> cluster.
>
> I think I should explain more carefully what this means. It means that if
> you give me a neighbourhood N of x, THEN I can find a cluster C that
> intersects N. Note that my choice of C is made AFTER you give me N, so I'm
> free to choose a different C for each different N you give me. It does not
> mean that there is a single cluster C such that C intersects every neighbour
> hood of x. The order of qualifiers is important!

But the real axis is a static, timeless object. There you have no hide
and seek play but a point either is disconnected from all points that
are smaller and larger, or it is not. I dont give you any N and no
quantifyer, but I ask whether the point x is single like the point 0
in the definition:
f(0) = 0 and f(x) = 1 for all x =/= 0.
If this is the case, then I know that the point 0 is a singleton and
interrupts a removed interval.

>
> > Intervals around rationals.
>
> You could say this of course, but it does not mean the singletons are
> "directly adjacent" to any clusters - there can be infinite sequences of
> clusters approaching closer and closer to x. [as suggested in my C_n, D_n
> example below]

And by this ecstasy of infinity we numbers get a bit muddled up and
countably many intervals are cut by uncountably many singletons
without getting uncountably many pieces. A fine result.

>
> If you don't think such a situation is possible, and that every x is the
> endpoint of a cluster then by all means provide a proof of this - but an
> assertion is not the same as a proof. And repeatedly stating "there must be
> something that makes it a singleton" is not a proof! :)

Any uncovered irrational is a singleton only when it interrupts an
interval.

> > > No, there needn't be a cluster lying to the right of x. Or rather, there
> > > needn't be any cluster directly adjacent to x.
>
> > Then x is not a singleton.
>
> Why not? This is exactly what you need to prove rather than just assert.

It is not necessary to prove it. It is the definition. A singleton is
a set of 1 element that is not covered, so it is 1 point, and there
must be an interval with measure > 0 at both sides that has been
covered. Remember, the real axis is a static object. There is no hide
and seek play adequate.

> No, this is your same mistake again: assuming that each singleton must be
> the endpoint of a cluster. [I see that several others have said the same
> thing to you in other threads, so this seems to be the key sticking point.]

If it is not the endpoint, then there is a point between it and the
endpoint, no?

>
> > And every interval is extended, i.e., it is not only a point.
>
> Well, most people count the closed interval [x,x] as an interval consisting
> of a single point.

That is correct. But all intervals that cover the rationals q_n have
the measure 10^-n. None of them is a point.

> Obviously such intervals aren't extended. However, all
> the intervals removed from the line (centred on the rationals) were by
> definition extended. What remains could be said to be singleton intervals
> (which aren't extended), or don't describe the {x} as intervals if you don't
> want [x,x] to be counted as an interval...

That's ok. They are counted as intervals. (Even the empty set is
counted as an interval.) But they are only then points, if they are
the only uncovered points in a neighbourhood where all other points
are covered.

So much for now.

Regards, WM

[Mike Terry]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:b425283f-00aa-4f1a...@d6g2000vbe.googlegroups.com...

> On 5 Jun., 20:11, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
news:457adda6-1640-4078...@b26g2000vbt.googlegroups.com...
> >
> > > On 5 Jun., 01:24, "Mike Terry"
> > > <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > > > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
> > > > This is wrong. Just as with Dedekind's construction of R as cuts in
Q,
> > the
> > > > fact that the set of Clusters is countable does not imply that the
> > "cuts" of
> > > > this set form a countable set.
> >
> > > Consider the unit interval bent to a circle such that 0 is the same
> > > point as 1.
> >
> > > Now define intervals, each by two points. Are you able to define
> > > countably many intervals while the complement contains uncountably
> > > many? How would you distinguish between intervals and intervals of the
> > > complement?
> >
> > Yes of course - the well known Cantor set is an example:
>
> No. There you have only uncountably many *points* in the remaining
> set. As there are also rationals like 1/4 and 3/10 remaining, that is
> not an obvious contradiction.

Each remaining point is isolated, and so forms a closed *interval* of length
0. Hence there are uncountably many intervals remaining in the complement -
AS REQUESTED. Nobody has claimed there is any contradiction in this!

Of course, technically whether [x,x] is called an interval is down to the
definition of interval. Most people count it as an interval. Do you? (I
will try to use whatever terminology you use - this does not alter the
maths.)

>
> But if in my example you happen, accidentally, to put all intervals of
> length 10^-n around all rational numbers q_n (that is very improbable
> but not impossible), then you have exactly countably many intervals
> with rationals and uncountably many intervals including singletons
> without rationals.
>
> Now let me hide the scale. How would you distinguish between the
> intervals?
> >
> >
> > I don't understand what you mean about "distinguish between intervals
and
> > intervals of the complement". Obviously they are distinguished by not
> > belonging to any of the intervals that were removed.
>
> There is nothing removed. All intervals remain where they are.

Duh - I'm not suggesting that the numbers in removed intervals have been
physically destroyed, or moved into a jam jar. OK, Let me reword: Obviously

they are distinguished by not belonging to any of the intervals that were

used to form the complement.

>
>
> > The intervals of the
> > complement are singleton intervals consisting of isolated single points
{x}.
>
> What isolates these points?

Are you really unable to answer this yourself? Look up the definition of
isolated, or read my explanations elsewhere. (I'm not going to repeat
explanations over and over for you.)

> >
> > Ah, maybe I see what you're asking. Perhaps you're pointing out that a
> > bunch of intervals were removed, and another bunch remain at the end.
The
> > intervals removed were countable, and an uncountable set remains! Well,
> > note that any set of disjoint intervals of non-zero extent will always
be
> > countable, because each contains a rational. However, this isn't the
case
> > with intervals of zero length (i.e. singleton sets {x}).
>
> What makes them having zero length?

They have zero length because there are "removed" points arbitrarily close
to them both above and below them.

>
> > Obviously we have
> > R = Union {{x}: x element of R}, and we have partitioned R into
uncountably
> > many disjoint "intervals" {x}. [Or perhaps you don't believe R is
> > uncountable, but that would be your problem, not ours...]
>
> I am open to every explanation. For instance if you distribute the
> same amount of intervals in another configuration, will there also be
> one kind countable and the other kind will not?

No. It depends on the configuration. E.g. if you bunch all the intervals
up so they are adjacent at the lower end of the interval, then their union
will be a solid block from 0 to 1/9. (1/9 being the total length of the
intervals). The complement will be the interval [1/9, 1].

> >
> > So perhaps the answer you're seeking is to point out the asymmetry
between
> > the intervals removed (which are of non-zero extent) and the intervals
that
> > remain (which are singleton sets)?
>
> As I said, nothing is removed. In addition you don't know the scale
> but put the intervals by accident.

(I've no idea what you mean by that.)

> >
> >
> > More to the point: if you believe only countably many intervals can
remain
> > then since it is YOU making claims here, it is YOU who needs to prove
this.
> > (Others don't need to prove anything or provide counter examples or what
> > not - the onus is on you to prove what you claim.)
>
> Consider a circle. Mark two pints: then you have two intervals. Mark
> another pair of points, this will again give you two intervals (not
> counting the interrelation with the first marked points). Mark another
> pair of points. That will give you again two intervals. Mark aleph_0
> pairs of points. That will give you, according to your belief, aleph_0
> intervals of the first kind and 2^aleph_0 intervals of the second
> kind. But which is the first and which is the second kind? If you
> can't answer this question, then I believe I have proved my claim.

Without going into any details, if you ask me to do something and I don't do
it for whatever reason, that in no way provides a proof of anything you
claim.


Regards,
Mike.

[WM]

On 5 Jun., 22:01, "Mike Terry"

<news.dead.person.sto...@darjeeling.plus.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>

> news:b425283f-00aa-4f1a...@d6g2000vbe.googlegroups.com...>

>
> Of course, technically whether [x,x] is called an interval is down to the
> definition of interval. Most people count it as an interval. Do you? (I
> will try to use whatever terminology you use - this does not alter the
> maths.)

I call [x,x] an interval.

> > There is nothing removed. All intervals remain where they are.
>
> Duh - I'm not suggesting that the numbers in removed intervals have been
> physically destroyed, or moved into a jam jar. OK, Let me reword: Obviously
> they are distinguished by not belonging to any of the intervals that were
> used to form the complement.

Yes. But at the beginning there is not at all clear what distinguishes
an interval from the complement. You just cut two notches into the
circle and have an interval and its complement. But apart from the
naming there is no asymmetry between them. So let us better say
intervals of the first kind and the second kind with no qualification.

>
> > I am open to every explanation. For instance if you distribute the
> > same amount of intervals in another configuration, will there also be
> > one kind countable and the other kind will not?
>
> No. It depends on the configuration. E.g. if you bunch all the intervals
> up so they are adjacent at the lower end of the interval, then their union
> will be a solid block from 0 to 1/9. (1/9 being the total length of the
> intervals). The complement will be the interval [1/9, 1].

Please let us wait a bit before we union anything. We have aleph_0
intervals of the first kind and how many of the second kind?

> > As I said, nothing is removed. In addition you don't know the scale
> > but put the intervals by accident.
>
> (I've no idea what you mean by that.)

We have cut aleph_0 notches into the circle and have produced aleph_0
intervals of the first kind. These notches have been cut at random.

>
>
>
>
>
>
>
>
>
> > > More to the point: if you believe only countably many intervals can
> remain
> > > then since it is YOU making claims here, it is YOU who needs to prove
> this.
> > > (Others don't need to prove anything or provide counter examples or what
> > > not - the onus is on you to prove what you claim.)
>
> > Consider a circle. Mark two pints: then you have two intervals. Mark
> > another pair of points, this will again give you two intervals (not
> > counting the interrelation with the first marked points). Mark another
> > pair of points. That will give you again two intervals. Mark aleph_0
> > pairs of points. That will give you, according to your belief, aleph_0
> > intervals of the first kind and 2^aleph_0 intervals of the second
> > kind. But which is the first and which is the second kind? If you
> > can't answer this question, then I believe I have proved my claim.
>
> Without going into any details, if you ask me to do something and I don't do
> it for whatever reason, that in no way provides a proof of anything you
> claim.

I see. Above you said the onus is on me to prove what I claim. But if
I tell you my proof and you don't listen for whatever reason, then
that in no way provides a proof of anything I claim.

Regards, WM

[Uergil]

In article
<457adda6-1640-4078...@b26g2000vbt.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 5 Jun., 01:24, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>
> > This is wrong.  Just as with Dedekind's construction of R as cuts in Q, the
> > fact that the set of Clusters is countable does not imply that the "cuts" of
> > this set form a countable set.
>
>
> Consider the unit interval bent to a circle such that 0 is the same
> point as 1.
>
> Now define intervals, each by two points. Are you able to define
> countably many intervals while the complement contains uncountably
> many? How would you distinguish between intervals and intervals of the
> complement?

In the sequence of intervals I_n for covering the rationals of [0,1]
there will be at least one contain 0 and at least one containing 1.
Add one more "interval" of their union to the front of the sequence so
that one cluster goes "round the corner".

Other than now having the combined 0-1 point in a cluster, this is
excatly the same as before being circled.

So WM's attempt to get "cute" again fails.

[Uergil]

In article
<6cb84549-1e6b-4efe...@j9g2000vbk.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 5 Jun., 01:24, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
> > news:fa443090-fc4c-4fe5...@ec4g2000vbb.googlegroups.com...
> >
> > > All rational numbers of the unit interval [0, 1] can be covered by
> > > countably many intervals, such that the n-th rational is covered by an
> > > interval of measure 1/10^n. There remain countably many complementary

> > > intervals of measure 8/9 ļæ½in total.

> >
> > I'm not sure if this post represents your latest thoughts after feedback
> > from others, but it seems to match what I found in a PDF file on your web

> > site. ļæ½If there's a more up to date version please let me know. ļæ½Anyway to

> > comment on this post...
> >
> > Not countably many - just like even though a tree has countably many nodes
> > there are uncountably many paths through the tree...
>
> Wrong. If the real axis contains countably many intervals, then there
> are at most countably many intervals in the complement.

While there are zero intervals in the compliment of the real axis, there
are uncountably many in the axis itself.

Not that if any real interval of positive length were countable, it
would have a cover of arbirtarily small measure.

>
> The Binary Tree has not uncountably many paths. Proof: Construct the
> complete Binary Tree using only all countably many finite paths each
> of being appended by an infinite tail of zeros (or, alternatively ones
> or some other fixed choice). From the result you cannot determine what
> kind of paths has been used and what paths are missing. Therefore you
> assertion is not anything that can be proved by mathematics of digit
> sequences.

The uncountability follows immediately from the original Cantor
Dialgonal proof that there are uncountalby many infinite binary
sequences.

> >

>
> >
> > For example, for an isolated irrational x, we could have two infinite
> > sequences of clusters C_1, C_2, C3,... and D_1, D_2, D3,... with

> > ļæ½ ļæ½ C_1 < C_2 < C_3 < ... < x < ... < D_3 < D_2 < D_1

> > and x is the least upper bound of Union(C_n) and the greatest lower bound of
> > Union(D_n)
> >
> > (I've introduced the obvious ordering relationship on the clusters and
> > isolated irrationals.)
> >
> > Note that x is not the end point of any cluster.
>
> Since x is a singleton, there must be something that makes it a
> singleton. And there is no other means than an interval or cluster.

How about a seguence of clusters (and uncovered irrationls) coverging to
it?

Works for me, and is workabloe for anyone other than WM.

> >
> > > (If they don't overlap, then the limits must come
> > > earlier, but in any case infinitely many endpoints are required to
> > > form a limit.) Such an infinite set of overlapping intervals is called
> > > a cluster. In principle, given a fixed enumeration of the rationals,
> > > we can calculate every cluster C_k and the limits of its union. Since
> > > two clusters are disjoint (by their limits), there are only countably
> > > many clusters (disjoint subsets of the countable set of intervals
> > > I_n).
> >
> > Yes.
> >
> > > Therefore, every irrational x can be put in bijection with the
> > > cluster lying right of it, say, between x and its next right
> > > neighbour.
> >

> > No, there needn't be a cluster lying to the right of x. ļæ½Or rather, there

> > needn't be any cluster directly adjacent to x.
>
> Then x is not a singleton.

I do not know, or care, what WM means by "singleton", but if it
prohibits an uncovered irrational from being the limit of a sequence of
uncovered irrationals, or the limit of a sequence o clusters, it is
wrong.
>
>
> ļæ½As with my example above,
> > there is no cluster *immediately* to the right of x. ļæ½Of course, all the D_n

> > are "to the right of x", but not *adjacent*, and the key point is you can't
> > choose one of them to be uniquely associated with x in a bijection...
>
> It is not necessary to choose an adjacent one. It is important that
> there *is* an adjacent one, because otherwise x is not a singleton.

Then there are, and must be, x's which are not singletons, i.e., are not
boundary points of clusters.

> And every interval is extended, i.e., it is not only a point.
> Therefore x is even isolated.

