Clusters and Cantor dust

[WM]

Let all rational numbers q_n of the interval (0, oo) be covered by intervals I_n of measure |I_n| = 2^−n, such that q_n is the centre of I:n. Then the endpoints are rational numbers. The irrational numbers x of the complement of infinite measure, not covered by the intervals I_n, form particles of a totally disconnected space, so-called "Cantor dust" C.

Every particle x of C must be separated from every particle y of C by at least one rational number q_n and hence by at least one interval I_n covering q_n. Since the end points 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 intervals don't overlap, then a limit must lie between them, but in any case infinitely many endpoints are required to establish a limit.)

Such an infinite set of overlapping intervals is called a cluster. In principle, given a fixed and complete enumeration of the rationals, we can calculate every cluster and the limits of its union. Therefore, every irrational x of C can be put in bijection with the infinite set of intervals converging to it from the right-hand side, say. There are countably many sets I_n and therefore not more than countably many disjoint clusters with limits and therefore not more limits. Where are the other irrational numbers of the complement?

Regards, WM

[George Greene]

On Monday, December 31, 2018 at 5:39:32 AM UTC-5, WM wrote:
> Let all rational numbers q_n of the interval (0, oo) be covered by intervals I_n of measure |I_n| = 2^−n, such that q_n is the centre of I:n. Then the endpoints are rational numbers. The irrational numbers x of the complement of infinite measure, not covered by the intervals I_n, form particles of a totally disconnected space, so-called "Cantor dust" C.
>
> Every particle x of C must be separated from every particle y of C by at least one rational number q_n and hence by at least one interval I_n covering q_n. Since the end points 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 intervals don't overlap, then a limit must lie between them, but in any case infinitely many endpoints are required to establish a limit.)
>
> Such an infinite set of overlapping intervals is called a cluster.
> In principle, given a fixed and complete enumeration of the rationals,
> we can calculate every cluster and the limits of its union.

The most problematic word in that sentence was "can".
Most of the clusters are going to be undefinable.
Their limit, once they exist, will be well-defined GIVEN the sequence&cluster,
but for most reals, you will NOT be able TO GIVE the sequence&cluster finitarily or according to any definable rule.
Given the rationals, the class of infinite-sequences-of-rationals-with-limits
IS VERY large, and, again, because of the higher infinity, most of these sequences will be undefinable. Nevertheless, if they exist and have a limit, well, THEY HAVE ONE, REGARDLESS of whether anyone "can" or
cannot "calculate" it.

> Therefore, every irrational x of C can be put in bijection with

wait for it.. it's not wrong yet...

> the infinite set of

SORRY, the *the* is wrong. There are MANY DIFFERENT sequences of rationals
converging to any given irrational real. You can't get a bijection unless
you have a rule for picking one of the many, and that's quite hopeless.
Even if you limit to the ones that match bit-strings (since you're using 2^-n), there will STILL be too many OF THOSE.

> intervals converging to it from the right-hand side, say.

Yes, if the point x is in the complement, any sequence of rationals converging to it from the right side must define a sequence of left-endpoints-of-those-associated intervals that also converge to the same limit. The left endpoints of these intervals of rationals-converging-from-the-right are in fact always closer to the limit than the converging rationals, and if you have rational endpoints, then that sequence (of left end points, not intervals) is ANOTHER sequence of rationals (in addition to the original sequence of rationals) that also converges to the same limit.

The clusters are not unique but they are all subsets of some larger convergent set of intervals.

> There are countably many sets I_n

right.

> and therefore not more than countably many disjoint

Disjointness is highly irrelevant. Many clusters that converge WILL OVERLAP.
Every convergent cluster overlaps with some other nearby convergent cluster, OBVIOUSLY. It is possible to define a cluster-converging-to-one-irrational such that none of its own elements overlap with each other, but getting them not to overlap with any elements of the cluster converging to a DIFFERENT nearby rational is harder (though I could explain how).

In any case, regardless of disjointness, your "countably"is just completely wrong. There are uncountably many limit-points to begin with so obviously
there have to be uncountably many clusters. All the clusters are countably infinite subsets of the countably infinite set of intervals SO OBVIOUSLY the class of infinite subsets of an infinite set is NOT countable.

> clusters with limits and therefore not more limits.

The number of clusters is also uncountable, even after you require them to be disjoint.

Given any two different reals x<y you can in fact always "find" (even if not *de*fine) two disjoint clusters converging to them simply by choosing an n so large that that the first q_n in each limit-sequence is no more than (y-x)/4 from its intended target, and all the subsequent intervals are smaller than (y-x)/8 -- indeed, if you are converging from the right to both different points then the clusters are guaranteed to be disjoint for that reason alone since every element of the cluster converging to y is greater than y, while every element of the cluster converging to x is less than y.

[WM]

Am Dienstag, 1. Januar 2019 03:53:06 UTC+1 schrieb George Greene:
> On Monday, December 31, 2018 at 5:39:32 AM UTC-5, WM wrote:


> > Such an infinite set of overlapping intervals is called a cluster.
> > In principle, given a fixed and complete enumeration of the rationals,
> > we can calculate every cluster and the limits of its union.
>
> The most problematic word in that sentence was "can".
> Most of the clusters are going to be undefinable.

Hardly. The clusters are disjoint subsets of the countable set of intervals I_n with centre q_n. Therefore there are countably many clusters.

> Their limit, once they exist, will be well-defined GIVEN the sequence&cluster,
> but for most reals, you will NOT be able TO GIVE the sequence&cluster finitarily or according to any definable rule.

The clusters as disjoint subsets of the countable set of intervals I_n are well-defined. Your argument shows that not more than countably many irrational numbers can exist as limits of clusters. Every cluster can define at most two limits, one at its left-hand side and one at its right-hand-side.

> Given the rationals, the class of infinite-sequences-of-rationals-with-limits

They are mostly irrelevant. Every cluster defines only two limits at its ends. Rational numbers belonging to different clusters do not define limits. As soon as a cluster ends, all its intervals have defined the limit at this end.

Superimpose intervals of lenght 1/2, 1/4, 1/8, .... such that every midpoint is the ednpoint of the preceding interval. Then there is a limit. What happens at the next cluster is irrelevant for the members of this cluster.

If you construct a converging sequence of points of infinitely many clusters then the limit either coincides with the limit of one cluster or it falls into an interval and is irrelevant for the points of the complement.

> IS VERY large, and, again, because of the higher infinity, most of these sequences will be undefinable. Nevertheless, if they exist and have a limit, well, THEY HAVE ONE, REGARDLESS of whether anyone "can" or
> cannot "calculate" it.

These sequences formed by points belonging to different clusters and their limits are irrelevant for the limits of clusters. Either these limits coincide with the limit of a single cluster or they fall into an interval and therefore are irrelevant for the points in the complement. In fact they are only imagined limits, not materialized by really overlapping intervals.

>
> > Therefore, every irrational x of C can be put in bijection with
>

> You can't get a bijection unless
> you have a rule for picking one of the many, and that's quite hopeless.

No, see above. Every cluster can have two irrational limits - not more.

Regards, WM

[Julio Di Egidio]

On Monday, 31 December 2018 11:39:32 UTC+1, WM wrote:

> Let all rational numbers q_n of the interval (0, oo) be covered
> by intervals I_n of measure |I_n| = 2^−n, such that q_n is the
> centre of I:n.

Which is the n-th rational number? (How do you map 1-to-1 rationals to
naturals?) Doesn't it matter? And how do you prove they are disjoint?

Julio

[Julio Di Egidio]

That is, how do you prove the intervals are disjoint?

Julio

[WM]

Am Dienstag, 1. Januar 2019 03:53:06 UTC+1 schrieb George Greene:

> On Monday, December 31, 2018 at 5:39:32 AM UTC-5, WM wrote:


> Yes, if the point x is in the complement, any sequence of rationals converging to it from the right side must define a sequence of left-endpoints-of-those-associated intervals that also converge to the same limit. The left endpoints of these intervals of rationals-converging-from-the-right are in fact always closer to the limit than the converging rationals, and if you have rational endpoints, then that sequence (of left end points, not intervals) is ANOTHER sequence of rationals (in addition to the original sequence of rationals) that also converges to the same limit.

The sequence of left endpoints s_n of intervals I_n = [s_n, t_n] is the decisive sequence defining the limit x in the complement.

>
> The clusters are not unique but they are all subsets of some larger convergent set of intervals.

The clusters are unique. x is not covered by a cluster but x is the endpoint of two clusters.