Consider the following irrational sequences in (0,1):

1/sqrt(2) - 1/(sqrt(13)*n)
and
1/sqrt(2) + 1/(sqrt(13)*n)

Their members are all irrational and could well be uncovered irrationals
in one of WM's constructions.

But note that 1/sqrt(2) is now separated by at least one rational and
one irrational from EACH sequence member, but is not separated by any
rational, or irrational, or anything else, from ALL sequence members or
all spaces between consecutive sequence members, as both sequences
converge to it.

Thus 1/sqrt(2) could itself be an uncovered irrational of the very sort
that WM claims cannot ever exist.

Thus demonstrating that what WM claims cannot occur can quite easily
occur.

>
> > > say, between x_a
> > > and its right neighbour x_b.
> >
> > No, x_a does not have a "right neighbour" x_b.
>
> Yes, because x_a = x_b.

In a rational world neighbors differ from each other, and are not clones.

>
>
> > > There are countably many disjoint sets
> > > like {t | t in (t_ )} of elements of the sequences (t_ ) converging to
> > > one of the x_a. By this bijection we get a countable set of not
> > > covered irrational numbers x_a.
> >

> > This is wrong. ļæ½Just as with Dedekind's construction of R as cuts in Q, the

> > fact that the set of Clusters is countable does not imply that the "cuts" of
> > this set form a countable set.
>
> Of course the cuts form a countable set.

WRONG!

>Every cut is defined by a finite word.

WRONG!

> The set of finite words is countable.

Irrelevant.

[Uergil]

In article
<80ab7315-3870-42f0...@q2g2000vbv.googlegroups.com>,

And you have not ever shown that this must be the case while several
people, have shown it need not be the case.

>
> > > > No, there needn't be a cluster lying to the right of x. Or rather, there
> > > > needn't be any cluster directly adjacent to x.
> >
> > > Then x is not a singleton.
> >
> > Why not? This is exactly what you need to prove rather than just assert.
>
> It is not necessary to prove it. It is the definition. A singleton is
> a set of 1 element that is not covered, so it is 1 point, and there
> must be an interval with measure > 0 at both sides that has been
> covered. Remember, the real axis is a static object. There is no hide
> and seek play adequate.

Consider the following irrational sequences in (0,1):

1/sqrt(2) - 1/(sqrt(13)*n)
and
1/sqrt(2) + 1/(sqrt(13)*n)

Their members are all irrational and could well be uncovered irrationals
in one of WM's constructions.

But note that 1/sqrt(2) is now separated by at least one rational and at
least one irrational from EACH real except itself, but is not separated

by any rational, or irrational, or anything else, from ALL sequence
members or all spaces between consecutive sequence members, as both
sequences converge to it.

Thus 1/sqrt(2) could itself be an uncovered irrational of the very sort

that WM claims ( but never has proved ) cannot ever exist.

Thus demonstrating that what WM forbids occur can quite easily be the
case.

>
>
> > No, this is your same mistake again: assuming that each singleton must be
> > the endpoint of a cluster. [I see that several others have said the same
> > thing to you in other threads, so this seems to be the key sticking point.]
>
> If it is not the endpoint, then there is a point between it and the
> endpoint, no?

In fact infinitely many points must exist between any two distinct
points.

And possibly infinitely many clusters between them as well.

> >
> > > And every interval is extended, i.e., it is not only a point.
> >
> > Well, most people count the closed interval [x,x] as an interval consisting
> > of a single point.
>
> That is correct. But all intervals that cover the rationals q_n have
> the measure 10^-n. None of them is a point.

Irrelevant.

>
>
> > Obviously such intervals aren't extended. However, all
> > the intervals removed from the line (centred on the rationals) were by
> > definition extended. What remains could be said to be singleton intervals
> > (which aren't extended), or don't describe the {x} as intervals if you don't
> > want [x,x] to be counted as an interval...
>
> That's ok. They are counted as intervals. (Even the empty set is
> counted as an interval.) But they are only then points, if they are
> the only uncovered points in a neighbourhood where all other points
> are covered.
>

So much nonsense from WM for now!

Consider the following irrational sequences in (0,1):

1/sqrt(2) - 1/(sqrt(13)*n)
and
1/sqrt(2) + 1/(sqrt(13)*n)

Their members are all irrational and could well be uncovered irrationals
in one of WM's constructions.

But note that 1/sqrt(2) is now separated by at least one rational and
one irrational from EACH sequence member, but is not separated by any
rational, or irrational, or anything else, from ALL sequence members or
all spaces between consecutive sequence members, as both sequences
converge to it.

Thus 1/sqrt(2) could itself be an uncovered irrational of the very sort
that WM claims cannot ever exist.

Thus demonstrating that what WM claims cannot occur can quite easily
occur.

And this argument WM cannot fault ( for at least has not fAULTED.

[Uergil]

In article
<b425283f-00aa-4f1a...@d6g2000vbe.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> then I believe I have proved my claim.

Wrong! Again!! As Usual!!!

[Uergil]

In article
<20898bb2-1e18-435c...@d6g2000vbe.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> Yes. But at the beginning there is not at all clear what distinguishes
> an interval from the complement. You just cut two notches into the
> circle and have an interval and its complement. But apart from the
> naming there is no asymmetry between them. So let us better say
> intervals of the first kind and the second kind with no qualification.

Pick any point of your circle as the 0 point and by measuring distances
as fractions of the circumference counterclockwise from 0, each point
becomes either a rational point or an irrational point with value from 0
up to but not including 1.

Take the set of intervals on [0,1] and wrap them onto that circle with
the 0 an 1 overlayed and any interval with a zero welded to an interval
with a 1 in it.

Then the original set of rationals are still rational, the original
irrationals are still irrational, there willneed to be some changes in
the I_n, but only one cluster will change, and no uncovered irrational
will change.

So all of the proofs that WM is abjectly wrong will remain totally valid.

[Mike Terry]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:80ab7315-3870-42f0...@q2g2000vbv.googlegroups.com...

> On 5 Jun., 19:48, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
>
> > Or put another way, every neighbourhood of x contains a point belonging
to a
> > cluster.
> >
> > I think I should explain more carefully what this means. It means that
if
> > you give me a neighbourhood N of x, THEN I can find a cluster C that
> > intersects N. Note that my choice of C is made AFTER you give me N, so
I'm
> > free to choose a different C for each different N you give me. It does
not
> > mean that there is a single cluster C such that C intersects every
neighbour
> > hood of x. The order of qualifiers is important!
>
> But the real axis is a static, timeless object. There you have no hide
> and seek play but a point either is disconnected from all points that
> are smaller and larger, or it is not. I dont give you any N and no
> quantifyer, but I ask whether the point x is single like the point 0
> in the definition:
> f(0) = 0 and f(x) = 1 for all x =/= 0.
> If this is the case, then I know that the point 0 is a singleton and
> interrupts a removed interval.

Right I've explained the exact sense in which x is a "singleton" several
times.

If you don't like that definition call it something else - a "separated"
point or whatever. Names do not change the truth of things.

If you like you could make your own definition of singleton. Let's see:

Def. x is a singleton if it's like the point 0 in the definition:

f(0) = 0 and f(x) = 1 for all x =/= 0.

Umm, no. Sorry, that's not going to do :-) Much to vague!

Hmmm, although I was being sarcastic [sorry], I've suddenly had an idea
about what you really might mean by "singleton"! How about this:

Def. x is a singleton if there is a punctured neighbourhood N' of x
such that N' is a subset of the union of all the clusters.

or, equivalently if you don't like the language of neighbourhoods etc. from
topology:

Def. x is a singleton if there is an interval measure > 0
containing x as an interior point such that all the
remaining points of the interval (other than x) belong
to clusters.

Yep, that makes some sense, and I believe normally we would describe such
points as "isolated". If x is "isolated", then it follows it is the end
point of two clusters, one above it and one below. (That seems to fit in
with what you've been saying?)


...So, let me summarise where we've got to!

We start with the interval [0,1] and "remove" intervals of length 10^-n
centred around the corresponding rationals {q_n}, and consider the
complement that remains.

a) The connected components of the complement consist of singleton
sets {x}, for certain irrational values of x. These points are
"separated" from each other, in the sense I've described
previously, but are not all "singletons interrupting an interval".

b) None the less, we will say the set with just one member {x} is a
"singleton set".
[Sorry, that's standard terminology in set theory that everybody
uses, and I see no value in changing that, even for you!]
And we might as well then admit to call the singleton set {x}
a singleton interval [x,x]. (But if you like we won't call
x a singleton point in any other context...)

c) SOME of the x are "isolated", and can be described as
"singletons interrupting an interval". These x have clusters
immediately above and below them. [I'd rather just call these
"isolated", and reserve the use of "singleton" to it's set
theory meaning.]

d) Other x are not adjacent to any clusters, but instead have infinite
sequences of clusters above and below them, the clusters converging
to the point x. So these x are not "isolated" and cannot be
described as "singletons interrupting an interval". If you
like we could avoid calling such x "singletons", although they
are unarguably "singleton sets", being sets consisting of
just one member!

You seem to doubt the existence of points mentioned in (d)? If so, you
should prove there are no points of this kind!

> >
> > > Intervals around rationals.
> >
> > You could say this of course, but it does not mean the singletons are
> > "directly adjacent" to any clusters - there can be infinite sequences of
> > clusters approaching closer and closer to x. [as suggested in my C_n,
D_n
> > example below]
>
> And by this ecstasy of infinity we numbers get a bit muddled up and
> countably many intervals are cut by uncountably many singletons
> without getting uncountably many pieces. A fine result.

Yeah, that's business as usual in the world of rationals and reals! I
remember thinking this was neat when I was about 12, but I guess we get used
to such results and lose our initial sense of wonder as we can just "see
it's true" and get on with it. Like, how can there be infinitely many
naturals, and yet...and yet... every natural number is finite!!!

Sometimes posters come to sci.math and they're genuinely confused about
simple things like this, having studied too much philosophy or something I
guess. After several posts trying to explain, I feel like saying "hey if
you can't just SEE this is the case, e.g. that AxEy is not the same thing as
EyAx, or whatever, then maths isn't going to be for you - it's only going to
get harder!!" :-)

> >
> > If you don't think such a situation is possible, and that every x is the
> > endpoint of a cluster then by all means provide a proof of this - but an
> > assertion is not the same as a proof. And repeatedly stating "there
must be
> > something that makes it a singleton" is not a proof! :)
>
> Any uncovered irrational is a singleton only when it interrupts an
> interval.

Maybe this difficulty has been cured with my terminology change above? SOME
of the uncovered irrationals are "isolated" and can be said to "interrupt an
interval".

Others are not "isolated", and do not "interrupt an interval". Neverless,
they are still in the complement set. And their connected components in the
complement are "singleton sets", but I'll try not to call the point x itself
a singleton point if you like...

>
> > > > No, there needn't be a cluster lying to the right of x. Or rather,
there
> > > > needn't be any cluster directly adjacent to x.
> >
> > > Then x is not a singleton.
> >
> > Why not? This is exactly what you need to prove rather than just
assert.
>
> It is not necessary to prove it. It is the definition. A singleton is
> a set of 1 element that is not covered, so it is 1 point, and there
> must be an interval with measure > 0 at both sides that has been
> covered. Remember, the real axis is a static object. There is no hide
> and seek play adequate.

Fair enough if you want to make this definition, but it doesn't change the
underlying maths! So with your definition I could agree that a singleton
{x} from the complement has a cluster immediately above and below it.
But... now the problem is that there are connected components of the
complement that are "sets with one element" {x}, but are lacking any

"interval with measure > 0 at both sides that has been covered".

So you still end up having to prove the same thing, but using different
definitions. :)

To summarise: you need to provide a proof that all x in the complement have

an interval with measure > 0 at both sides that has been covered.

>
>

> > No, this is your same mistake again: assuming that each singleton must
be
> > the endpoint of a cluster. [I see that several others have said the
same
> > thing to you in other threads, so this seems to be the key sticking
point.]
>
> If it is not the endpoint, then there is a point between it and the
> endpoint, no?

Here is my description of the situation from an earlier post:

We have infinite sequences of clusters

C_1, C_2, C3,... and D_1, D_2, D3

We have the "uncovered point x"

We have C_1 < C_2 < C_3 < ... < x < ... < D_3 < D_2 < D_1

x is the least upper bound of Union(C_n) and

x is the greatest lower bound of Union(D_n)

So x is firmly "sandwiched" between the sequences of clusters above and
below it.

Now let's try to answer your question: "If x is not the endpoint, then

there is a point between it and the endpoint, no?"

The obvious problem with your question is "between it and WHAT endpoint?"
There is no "the" endpoint so your question doesn't make sense...

Well, in my diagram, x is not the endpoint of D_3 - let's focus on that.
For sure there is a point between D_3 and x. In fact, between D_3 and x
there are both entire clusters [D_4, D_5, etc.] and there are further
"uncovered" points, viz the endpoints of the clusters D_4, D_5, etc..

So do you want me to answer "yes" to your question for this reason?

Otherwise you need to think more carefully what you're asking. Specifically
when you say "between x and the interval", I would like to know WHAT
interval you're referring to.

> >
> > > And every interval is extended, i.e., it is not only a point.
> >
> > Well, most people count the closed interval [x,x] as an interval
consisting
> > of a single point.
>
> That is correct. But all intervals that cover the rationals q_n have
> the measure 10^-n. None of them is a point.

Sure - all the "covering" or as I sometimes call them "removed" intervals
have length > 0.

However every "remaining" interval in the complement has length = 0, i.e.
the interval is a singleton set (as the term is used in set theory, viz a
set with just one element).

>
>
> > Obviously such intervals aren't extended. However, all
> > the intervals removed from the line (centred on the rationals) were by
> > definition extended. What remains could be said to be singleton
intervals
> > (which aren't extended), or don't describe the {x} as intervals if you
don't
> > want [x,x] to be counted as an interval...
>
> That's ok. They are counted as intervals. (Even the empty set is
> counted as an interval.) But they are only then points, if they are
> the only uncovered points in a neighbourhood where all other points
> are covered.

Yeah - but this is the case for ALL the remaining points in the complement
set!

At least I know it's OK for me to call them intervals, although there's some
remaining doubt about the terminology "singleton". :(

[Mike Terry]

"WM" <muec...@rz.fh-augsburg.de> wrote in message

news:20898bb2-1e18-435c...@d6g2000vbe.googlegroups.com...

> On 5 Jun., 22:01, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> wrote in message
> >
> >
news:b425283f-00aa-4f1a...@d6g2000vbe.googlegroups.com...>
>
> >
> > Of course, technically whether [x,x] is called an interval is down to
the
> > definition of interval. Most people count it as an interval. Do you?
(I
> > will try to use whatever terminology you use - this does not alter the
> > maths.)
>
> I call [x,x] an interval.
>
> > > There is nothing removed. All intervals remain where they are.
> >
> > Duh - I'm not suggesting that the numbers in removed intervals have been
> > physically destroyed, or moved into a jam jar. OK, Let me reword:
Obviously
> > they are distinguished by not belonging to any of the intervals that
were
> > used to form the complement.
>
> Yes. But at the beginning there is not at all clear what distinguishes
> an interval from the complement. You just cut two notches into the
> circle and have an interval and its complement. But apart from the
> naming there is no asymmetry between them. So let us better say
> intervals of the first kind and the second kind with no qualification.

OK I think I see what you're saying:

You want me to simply "mark the ends of potential intervals" by selecting
just endpoints. Then continue selecting endpoints until I've got countably
infinitely many of them. Then I can start looking at what intervals I can
make?

IOW at each stage n we are just removing 2 points from the circle, not an
interval.

This is sufficiently different from what we originally were discussing for
me not to want to consider it at this point. Maybe make it a new thread and
see if anyone is interested? (If you do, try to be as explicit as you can
about the exact process going on, and what your question is at the end!)

....

> >
> > Without going into any details, if you ask me to do something and I
don't do
> > it for whatever reason, that in no way provides a proof of anything you
> > claim.
>
> I see. Above you said the onus is on me to prove what I claim. But if
> I tell you my proof and you don't listen for whatever reason, then
> that in no way provides a proof of anything I claim.

That's not what I said - and I'm sure you understood what I said, so you are
misrepresenting me.

If you "ask me to do something" that is not "telling me your proof". Proofs
do not consist of asking other people to do things! [Except in the
metaphorical sense we use language during a proof: "...Let x be ....
Consider the set of ...." etc..]

So let's get real. If you offer what you consider a proof, it could be that
I disagree with it - or maybe everybody in sci.math disagrees. Well that's
life in the big city! Mathematics works like that - you could consider
instead publishing your result for peer review. Or you could just accept
that sci.mathers don't agree with your proof even though (in your opinion)
it is correct. Meanwhile others will live the rest of their lives believing
your proof is duff in the ways they pointed out to you and you ignored.
Stalemate! [I'm sure JSH to this day believes he proved Fermat's Last
Theorem and has been unfairly denied his due glory by an evil cabal of
sci.mathers! :)]

Regards,
Mike.

[Uergil]

In article <UIednSvbW71KD1PS...@brightview.co.uk>,

"Mike Terry" <news.dead.p...@darjeeling.plus.com> wrote:

> So let's get real. If you offer what you consider a proof, it could be that
> I disagree with it - or maybe everybody in sci.math disagrees. Well that's
> life in the big city! Mathematics works like that - you could consider
> instead publishing your result for peer review. Or you could just accept
> that sci.mathers don't agree with your proof even though (in your opinion)
> it is correct. Meanwhile others will live the rest of their lives believing
> your proof is duff in the ways they pointed out to you and you ignored.
> Stalemate! [I'm sure JSH to this day believes he proved Fermat's Last
> Theorem and has been unfairly denied his due glory by an evil cabal of
> sci.mathers! :)]
>
> Regards,
> Mike.

There are a bunch of similarities between JSH postings and WM postings

[WM]

On 6 Jun., 01:14, "Mike Terry"

<news.dead.person.sto...@darjeeling.plus.com> wrote:
> You want me to simply "mark the ends of potential intervals" by selecting
> just endpoints.  Then continue selecting endpoints until I've got countably
> infinitely many of them.  Then I can start looking at what intervals I can
> make?
>
> IOW at each stage n we are just removing 2 points from the circle, not an
> interval.

Two points mark two intervals, one of the first kind, and one of the
second kind.

>
> This is sufficiently different from what we originally were discussing for
> me not to want to consider it at this point.

No, it is not different as long as we are able to let the notches
slide over the circle from the random position to the position of our
intervals covering the rationals or from that to a random position or
even to a position where every pair splits the circle into two halves.

If the real line is continuous, this should be allowed. But I see that
(and why) you are not willing to discuss this eaxmple but prefer to
stick to points that are not separated but not connected either.

Regards, WM

[WM]

Am Mittwoch, 6. Juni 2012 00:43:20 UTC+2 schrieb Mike Terry:
> "WM" <muec...@rz.fh-augsburg.de> wrote in message
> news:80ab7315-3870-42f0...@q2g2000vbv.googlegroups.com...

> Def. x is a singleton if there is an interval measure > 0
> containing x as an interior point such that all the
> remaining points of the interval (other than x) belong
> to clusters.
>
> Yep, that makes some sense, and I believe normally we would describe such
> points as "isolated". If x is "isolated", then it follows it is the end
> point of two clusters, one above it and one below. (That seems to fit in
> with what you've been saying?)
>
>
> ...So, let me summarise where we've got to!
>
> We start with the interval [0,1] and "remove" intervals of length 10^-n
> centred around the corresponding rationals {q_n}, and consider the
> complement that remains.
>
> a) The connected components of the complement consist of singleton
> sets {x}, for certain irrational values of x. These points are
> "separated" from each other, in the sense I've described
> previously, but are not all "singletons interrupting an interval".

That is wrong. Every interval has measure > 0, so it is an interval. And between two irrationals there has to be an interval.

>
> b) None the less, we will say the set with just one member {x} is a
> "singleton set".

Of course. But it is isolated by two intervals of measure > 0.

> c) SOME of the x are "isolated", and can be described as
> "singletons interrupting an interval". These x have clusters
> immediately above and below them. [I'd rather just call these
> "isolated", and reserve the use of "singleton" to it's set
> theory meaning.]
>
> d) Other x are not adjacent to any clusters, but instead have infinite
> sequences of clusters above and below them, the clusters converging
> to the point x. So these x are not "isolated" and cannot be
> described as "singletons interrupting an interval". If you
> like we could avoid calling such x "singletons", although they
> are unarguably "singleton sets", being sets consisting of
> just one member!

Wrong. The infinite sequences are nothing but a red herring. For the topology of the real line it makes no difference, whether you point to 0 or point to 1 - 0.999... or use the sequence that dilettante constructed. The only necessity is that 0 is a point that is not connected to any other point, i.e. it is isolated by intervals of measure > 0.

>
> You seem to doubt the existence of points mentioned in (d)? If so, you
> should prove there are no points of this kind!

That's simple. There is no pair of irrationals without a rational between them, hence without an extended interval (in your opinion even containing uncountably many points) between them. If not yet convinced, look into my last post with the "circular" argument.

Regards, WM

[WM]

Am Mittwoch, 6. Juni 2012 00:43:20 UTC+2 schrieb Mike Terry:

> Like, how can there be infinitely many
> naturals, and yet...and yet... every natural number is finite!!!

There are not infinitely many. Unfortunaletly potential and actual infinity are intermingled here. For every n there is a larger one. That does not mean that "there are" (where, by the way) all.

Here only a simple quantifier exchange has been done. We can enumerate every initial segment of rationals, but not all. It is a bit ridiculous that logicans have been trapped by a simple error like: Every US citicen lives in America. <==> Every American lives in the USA.

Regards, WM

[WM]

On 5 Jun., 22:34, Uergil <Uer...@uer.net> wrote:
>
> In the sequence of intervals I_n for  covering the rationals of [0,1]
> there will be at least one contain 0 and at least one containing 1.

Of all pairs of intervals that are simultaneously created, there will
allways one interval contain the point 0 = 1, unless 0 = 1 itself is
the cus.

> Add one more "interval" of their union to the front of the sequence so
> that one cluster goes "round the corner".

Beforehand there is no union at all. There are aleph_0 intervals, each
interval I_n stretching from c_n_1 to c_n_2. For the beginning you can
always cut the circle into two equal parts.

>
> Other than now having the combined 0-1 point in a cluster, this is
> excatly the same as before being circled.

Will you claim that after application of aleph_0 cuts the intervals
containing the point 0 = 1 make up an uncountable set?

Regards, WM

[WM]

On 5 Jun., 22:54, Uergil <Uer...@uer.net> wrote:

>
> The uncountability  follows immediately from the original Cantor
> Dialgonal proof that there are uncountalby many infinite  binary
> sequences.

This argument is invalid. If the Cantor-list is not defined, then
there is not a defined diagonal number, i.e. no diagonal number. If
the Cantor-list is completely defined, then itself and its diagonal
number belong to a countable set.

>
>
> >Every cut is defined by a finite word.
>
> WRONG!

Define a cut that is not.

>
> > The set of finite words is countable.
>
> Irrelevant.

Not as irrelevant as transfinity.

Regards, WM

[Uergil]

In article
<8532fc05-23b8-4576...@n33g2000vbi.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 5 Jun., 22:54, Uergil <Uer...@uer.net> wrote:
>
> >
> > The uncountability  follows immediately from the original Cantor
> > Dialgonal proof that there are uncountalby many infinite  binary
> > sequences.
>
> This argument is invalid. If the Cantor-list is not defined, then
> there is not a defined diagonal number, i.e. no diagonal number. If
> the Cantor-list is completely defined, then itself and its diagonal
> number belong to a countable set.

For which set there is a new non-member, and so on ad infinitum.

But no countable set of binary sequences can be a complete set of ALL
binary sequences because for every such countable set there is at least
one non-member binary sequence.

Cantor proved that for any countable set of binary sequences there is at
least one binary sequence not in it.

Thus no complete and simultaneously countable set of them is possible.

[Uergil]

In article
<ec7c31a0-adea-4aef...@j9g2000vbk.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 5 Jun., 22:34, Uergil <Uer...@uer.net> wrote:
> >
> > In the sequence of intervals I_n for  covering the rationals of [0,1]
> > there will be at least one contain 0 and at least one containing 1.
>
> Of all pairs of intervals that are simultaneously created, there will
> allways one interval contain the point 0 = 1, unless 0 = 1 itself is
> the cus.

WM claimed a list of closed intervals, I_n, whose sum of lengths was
considerably less that 1 and whose endpoints were rationals.
I was not aware that "0 = 1" was a member of one of them

>
> > Add one more "interval" of their union to the front of the sequence so
> > that one cluster goes "round the corner".
>
> Beforehand there is no union at all. There are aleph_0 intervals, each
> interval I_n stretching from c_n_1 to c_n_2. For the beginning you can
> always cut the circle into two equal parts.
> >
> > Other than now having the combined 0-1 point in a cluster, this is

> > exactly the same as before being circled.

>
> Will you claim that after application of aleph_0 cuts the intervals
> containing the point 0 = 1 make up an uncountable set

If those "cuts" are the only endpoints, then the number of two point
subsets of your countable set of "cuts" is itself countable, thus so is
the set of intervals which can be made from them.

Note that WM's attempt to impose is corrupt matheological version of
mathematics on sci.math has again fallen flat on its ugly face.

[WM]

On 6 Jun., 22:09, Uergil <Uer...@uer.net> wrote:
> In article
> <8532fc05-23b8-4576-b28f-734a259df...@n33g2000vbi.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 5 Jun., 22:54, Uergil <Uer...@uer.net> wrote:
>
> > > The uncountability follows immediately from the original Cantor
> > > Dialgonal proof that there are uncountalby many infinite binary
> > > sequences.
>
> > This argument is invalid. If the Cantor-list is not defined, then
> > there is not a defined diagonal number, i.e. no diagonal number. If
> > the Cantor-list is completely defined, then itself and its diagonal
> > number belong to a countable set.
>
> For which set there is a new non-member, and so on ad infinitum.

That holds for every infinite set. But that does not make any set
larger than an infinite set. Some elements simply do not exist. If the
diagonal existed in advance of the list construction, why would it be
left out?

>
> But no countable set of binary sequences can be a complete set of ALL
> binary sequences because for every such countable set there is at least
> one non-member binary sequence.

And for every set of natural numbers there is at least a non-member
natural number.

>
> Cantor proved that for any countable set of binary sequences there is at
> least one binary sequence not in it.
>
> Thus no complete and simultaneously countable set of them is possible.

No complete and simultaneously incomplete (infinite) set if possible.

Regards, WM

[Uergil]

In article
<b89eadf0-f174-4b1a...@p27g2000vbl.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 6 Jun., 01:14, "Mike Terry"
> <news.dead.person.sto...@darjeeling.plus.com> wrote:
> > You want me to simply "mark the ends of potential intervals" by selecting
> > just endpoints.  Then continue selecting endpoints until I've got countably
> > infinitely many of them.  Then I can start looking at what intervals I can
> > make?
> >
> > IOW at each stage n we are just removing 2 points from the circle, not an
> > interval.
>
> Two points mark two intervals, one of the first kind, and one of the
> second kind.

Other than containing or not containing the original endpoints of the
interval which were identified with each other to form your circle, how
do the two intervals created by such a pair of distinct points differ in
"kind"?

> >
> > This is sufficiently different from what we originally were discussing for
> > me not to want to consider it at this point.
>
> No, it is not different as long as we are able to let the notches
> slide over the circle from the random position to the position of our
> intervals covering the rationals or from that to a random position or
> even to a position where every pair splits the circle into two halves.

So that WM claims that a closed interval and a circle are no different?

WM is getting desperate. Each of his attempts to foist his matheologal
mish-mashes onto the world has again failed and he is running out of
ideas.

>
> If the real line is continuous, this should be allowed. But I see that
> (and why) you are not willing to discuss this eaxmple but prefer to
> stick to points that are not separated but not connected either.

It was WM's, model proposed by WM himself, and argued for and supported
by WM, until it was shown to everyone that WM's own model contradicted
WM's matheoogical own claims. Now WM wants that model forgot in favor of
another, which will ultimately contradict his matheoogy quite as
thoroughly as his last one did.

[WM]

On 6 Jun., 22:19, Uergil <Uer...@uer.net> wrote:

> > > Add one more "interval" of their union to the front of the sequence so
> > > that one cluster goes "round the corner".
>
> > Beforehand there is no union at all. There are aleph_0 intervals, each
> > interval I_n stretching from c_n_1 to c_n_2. For the beginning you can
> > always cut the circle into two equal parts.
>
> > > Other than now having the combined 0-1 point in a cluster, this is
> > > exactly the same as before being circled.
>
> > Will you claim that after application of aleph_0 cuts the intervals

> > containing the point 0 = 1 make up an uncountable set?

>
> If those "cuts" are the only endpoints, then the number of two point
> subsets of your countable set of "cuts" is itself countable, thus so is
> the set of intervals which can be made from them.

But when the cuts slip arount the circle such that the nth pair in a
relative distance of 10^-n includes the rational q_n, then the set of
intervals in the complement becomes uncountable?

Regards, WM

[Uergil]

In article <2076caf9-b44b-4717...@googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> Am Mittwoch, 6. Juni 2012 00:43:20 UTC+2 schrieb Mike Terry:
>
> > Like, how can there be infinitely many
> > naturals, and yet...and yet... every natural number is finite!!!
>
> There are not infinitely many. Unfortunaletly potential and actual infinity
> are intermingled here. For every n there is a larger one. That does not mean
> that "there are" (where, by the way) all.

There are certainly more than any finite number of them.

>
> Here only a simple quantifier exchange has been done. We can enumerate every
> initial segment of rationals, but not all.