>
> > There are countably many sets I_n
>
> right.
>
> > and therefore not more than countably many disjoint
>
> Disjointness is highly irrelevant.

It is fact however. Between two clusters there is a limit. Since all rationals are within clusters, this limit is irrational and is in the complement.

> Many clusters that converge WILL OVERLAP.

None will overlap an irrational x in the complement. By definition, a cluster is a complete set of overlapping intervals.

> Every convergent cluster overlaps with some other nearby convergent cluster, OBVIOUSLY.

A cluster is an infinite set of overlapping intervals with two limits. These limits are the boundaries of that cluster.

> It is possible to define a cluster-converging-to-one-irrational such that none of its own elements overlap with each other

That is possible but useless. The limit x is either within another cluster or it is ib the complement. Then it is necessarily a limit of that cluster that has x as one of its two limits.

Imagine three clusters

...123...x_1...456...x_2...789...

and let the points 2, 5, 6, 7, ... converge to an irrational limit in the complement, then this limit is also the limit of a cluster with endpoints of intervals converging to it.

>, but getting them not to overlap with any elements of the cluster converging to a DIFFERENT nearby rational is harder (though I could explain how).

There is no possibility to explain this because it is nonsense.

>
> In any case, regardless of disjointness, your "countably"is just completely wrong. There are uncountably many limit-points

And as you have seen, that is impossible. Uncountability is a notion of nonsense.

> to begin with so obviously
> there have to be uncountably many clusters.

But they cannot exist. A cluster covers all its points and ends only at a limit. It is impossible that one of these few limits is covered by another cluster, because then it would disappear.

>
> The number of clusters is also uncountable,

That is impossible. Clusters are sets of intervals, in fact infinite sets, and cover each of their points. Clusters by definition do not overlap with other clusters. Therefore their number is countable.

> Given any two different reals x<y you can in fact always "find" (even if not *de*fine) two disjoint clusters

That is true but irrelevant, as I have frequently shown. It is impossible to find two clusters converging to x and y where x and y are not separated by at least another cluster the limits of which are converging to x and / or y.

Uncountably many different sequences converge to at most countably many limits of clusters - if uncountable and countable are sensible notions.

Regards, WM

[WM]

Am Dienstag, 1. Januar 2019 13:02:39 UTC+1 schrieb Julio Di Egidio:
> On Tuesday, 1 January 2019 13:00:43 UTC+1, Julio Di Egidio wrote:
> > On Monday, 31 December 2018 11:39:32 UTC+1, WM wrote:
> >
> > > Let all rational numbers q_n of the interval (0, oo) be covered
> > > by intervals I_n of measure |I_n| = 2^−n, such that q_n is the
> > > centre of I:n.
> >
> > Which is the n-th rational number? (How do you map 1-to-1 rationals to
> > naturals?)

I use Cantors example.

> Doesn't it matter?

No. Every enumeration shows the same contradiction.

> And how do you prove they are disjoint?
>
> That is, how do you prove the intervals are disjoint?

The intervals are not disjoint in general, because the endpoints of each interval are rational and therefore also covered by intervals.

The clusters are disjoint by definition- I said: Such an infinite set of overlapping intervals is called a cluster.

Therefore the clusters are dijoint by limits which are irrational and lie in the complement.

Regards, WM

[Julio Di Egidio]

On Tuesday, 1 January 2019 14:34:32 UTC+1, WM wrote:
> Am Dienstag, 1. Januar 2019 13:02:39 UTC+1 schrieb Julio Di Egidio:
> > On Tuesday, 1 January 2019 13:00:43 UTC+1, Julio Di Egidio wrote:
> > > On Monday, 31 December 2018 11:39:32 UTC+1, WM wrote:
> > >
> > > > Let all rational numbers q_n of the interval (0, oo) be covered
> > > > by intervals I_n of measure |I_n| = 2^−n, such that q_n is the
> > > > centre of I:n.
> > >
> > > Which is the n-th rational number? (How do you map 1-to-1 rationals to
> > > naturals?)
>
> I use Cantors example.

(q_n) = (1/1, 1/2, 2/1, 1/3, ...)

|I_n|/2 = (2^(-n))/2 = 2^(-n-1) = 1/2^(n+1)

I_0 = 1/1 -+ 1/2^1 = (2 -+ 1)/2 = [1/2,3/2] = [0.5,1.5]

I_1 = 1/2 -+ 1/2^2 = (2 -+ 1)/4 = [1/4,3/4] = [0.25,0.75]

I_2 = 2/1 -+ 1/2^3 = (8 -+ 1)/8 = [7/8,9/8] = [0.875,1.125]

I_3 = 1/3 -+ 1/2^4 = (16 -+ 3)/48 = [13/48,19/48] = [0.2708(3),0.3958(3)]

...

> > Doesn't it matter?
>
> No. Every enumeration shows the same contradiction.
>
> > And how do you prove they are disjoint?
> >
> > That is, how do you prove the intervals are disjoint?
>
> The intervals are not disjoint in general, because the endpoints of each
> interval are rational and therefore also covered by intervals.
>
> The clusters are disjoint by definition- I said: Such an infinite set of
> overlapping intervals is called a cluster.
>
> Therefore the clusters are dijoint by limits which are irrational and lie
> in the complement.

I do not understand which clusters you are talking about. Those intervals
indeed overlap all over the place. So, where is the "complement of infinite
measure, not covered by the intervals I_n"?

Julio

[jvr]

On Monday, December 31, 2018 at 11:39:32 AM UTC+1, WM wrote:
> Let all rational numbers q_n of the interval (0, oo) be covered by intervals I_n of measure |I_n| = 2^−n, such that q_n is the centre of I:n. Then the endpoints are rational numbers. The irrational numbers x of the complement of infinite measure, not covered by the intervals I_n, form particles of a totally disconnected space, so-called "Cantor dust" C.
>
> Every particle x of C must be separated from every particle y of C by at least one rational number q_n and hence by at least one interval I_n covering q_n. Since the end points of the I_n are rational numbers too, also being covered by their own intervals,

Up to this point everything is ok. Your "clusters" are intervals - so C is the complement of a countably infinite set of disjoint intervals with finite total measure covering the rationals.

> the particles of Cantor dust can only be limits of infinite sequences of endpoints of overlapping intervals I_n.

This is an error. The counter-intuitive fact is that C does not consist entirely of endpoints of the intervals in its complement. You are claiming that C is countable, which contradicts the fact that the Lebesgue measure of C in non-zero.

(If intervals don't overlap, then a limit must lie between them, but in any case infinitely many endpoints are required to establish a limit.)
>
> Such an infinite set of overlapping intervals is called a cluster. In principle, given a fixed and complete enumeration of the rationals, we can calculate every cluster and the limits of its union. Therefore, every irrational x of C can be put in bijection with the infinite set of intervals converging to it from the right-hand side, say. There are countably many sets I_n and therefore not more than countably many disjoint clusters with limits and therefore not more limits. Where are the other irrational numbers of the complement?

This just repeats the misconception noted above. There is no such bijection.

Despite the error I am positively surprised by the fact that you have succeeded in writing several sentences that are both unambiguous and free of gross mistakes.

[WM]

Am Dienstag, 1. Januar 2019 15:34:32 UTC+1 schrieb Julio Di Egidio:
> On Tuesday, 1 January 2019 14:34:32 UTC+1, WM wrote:


> > I use Cantors example.
>
> (q_n) = (1/1, 1/2, 2/1, 1/3, ...)

I_n = [s_n, t_n] has the centre q_n and measure |I_n| = 1/2^n.

>
> I_0 = 1/1 -+ 1/2^1 = (2 -+ 1)/2 = [1/2,3/2] = [0.5,1.5]

Okay, when you start with n = 0. It does not matter.

>
> I_1 = 1/2 -+ 1/2^2 = (2 -+ 1)/4 = [1/4,3/4] = [0.25,0.75]
>
> I_2 = 2/1 -+ 1/2^3 = (8 -+ 1)/8 = [7/8,9/8] = [0.875,1.125]

Here is an error: [15/8, 17/8]
>

> > Therefore the clusters are disjoint by limits which are irrational and lie

> > in the complement.
>
> I do not understand which clusters you are talking about. Those intervals
> indeed overlap all over the place.

That is impossible because their total measure is 1 (or 2, if you start with 0) but not infinite.

The clusters are constructed as follows: If we start with [1/2,3/2], then 1/2 is in the centre of [1/4,3/4] and also 3/2 will appear in Cantor's enumeration. But also 1/4 will be in the centre of an interval and so on.

There are irrational numbers provably not covered. See for instance Liouville's theorem for an easy approach: https://www.hs-augsburg.de/~mueckenh/HI/HI10.PPT

Regards, WM

[Julio Di Egidio]

On Tuesday, 1 January 2019 16:53:42 UTC+1, WM wrote:
> Am Dienstag, 1. Januar 2019 15:34:32 UTC+1 schrieb Julio Di Egidio:
> > On Tuesday, 1 January 2019 14:34:32 UTC+1, WM wrote:
>
> > > I use Cantors example.
> >
> > (q_n) = (1/1, 1/2, 2/1, 1/3, ...)
>
> I_n = [s_n, t_n] has the centre q_n and measure |I_n| = 1/2^n.
> >
> > I_0 = 1/1 -+ 1/2^1 = (2 -+ 1)/2 = [1/2,3/2] = [0.5,1.5]
>
> Okay, when you start with n = 0. It does not matter.
> >
> > I_1 = 1/2 -+ 1/2^2 = (2 -+ 1)/4 = [1/4,3/4] = [0.25,0.75]
> >
> > I_2 = 2/1 -+ 1/2^3 = (8 -+ 1)/8 = [7/8,9/8] = [0.875,1.125]
>
> Here is an error: [15/8, 17/8]

My bad, but that doesn't change anything.

> > > Therefore the clusters are disjoint by limits which are irrational and lie
> > > in the complement.
> >
> > I do not understand which clusters you are talking about. Those intervals
> > indeed overlap all over the place.
>
> That is impossible because their total measure is 1 (or 2, if you start with 0) but not infinite.
>
> The clusters are constructed as follows: If we start with [1/2,3/2], then 1/2 is in the centre of [1/4,3/4] and also 3/2 will appear in Cantor's enumeration. But also 1/4 will be in the centre of an interval and so on.
>
> There are irrational numbers provably not covered.

Despite the size of the intervals tends to vanish, it never actually vanishes,
so I still cannot see how that sequence is not covering the whole (0,oo) (nor,
where are you taking any limits). But it might be me...

Julio

[WM]

Am Dienstag, 1. Januar 2019 16:39:42 UTC+1 schrieb jvr:
> On Monday, December 31, 2018 at 11:39:32 AM UTC+1, WM wrote:
> > Let all rational numbers q_n of the interval (0, oo) be covered by intervals I_n of measure |I_n| = 2^−n, such that q_n is the centre of I:n. Then the endpoints are rational numbers. The irrational numbers x of the complement of infinite measure, not covered by the intervals I_n, form particles of a totally disconnected space, so-called "Cantor dust" C.
> >
> > Every particle x of C must be separated from every particle y of C by at least one rational number q_n and hence by at least one interval I_n covering q_n. Since the end points of the I_n are rational numbers too, also being covered by their own intervals,
>
> Up to this point everything is ok. Your "clusters" are intervals - so C is the complement of a countably infinite set of disjoint intervals with finite total measure covering the rationals.

Yes.

>
> > the particles of Cantor dust can only be limits of infinite sequences of endpoints of overlapping intervals I_n.
>
> This is an error. The counter-intuitive fact is that C does not consist entirely of endpoints of the intervals in its complement.

Limits of clusters, to be precise.

> You are claiming that C is countable, which contradicts the fact that the Lebesgue measure of C is non-zero.

That shows the contradiction.

There is only the the wisdom that I quoted already and that is true. Infinite sets behave differently from finite sets. For instance: If we remove N intervals from the real axis, then N+1 intervals will remain. But if we remove countably many intervals, then uncountably many atoms or degenerate intervals will remain. But the most important application is rarely mentioned: If we biject finite sets, then we get information about their relative cardinality. For infinite sets this method fails. Same with the diagonal-argument. There is no information obtained. Otherwise the intervals or clusters could be used as an enumeration of the irrationals they converge to. Their number does not change when shuffled. But the method itself is invalid - since infinite sets behave differently.+

Have you never wondered why the bijection between every q and every n is constructed for every pair (q, n) in a finite set (since every n belongs to a finite initial segment) but is claimed to be valid for the infinite sets? Injectivity and surjectivity can be proven for every q and n. But all belong to a finite initial segment. Nothing is proven and nothing can be proven for the whole infinite sets. We know: Infinite sets behave differently!

>
> (If intervals don't overlap, then a limit must lie between them, but in any case infinitely many endpoints are required to establish a limit.)
> >
> > Such an infinite set of overlapping intervals is called a cluster. In principle, given a fixed and complete enumeration of the rationals, we can calculate every cluster and the limits of its union. Therefore, every irrational x of C can be put in bijection with the infinite set of intervals converging to it from the right-hand side, say. There are countably many sets I_n and therefore not more than countably many disjoint clusters with limits and therefore not more limits. Where are the other irrational numbers of the complement?
>
> This just repeats the misconception noted above. There is no such bijection.

Every cluster can be mapped on the pair of its limits.

Regards, WM

[George Greene]

On Tuesday, January 1, 2019 at 7:00:43 AM UTC-5, Julio Di Egidio wrote:
> Which is the n-th rational number? (How do you map 1-to-1 rationals to
> naturals?) Doesn't it matter?

You begin by assuming that that's possible.
WM doesn't believe it is so he is going to try to derive a contradiction
from this is assumption. Since he is arguing *against*us*, WE don't GET
to object to the existence of this bijection between N and Q -- we're the
side *defending* the existence of this bijection (we're locally calling it q_n).
We're taking this bijection as given; we are giving it to him.

> And how do you prove they are disjoint?

Again, WE *GIVE* to him the fact that every IRrational real has an infinite sequence of RATIONAL reals with that real as its limit. In other words, it has
a subsequence of q_n whose limit is that irrational. Any infnite subsequence OF that subsequence WILL ALSO have the same limit, so, basically, you just keep taking later and later subsequences until the n's get big enough to keep the intervals small enough that they are disjoint.

[Julio Di Egidio]

On Tuesday, 1 January 2019 17:18:10 UTC+1, George Greene wrote:
> On Tuesday, January 1, 2019 at 7:00:43 AM UTC-5, Julio Di Egidio wrote:
> > Which is the n-th rational number? (How do you map 1-to-1 rationals to
> > naturals?) Doesn't it matter?
>
> You begin by assuming that that's possible.

*He* begins by assuming that, the very moment he introduces q_n!

> WM doesn't believe it is so he is going to try to derive a contradiction
> from this is assumption. Since he is arguing *against*us*, WE don't GET
> to object to the existence of this bijection between N and Q -- we're the
> side *defending* the existence of this bijection (we're locally calling it q_n).
> We're taking this bijection as given; we are giving it to him.
>
> > And how do you prove they are disjoint?
>
> Again, WE *GIVE* to him the fact that every IRrational real has an infinite sequence of RATIONAL reals with that real as its limit. In other words, it has
> a subsequence of q_n whose limit is that irrational. Any infnite subsequence OF that subsequence WILL ALSO have the same limit, so, basically, you just keep taking later and later subsequences until the n's get big enough to keep the intervals small enough that they are disjoint.

Thank you, I start getting what kind of "clusters" you guys are talking about...

Julio

[WM]

Am Dienstag, 1. Januar 2019 17:11:56 UTC+1 schrieb George Greene:

> On Tuesday, January 1, 2019 at 6:25:28 AM UTC-5, WM wrote:
> > No, see above. Every cluster can have two irrational limits - not more.
>

> Not exactly "wrong", but the POINT is, every cluster has *ONE* irrational limit, not more

That would not change the countability of the limits. However, every cluster is confined by parts of the complement. And the complement has no intervals.

>, because the left and right endpoints get arbitrarily close TO EACH OTHER, so the limit of the left endpoints, the limit of the center rationals, and the limit of the right endpoints, ARE ALL ALWAYS THE SAME.

That is not excluded (but does not increase the number of limits). By the "two" limits I mean that on the left-hand side and that on the right-hand side of the cluster.

Regards, WM

[WM]

Am Dienstag, 1. Januar 2019 17:53:13 UTC+1 schrieb George Greene:

> On Tuesday, January 1, 2019 at 7:28:50 AM UTC-5, WM wrote:
>
> > The sequence of left endpoints s_n of
> > intervals I_n = [s_n, t_n] is the decisive sequence defining
> > the limit x in the complement.
>

> That is absolutely true, but what you seem to be missing is that
> the sequence q_n of the centers and the sequence t_n of the right
> endpoints ALSO HAVE THAT EXACT SAME x as THEIR limit.