Actually there are numerous formulae which "enumerate" all of the
rationals in the sense of pairing them off with positive integers with
none of either left over.

> It is a bit ridiculous that
> logicans have been trapped by a simple error like: Every US citicen lives in
> America. <==> Every American lives in the USA.

However
"Every US citicen lives in America" even if "citicen" means "citizen"
and

"Every American lives in the USA"

are both false..

[Uergil]

In article <226a3f41-9f92-48ca...@googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> Am Mittwoch, 6. Juni 2012 00:43:20 UTC+2 schrieb Mike Terry:
> > "WM" <muec...@rz.fh-augsburg.de> wrote in message
> > news:80ab7315-3870-42f0...@q2g2000vbv.googlegroups.com...
>
>
> > Def. x is a singleton if there is an interval measure > 0
> > containing x as an interior point such that all the
> > remaining points of the interval (other than x) belong
> > to clusters.
> >
> > Yep, that makes some sense, and I believe normally we would describe such
> > points as "isolated". If x is "isolated", then it follows it is the end
> > point of two clusters, one above it and one below. (That seems to fit in
> > with what you've been saying?)
> >
> >
> > ...So, let me summarise where we've got to!
> >
> > We start with the interval [0,1] and "remove" intervals of length 10^-n
> > centred around the corresponding rationals {q_n}, and consider the
> > complement that remains.
> >
> > a) The connected components of the complement consist of singleton
> > sets {x}, for certain irrational values of x. These points are
> > "separated" from each other, in the sense I've described
> > previously, but are not all "singletons interrupting an interval".
>
> That is wrong. Every interval has measure > 0, so it is an interval. And
> between two irrationals there has to be an interval.

Between two uncovered irrationals there may well be infinitely many
cluster intervals, in which case there may also be uncovered irrationals
which are not endpoints of any cluster.

Consider the following irrational sequences in (0,1):

1/sqrt(2) - 1/(sqrt(13)*n)
and
1/sqrt(2) + 1/(sqrt(13)*n)

Their members are all irrational and could well be uncovered irrationals
in one of WM's constructions.

But note that 1/sqrt(2) is now separated by at least one rational and
one irrational from EACH sequence member, but is not separated by any
rational, or irrational, or anything else, from ALL sequence members or
all spaces between consecutive sequence members, as both sequences
converge to it.

Thus 1/sqrt(2) could itself be an uncovered irrational of the very sort
that WM claims cannot ever exist.

Thus demonstrating that what WM claims cannot occur can quite easily
occur.

> >

> > b) None the less, we will say the set with just one member {x} is a
> > "singleton set".
>
> Of course. But it is isolated by two intervals of measure > 0.

But it may also be that it is isolated from every cluster by a positive
distance, as in my example above.

>
>
> > c) SOME of the x are "isolated", and can be described as
> > "singletons interrupting an interval". These x have clusters
> > immediately above and below them. [I'd rather just call these
> > "isolated", and reserve the use of "singleton" to it's set
> > theory meaning.]
> >
> > d) Other x are not adjacent to any clusters, but instead have infinite
> > sequences of clusters above and below them, the clusters converging
> > to the point x. So these x are not "isolated" and cannot be
> > described as "singletons interrupting an interval". If you
> > like we could avoid calling such x "singletons", although they
> > are unarguably "singleton sets", being sets consisting of
> > just one member!
>
> Wrong. The infinite sequences are nothing but a red herring. For the topology
> of the real line it makes no difference, whether you point to 0 or point to 1
> - 0.999... or use the sequence that dilettante constructed. The only
> necessity is that 0 is a point that is not connected to any other point, i.e.
> it is isolated by intervals of measure > 0.

0 is a rational, so, by WM's original construction, must be an interior
point of some cluster

Otherwise one may repeat WM's original I_n on the open interval (0,1),
and not have to worry about the oddities of 0 and 1 in WM's orginal
construction.

> >
> > You seem to doubt the existence of points mentioned in (d)? If so, you
> > should prove there are no points of this kind!
>
> That's simple. There is no pair of irrationals without a rational between
> them, hence without an extended interval (in your opinion even containing
> uncountably many points) between them.

That does not prohibit there being infinitely many clusters between
them, as must sometimes occur.

And whenever it does one can have an uncovered irrational which NOT an
endpoint of any cluster, but a limit point of both clusters and other
uncovered irrationals.

The example I have given above shows how that can occur, and it must
occur unless the outer measure on the unit interval is zero.

> If not yet convinced, look into my
> last post with the "circular" argument.

I have looked at all of WM's posts and have not found in any of them any
reason why my example cannot occur, i.e., having uncovered irrationals
which are NOT the endpoint of ANY cluster.


WM must be both blind and desperate to reject something so obvious.

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag
news:d83ca3eb-f19d-4ed1...@d6g2000vbe.googlegroups.com...

> On 6 Jun., 22:09, Uergil <Uer...@uer.net> wrote:
>> In article
>> <8532fc05-23b8-4576-b28f-734a259df...@n33g2000vbi.googlegroups.com>,
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:
>> > On 5 Jun., 22:54, Uergil <Uer...@uer.net> wrote:
>>
>> > > The uncountability follows immediately from the original Cantor
>> > > Dialgonal proof that there are uncountalby many infinite binary
>> > > sequences.
>>
>> > This argument is invalid. If the Cantor-list is not defined, then
>> > there is not a defined diagonal number, i.e. no diagonal number. If
>> > the Cantor-list is completely defined, then itself and its diagonal
>> > number belong to a countable set.
>>
>> For which set there is a new non-member, and so on ad infinitum.
>
> That holds for every infinite set. But that does not make any set
> larger than an infinite set. Some elements simply do not exist. If the
> diagonal existed in advance of the list construction, why would it be
> left out?

Bravo! That's it! Yet another proof - as though we needed more than one -
that the reals are countable. God almighty! How dense these matheologians
are!

Again (please listen closely, you deaf matheologian):
Let L(n) be the list of all those reals that exist. There exists,
indisputably, a real number d not in the list, by virtue of the
diagonal argument. Therefore, either the list was incomplete or
d did not exist when the list was made. So obviously d
came into existence after the list was made. QED

That was easy.

[Ross A. Finlayson]

Heh, sounds much like, the state of the argument before. Hell and
last week I'm working up counterpoint to Rice's theorem. I suggest
you find some applications.

The workup of the counterpoint to the ideal for Rice's theorem then
has the encoding problem, the format of the program and its data is
defined, then deterministically that builds Halts(). Relevant to,
"matheology"? Yes, relevant in various senses.

Thanks no that's last week.

That's where I suggest you either denote the novel parts or work up
the summary.

If you're reaching for (or to) a point please at least make it a
statement.

Regards,

Ross Finlayson

[Uergil]

In article
<6d8da7ed-a736-41d3...@z19g2000vbe.googlegroups.com>,

The set of intervals in any open real interval of positive measure is
uncountable. Unless the endpoints of those intervals are restricted to a
countable set, in which case there are only countably many of them.

So countable many "cuts" produce countably many intervals, and
uncountably any "cuts" produce uncountably many intervals.

[Uergil]

In article
<d83ca3eb-f19d-4ed1...@d6g2000vbe.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 6 Jun., 22:09, Uergil <Uer...@uer.net> wrote:
> > In article
> > <8532fc05-23b8-4576-b28f-734a259df...@n33g2000vbi.googlegroups.com>,
> >
> >  WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 5 Jun., 22:54, Uergil <Uer...@uer.net> wrote:
> >
> > > > The uncountability follows immediately from the original Cantor
> > > > Dialgonal proof that there are uncountalby many infinite binary
> > > > sequences.
> >
> > > This argument is invalid. If the Cantor-list is not defined, then
> > > there is not a defined diagonal number, i.e. no diagonal number. If
> > > the Cantor-list is completely defined, then itself and its diagonal
> > > number belong to a countable set.
> >
> > For which set there is a new non-member, and so on ad infinitum.
>
> That holds for every infinite set. But that does not make any set
> larger than an infinite set. Some elements simply do not exist. If the
> diagonal existed in advance of the list construction, why would it be
> left out?

Put it in the list and the new list leaves a different sequence out.

(Actually, for any such listing, a slight modification of Cantor's
argument shows that there are as many left out as listed)

The point is that WHATEVER list one has, THAT list leaves something out.

So the very constructing a list AUTOMATICALLY leaves something out.
Changing a list changes what gets left out, but there is always
something.

> >
> > But no countable set of binary sequences can be a complete set of ALL
> > binary sequences because for every such countable set there is at least
> > one non-member binary sequence.
>
> And for every set of natural numbers there is at least a non-member
> natural number.

The Cantor argument has been proven, yours claim has not, and cannot be.

> >
> > Cantor proved that for any countable set of binary sequences there is at
> > least one binary sequence not in it.
> >
> > Thus no complete and simultaneously countable set of them is possible.
>
> No complete and simultaneously incomplete (infinite) set if possible.

No complete and simultaneously complete and incomplete anything can
exist, and only WM's matheology would even suggest otherwise.

[WM]

On 6 Jun., 23:11, Uergil <Uer...@uer.net> wrote:

> I have looked at all of WM's posts and have not found in any of them any
> reason why my example cannot occur, i.e., having uncovered irrationals
> which are NOT the endpoint of ANY cluster.

Take one of those uncovered points x_0There are two possibilities:
Either x_0s adjacent to a covered point (i.e. a point of an interval)
or to a not covered point, i.e. an uncovered irrational. The latter
possibility is a contradiction. Hence only the first is really
possible.

You may try to escape by saying there is no adjacent point at all. But
that is false as the real axis is continuous. You can define the
function
f(x) = 1 for covered x
and
f(x) = 0 for uncovered x.
Then the function we look for is
f(x) = 1 for a < x < x_0
f(x) = 0 for x = x_0
f(x) = 1 for x_0 < x < b.

As every interval covering a rational has measure > 0, there is in
every case an a and a b, even when we cannot define them.

Regards, WM

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag

news:59f5c255-76ae-4a6e...@d6g2000vbe.googlegroups.com...

> On 6 Jun., 23:11, Uergil <Uer...@uer.net> wrote:
>
>> I have looked at all of WM's posts and have not found in any of them any
>> reason why my example cannot occur, i.e., having uncovered irrationals
>> which are NOT the endpoint of ANY cluster.
>
> Take one of those uncovered points x_0There are two possibilities:
> Either x_0s adjacent to a covered point (i.e. a point of an interval)
> or to a not covered point, i.e. an uncovered irrational. The latter
> possibility is a contradiction. Hence only the first is really
> possible.

Dear Professor Dr. Mückenheim,

As I have previously admitted, you have convinced me by your
wonderful proofs that all of your assertions are correct and
that you are being unjustly persecuted by the matheologians.

However, I sometimes have difficulty understanding your
terminology. Needless to say it is necessary to introduce
new concepts when new facts are discovered. May I ask you
to define these new concepts when they are introduced?

In the particular case you are using the notion "adjacent",
which appears to be undefined in the present context.

Very sincerely Yours,

Jürgen R.

[WM]

On 7 Jun., 14:57, Jürgen R. <jurg...@arcor.de> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> schrieb im Newsbeitragnews:59f5c255-76ae-4a6e...@d6g2000vbe.googlegroups.com...

>
> > On 6 Jun., 23:11, Uergil <Uer...@uer.net> wrote:
>
> >> I have looked at all of WM's posts and have not found in any of them any
> >> reason why my example cannot occur, i.e., having uncovered irrationals
> >> which are NOT the endpoint of ANY cluster.
>
> > Take one of those uncovered points x_0There are two possibilities:
> > Either x_0s adjacent to a covered point (i.e. a point of an interval)
> > or to a not covered point, i.e. an uncovered irrational. The latter
> > possibility is a contradiction. Hence only the first is really
> > possible.
>
> Dear Professor Dr. Mückenheim,
>
> As I have previously admitted, you have convinced me by your
> wonderful proofs that all of your assertions are correct and
> that  you are being unjustly persecuted by the matheologians.
>
> However, I sometimes have difficulty understanding your
> terminology. Needless to say it is necessary to introduce
> new concepts when new facts are discovered. May I ask you
> to define these new concepts when they are introduced?
>
> In the particular case you are using the notion "adjacent",
> which appears to be undefined in the present context.

Adjacent means in German angrenzend.

If f(x) = 0 for x = 0 and f(x) = 1 for x > 0, then the interval (0,
oo) is adjacent to the point 0. That means there is no real number
between 0 and the interval (0, oo). In particular there is no real
number y with f(y) = 5.

Regards, WM

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag

news:5cae398f-140e-4438...@eh4g2000vbb.googlegroups.com...

Please be patient because this kind of thinking is
very new to me. You meant, in other words, that x_0 is the
endpoint of of the interval (0,a)? Then why didn't
you say that?

But what did you mean when you said that x_0 might be
adjacent to a "covered point"? When are two points
adjacent?

[Uergil]

In article
<59f5c255-76ae-4a6e...@d6g2000vbe.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 6 Jun., 23:11, Uergil <Uer...@uer.net> wrote:
>
> > I have looked at all of WM's posts and have not found in any of them any
> > reason why my example cannot occur, i.e., having uncovered irrationals
> > which are NOT the endpoint of ANY cluster.
>
> Take one of those uncovered points x_0There are two possibilities:
> Either x_0s adjacent to a covered point (i.e. a point of an interval)
> or to a not covered point, i.e. an uncovered irrational.

WM shows his abysmal ignorance of the real line, or any densely ordered
set, by claiming the there can be anything like "adjacent" members.

> The latter
> possibility is a contradiction. Hence only the first is really
> possible.

The claim of adjacent points is the only contradiction I see, and that
entirely buggers WM's argument!

>
> You may try to escape by saying there is no adjacent point at all. But
> that is false as the real axis is continuous.

That very continuity REQUIRES non-adjacency!

Given any x and y on the real line, they are not adjacent because
(x+y)/2 is strictly between them.

You can define the
> function
> f(x) = 1 for covered x
> and
> f(x) = 0 for uncovered x.
> Then the function we look for is
> f(x) = 1 for a < x < x_0
> f(x) = 0 for x = x_0
> f(x) = 1 for x_0 < x < b.

NOPE! While there will be isolated uncovered points, since there will
have to be infinitely many of them in in finite interval, not all of
them can be isolated.

Consider the following irrational sequences in (0,1):

1/sqrt(2) - 1/(sqrt(13)*n)
and
1/sqrt(2) + 1/(sqrt(13)*n)

Their members are all irrational and could well be uncovered irrationals
in one of WM's constructions.

But note that 1/sqrt(2) is now separated by at least one rational and
one irrational from EACH sequence member, but is not separated by any
rational, or irrational, or anything else, from ALL sequence members or
all spaces between consecutive sequence members, as both sequences
converge to it.

Thus 1/sqrt(2) could itself be an uncovered irrational of the very sort
that WM claims cannot ever exist.