No, I am not missing that. But it is highly irrelevant. Concerning the limit of the cluster at the right-hand side only the sequence formed by the right-hand endpoints of the interval is relevant. All other sequences either have the same limit or other limits. Wherever the latter may lie, they are irrelevant for finding points of the complement.

Regards, WM

[WM]

Am Dienstag, 1. Januar 2019 17:51:45 UTC+1 schrieb George Greene:
> On Tuesday, January 1, 2019 at 6:25:28 AM UTC-5, WM wrote:

> > Am Dienstag, 1. Januar 2019 03:53:06 UTC+1 schrieb George Greene:
> > > On Monday, December 31, 2018 at 5:39:32 AM UTC-5, WM wrote:
> >
> >
> > > > Such an infinite set of overlapping intervals is called a cluster.
> > > > In principle, given a fixed and complete enumeration of the rationals,
> > > > we can calculate every cluster and the limits of its union.
> > >
> > > The most problematic word in that sentence was "can".
> > > Most of the clusters are going to be undefinable.
> >
> > Hardly. The clusters are disjoint
>

> Most clusters ARE NOT disjoint,

Then they are one and the same cluster.

> Every irrational has
> A GREAT MANY DIFFERENT clusters converging to it.

No. The irrtaional x may have many sequences of real numbers converging to it. But a cluster is defined by two limits confining it from both sides.

> *A*N*Y* sequence of rational numbers converging to a given irrational
> x defines a cluster.

No.

> Some of them are going to be overlapping

Then they are one and the same cluster. They define one and the same limit in the exterior, but none in their interior, at least none of interest here.

> and others
> are going to be disjoint, WITHIN the cluster (some clusters have the property
> that every interval in them is disjoint from every other interval in them).

Then there is a point x (not covered and therefore belonging to the complement) dividing the cluster in two different clusters.

>
> > subsets of the countable set of intervals I_n with centre q_n.
>

> Yes, they are all that.

And they are ovrlapping!

>
> > Therefore there are countably many clusters.
>

> For any GIVEN ONE n, there
> is only ONE interval I_n with only ONE center q_n. There are not
> multiple overlapping intervals all with center q_n.

Of course not. Who said so?

> It is the q_n's (as n gets bigger and bigger) that converge to the irrational
> x.

That is irrelevant. Relevant are only the endpoints in direction to the limit.

> The intervals are actually completely irrelevant.

That is a mistake. The intervals define the cluster by overlapping each other.

> For large enough n
> their widths are small enough that said widths just don't matter at all -- you
> might as well be dealing with the q_n's *WITHOUT* the intervals.

Without intervals, the irrational limits of sequences could lie everywhere. By overlapping the intervals decrease the possible sites in the complement to limits of clusters.

Regards, WM

[WM]

Am Dienstag, 1. Januar 2019 17:28:34 UTC+1 schrieb Julio Di Egidio:


> Thank you, I start getting what kind of "clusters" you guys are talking about...
>

---------------
-----------------
------------
-------

Regards, WM

[jvr]

On Tuesday, January 1, 2019 at 5:02:02 PM UTC+1, WM wrote:
> Am Dienstag, 1. Januar 2019 16:39:42 UTC+1 schrieb jvr:
> > On Monday, December 31, 2018 at 11:39:32 AM UTC+1, WM wrote:
> > > Let all rational numbers q_n of the interval (0, oo) be covered by intervals I_n of measure |I_n| = 2^−n, such that q_n is the centre of I:n. Then the endpoints are rational numbers. The irrational numbers x of the complement of infinite measure, not covered by the intervals I_n, form particles of a totally disconnected space, so-called "Cantor dust" C.
> > >
> > > Every particle x of C must be separated from every particle y of C by at least one rational number q_n and hence by at least one interval I_n covering q_n. Since the end points of the I_n are rational numbers too, also being covered by their own intervals,
> >
> > Up to this point everything is ok. Your "clusters" are intervals - so C is the complement of a countably infinite set of disjoint intervals with finite total measure covering the rationals.
>
> Yes.
> >
> > > the particles of Cantor dust can only be limits of infinite sequences of endpoints of overlapping intervals I_n.
> >
> > This is an error. The counter-intuitive fact is that C does not consist entirely of endpoints of the intervals in its complement.
>
> Limits of clusters, to be precise.
>
> > You are claiming that C is countable, which contradicts the fact that the Lebesgue measure of C is non-zero.
>
> That shows the contradiction.

Right. There are proven mathematical facts, e.g. that Lebesgue measure is additive; and there is your vague intuition and your unproven claims, e.g. that there is a 1-1 mapping between a certain set of reals and a vaguely defined set of vaguely defined sequences of sets.

The former is right and the latter provably is wrong. That's the contradiction.

[Mike Terry]

On 01/01/2019 16:21, George Greene wrote:
> On Tuesday, January 1, 2019 at 6:25:28 AM UTC-5, WM wrote:

>> Hardly. The clusters are disjoint subsets of the countable
>> set of intervals I_n with centre q_n.
>

> The clusters IN GENERAL *ARE*NOT* disjoint, DUMBASS.
> Even if they were, they still wouldn't be countable.
>
>> Therefore
>
> You still do not know the meaning of this word.
> The clusters are countably INFINITE subsets of the countably INFINITE
> set of intervals I_n.

>
>> there are countably many clusters.

> No, dumbass, the class of infinite subsets of an infinite set IS UNcountable.
> If you do this with bit-strings then you can map each susbet to a countably-
> infinitely-wide bit-string with an 0 or a 1 in the spot for each element of
> the original underlying set. The cardinality of the class of all these bit-strings is 2^aleph_0, i.e., IS UNcountable.
>

George - I think you've misunderstood how the term "cluster" is being
used. Clusters are not simply arbitrary unions of arbitrary intervals,
or even arbitrary unions of WM's countable set of intervals I_n centred
on the rationals.

Take I_n as the interval centred on q_n as defined elsewhere, then lets
define:

D = Union {I_n: n in |N}
so C = D' (complement of D)


a cluster is a maximal path-connected component of D.

[Well, strictly this is my interpretation, but I'm pretty sure it's what
WM would say if he could express himself better mathematically speaking.]


Since each I_n is path connected and is a subset of D, each I_n belongs
to a unique cluster. Also we can say each cluster is the union of the
intervals I_n which are a subset of that cluster.

Some properties of the clusters:

- each cluster is an interval (but not one of the I_n intervals).
This is just because path-connected components of any subset
of R (reals) are always (possibly degenerate) intervals.
Clusters are never degenerate (single point) intervals as
each cluster contains some I_n.

- each cluster is disjoint.
This is just because if two clusters overlapped, their
union would be a larger path-connected subset of D, but
clusters are maximal path-connected subsets of D.

- When WM says "the limits of a cluster" he just means one
of the two end-points of the cluster interval.

- the endpoints of each cluster interval must be irrational.
This is because otherwise the endpoint would be some q_n,
and so the interval I_n if added to the cluster would
extend it and still be path connected.

- the endpoints of each cluster interval do not belong to
the cluster, i.e. the clusters are open.
This is because if c is the endpoint of a cluster, and
c belonged to the cluster, then by definition of cluster,
c belongs to set D, and so is in some interval I_n. But
I_n has rational endpoints, so c is in the interior of
I_n, which, as in the previous paragraph, can't be the
case. (The cluster would not be maximal path connected.)

- the set of clusters is countable, since each cluster
contains some I_n, and that I_n belongs only to that
cluster and no other.
So also the set of "cluster endpoints" is countable.

- although all cluster endpoints are in C, not every
element of C is a cluster endpoint. (This is the bit
WM claims but never proves.)
It's true that for each x in C, either x is an
endpoint of some cluster, or there is an infinite
sequence of clusters that approach x arbitrarily
closely on each side of x.

HTH,
Mike.

[WM]

Am Dienstag, 1. Januar 2019 19:04:42 UTC+1 schrieb jvr:
> On Tuesday, January 1, 2019 at 5:02:02 PM UTC+1, WM wrote:


> > > You are claiming that C is countable,

I have proved it.

> > > which contradicts the fact that the Lebesgue measure of C is non-zero.
> >
> > That shows the contradiction.
>
> Right. There are proven mathematical facts,

e.g., that infinite sets behave differently from finite sets and therefore infinite bijections behave differently from finite bijections.

> e.g. that Lebesgue measure is additive;

showing that the rational numbers are not in one-to-one correspondence with the natural numbers. In fact if they were, then every reordering would prove it. On the contrary, Cantor's enumeration fails always somewhere in the finite realm. Bad intuition extrapolates it into the infinite.

If we apply really infinite sets, for instance all rationals Q_n between n and n+1 and relate them with n, then we have the proof that the relation between Q and |N is oo to 1 and not 1 to 1.

Why do you believe stronger in Cantor's insufficient process than in the really infinite sets Q_n? It is a matter of intellectual inertia.

> and there is your vague intuition and your unproven claims, e.g. that there is a 1-1 mapping between a certain set of reals and a vaguely defined set of vaguely defined sequences of sets.

That is mathematics. There cannot exist two irrational numbers in the complement without a cluster between them. Therefor we have a bijection between clusters and their limits. Clear cut mathematics. No intuition.

Regards, WM

[Jim Burns]

On 12/31/2018 5:39 AM, WM wrote:

> Let all rational numbers q_n of the interval (0, oo) be
> covered by intervals I_n of measure |I_n| = 2^−n, such
> that q_n is the centre of I:n. Then the endpoints are
> rational numbers. The irrational numbers x of the complement
> of infinite measure, not covered by the intervals I_n,
> form particles of a totally disconnected space, so-called
> "Cantor dust" C.
>
> Every particle x of C must be separated from every particle y
> of C by at least one rational number q_n and hence by at
> least one interval I_n covering q_n. Since the end points 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 intervals don't overlap, then a limit must lie

> between them, but in any case infinitely many endpoints are
> required to establish a limit.)
>

> Such an infinite set of overlapping intervals is called a
> cluster. In principle, given a fixed and complete enumeration
> of the rationals, we can calculate every cluster and the

> limits of its union. Therefore, every irrational x of C can
> be put in bijection with the infinite set of intervals
> converging to it from the right-hand side, say. There are
> countably many sets I_n and therefore not more than countably
> many disjoint clusters with limits and therefore not more
> limits. Where are the other irrational numbers of the
> complement?

I'm trying to clean up your description in order to make sense
of what you are saying. Here goes:

We have a list q_n of all of the rationals in (0,oo).
We have a list of intervals I_n (which I'll say are open
intervals) of measure |I_n| = 2^-n, centered on q_n.

As you point out, the endpoints of each interval I_n are
in the interior of some other intervals.

Instead of talking about clusters, I will partition the
intervals I_n, so that all the intervals in a given
partition overlap (in a certain sense), and do not overlap
with other partitions.

Define
I_k overlap I_m
if and only if
there are points x in I_k and y in I_m such that
every point z between x and y is in some interval I_j
(that is, z is in UNION{ I_n | n e N })

The relation I_k overlap I_m is reflexive, symmetric, and
transitive, so there is some collection of sets of
intervals I_n with each interval in exactly one set
(that is, a partition of { I_n | n e N }), with the _union_
of each set of intervals being s single open interval.
By construction, for each such union, every point is in
its interior (it's open) and there are no gaps (it's
an interval).

Let [ I_n ] be the set of all intervals that overlap I_n,
in the sense I used above.

For all n, there are some a_n, b_n such that
UNION[ I_n ] = (a_n,b_n)

There are no more than countably many of these
union-of-partition intervals. We can index each one, for
example, with the index of the first I_n in a given partition.

I _think_ that what you're calling a cluster is one of
the partitions [ I_n ] of intervals, and that the limits
you are focusing on are the endpoints of the union of
a partition a_n and b_n.

Because of how we constructed the intervals I_n, we know
that a_n and b_n must be irrational. And there are no more
than countably many a_n and b_n.

However, those endpoints _are not_ the only points in
the complement C of the union of intervals I_n. The
argument you make that they _are_ the only points has really
been unchanged through this whole discussion, just given
a new coat of paint and sent out again. And again, and again.

You depend upon there being some partition of intervals
_right next_ to every point in the complement C. And it
it the (alleged) right-next-to partition that (allegedly)
gives you your bijection and so the countability of the
the reals (allegedly).

That would be fine if these partitions of intervals
(probably your "clusters") were only finitely many.
Then you would have a well-order, and everything would
work out nicely. *However* there are infinitely many
of these partitions of intervals I_n. So, there would
usually be no "right-next-to" partition of intervals.

Just as earlier there would be no "right-next-to" I_n, and
just as, before that, there would be no "right-next-to" q_n.

What you call "simple logic" is just recycling the same
old intuition about _finite_ sets that you started with.

----
The argument for missing most of the real numbers
doesn't change either. It's a variation on Cantor's
nested intervals.

We list all of the q_n or all of the I_n or all of
the UNION[ I_n ] = ( a_n, b_n ). We sort them out, all of them
either on one side of ...somewhere... or the other side --
and, by Dedekind completeness, by _what we mean by continuum_ ,
there must be something in the somewhere between the one side
and the other. These between-points are the unaccounted-for
real numbers.

[jvr]

On Tuesday, January 1, 2019 at 7:27:56 PM UTC+1, WM wrote:
> Am Dienstag, 1. Januar 2019 19:04:42 UTC+1 schrieb jvr:
> > On Tuesday, January 1, 2019 at 5:02:02 PM UTC+1, WM wrote:
>
>
> > > > You are claiming that C is countable,
>
> I have proved it.
>

You have proved no such thing.

[WM]

Am Dienstag, 1. Januar 2019 19:18:02 UTC+1 schrieb Mike Terry:
> On 01/01/2019 16:21, George Greene wrote:
> > On Tuesday, January 1, 2019 at 6:25:28 AM UTC-5, WM wrote:


> a cluster is a maximal path-connected component of D.
>
> [Well, strictly this is my interpretation, but I'm pretty sure it's what
> WM would say if he could express himself better mathematically speaking.]
>

Perhaps you have overlooked the definition in the OP: "Such an infinite set of overlapping intervals is called a cluster."

> Some properties of the clusters:
>

> Clusters are never degenerate (single point) intervals as
> each cluster contains some I_n.

Infinitely many I_n, in order to get a limit.

>
> - each cluster is disjoint.
> This is just because if two clusters overlapped, their
> union would be a larger path-connected subset of D, but
> clusters are maximal path-connected subsets of D.

Yes.

>
> - When WM says "the limits of a cluster" he just means one
> of the two end-points of the cluster interval.

Well, the I_n are closed but the clusters are open. Their limits are the endpoints belonging to the complement.

>
> - the endpoints of each cluster interval must be irrational.
> This is because otherwise the endpoint would be some q_n,
> and so the interval I_n if added to the cluster would
> extend it and still be path connected.

Yes.

>
> - the endpoints of each cluster interval do not belong to
> the cluster, i.e. the clusters are open.
> This is because if c is the endpoint of a cluster, and
> c belonged to the cluster, then by definition of cluster,
> c belongs to set D, and so is in some interval I_n. But
> I_n has rational endpoints, so c is in the interior of
> I_n, which, as in the previous paragraph, can't be the
> case. (The cluster would not be maximal path connected.)

Yes.

>
> - the set of clusters is countable, since each cluster
> contains some I_n, and that I_n belongs only to that
> cluster and no other.
> So also the set of "cluster endpoints" is countable.

Yes.

>
> - although all cluster endpoints are in C, not every
> element of C is a cluster endpoint. (This is the bit
> WM claims but never proves.)

That is proved by assuming such a point y. It cannot belong to a cluster because it is in C. If it is not an endpoint, then it must be disconnected from every endpoint by at least one cluster. But that makes y an endpoint of a cluster.

> It's true that for each x in C, either x is an
> endpoint of some cluster, or there is an infinite
> sequence of clusters that approach x arbitrarily
> closely on each side of x.

An infinite sequence of clusters? Clusters consist of intervals. If clusters approach x arbitrarily closely then in fact their intervals approach x arbitrarily closely and we have the same situation as before. Then x is the limit of intervals and as such an endpoint of a cluster.

But if you claim another limit, say y, of a sequence of clusters or of any arbitrary sequence of points (can you tell me what should be the merits of a sequence of clusters here?), then y cannot float in free space. If it is not an endpoint x of a cluster, then it must be disconnected from every endpoint x by at least one cluster. In every neighbourhood there must be points of a cluster at the right-hand side and of a clusters at the left-hand side. But that makes y an endpoint of two clusters, contrary to the assumption.

Regards, WM

[WM]

Am Dienstag, 1. Januar 2019 19:42:52 UTC+1 schrieb Jim Burns:


> You depend upon there being some partition of intervals
> _right next_ to every point in the complement C. And it
> it the (alleged) right-next-to partition that (allegedly)
> gives you your bijection and so the countability of the
> the reals (allegedly).