Thus demonstrating that what WM claims cannot occur can quite easily
occur.

>

> As every interval covering a rational has measure > 0, there is in
> every case an a and a b, even when we cannot define them.

Consider the set of irrationals in my example above.

That sort of situation is totally compatible with your own model.

To deny it is to reject your own model and admit its failure.

[Uergil]

In article
<5cae398f-140e-4438...@eh4g2000vbb.googlegroups.com>,

So, as far as WM has defined it, adjacency in the set of reals can only
occur between a point and a not closed set of points such as an open
inteval.

But WM has NOT defined what it means for his x_0 to be adjacent to a
point, whether a covered or a not covered irrational point, even though
that is what he is trying to claim.

Note that the ordered set of reals and every subinterval of the reals,
has the property of density, which means, among other things, that no
point is ever "adjacent" to any other point, even though WM may insist
otherwise..

According to WM's model, one has in the open interval (0,1) each
rational contained in an open interval such that the cumulative lengths
of those open intervals is less than one but greater than zero, leaving
abounded but infinite set of uncovered irrationals (of necessarily
positive outer measure) which he claims cannot contain a single cluster
point.

And WM is careful to ignore all posted proofs that his model does not
behave as he claims it must.

[WM]

On 7 Jun., 18:00, Jürgen R. <jurg...@arcor.de> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> schrieb im Newsbeitragnews:5cae398f-140e-4438...@eh4g2000vbb.googlegroups.com...

>
>
>
>
>
> > On 7 Jun., 14:57, J rgen R. <jurg...@arcor.de> wrote:
> >> "WM" <mueck...@rz.fh-augsburg.de> schrieb im
> >> Newsbeitragnews:59f5c255-76ae-4a6e...@d6g2000vbe.googlegroups.com...
>
> >> > On 6 Jun., 23:11, Uergil <Uer...@uer.net> wrote:
>
> >> >> I have looked at all of WM's posts and have not found in any of them
> >> >> any
> >> >> reason why my example cannot occur, i.e., having uncovered irrationals
> >> >> which are NOT the endpoint of ANY cluster.
>
> >> > Take one of those uncovered points x_0There are two possibilities:
> >> > Either x_0s adjacent to a covered point (i.e. a point of an interval)
> >> > or to a not covered point, i.e. an uncovered irrational. The latter
> >> > possibility is a contradiction. Hence only the first is really
> >> > possible.
>

> >> Dear Professor Dr. M ckenheim,

>
> >> As I have previously admitted, you have convinced me by your
> >> wonderful proofs that all of your assertions are correct and
> >> that  you are being unjustly persecuted by the matheologians.
>
> >> However, I sometimes have difficulty understanding your
> >> terminology. Needless to say it is necessary to introduce
> >> new concepts when new facts are discovered. May I ask you
> >> to define these new concepts when they are introduced?
>
> >> In the particular case you are using the notion "adjacent",
> >> which appears to be undefined in the present context.
>
> > Adjacent means in German angrenzend.
>
> > If f(x) = 0 for x = 0 and f(x) = 1 for x > 0, then the interval (0,
> > oo) is adjacent to the point 0. That means there is no real number
> > between 0 and the interval (0, oo). In particular there is no real
> > number y with f(y) = 5.
>
> > Regards, WM
>
> Please be patient because this kind of thinking is
> very new to me. You meant, in other words, that x_0 is the
> endpoint of of the interval (0,a)? Then why didn't
> you say that?

x_0 is the endpoint of an interval or of a cluster, i.e. of a
connected component,

>
> But what did you mean when you said that x_0 might be
> adjacent to a "covered point"? When are two points

> adjacent?-

There is no single point that could be identified as a point next to
x_0. But we can prove that an interval or cluster next to x_0 consists
of covered points only.

Regards, WM

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag

news:74175e0f-0c65-4861...@d17g2000vbv.googlegroups.com...

I can't say that this really makes it clear to me what you meant by
saying

"Either x_0s adjacent to a covered point (i.e. a point of an interval)
or to a not covered point"

in English this means that x_0 is always adjacent to some
point. Perhaps this is not what you meant to say?

>
> Regards, WM

[Uergil]

In article
<74175e0f-0c65-4861...@d17g2000vbv.googlegroups.com>,

If that means that every uncovered point is in the closure of a cluster,
that is obviously false for any number of reasons.

For one thing, that would require the closure of the union of open
cluster intervals to be the entire interval of length 1 even though one
can make sure, by chosing the original I_n appropriately, that the set
of open cluster intervals has outer measure less than any positive
epsilon.

[WM]

On 7 Jun., 23:13, Jürgen R. <jurg...@arcor.de> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> schrieb im

> >> Please be patient because this kind of thinking is
> >> very new to me. You meant, in other words, that x_0 is the
> >> endpoint of of the interval (0,a)? Then why didn't
> >> you say that?
>
> > x_0 is the endpoint of an interval or of a cluster, i.e. of a
> > connected component,
>
> >> But what did you mean when you said that x_0 might be
> >> adjacent to a "covered point"? When are two points
> >> adjacent?-
>
> > There is no single point that could be identified as a point next to
> > x_0. But we can prove that an interval or cluster next to x_0 consists
> > of covered points only.
>
> I can't say that this really makes it clear to me what you meant by
> saying
>
> "Either x_0s adjacent to a covered point (i.e. a point of an interval)
> or to a not covered point"
>
> in English this means that x_0 is always adjacent to some
> point. Perhaps this is not what you meant to say?

I said: x_0 is the endpoint of an interval or of a cluster, i.e. of a
connected component. Sorry, I don't see how this could not be
understood by a mathematician.

But it is not so important for the result. Simply start with a circle
of measure 1 and cut it into aleph_0 intervals, each of measure 1/2.
Can you imagine that you have 2^aleph_0 intervals? I dare to guess you
cannot.

Then slide the cuts around the circle in an arbitrary way. Can you
imagine that you get 2^aleph_0 intervals?

Regards, WM

[WM]

On 7 Jun., 23:52, Uergil <Uer...@uer.net> wrote:
> In article

> For one thing, that would require the closure of the union of open
> cluster intervals to be the entire interval of length 1 even though one
> can make sure, by chosing the original I_n appropriately, that the set
> of open cluster intervals has outer measure less than any positive
> epsilon.

If the notion of countability had the meaning of finishing an infinite
set. But it hasn't.

Regards, WM

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag

news:babc6e33-f0ea-42b5...@5g2000vbf.googlegroups.com...

> On 7 Jun., 23:13, Jürgen R. <jurg...@arcor.de> wrote:
>> "WM" <mueck...@rz.fh-augsburg.de> schrieb im
>
>> >> Please be patient because this kind of thinking is
>> >> very new to me. You meant, in other words, that x_0 is the
>> >> endpoint of of the interval (0,a)? Then why didn't
>> >> you say that?
>>
>> > x_0 is the endpoint of an interval or of a cluster, i.e. of a
>> > connected component,
>>
>> >> But what did you mean when you said that x_0 might be
>> >> adjacent to a "covered point"? When are two points
>> >> adjacent?-
>>
>> > There is no single point that could be identified as a point next to
>> > x_0. But we can prove that an interval or cluster next to x_0 consists
>> > of covered points only.
>>
>> I can't say that this really makes it clear to me what you meant by
>> saying
>>
>> "Either x_0s adjacent to a covered point (i.e. a point of an interval)
>> or to a not covered point"
>>
>> in English this means that x_0 is always adjacent to some
>> point. Perhaps this is not what you meant to say?
>
> I said: x_0 is the endpoint of an interval or of a cluster, i.e. of a
> connected component. Sorry, I don't see how this could not be
> understood by a mathematician.

I know I must be trying your patience, but I really would like
to understand your reasoning. I thought you were proving that x_0
is an endpoint. Now it seems you were assuming what needs to be
proven - and I am sure that can't be what you meant.

The problematical part was this:

"Take one of those uncovered points x_0There are two possibilities:

Either x_0s adjacent to a covered point (i.e. a point of an interval)

or to a not covered point, i.e. an uncovered irrational. The latter
possibility is a contradiction. Hence only the first is really
possible."

And I still am unclear what you mean by saying two points are
"adjacent".

>
> But it is not so important for the result. Simply start with a circle
> of measure 1 and cut it into aleph_0 intervals, each of measure 1/2.
> Can you imagine that you have 2^aleph_0 intervals? I dare to guess you
> cannot.

You are completely right! The number of intervals certainly
certainly is countable.

>
> Then slide the cuts around the circle in an arbitrary way. Can you
> imagine that you get 2^aleph_0 intervals?

Absolutely not! There are countably many intervals,
although there must be uncountably many points in those
intervals.
Again I must apologize for my difficulties. I cannot see the
relationship between this example and the original construction,
i.e. the ternary set type of construction. Perhaps you could
transfer this to the circle?

>
> Regards, WM
>

[WM]

On 8 Jun., 12:19, Jürgen R. <jurg...@arcor.de> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> schrieb im

> > But it is not so important for the result. Simply start with a circle
> > of measure 1 and cut it into aleph_0 intervals, each of measure 1/2.
> > Can you imagine that you have 2^aleph_0 intervals? I dare to guess you
> > cannot.
>
> You are completely right! The number of intervals certainly
> certainly is countable.
>

> > Then slide the cuts around the circle in an arbitrary way. Can you
> > imagine that you get 2^aleph_0 intervals?
>
> Absolutely not! There are countably many intervals,
> although there must be uncountably many points in those
> intervals.

In the first kind of intervals, there can be uncountably many points.
In the second kind, that it not possible, because there are absolutely
no rational numbers available.

> Again I must apologize for my difficulties. I cannot see the
> relationship between this example and the original construction,
> i.e. the ternary set type of construction. Perhaps you could
> transfer this to the circle?
>

No, Cantor's example does not eliminate all rational numbers. There
remain rationals like 1/4 or 3/10. But what should this special
example in ternary be good for? My example works in every basis.

Regards, WM

[Uergil]

In article
<c70fb203-734c-4136...@q2g2000vbv.googlegroups.com>,

One would not have to actually finish anything to show that the outer
measure is effectively zero.

And WM has already conceded the ability to finish infinite processes by
creating his model, based on the I_n's, without being able to prove
that model's own inconsistency.

[Uergil]

In article
<babc6e33-f0ea-42b5...@5g2000vbf.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 7 Jun., 23:13, Jürgen R. <jurg...@arcor.de> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> schrieb im
>
> > >> Please be patient because this kind of thinking is
> > >> very new to me. You meant, in other words, that x_0 is the
> > >> endpoint of of the interval (0,a)? Then why didn't
> > >> you say that?
> >
> > > x_0 is the endpoint of an interval or of a cluster, i.e. of a
> > > connected component,
> >
> > >> But what did you mean when you said that x_0 might be
> > >> adjacent to a "covered point"? When are two points
> > >> adjacent?-
> >
> > > There is no single point that could be identified as a point next to
> > > x_0. But we can prove that an interval or cluster next to x_0 consists
> > > of covered points only.
> >
> > I can't say that this really makes it clear to me what you meant by
> > saying
> >
> > "Either x_0s adjacent to a covered point (i.e. a point of an interval)
> > or to a not covered point"
> >
> > in English this means that x_0 is always adjacent to some
> > point. Perhaps this is not what you meant to say?
>
> I said: x_0 is the endpoint of an interval or of a cluster, i.e. of a
> connected component. Sorry, I don't see how this could not be
> understood by a mathematician.

What you actually said, at least in standard English, does not convey
that, and if that is what you meant to convey, it is wrong, since there
can quite easily be uncovered irrational points which are not endpoints
of any cluster, but any neighborhood of such a point meets infinitely
many clusters.

as in my example:

Consider the following irrational sequences in (0,1):

1/sqrt(2) - 1/(sqrt(13)*n)
and
1/sqrt(2) + 1/(sqrt(13)*n)

Their members are all irrational and could well be uncovered irrationals

and endpoints of clusters in one of WM's I_n constructions.

But note that 1/sqrt(2) is now separated by at least one rational and

one irrational from EACH sequence member and each such cluster, but is

not separated by any rational, or irrational, or anything else, from ALL

sequence members or all clusters, as both sequences converge to it.

Thus 1/sqrt(2) could itself be an uncovered irrational of the very sort
that WM claims cannot ever exist.

Thus demonstrating that what WM claims cannot occur can quite easily
occur.

>

> But it is not so important for the result. Simply start with a circle
> of measure 1 and cut it into aleph_0 intervals, each of measure 1/2.
> Can you imagine that you have 2^aleph_0 intervals? I dare to guess you
> cannot.

Why should anyone, except nuts like WM, conflate aleph_0 with 2^aleph_0?

>
> Then slide the cuts around the circle in an arbitrary way. Can you
> imagine that you get 2^aleph_0 intervals?

That is not at all the same as removing from the circle countably many
non-overlapping open intervals whose lengths add up to considerably less
that the circumference of that circle.

And it is only that which is at all analogous to your original
construction on [0,1], and not your circles.

[Uergil]

In article
<564cd36f-e5cd-4c55...@a16g2000vby.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 8 Jun., 12:19, Jürgen R. <jurg...@arcor.de> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> schrieb im
>
> > > But it is not so important for the result. Simply start with a circle
> > > of measure 1 and cut it into aleph_0 intervals, each of measure 1/2.
> > > Can you imagine that you have 2^aleph_0 intervals? I dare to guess you
> > > cannot.
> >
> > You are completely right! The number of intervals certainly
> > certainly is countable.
> >
>
> > > Then slide the cuts around the circle in an arbitrary way. Can you
> > > imagine that you get 2^aleph_0 intervals?
> >
> > Absolutely not! There are countably many intervals,
> > although there must be uncountably many points in those
> > intervals.
>
> In the first kind of intervals, there can be uncountably many points.
> In the second kind, that it not possible, because there are absolutely
> no rational numbers available.

As far as I can tell, sliding those intervals around does not change any
of the rational points into irrationals or vice versa, so where did all
those rational points disappear to?

>
> > Again I must apologize for my difficulties. I cannot see the
> > relationship between this example and the original construction,
> > i.e. the ternary set type of construction. Perhaps you could
> > transfer this to the circle?
> >
> No, Cantor's example does not eliminate all rational numbers. There
> remain rationals like 1/4 or 3/10. But what should this special
> example in ternary be good for?

It shows that WM's matheology does not hold in real interval or real
circles.

> My example works in every basis.

Like so many of WM's matheological mysteries, that example, while often
enough claimed by WM, it has yet to be proved, every alleged proof
having been shown to be erroneous.