We do not need that. We can safely say that every cluster has two endpoints (counting each one twice, but that does not matter).

>
> That would be fine if these partitions of intervals
> (probably your "clusters") were only finitely many.

And here it comes again. That argument is correct but to be applied to infinite bijections: Infinite sets behave differently from finite sets. For instance: If we remove N intervals from the real axis, then N+1 intervals will remain. But if we remove countably many intervals, then uncountably many atoms or degenerate intervals will remain. But the most important application is rarely mentioned: If we biject finite sets, then we get information about their relative cardinality. For infinite sets this method fails. Same with the diagonal-argument. There is no information obtained. Otherwise the intervals or clusters could be used as an enumeration of the irrationals they converge to. Their number does not change when shuffled. But the method itself is invalid - since infinite sets behave differently.

You must confess that Cantor bijects only finite sets. Have you never wondered why the bijection between every q and every n is constructed for every pair (q, n) in a finite set (since every n belongs to a finite initial segment) but is claimed to be valid for the infinite sets? Injectivity and surjectivity can be proven for every q and n. But all belong to a finite initial segment. Nothing is proven and nothing can be proven for the whole infinite sets. We know: Infinite sets behave differently.

If you want to apply the really infinite sets, then every n has to be related to the rational numbers between bn and n+1. Then you get the correct result: Q to |N is oo to 1.

And plase note: Clusters have two endpoints, whether they are well-ordered or not.

Regards, WM

[Jim Burns]

On 1/1/2019 2:12 PM, WM wrote:
> Am Dienstag, 1. Januar 2019 19:18:02 UTC+1 schrieb Mike Terry:

>> a cluster is a maximal path-connected component of D.
>>
>> [Well, strictly this is my interpretation, but I'm pretty
>> sure it's what WM would say if he could express himself
>> better mathematically speaking.]
>
> Perhaps you have overlooked the definition in the OP:
> "Such an infinite set of overlapping intervals is called
> a cluster."

That doesn't seem likely.

Suppose that, for three intervals I_i, I_j, I_k,
I_i overlaps (has a non-empty intersection with) I_j,
I_j overlaps I_k, and I_i DOES NOT overlap I_k.

The you would say that I_i and I_k are in the same
"cluster", isn't that correct?

That's what Mike Terry said -- or its equivalent.

[Jim Burns]

On 1/1/2019 2:25 PM, WM wrote:
> Am Dienstag, 1. Januar 2019 19:42:52 UTC+1
> schrieb Jim Burns:

>> You depend upon there being some partition of intervals
>> _right next_ to every point in the complement C. And it
>> it the (alleged) right-next-to partition that (allegedly)
>> gives you your bijection and so the countability of the
>> the reals (allegedly).
>
> We do not need that. We can safely say that every cluster
> has two endpoints (counting each one twice, but that does
> not matter).

That would prove that there are _at least_ countably many
endpoints (points in The Complement).

What you want is that there are _at most_ countably many
endpoints.

You're not going to be able to prove the second.
The points in the complement are not required to be
endpoints of any of the cluster/union-of-path-connected-
-intervals. That they are not so required is the point
of my example of
UNION{ [1/4,3/4], [1/8,7/8], 1/16,15/16], ... } = (0,1)

>> That would be fine if these partitions of intervals
>> (probably your "clusters") were only finitely many.
>
> And here it comes again. That argument is correct but to
> be applied to infinite bijections: Infinite sets behave

> differently from finite sets. [...]

This is a whole other can of worms. It needs a post of its own.

In short, though, you are trying to show that our ideas
about infinite sets don't work, by producing contradictions
from them. And then, when we show how you _haven't_
produced a contradiction from our ideas, you excuse this
by saying that our ideas about infinity are bad, so the
absence of a contradiction doesn't matter. You just chase
around and around in a circle.

I've been back and forth over this ground enough times that
I think I can put all of the reasoning in a single post,
from the formalization of our 'forall' and 'exists' to
reasonably important and non-obvious results, and then
defy you to find a problem with the whole bunch. But that's
for another post (or possibly more than one).

[WM]

Am Dienstag, 1. Januar 2019 21:59:53 UTC+1 schrieb Jim Burns:
> On 1/1/2019 2:25 PM, WM wrote:
> > Am Dienstag, 1. Januar 2019 19:42:52 UTC+1
> > schrieb Jim Burns:
>
> >> You depend upon there being some partition of intervals
> >> _right next_ to every point in the complement C. And it
> >> it the (alleged) right-next-to partition that (allegedly)
> >> gives you your bijection and so the countability of the
> >> the reals (allegedly).
> >
> > We do not need that. We can safely say that every cluster
> > has two endpoints (counting each one twice, but that does
> > not matter).
>
> That would prove that there are _at least_ countably many
> endpoints (points in The Complement).

And we can say that every point has two clusters around it.

>
> What you want is that there are _at most_ countably many
> endpoints.
>
> You're not going to be able to prove the second.

We can say that every point is confined between two clusters.

> The points in the complement are not required to be
> endpoints of any of the cluster/union-of-path-connected-
> -intervals. That they are not so required is the point
> of my example of
> UNION{ [1/4,3/4], [1/8,7/8], 1/16,15/16], ... } = (0,1)

Your example is irrelevant. (Of course it is the same nonsense as the claim that there are points not confined between two clusters.) Why do yoz repeat it? You main argument is that infinite sets differ from finite sets. And you are right! But you remember it some decades too late. It had been the correct answer to Cantor's claim that finite bijections can be exrapolated to infinite sets.

Better use these infinite sets: Every infinite set of rational numbers between n and n+1 may be correlated to n. Then you have at least the complete infinite rational sets.

>

> > And here it comes again. That argument is correct but to
> > be applied to infinite bijections: Infinite sets behave
> > differently from finite sets. [...]
>
> This is a whole other can of worms. It needs a post of its own.
>
> In short, though, you are trying to show that our ideas
> about infinite sets don't work, by producing contradictions
> from them. And then, when we show how you _haven't_
> produced a contradiction from our ideas

You believe in extrapolations from the finite into the infinite, but only in some cases. I have clear mathematical facts: Every cluster has two endpoints and between every two points of the complement there is at least a cluster. Nothing can be more basic mathematics.

Regards, WM

[WM]

Am Dienstag, 1. Januar 2019 21:25:50 UTC+1 schrieb Jim Burns:
> On 1/1/2019 2:12 PM, WM wrote:
> > Am Dienstag, 1. Januar 2019 19:18:02 UTC+1 schrieb Mike Terry:
>
> >> a cluster is a maximal path-connected component of D.
> >>
> >> [Well, strictly this is my interpretation, but I'm pretty
> >> sure it's what WM would say if he could express himself
> >> better mathematically speaking.]
> >
> > Perhaps you have overlooked the definition in the OP:
> > "Such an infinite set of overlapping intervals is called
> > a cluster."
>
> That doesn't seem likely.

That is the definition used in this thread.

>
> Suppose that, for three intervals I_i, I_j, I_k,
> I_i overlaps (has a non-empty intersection with) I_j,
> I_j overlaps I_k, and I_i DOES NOT overlap I_k.
>
> The you would say that I_i and I_k are in the same
> "cluster", isn't that correct?

Yes, of course.
>
This, projected to the bottom, is a cluster consistung of four intervals.

[WM]

I have proved the following:

If every cluster has two endpoints in C and if every point in C is separated by at least one cluster from every other point in C, then there are countably many points in C.

Regards, WM

[jvr]

Your "proof" is incorrect, the assertion is false, and you have been given counter-examples.

[Jim Burns]

On 1/1/2019 4:19 PM, WM wrote:
> Am Dienstag, 1. Januar 2019 21:25:50 UTC+1
> schrieb Jim Burns:
>> On 1/1/2019 2:12 PM, WM wrote:
>>> Am Dienstag, 1. Januar 2019 19:18:02 UTC+1
>>> schrieb Mike Terry:

>>>> a cluster is a maximal path-connected component of D.
>>>>
>>>> [Well, strictly this is my interpretation, but I'm pretty
>>>> sure it's what WM would say if he could express himself
>>>> better mathematically speaking.]
>>>
>>> Perhaps you have overlooked the definition in the OP:
>>> "Such an infinite set of overlapping intervals is called
>>> a cluster."
>>
>> That doesn't seem likely.
>
> That is the definition used in this thread.

What I am calling unlikely is your theory that Mike Terry
overlooked that definition. He told you rather a lot about
the consequences of your definition.

Perhaps you overlooked that.

>> Suppose that, for three intervals I_i, I_j, I_k,
>> I_i overlaps (has a non-empty intersection with) I_j,
>> I_j overlaps I_k, and I_i DOES NOT overlap I_k.
>>
>> The you would say that I_i and I_k are in the same
>> "cluster", isn't that correct?
>
> Yes, of course.
>

[Jim Burns]

On 1/1/2019 4:14 PM, WM wrote:
> Am Dienstag, 1. Januar 2019 21:59:53 UTC+1
> schrieb Jim Burns:
>> On 1/1/2019 2:25 PM, WM wrote:
>>> Am Dienstag, 1. Januar 2019 19:42:52 UTC+1
>>> schrieb Jim Burns:

>>>> You depend upon there being some partition of intervals
>>>> _right next_ to every point in the complement C. And it
>>>> it the (alleged) right-next-to partition that (allegedly)
>>>> gives you your bijection and so the countability of the
>>>> the reals (allegedly).
>>>
>>> We do not need that. We can safely say that every cluster
>>> has two endpoints (counting each one twice, but that does
>>> not matter).
>>
>> That would prove that there are _at least_ countably many
>> endpoints (points in The Complement).
>
> And we can say that every point has two clusters around it.

We can say that there are _no_ clusters _next to_ any
given point in The Complement. Between any point and
a cluster will be at least another cluster.

This (alleged) bijection: how are you selecting the
cluster to which a complement-point gets matched? Can you
show that different complement points are matched to
different clusters? That is what you need to show.

>> What you want is that there are _at most_ countably many
>> endpoints.
>>
>> You're not going to be able to prove the second.
>
> We can say that every point is confined between two clusters.

How many other points are between those two clusters?
Can you prove your answer?

>> The points in the complement are not required to be
>> endpoints of any of the cluster/union-of-path-connected-
>> -intervals. That they are not so required is the point
>> of my example of
>> UNION{ [1/4,3/4], [1/8,7/8], 1/16,15/16], ... } = (0,1)
>
> Your example is irrelevant.

My example shows that your obsession with endpoints
is irrelevant.

[Mike Terry]

On 01/01/2019 19:12, WM wrote:
> Am Dienstag, 1. Januar 2019 19:18:02 UTC+1 schrieb Mike Terry:
>> On 01/01/2019 16:21, George Greene wrote:
>>> On Tuesday, January 1, 2019 at 6:25:28 AM UTC-5, WM wrote:
>
>
>> a cluster is a maximal path-connected component of D.
>>
>> [Well, strictly this is my interpretation, but I'm pretty sure it's what
>> WM would say if he could express himself better mathematically speaking.]
>>
> Perhaps you have overlooked the definition in the OP: "Such an infinite set of overlapping intervals is called a cluster."
>

No, I gave a simple and rigorous definition for the term.

Jim tried heroically to make an actual definition out of your words, and
I would say he succeeded, but in the end his definition amounts to the
same as my "path-connected" definition.)

Still, I think we've both done an OK job, which covers what you intended.

No it doesn't. That's just a wrong claim, not a proof of anything. As
this is the key sticking point for the thread, you need to supply a
(rigorous) proof skipping the vague terms and unsupported claims.

OK, I'll help you in turning it into a proper proof...


Suppose there exists a point of C which is not an endpoint
of any cluster. Let y be such a point.

y does not belong to any cluster, since every cluster
is a subset of D, and y is in C (the complement of D).

Let x be an endpoint of some cluster F. So x != y. WLOG we
can consider the casex > y.
(Just to aid stating comparisons below)

Between x and y there is some rational q_n. Its
corresponding interval I_n belongs to a unique cluster which
we'll label

F_1 = (f_1, g_1). (F_1 being an open interval...)

Since x is not in F_1, and y < q_n, we have

y ≤ f_1 < g_1 ≤ x.

(since otherwise f1 < y < q_n, which would imply y ∈ F_1)

No problem so far - you've basically deduced that there is a cluster F_1
to the right of y. You have not shown that y is an endpoint of F_1 or
reached any contradiction.

So what next? Let me have a guess...

If y != f_1, we can repeat the above argument, using f_1 in place of y,
and we deduce there exists a second cluster

F_2 = (f_2, g_2)

satisfying

y ≤ f2 < g_2 ≤ f_1 < g_1 ≤ x

Well, that's not helped - now you've the same problem with y and f_2,
and you can repeat this as often as you like without reaching any
contradiction, or finding a cluster having y as an endpoint...

So over to you! How to finish the proof? Or start afresh, but it needs
to be a proper proof along the lines of how I started, not wishy-washy
phrases and unjustified claims...

>
>> It's true that for each x in C, either x is an
>> endpoint of some cluster, or there is an infinite
>> sequence of clusters that approach x arbitrarily
>> closely on each side of x.
>
> An infinite sequence of clusters? Clusters consist of intervals. If clusters approach x arbitrarily closely

of course clusters approach x arbitrarily closely. x has rationals in
each of its neighbourhoods, and those rationals all belong to various
clusters.

> then in fact their intervals approach x arbitrarily closely and we
have the same situation as before. Then x is the limit of intervals and
as such an endpoint of a cluster.

No that doesn't work, because these are a sequence of disjoint clusters.
There is no unifying "overlapping set of intervals" which extends all
the way up to x. Instead all you have is a SEQUENCE of clusters that
get closer and closer to x, but each terminates short of x so that x is
never an endpoint of any of them. Not at all the "same situation as
before".

In short, the union of a sequence of clusters "approaching x arbitrarily
closely" is NOT A CLUSTER. Same for intervals belonging to the
clusters, because those intervals DO NOT OVERLAP - there are infinitely
many points of C keeping them as separate clusters.

Still, the proof of the puding will be when you supply the missing proof
e.g. by completing the proof I started above (or otherwise).

>
> But if you claim another limit, say y, of a sequence of clusters or of any arbitrary sequence of points (can you tell me what should be the merits of a sequence of clusters here?), then y cannot float in free space.If it is not an endpoint x of a cluster, then it must be disconnected

from every endpoint x by at least one cluster. In every neighbourhood
there must be points of a cluster at the right-hand side and of a
clusters at the left-hand side.

Yes of course - in the following sense: If N is any neighbourhood of y,
then there exists a point x ∈ N, with x > y and x is the endpoint of
some cluster. [Similarly with x < y].

NOTE: neither of the clusters identified above necessarily have y as an
endpoint.

> But that makes y an endpoint of two clusters, contrary to the assumption.

It certainly does not - which two clusters exactly have y as an
endpoint? You're muddling quantifiers again - taking

"for every neighbourhood there must be points of a cluster.."

as

"there is a cluster with points in every neighbourhood"

Two VERY different statements, but only the second leads to a conclusion
that y is an endpoint of a cluster....


Regards,
Mike.

[George Greene]

On Monday, December 31, 2018 at 5:39:32 AM UTC-5, WM wrote:
> Let all rational numbers q_n of the interval (0, oo)

No, dammit. The relevant interval is [0,1].

The structure that occurs in this interval IS JUST REPEATED aleph_0 times
between EVERY successive pair of adjacent integers. NOTHING NEW HAPPENS above/beyond/to-the-right-of 1.

> be covered by intervals I_n of measure |I_n| = 2^−n, such

> that q_n is the centre of I_n.

If you start at n=2 then the total sum of the measures of these intervals
will be 1/2. This is good. If you deem this starting point unnatural then
by all means define I_0 and I_1 as degenerate (I_0=[0,0] and I_1=[1,1]) and
continue FROM THERE.

> Then the endpoints are rational numbers.

Specifically and explicitly, I_n = [ q_n-(2^-(n+1)) , q_n+(2^-(n+1)) ].
These endpoints must indeed be rational since the q's are and the (negative) powers of 2 are.

> The irrational numbers x of the complement of infinite measure,

THAT infinity is NOT relevant, so since your actual goal is to argue against
infinities, it does not help for you to introduce irrelevant ones. In the relevant unit interval, the complement will have measure at least 1/2 (though less than 1) and the intervals will have measure at most 1/2.

> not covered by the intervals I_n, form particles of a
> totally disconnected space, so-called "Cantor dust" C.
>
> Every particle x of C must be separated from every particle y of C by
> at least one rational number q_n

Every real number is separated from every other real number by countably INFINITELY many rationals, NOT JUST ONE, AND by continuum-many IRrationals.
The fact that you are choosing *1* here when the ACTUAL number IS INFINITY is just laughable. Unlike the infinitely long right half of the real line, THIS is an infinity THAT IS relevant and yet you are hiding it. How Did You Get To Be So BAD at this??