[WM]

On 8 Jun., 21:34, Uergil <Uer...@uer.net> wrote:
> In article

> <564cd36f-e5cd-4c55-bd90-2fc440d7d...@a16g2000vby.googlegroups.com>,

>
>
>
>
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 8 Jun., 12:19, Jürgen R. <jurg...@arcor.de> wrote:
> > > "WM" <mueck...@rz.fh-augsburg.de> schrieb im
>
> > > > But it is not so important for the result. Simply start with a circle
> > > > of measure 1 and cut it into aleph_0 intervals, each of measure 1/2.
> > > > Can you imagine that you have 2^aleph_0 intervals? I dare to guess you
> > > > cannot.
>
> > > You are completely right! The number of intervals certainly
> > > certainly is countable.
>
> > > > Then slide the cuts around the circle in an arbitrary way. Can you
> > > > imagine that you get 2^aleph_0 intervals?
>
> > > Absolutely not! There are countably many intervals,
> > > although there must be uncountably many points in those
> > > intervals.
>
> > In the first kind of intervals, there can be uncountably many points.
> > In the second kind, that it not possible, because there are absolutely
> > no rational numbers available.
>
> As far as I can tell, sliding those intervals around does not change any
> of the rational points into irrationals or vice versa, so where did all
> those rational points disappear to?

Each rational q_n is now in an interval I_n of measure or length 10^-
n.

Regards, WM

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag

news:564cd36f-e5cd-4c55...@a16g2000vby.googlegroups.com...

> On 8 Jun., 12:19, Jürgen R. <jurg...@arcor.de> wrote:
>> "WM" <mueck...@rz.fh-augsburg.de> schrieb im
>

I am afraid you forget to rrespond to the main point - indeed, you
must have inadvertantly deleted the question -

What did you mean when you called two points "adjacent"?

You undoubtedly understand that your otherwise fascinating and
important argument suffers from the use of undefined concepts.

>> > But it is not so important for the result. Simply start with a circle
>> > of measure 1 and cut it into aleph_0 intervals, each of measure 1/2.
>> > Can you imagine that you have 2^aleph_0 intervals? I dare to guess you
>> > cannot.
>>
>> You are completely right! The number of intervals certainly
>> certainly is countable.
>>
>
>> > Then slide the cuts around the circle in an arbitrary way. Can you
>> > imagine that you get 2^aleph_0 intervals?
>>
>> Absolutely not! There are countably many intervals,
>> although there must be uncountably many points in those
>> intervals.
>
> In the first kind of intervals, there can be uncountably many points.
> In the second kind, that it not possible, because there are absolutely
> no rational numbers available.

Sorry - you have lost me - first kind, second kind, rational
numbers all used up? I don't follow.

>
>> Again I must apologize for my difficulties. I cannot see the
>> relationship between this example and the original construction,
>> i.e. the ternary set type of construction. Perhaps you could
>> transfer this to the circle?
>>
> No, Cantor's example does not eliminate all rational numbers. There
> remain rationals like 1/4 or 3/10. But what should this special
> example in ternary be good for? My example works in every basis.

Actually it is an excellent example: It removes a countable set of points -
a subset of the rationals, namely the centers of the middle third
intervals, and a neighborhood of each.

And it allows every assertion concerning the properties of the
residual set to be answered explicitly and numerically.

But, of course, the latter may not be desirable for your purposes.

>
> Regards, WM

[WM]

On 9 Jun., 10:53, Jürgen R. <jurg...@arcor.de> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> schrieb im Newsbeitragnews:564cd36f-e5cd-4c55...@a16g2000vby.googlegroups.com...

>
> > On 8 Jun., 12:19, Jürgen R. <jurg...@arcor.de> wrote:
> >> "WM" <mueck...@rz.fh-augsburg.de> schrieb im
>
> I am afraid you forget to rrespond to the main point - indeed, you
> must have inadvertantly deleted the question -
>
> What did you mean when you called two points "adjacent"?
>
> You undoubtedly understand that your otherwise fascinating and
> important argument suffers from the use of undefined concepts.

You know that there are no two real numbers that are really adjadent
on the real axis. Therefore I have introduced intervals (and clusters
which also form inervals). An endpoint x is adjacent to its interval.
That means between x and the interval there are no further points.

If we define: two points are adjacent* if no uncovered irrational
numbers lie between them, then (and only in this respect: there are no
uncovered irrational numbers between x and the interval) the endpoint
x can be said to be adjacent to every point of the interval.

>
> Sorry - you have lost me - first kind, second kind, rational
> numbers all used up? I don't follow.
>

First kind: After sliding to the final position containing q_n and
having length 10^-n. Second kind: the complements.

>
>
> >> Again I must apologize for my difficulties. I cannot see the
> >> relationship between this example and the original construction,
> >> i.e. the ternary set type of construction. Perhaps you could
> >> transfer this to the circle?
>
> > No, Cantor's example does not eliminate all rational numbers. There
> > remain rationals like 1/4 or 3/10. But what should this special
> > example in ternary be good for? My example works in every basis.
>
> Actually it is an excellent example: It removes a countable set of points -
> a subset of the rationals, namely the centers of the middle third
> intervals, and a neighborhood of each.
>
> And it allows every assertion concerning the properties of the
> residual set to be answered explicitly and numerically.
>
> But, of course, the latter may not be desirable for your purposes.

If a theory is contradictory, it is of no use to discuss only those
examples which do not obviously contain contradictions.

Regards, WM

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag

news:189c8d52-3de5-4d7c...@w24g2000vby.googlegroups.com...

> On 9 Jun., 10:53, Jürgen R. <jurg...@arcor.de> wrote:
>> "WM" <mueck...@rz.fh-augsburg.de> schrieb im
>> Newsbeitragnews:564cd36f-e5cd-4c55...@a16g2000vby.googlegroups.com...
>>
>> > On 8 Jun., 12:19, Jürgen R. <jurg...@arcor.de> wrote:
>> >> "WM" <mueck...@rz.fh-augsburg.de> schrieb im
>>
>> I am afraid you forget to rrespond to the main point - indeed, you
>> must have inadvertantly deleted the question -
>>
>> What did you mean when you called two points "adjacent"?
>>
>> You undoubtedly understand that your otherwise fascinating and
>> important argument suffers from the use of undefined concepts.
>
> You know that there are no two real numbers that are really adjadent
> on the real axis. Therefore I have introduced intervals (and clusters
> which also form inervals). An endpoint x is adjacent to its interval.
> That means between x and the interval there are no further points.
>
> If we define: two points are adjacent* if no uncovered irrational
> numbers lie between them, then (and only in this respect: there are no
> uncovered irrational numbers between x and the interval) the endpoint
> x can be said to be adjacent to every point of the interval.

So that means that two points are "adjacent" if and only if
they belong to the closure of one of the connected components
if the covering.

That's a good clear simple definition. But I still have a problem
with the original claim, which was:

"Take one of those uncovered points x_0There are two possibilities:
Either x_0s adjacent to a covered point (i.e. a point of an interval)
or to a not covered point, i.e. an uncovered irrational. The latter
possibility is a contradiction. Hence only the first is really
possible."

I am sorry, but this statement seems to make no sense at all with
the new definition of adjacent, viz. it now says every
uncovered point is an endpoint of a component because otherwise
it would be in the same component with another uncovered point.

Have I perhaps misunderstood something. I am almost certain that
must be the case.

>
>>
>> Sorry - you have lost me - first kind, second kind, rational
>> numbers all used up? I don't follow.
>>
> First kind: After sliding to the final position containing q_n and
> having length 10^-n. Second kind: the complements.

But how did all the rationals get used up? You mean there are no
rationals in the complementary intervals?

>>
>>
>> >> Again I must apologize for my difficulties. I cannot see the
>> >> relationship between this example and the original construction,
>> >> i.e. the ternary set type of construction. Perhaps you could
>> >> transfer this to the circle?
>>
>> > No, Cantor's example does not eliminate all rational numbers. There
>> > remain rationals like 1/4 or 3/10. But what should this special
>> > example in ternary be good for? My example works in every basis.
>>
>> Actually it is an excellent example: It removes a countable set of
>> points -
>> a subset of the rationals, namely the centers of the middle third
>> intervals, and a neighborhood of each.
>>
>> And it allows every assertion concerning the properties of the
>> residual set to be answered explicitly and numerically.
>>
>> But, of course, the latter may not be desirable for your purposes.
>
> If a theory is contradictory, it is of no use to discuss only those
> examples which do not obviously contain contradictions.

Precisely - what one needs to do is deduce a contradiction. But
unfortunately the deduction needs to be unassailable. Otherwise
the dicotomy "either my result is false or the axioms are
inconsistent" will come to haunt you.
>
> Regards, WM

[WM]

On 9 Jun., 12:03, Jürgen R. <jurg...@arcor.de> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> schrieb im Newsbeitragnews:189c8d52-3de5-4d7c...@w24g2000vby.googlegroups.com...

> >> I am afraid you forget to rrespond to the main point - indeed, you
> >> must have inadvertantly deleted the question -
>
> >> What did you mean when you called two points "adjacent"?
>
> >> You undoubtedly understand that your otherwise fascinating and
> >> important argument suffers from the use of undefined concepts.
>
> > You know that there are no two real numbers that are really adjadent
> > on the real axis. Therefore I have introduced intervals (and clusters
> > which also form inervals). An endpoint x is adjacent to its interval.
> > That means between x and the interval there are no further points.
>
> > If we define: two points are adjacent* if no uncovered irrational
> > numbers lie between them, then (and only in this respect: there are no
> > uncovered irrational numbers between x and the interval) the endpoint
> > x can be said to be adjacent to every point of the interval.
>
> So that means that two points are "adjacent" if and only if
> they belong to the closure of one of the connected components
> if the covering.

Then they are adjacent*. The asterisk is important.

>
> That's a good clear simple definition. But I still have a problem
> with the original claim, which was:
>
> "Take one of those uncovered points x_0There are two possibilities:
> Either x_0s adjacent to a covered point (i.e. a point of an interval)
> or to a not covered point, i.e. an uncovered irrational. The latter
> possibility is a contradiction. Hence only the first is really
> possible."
>
> I am sorry, but this statement seems to make no sense at all with
> the new definition of adjacent, viz. it now says every
> uncovered point is an endpoint of a component

No. Please think more carefully. Every x_0 is an endpoint of a /
covered/ interval.

> because otherwise
> it would be in the same component with another uncovered point.

And where is your problem? There are no uncovered intervals because
there are no uncovered rationals and there are no irrationals without
rationals between them.

> Have I perhaps misunderstood something. I am almost certain that
> must be the case.

See above.

>
>
>
> >> Sorry - you have lost me - first kind, second kind, rational
> >> numbers all used up? I don't follow.
>
> > First kind: After sliding to the final position containing q_n and
> > having length 10^-n. Second kind: the complements.
>
> But how did all the rationals get used up? You mean there are no
> rationals in the complementary intervals?

There are many rationals in the compelementary intervals which have
measure 1 - 10^-n. But every rational is covered by at least one
interval of the first kind.

>
> > If a theory is contradictory, it is of no use to discuss only those
> > examples which do not obviously contain contradictions.
>
> Precisely - what one needs to do is deduce a contradiction. But
> unfortunately the deduction needs to be unassailable.

There is no unassailable deduction. If tolerance towards humbug is
only sufficiently great, then even undefinable definitions,
unaccessible real numbers and time after never get accepted. And, as
we see here, disperging an interval into uncountably many parts by
using countably many cuts.

Real die-hard matheologians will never give up their dogmatic view.
They must die out.

Regards, WM

[Uergil]

In article
<03420584-76c2-4807...@n5g2000vbb.googlegroups.com>,

In each of which, having positive length, contains uncountably any
irrationals and outside of which there also uncountably any irrationals.
And these since these intervals have sum of lengths less that that of
intreval (0,10), uncountably many points not included in them.


And infinitely many rationals in any I_n are of form q_m with m > n, and
quite possibly some with I_m a subset of I_n.

[Uergil]

In article
<189c8d52-3de5-4d7c...@w24g2000vby.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 9 Jun., 10:53, Jürgen R. <jurg...@arcor.de> wrote:
> > "WM" <mueck...@rz.fh-augsburg.de> schrieb im
> > Newsbeitragnews:564cd36f-e5cd-4c55...@a16g2000vby.googlegroup
> > s.com...
> >
> > > On 8 Jun., 12:19, Jürgen R. <jurg...@arcor.de> wrote:
> > >> "WM" <mueck...@rz.fh-augsburg.de> schrieb im
> >
> > I am afraid you forget to rrespond to the main point - indeed, you
> > must have inadvertantly deleted the question -
> >
> > What did you mean when you called two points "adjacent"?
> >
> > You undoubtedly understand that your otherwise fascinating and
> > important argument suffers from the use of undefined concepts.
>
> You know that there are no two real numbers that are really adjadent
> on the real axis. Therefore I have introduced intervals (and clusters
> which also form inervals). An endpoint x is adjacent to its interval.
> That means between x and the interval there are no further points.

That assumes, contrary to fact, that every irrational is a boundary
point of some cluster. But it is quite possible, and even necessary,
for every neighborhood of such an uncovered irrational to contain
infinitely many clusters and infinitely many uncovered irrationals.

>
> If we define: two points are adjacent* if no uncovered irrational
> numbers lie between them, then (and only in this respect: there are no
> uncovered irrational numbers between x and the interval) the endpoint
> x can be said to be adjacent to every point of the interval.

That assumes, contrary to fact, that every irrational is a boundary
point of some cluster. But it is quite possible, and even necessary,
for every neighborhood of such an uncovered irrational to contain
infinitely many clusters and infinitely many uncovered irrationals.

>
> >
> > Sorry - you have lost me - first kind, second kind, rational
> > numbers all used up? I don't follow.
> >
> First kind: After sliding to the final position containing q_n and
> having length 10^-n. Second kind: the complements.

> If a theory is contradictory, it is of no use to discuss only those
> examples which do not obviously contain contradictions.

And WM's theories, at least the ones he posts here, contain
contradictions.

[Alan Smaill]

Uergil <Uer...@uer.net> writes:

> And WM has already conceded the ability to finish infinite processes by
> creating his model, based on the I_n's, without being able to prove
> that model's own inconsistency.

Of course, it's an entirely legitimate way to go about showing
the inconsistency of a formal system (eg classical set theory)
by proving its inconsistency in its own terms.

Not that he has succeeded in doing that yet, but for someone who
thinks some of the presuppositions are false, it's just like
Russell's letter to Frege -- it doesn't mean "conceding the
ability to finish infinite processes" outright, but only for
sake of argument.

Perhaps this much is beyond WM, since he refuses to pin down any
one such "proof", and instead insists that many such proofs
are necessary. Goodness knows why -- it's his loss (and ours).


--
Alan Smaill

[Alan Smaill]

WM <muec...@rz.fh-augsburg.de> writes:

> There is no unassailable deduction.

A proof that fits with underlying set theory would be unassailable;
there are large scale proofs that have been checked that way.
And your proofs are "simple and obvious", you say -- there is nothing
particularly deep in them, properly presented. So, go and do the
work!

> If tolerance towards humbug is
> only sufficiently great, then even undefinable definitions,
> unaccessible real numbers and time after never get accepted. And, as
> we see here, disperging an interval into uncountably many parts by
> using countably many cuts.

Where's the problem -- you can use any definition you want,
definable in set theory, to bring it down. Just pin down *one* proof
of contradiction.

>
> Regards, WM

--
Alan Smaill

[Uergil]

In article
<a0b12b06-0734-4280...@8g2000vbu.googlegroups.com>,

There are certainly uncovered irrationals that need not be limit points
of any clusters, unless those U_n intervals of cumulative length 1/9 can
cover the interval (0,1).

>
> > because otherwise
> > it would be in the same component with another uncovered point.
>
> And where is your problem? There are no uncovered intervals because
> there are no uncovered rationals and there are no irrationals without
> rationals between them.

The problem is that if each uncovered irrational is the common endpoint
of two clusters, then they are as countable as the rationals and one can
then cover the unit interval using closed intervals of cumulative length
less that 1.

Which I find self contradictory.

>
>
> > Have I perhaps misunderstood something. I am almost certain that
> > must be the case.
>
> See above.
> >
> >
> >
> > >> Sorry - you have lost me - first kind, second kind, rational
> > >> numbers all used up? I don't follow.
> >
> > > First kind: After sliding to the final position containing q_n and
> > > having length 10^-n. Second kind: the complements.
> >
> > But how did all the rationals get used up? You mean there are no
> > rationals in the complementary intervals?
>
> There are many rationals in the compelementary intervals which have
> measure 1 - 10^-n. But every rational is covered by at least one
> interval of the first kind.
> >
> > > If a theory is contradictory, it is of no use to discuss only those
> > > examples which do not obviously contain contradictions.
> >
> > Precisely - what one needs to do is deduce a contradiction. But
> > unfortunately the deduction needs to be unassailable.
>
> There is no unassailable deduction. If tolerance towards humbug is
> only sufficiently great, then even undefinable definitions

WM is the one who is proposing humbug. He is requiring a covering of the
unit interval by a list of intervals whose sum of lengths can be made
less that any positive number, thus WM is claiming that the unit
interval has outer measure zero.

> unaccessible real numbers and time after never get accepted. And, as
> we see here, disperging an interval into uncountably many parts by
> using countably many cuts.

Note that both the rationals and the irrationals, of different
cardinalities, are both dense in the ordered set of reals, so consider
the ordered set of uncovered irrationals and clusters.

WM claims, but has not proved, that in the resulting totally ordered
set, the uncovered irrationals are not dense in the total or that the
clusters are not dense in the total.

And without one or the other proof, his claims all fail.

[WM]

One is enough. Prove that you can use infinite binary strings (paths
in the Binary Tree) without having a finite definition.

Regards, WM

[WM]

On 9 Jun., 23:55, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

Cut the circle of length 1 aleph_0 times in two halves. That will
yield aleph_0 intervals of length 1/2 and aleph_0 intervals of length
1/2 in the complement. Observe that there are also aleph_0 intervals
between the cuts. Now let the cuts slide such that you get aleph_0
intervals I_n of length 10^-n centered around the rational q_n. How
can 2^aleph_0 intervals be created in the complement by this
continuous prodedure?

Regards, WM

[Uergil]

In article
<2918353b-ca95-49ec...@x21g2000vbc.googlegroups.com>,

Only WM claims that it can be.

This circular model is in no way related to WM's model on the unit
interval

WM claimed intervals I_n of length 1/10^n, thus of cumulative length
1/9, able to cover all of [0,1] but a set of outer measure 0.

Note that when one replaces each one of WM's clusters by its midpoint,
one must still have a "real" interval of length >= 9/10.

[Uergil]

In article
<f001d4f4-5ac7-4e7e...@fr28g2000vbb.googlegroups.com>,

Cantor's antidiagonal argument does not require any binary strings to
actually be represented, but still shows that any complete listing of
them is impossible.

[Alan Smaill]

You have this upside down --
you are the one who has to provide a proof of contradiction.

Show that a contradiction arises in a theory with
only countable many names that claims there are uncountably many
reals. You may use the axiom of choice.

[WM]

On 13 Jun., 11:31, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> WM <mueck...@rz.fh-augsburg.de> writes:

> > One is enough. Prove that you can use infinite binary strings (paths
> > in the Binary Tree) without having a finite definition.
>
> You have this upside down --
> you are the one who has to provide a proof of contradiction.
>
> Show that a contradiction arises in a theory with
> only countable many names that claims there are uncountably many
> reals.  You may use the axiom of choice.

If two elements of a set cannot be distinguished, then they are not
different. Even the axiom of choice will not help. Extensionality is
violated if you cannot distinguish the elements. Therefore it was
said, and thought, that most real numbers, although they cannot be
distinguished by finite definitions, are distinct by their infinite
representations. I have shown all of you that you are wrong. You
cannot distinguish what (countably many) real numbers I use to
construct a set that seems to contains all real numbers.

Regards, WM

[Alan Smaill]

WM <muec...@rz.fh-augsburg.de> writes:

> On 13 Jun., 11:31, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>
>> > One is enough. Prove that you can use infinite binary strings (paths
>> > in the Binary Tree) without having a finite definition.
>>
>> You have this upside down --
>> you are the one who has to provide a proof of contradiction.
>>
>> Show that a contradiction arises in a theory with
>> only countable many names that claims there are uncountably many
>> reals.  You may use the axiom of choice.
>
> If two elements of a set cannot be distinguished, then they are not
> different.

If you mean "effectively distinguished", then say so.
If you mean that if two sets have the same elements, then they
are identical, that is extensionality. (The elements themselves are sets
in the set theoretic foundation.)

> Even the axiom of choice will not help.

Giving you the chance to use AC makes your task in finding
a contradiction easier.

> Extensionality is
> violated if you cannot distinguish the elements.

Same comment as above.

> Therefore it was
> said, and thought, that most real numbers, although they cannot be
> distinguished by finite definitions, are distinct by their infinite
> representations.

Whatever that means, it is clear that in the usual exposition most
reals do not have definitions/names/labels.

> I have shown all of you that you are wrong. You
> cannot distinguish what (countably many) real numbers I use to
> construct a set that seems to contains all real numbers.

you "prove" that there's a set which *"seems"* to contain all reals?

In general, yes, most real numbers lack names, and we cannot effectively
distinguish them (in the usual story).

But where's the contradiction in all this?
(As opposed to WM's personal credo concerning the "facts".)

[WM]

On 13 Jun., 17:54, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> Whatever that means, it is clear that in the usual exposition most
> reals do not have definitions/names/labels.

Why do you call them distinct? By what do they differ?

>
> > I have shown all of you that you are wrong. You
> > cannot distinguish what (countably many) real numbers I use to
> > construct a set that seems to contains all real numbers.
>
> you "prove" that there's a set which *"seems"* to contain all reals?

If there are all reals of the unit interval, then the Binary Tree is a
representation of that set because it contains all of them in form of
infinite paths.

>
> In general, yes, most real numbers lack names, and we cannot effectively
> distinguish them (in the usual story).

Do you think that you can distinguish them ineffectively? And what do
you mean by that expression?

>
> But where's the contradiction in all this?

Numbers are used to distinguish objects (Dedekind). But if not even
the numbers can be distinguished, what are they good for?

Such numbers are only good for set theorists to play their silly
nonsense games. You know that there is not the least chance for set
theory to be useful in any respect? Even the claim it would be
required to complete mathematics, to make the real axis continuous, to
guarantee that every polynomial has its zeros, is nonsense, if you
cannot distinguish the real numbers - effectively or ineffectively.

All real numbers that ever can appear in mathematical calculations
have finite names and belong to a countable set. Therefore your
asserted uncountably many unreal reals are good for nothing like all
your asserted "mathematics".

Regards, WM

[WM]

On 10 Jun., 20:52, Uergil <Uer...@uer.net> wrote:

>
> > Cut the circle of length 1 aleph_0 times in two halves. That will
> > yield aleph_0 intervals of length 1/2 and aleph_0 intervals of length
> > 1/2 in the complement. Observe that there are also aleph_0 intervals
> > between the cuts. Now let the cuts slide such that you get aleph_0
> > intervals I_n of length 10^-n centered around the rational q_n. How
> > can 2^aleph_0 intervals be created in the complement by this
> > continuous prodedure?
>
> Only WM claims that it can be.
>
> This circular model is in no way related to WM's model on the unit
> interval

It is precisely the same. First we get aleph_0 intervals and aleph_0
complementary intervals. But we can definie complementary in an
arbitrary way. So there cannot be more intervals than complemtary
intervals. Then we let the endpoints slice continuously until the
intervals I_n have length 10^-n. What is impossible? What is different
in the result from the linear unit interval?

>
> WM claimed intervals I_n of length 1/10^n, thus of cumulative length
> 1/9,

Only in case countable was a meaningful notion!

However, a totally disconnected space is not disconnected by God but
by rationals, in my example by intervals of finite length.

Let two points be called adjacent, when there is no uncovered
irrational between them.
Then every uncovered irrational x is adjacent to all points of two
intervals, namely one below and one above. There is no chance to
realize the totally disconnected space in another way.

Regards, WM

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag

news:50130530-fdad-48b3...@f30g2000vbz.googlegroups.com...

Well, then tell me, Herr Professor Doktor Mueckenheim, how do you solve
the equation

ih partial(du/dt) = H(u)

where i = sqrt(-1), h Planck's constant and H the Hamiltonian
operator, u = u(x,t), in your discrete space?

>
> Regards, WM
>

[WM]

On 13 Jun., 20:09, Jürgen R. <jurg...@arcor.de> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> schrieb im Newsbeitragnews:50130530-fdad-48b3...@f30g2000vbz.googlegroups.com...

You would like to get a real number as a "solution" that cannot be
distinguished from uncountably many others? How will you distinguish
them? Or will you accept uncountably many different (differing by what
property) solutions?

Regards, WM

[Uergil]

In article
<e598701b-2021-4dd4...@n5g2000vbb.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 10 Jun., 20:52, Uergil <Uer...@uer.net> wrote:
>
> >
> > > Cut the circle of length 1 aleph_0 times in two halves. That will
> > > yield aleph_0 intervals of length 1/2 and aleph_0 intervals of length
> > > 1/2 in the complement. Observe that there are also aleph_0 intervals
> > > between the cuts. Now let the cuts slide such that you get aleph_0
> > > intervals I_n of length 10^-n centered around the rational q_n. How
> > > can 2^aleph_0 intervals be created in the complement by this
> > > continuous prodedure?
> >
> > Only WM claims that it can be.
> >
> > This circular model is in no way related to WM's model on the unit
> > interval
>
> It is precisely the same.

WM's model on the unit interval involved clusters, a cluster being an
open interval containing the closed I_n interval centered on any of its
rational members.

There is no way to do tis on a circle when one includes both closed
intervals having any particular pair of endpoints.

First we get aleph_0 intervals and aleph_0
> complementary intervals. But we can definie complementary in an
> arbitrary way. So there cannot be more intervals than complemtary
> intervals. Then we let the endpoints slice continuously until the
> intervals I_n have length 10^-n. What is impossible? What is different
> in the result from the linear unit interval?

No clusters!

>
> >
> > WM claimed intervals I_n of length 1/10^n, thus of cumulative length
> > 1/9,
>
> Only in case countable was a meaningful notion!

And WM has still not proved it otherwise

>
> However, a totally disconnected space is not disconnected by God but
> by rationals, in my example by intervals of finite length.
>
> Let two points be called adjacent, when there is no uncovered
> irrational between them.
> Then every uncovered irrational x is adjacent to all points of two
> intervals, namely one below and one above.

Often claimed but never proven and, in fact, not true.

Suppose in the following that the named irrationals are uncovered and
have no other uncovered irrationals between any two of them"

Consider the following irrational sequences in (0,1):

1/sqrt(2) - 1/(sqrt(13)*n)
and
1/sqrt(2) + 1/(sqrt(13)*n)

Their members are all irrational and could well be uncovered irrationals
in one of WM's constructions.
And the irrational 1/sqrt(2).

But note that 1/sqrt(2) is now separated by at least one rational and

one irrational from EACH sequence member, but is not separated by any

rational, or irrational, or anything else, from ALL sequence members or

all spaces between consecutive sequence members, as both sequences

converge to it.

Thus 1/sqrt(2) could itself be an uncovered irrational of the very sort
that WM claims cannot ever exist.

Thus demonstrating that what WM claims cannot occur can quite easily
occur.

> There is no chance to
> realize the totally disconnected space in another way.

In fact, if each cluster is replaced by one of its rationals one will
have something order isomorphic to a real interval.

[Uergil]

In article
<3b632643-0685-4c7b...@m3g2000vbl.googlegroups.com>,

But if all of your countable set have n-ary expansions for some n >= 2
then the Cantor diagonal argument can find one not included.

[Uergil]

In article
<50130530-fdad-48b3...@f30g2000vbz.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 13 Jun., 17:54, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> >
> > Whatever that means, it is clear that in the usual exposition most
> > reals do not have definitions/names/labels.
>
> Why do you call them distinct? By what do they differ?

Since they are dense, one may be able to find a rational between two of
them, tus showing them to be different.

> >
> > > I have shown all of you that you are wrong. You
> > > cannot distinguish what (countably many) real numbers I use to
> > > construct a set that seems to contains all real numbers.
> >
> > you "prove" that there's a set which *"seems"* to contain all reals?
>
> If there are all reals of the unit interval, then the Binary Tree is a
> representation of that set because it contains all of them in form of
> infinite paths.

Only a COMPLETE infinite binary tree need contain all of them (and
infinitely many of them twice) ,and WM's tree is always INcomplete.

[WM]

On 13 Jun., 22:37, Uergil <Uer...@uer.net> wrote:

> There is no way to do tis on a circle when one includes both closed
> intervals having any particular pair of endpoints.
>
>  First we get aleph_0 intervals and aleph_0
>
> > complementary intervals. But we can definie complementary in an
> > arbitrary way. So there cannot be more intervals than complemtary
> > intervals. Then we let the endpoints slice continuously until the
> > intervals I_n have length 10^-n. What is impossible? What is different
> > in the result from the linear unit interval?
>
> No clusters!

No? Even if at the end all intervals have the same positions, namely
all rational numbers as their centers? There must be something
different.

> > There is no chance to
> > realize the totally disconnected space in another way.
>
> In fact, if each cluster is replaced by one of its rationals one will
> have something order isomorphic to a real interval.

We need not "something so and so" but an explanation how aleph_0
intervals by continuous deformation become uncountaby many intervals.

Regards, WM

[WM]

On 13 Jun., 22:46, Uergil <Uer...@uer.net> wrote:
> In article

> <50130530-fdad-48b3-b642-844ef6e99...@f30g2000vbz.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 13 Jun., 17:54, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > > Whatever that means, it is clear that in the usual exposition most
> > > reals do not have definitions/names/labels.
>
> > Why do you call them distinct? By what do they differ?
>
> Since they are dense, one may be able to find a rational between two of
> them, tus showing them to be different.

But there are only countably many rationals. How can they distinguish
uncountably many reals?

>
>
>
> > > > I have shown all of you that you are wrong. You
> > > > cannot distinguish what (countably many) real numbers I use to
> > > > construct a set that seems to contains all real numbers.
>
> > > you "prove" that there's a set which *"seems"* to contain all reals?
>
> > If there are all reals of the unit interval, then the Binary Tree is a
> > representation of that set because it contains all of them in form of
> > infinite paths.
>
> Only a COMPLETE infinite binary tree need contain all of them (and
> infinitely many of them twice) ,and WM's tree is always INcomplete.

I have constructed a complete Binary Tree by means of countably many
paths. There is no single node missing. And you cannot find which path
is missing. If there are more paths claimed, then they must be
identified and, in addition, they must consist of more than nodes.

Regards, WM

[Jürgen R.]

"WM" <muec...@rz.fh-augsburg.de> schrieb im Newsbeitrag

news:c93843a7-2039-42e7...@fr28g2000vbb.googlegroups.com...

In mathematics and in physics solutions of differential equations
are not "numbers", nor are they "names".

In your world there is no derivative and all integrals
are zero. Or have I again misunderstood you?

Why don't you tell us a bit about how calculus
is done by non-matheological methods?

>
> Regards, WM

[Ross A. Finlayson]

On Jun 13, 1:41 pm, Uergil <Uer...@uer.net> wrote:
> In article

> <3b632643-0685-4c7b-b4c8-c1dc01e87...@m3g2000vbl.googlegroups.com>,

Nah, we already worked through that the only way to get that is to
define it on the range of EF, any expansion resolves to binary and
there's exactly one anti-diagonal. And, this function EF can be built
standardly, modeled by real functions as not-a-real-function, along
similar lines as the unit impulse or step functions.

Basically it's like a pasture where it is green on the other side of
the fence, with a sign saying "the other side doesn't exist".

So, look around.

Regards,

Ross Finlayson

[Ross A. Finlayson]

On Jun 13, 2:18 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:

EF (scaled) and the reverse EF or REF are the only functions, or here
one function with a real scalar that defaults to one, that work this
way.

And the integral of EF_x is x, where EF_x is xn/d as d -> oo and n ->
d; n, d E N, x E R.

Basically if you split the square evenly into infinitely many lines,
its area doubles, because each line has two sides, where when they are
re-integrated it's the square. This led to definitions of iota-sums
and iota-multiples as working features of these polydimensional
quantities.

Linearization introduces error terms and normalization is de-re-
normalization. Now, we know that higher energy experiments see the
universe larger and fundamental particles smaller, not more bounded
but more unbounded, in increasing precision.

Then, where iota is simple a least real or scalar quantity in a model
of the reals as partially ordered ring with rather restricted transfer
principle along the constructible lines of Schmieden and Laugwitz, it
seems a reasonable term.

Nothing to see here? Move along.

Regards,

Ross Finlayson

[Uergil]

In article
<aa169619-3b9b-4cdf...@s9g2000vbg.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 13 Jun., 22:46, Uergil <Uer...@uer.net> wrote:
> > In article
> > <50130530-fdad-48b3-b642-844ef6e99...@f30g2000vbz.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 13 Jun., 17:54, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> >
> > > > Whatever that means, it is clear that in the usual exposition most
> > > > reals do not have definitions/names/labels.
> >
> > > Why do you call them distinct? By what do they differ?
> >
> > Since they are dense, one may be able to find a rational between two of

> > them, thus showing them to be different.

>
> But there are only countably many rationals. How can they distinguish
> uncountably many reals?

All I said was between any two irrationals one ought to be able to find
at least one rational, which is enough to distinguish these two
irrationals from each other.

> >
> >
> >
> > > > > I have shown all of you that you are wrong. You
> > > > > cannot distinguish what (countably many) real numbers I use to
> > > > > construct a set that seems to contains all real numbers.
> >
> > > > you "prove" that there's a set which *"seems"* to contain all reals?
> >
> > > If there are all reals of the unit interval, then the Binary Tree is a
> > > representation of that set because it contains all of them in form of
> > > infinite paths.
> >
> > Only a COMPLETE infinite binary tree need contain all of them (and
> > infinitely many of them twice) ,and WM's tree is always INcomplete.
>
> I have constructed a complete Binary Tree by means of countably many
> paths.

You keep claiming to have done so , but as it is provable that no
countable set of paths can exhaust the set of paths in such a tree you
have failed.

> There is no single node missing.

But from any of WM's trees there are sets of nodes forming paths that
are missing.

There is an easy bijection between the set of paths of a truly complete
infinite binary tree and the set of all subsets (the power set) of N.
For each path, there is the subset of N of level at which the path
branches left.

For WM's trees no such bijection exists. In his tree one needs only the
finite subsets of N, those node levels before the common tail of all
paths occurs.

> And you cannot find which path
> is missing.

As soon as you let me find out the common tail (or up to countably many
tails) of all your paths, I can find a path not in your tree.

> If there are more paths claimed, then they must be
> identified and, in addition, they must consist of more than nodes.

I claim one path, P, for every subset, S, of N, P consisting of the
root node and for every n in S, the left child of the nth node of P and
for every n not in S, the right child of the nth node of P.

Since paths never consist of anything other than nodes, why must my
paths consist of more when yours don't?

[Uergil]

In article
<cdde27b2-afaa-4c29...@eh4g2000vbb.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 13 Jun., 22:37, Uergil <Uer...@uer.net> wrote:
>

> > There is no way to do it on a circle when one includes both closed

> > intervals having any particular pair of endpoints.
> >
> >  First we get aleph_0 intervals and aleph_0
> >
> > > complementary intervals. But we can definie complementary in an
> > > arbitrary way. So there cannot be more intervals than complemtary
> > > intervals. Then we let the endpoints slice continuously until the
> > > intervals I_n have length 10^-n. What is impossible? What is different
> > > in the result from the linear unit interval?
> >
> > No clusters!
>
> No? Even if at the end all intervals have the same positions, namely
> all rational numbers as their centers? There must be something
> different.

Also, for each interval of length 10^(-n)m you also now have an interval
of length 1 - 10^(-n), so you may well end up with only one cluster of
of the whole circle.

>
> > > There is no chance to
> > > realize the totally disconnected space in another way.
> >
> > In fact, if each cluster is replaced by one of its rationals one will
> > have something order isomorphic to a real interval.
>
> We need not "something so and so" but an explanation how aleph_0
> intervals by continuous deformation become uncountaby many intervals.

You circle example seem more likely to end up not with any "intervals"
at all , but with the whole circle.

[Uergil]

In article
<c93843a7-2039-42e7...@fr28g2000vbb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

Non-anwers are copouts!

[Ross A. Finlayson]

http://www.tiki-lounge.com/~raf/finlayson_injectrationals.pdf

[Uergil]

In article
<5bdc7e6d-0a46-4e08...@h10g2000pbi.googlegroups.com>,
"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

> http://www.tiki-lounge.com/~raf/finlayson_injectrationals.pdf

Claimed Proof flawed!
Every positive irrational is the limit of a strictly increasing sequence
of rationals and an increasing sequence of irrationals, which alows your
Q_(/>i) , which you claim to be not empty, to be empty.

[Ross A. Finlayson]

On Jun 13, 5:38 pm, Uergil <Uer...@uer.net> wrote:
> In article

> <5bdc7e6d-0a46-4e08-a69c-bf5491677...@h10g2000pbi.googlegroups.com>,

>  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
> >http://www.tiki-lounge.com/~raf/finlayson_injectrationals.pdf
>
> Claimed Proof flawed!
> Every positive irrational is the limit of a strictly increasing sequence
> of rationals and an increasing sequence of irrationals, which alows your
> Q_(/>i) , which you claim to be not empty, to be empty.
> --
> "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

Keep reading about the contrapositive bit, there.

What you don't have density of the rationals in the reals? And the
irrationals in the reals?

No, thank you, that's addressed there, in terms of the contradiction
that otherwise follows.

You'll notice there's no mention of EF or any other features of the
real numbers at all, it's quite self-contained following definitions
of the rationals, the reals, and the rationals' complement in the
reals.

No, you don't have that a non-empty interval contains no rationals,
except insofar as perhaps your interval is degenerate, where here it
isn't.
.
Regards,

Ross Finlayson

[Uergil]

In article
<a1a8c91b-72da-4594...@t1g2000pbl.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

> On Jun 13, 5:38 pm, Uergil <Uer...@uer.net> wrote:
> > In article
> > <5bdc7e6d-0a46-4e08-a69c-bf5491677...@h10g2000pbi.googlegroups.com>,
> >  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
> >
> > >http://www.tiki-lounge.com/~raf/finlayson_injectrationals.pdf
> >
> > Claimed Proof flawed!
> > Every positive irrational is the limit of a strictly increasing sequence
> > of rationals and an increasing sequence of irrationals, which alows your
> > Q_(/>i) , which you claim to be not empty, to be empty.
> > --
> > "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
>
> Keep reading about the contrapositive bit, there.

One error is enough!

[Ross A. Finlayson]

On Jun 13, 8:21 pm, Uergil <Uer...@uer.net> wrote:
> In article

> <a1a8c91b-72da-4594-82ee-5feb6cb0f...@t1g2000pbl.googlegroups.com>,

>  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
>
>
>
>
> > On Jun 13, 5:38 pm, Uergil <Uer...@uer.net> wrote:
> > > In article
> > > <5bdc7e6d-0a46-4e08-a69c-bf5491677...@h10g2000pbi.googlegroups.com>,
> > >  "Ross A. Finlayson" <ross.finlay...@gmail.com> wrote:
>
> > > >http://www.tiki-lounge.com/~raf/finlayson_injectrationals.pdf
>
> > > Claimed Proof flawed!
> > > Every positive irrational is the limit of a strictly increasing sequence
> > > of rationals and an increasing sequence of irrationals, which alows your
> > > Q_(/>i) , which you claim to be not empty, to be empty.
> > > --
> > > "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
>
> > Keep reading about the contrapositive bit, there.
>
> One error is enough!
> --
> "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

Sorry, bud, I want the whole schmeer.

What you'll notice is that either follows. Either is as fundamental
as the other.

Yeah, one error is enough.

Eudoxus/Cauchy/Dedekind: insufficient to fully represent the real
numbers.

And if you poll instructors of topology, you'll find them either
agreeing there are vagaries in the definitions, or incompetent, or
maybe, just as it were, single-minded and not very familiar with the
panoply of surrounding arguments.

Then, I agree, it would require a function from, say, the naturals,
to, say, the unit interval of reals, to really see how this carries
out.

Regards,

Ross Finlayson

[Uergil]

In article
<58d82e88-3fda-4dce...@re8g2000pbc.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

There are lots of such functions, just no surjectons.

Or do you reject both of Cantor's relevant theorems?

Both of which establish the impossibility of such surjections.
*********************************************
A PROOF OF THE UNCOUNTABILITY OF THE REALS

This proof is based on the difference between the order properties of
the real line and the order properties of the set of naturals numbers.

A strictly increasing sequence of naturals cannot have a natural as its
limit,
A strictly increasing but bounded sequence of reals has a real limit,
its least upper bound, different from every member of the sequence.

A strictly decreasing but bounded sequence of reals has a real limit,
its greatest lower bound, different from every member of the sequence.

Proof:

If the reals are countable then we may assume each real has been paired
with a natural so that different reals are paired with different
naturals with none of either left out.

Assuming this has been done, take the two reals with the lowest naturals
as endpoints of a real interval.
It is clear that all the interior points of this real interval must be
paired with naturals larger that those of its endpoints.

Now take the two reals interior to that interval with the lowest
naturals to be the endpoints of a subinterval of that interval.

It is clear that the interior points of this real interval must be
paired with naturals larger that those of the endpoints.

By repeating this process one generates a decreasing, but never empty,
sequence of closed real intervals each of which contains only points
with higher attached naturals than its endpoints have.

The intersection of such a nested sequence of closed intervals is not
empty, but the natural associated with any of its members is necessarily
larger than all of the infinitely many natural numbers associated with
those infinitely many endpoints.

But there is no natural number larger thatn infinitel many different
natural numbers.

This is a contradiction which can only have been caused by our original
assumption that the reals were countable, so it proves they are not
countable.

While I have never seen this particular form of proof in the literature
of countabiity, it is so obvious that I doubt that it is original with
me.

[Ross A. Finlayson]

On Jun 13, 9:33 pm, Uergil <Uer...@uer.net> wrote:
> In article

> <58d82e88-3fda-4dce-aafb-7ddd14a49...@re8g2000pbc.googlegroups.com>,

This doesn't hold for EF, whatever its range is. You paraphrase
Cantor's nested intervals result, which sees EF uniquely a special
function, here of a sort where the special functions are generally a
particular category of real and complex valued functions. EF is
unique among functions in this regard.

> By repeating this process one generates a decreasing, but never empty,
> sequence of closed real intervals each of which contains only points
> with higher attached naturals than its endpoints have.
>
> The intersection of such a nested sequence of closed intervals  is not
> empty, but the natural associated with any of its members is necessarily
> larger than all of the infinitely many natural numbers associated with
> those infinitely many endpoints.
>
> But there is no natural number larger thatn infinitel many different
> natural numbers.
>
> This is a contradiction which can only have been caused by our original
> assumption that the reals were countable, so it  proves they are not
> countable.
>
> While I have never seen this particular form of proof in the literature
> of countabiity, it is so obvious that I doubt that it is original with
> me.
> --
> "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

Nope, doesn't follow.

Uergil, you're Virgil.

Regards,

Ross F,

[Uergil]

In article
<80c273b0-1571-44ae...@oo8g2000pbc.googlegroups.com>,

"Ross A. Finlayson" <ross.fi...@gmail.com> wrote:

As my argument does not rely on any EF, whatever EF may mean to Ross,
Ross' objection fails.

> You paraphrase
> Cantor's nested intervals result,

Certainly a similar argument, but not identical.

> which sees EF uniquely a special
> function, here of a sort where the special functions are generally a
> particular category of real and complex valued functions. EF is
> unique among functions in this regard.

Since your 'EF' remains unidentified, your argument remains failed.

>
> > By repeating this process one generates a decreasing, but never empty,
> > sequence of closed real intervals each of which contains only points
> > with higher attached naturals than its endpoints have.
> >
> > The intersection of such a nested sequence of closed intervals �is not
> > empty, but the natural associated with any of its members is necessarily
> > larger than all of the infinitely many natural numbers associated with
> > those infinitely many endpoints.
> >

> > But there is no natural number larger than infinitely many different

> > natural numbers.
> >
> > This is a contradiction which can only have been caused by our original
> > assumption that the reals were countable, so it �proves they are not
> > countable.
> >
> > While I have never seen this particular form of proof in the literature
> > of countabiity, it is so obvious that I doubt that it is original with
> > me.
> > --
> > "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
>
> Nope, doesn't follow.
>

It follows for those who follow logic.
What does not follow is your objections to it.
>
> Regards,
>
> Ross F,