> and hence by at least one interval I_n covering q_n.

Again, AND BY INFINITELY MANY such intervals, since THERE HAVE TO EXIST MANY SEQUENCES of rationals CONVERGING TO x, and if x is in the complement, the intervals around the members of this sequence ARE NOT big enough to reach x.

> Since the end points of the I_n are rational numbers too,
> also being covered by their own intervals, the particles of Cantor dust can

wait for it ... it's not wrong yet.

> only

NOW it's wrong. ANY AND EVERY real number can be AND IS the limit of myriad
different sequences of rationals. Given any ONE sequence whose limit is x,
you can create UNCOUNTABLY MANY OTHER such sequences JUST by taking SUBSEQUENCES of the given one!!

> be limits of infinite sequences of endpoints

Wrong again; the CENTERS can also converge to a point x in the complement, OR TO ANY real number -- the complement is NOT SPECIAL in that regard. More to the point, IF the centers converge to x, then the left endpoints of the associated intervals AND the right endpoints of the associated intervals MUST ALSO converge to x. So you CAN'T SAY that x can "only" be the limit OF ANYthing -- it can be the limit OF MYRIAD DIFFERENT things. In particular, the IRrationals IN the complement ARE ALL limits of sequences of RATIONALS NOT IN the complement.

> of overlapping intervals I_n.

YOU DON'T KNOW whether the intervals overlap OR NOT! Sequences-converging-to-x-where-the-intervals-overlap AND sequences-converging-to-x-where-the-intervals (within the sequence) ARE DISJOINT are BOTH definable, USUALLY!
Indeed, the disjoint sequences ARE ALWAYS definable just by going up to high enough n's to make the intervals small enough! Unless you intentionally defined the q_n to thwart that for some chosen family of irrationals. You canNOT thwart it for MOST of them.

> (If intervals don't overlap, then a limit must lie between them,

This is very blatantly false.

> but in any case infinitely many endpoints are required to establish a limit.)

This is even MORE BLATANTLY false -- *E*V*E*R*Y* irrational real number is the limit of some sequence OF RATIONAL numbers -- of THE CENTERS of these intervals, NOT their endpoints (though the endpoints are rational too -- every endpoint is therefore also a center) -- but that DOES NOT EVEN MATTER since FOR ANY sequence of I_n where THE CENTERS converge to a given irrational number x, the sequence of their left endpoints and the sequence of their RIGHT endpoints MUST BOTH ALSO CONVERGE TO x!! NO ENDPOINTS AT ALL are required to establish a limit!

[WM]

The assertion is the non-existence of irrational numbers without clusters between them.

> and you have been given counter-examples.

The "counter example", the Cantor set, is is based on the same fallacies of set theor< but not as easy to disprove because it does not have irrational numbers without rationals between them and vice versa.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 00:35:57 UTC+1 schrieb George Greene:


> > > The clusters are not unique but they are all subsets of some larger convergent set of intervals.
> >
> > The clusters are unique
>
> No, they are not.

Given a fixed enumeration of the rational numbers the clusters are unique and reach from their lower limit x to their upper limit y, i.e., they are a set of overlapping intervals I_n, confined by two points, x and y, that do not belong to any interval but to the complement. That is so by definition.

Further we have for every point x of the complement, that it is separated from every other point of the complement by at least one cluster. From the point y dfined above, it is separated by precisely one cluster.
>
> >. x is not covered by a cluster but x is the endpoint of two clusters.
>
> x is the limit OF EVERY cluster arising FROM ANY sequence of rationals that
> converges to x.

That is irrelevant. Relevant is that, given a complete enumeration of the rational numbers, there is one cluster that belongs to every neighbourhood of of x (and a second one on the other side).

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 02:47:20 UTC+1 schrieb Mike Terry:


> Suppose there exists a point of C which is not an endpoint
> of any cluster. Let y be such a point.
>
> y does not belong to any cluster, since every cluster
> is a subset of D, and y is in C (the complement of D).
>
> Let x be an endpoint of some cluster F. So x != y. WLOG we

> can consider the case x > y.

> (Just to aid stating comparisons below)
>
> Between x and y there is some rational q_n. Its
> corresponding interval I_n belongs to a unique cluster which
> we'll label
>
> F_1 = (f_1, g_1). (F_1 being an open interval...)
>
> Since x is not in F_1, and y < q_n, we have
>
> y ≤ f_1 < g_1 ≤ x.
>
> (since otherwise f1 < y < q_n, which would imply y ∈ F_1)
>
> No problem so far - you've basically deduced that there is a cluster F_1
> to the right of y. You have not shown that y is an endpoint of F_1 or
> reached any contradiction.
>
> So what next?

If y is not an endpoint of a cluster (on the right-hand side), then there is a neighbourhood of y (on the right-hand side) that does not contain points of D. These points are irrational.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 01:42:11 UTC+1 schrieb Jim Burns:
> On 1/1/2019 4:14 PM, WM wrote:

> > And we can say that every point has two clusters around it.
>
> We can say that there are _no_ clusters _next to_ any
> given point in The Complement. Between any point and
> a cluster will be at least another cluster.

No. Maybe you confuse clusters and intervals. Between any interval I_n with rational endpoint and a point of the complement, there are infinitely many more intervals I_m with rational endpoints. But between an endpoint of a cluster and the cluster having that endpoint, there is no cluster.

>
> This (alleged) bijection: how are you selecting the
> cluster to which a complement-point gets matched?

I calculate its limit.

> Can you
> show that different complement points are matched to
> different clusters? That is what you need to show.

When I have the endpoints x any y of a cluster, then every other point z of the complement is not an endpoint of this cluster. Either it is endpoint of another cluster, or it is has no cluster in every neighbourhood. Then it is a forbidden point - I mean forbidden by mathematics.

>
> > We can say that every point is confined between two clusters.
>
> How many other points are between those two clusters?
> Can you prove your answer?

Between two given clusters, there are infinitely many points. But two points not separated by a cluster would contradict mathematics.

>
> >> The points in the complement are not required to be
> >> endpoints of any of the cluster/union-of-path-connected-
> >> -intervals. That they are not so required is the point
> >> of my example of
> >> UNION{ [1/4,3/4], [1/8,7/8], 1/16,15/16], ... } = (0,1)
> >
> > Your example is irrelevant.
>
> My example shows that your obsession with endpoints
> is irrelevant.

While Cantor-sets are based on the same fallacy of set theory, namely the unjustified extrapolation from finite to infinite bijections, they are not as easily to refute because there is not the crucial condition that irrational points must be separated by clusters of finite measure.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 01:25:20 UTC+1 schrieb Jim Burns:
> On 1/1/2019 4:19 PM, WM wrote:


> > That is the definition used in this thread.
>
> What I am calling unlikely is your theory that Mike Terry
> overlooked that definition.

It is in fact unlikely. More likely is that he likes to be patronizing. But I didn't want to insult him, because he has at least grasped the essential parts of my argument.

> He told you rather a lot about
> the consequences of your definition.
>
> Perhaps you overlooked that.

Most of his arguing is correct and I agreed. There is only a small point. But I am confident that you and he are going to understand it:

If a sequence of whatever chosen points has a limit x in the complement, then this point is the endpoint of a cluster because there are two constraint:

(1) By definition, every cluster is confined its two endpoints x and y. This cluster cannot reach farther and it cannot end before. That means there are points, x and y, that are separated by precisely one cluster.

(2) Every point of the complement is the endpoint of two clusters. Two points, x and y, not separated by any cluster are contradicting mathematics.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 02:47:20 UTC+1 schrieb Mike Terry:

> On 01/01/2019 19:12, WM wrote:

> > But if you claim another limit, say y, of a sequence of clusters or of any arbitrary sequence of points (can you tell me what should be the merits of a sequence of clusters here?), then y cannot float in free space.If it is not an endpoint x of a cluster, then it must be disconnected
> from every endpoint x by at least one cluster. In every neighbourhood
> there must be points of a cluster at the right-hand side and of a
> clusters at the left-hand side.
>
> Yes of course - in the following sense: If N is any neighbourhood of y,
> then there exists a point x ∈ N, with x > y and x is the endpoint of
> some cluster. [Similarly with x < y].
>
> NOTE: neither of the clusters identified above necessarily have y as an
> endpoint.

If y is not an endpoint of any cluster, then y has a neighbourhood not containing points of D. This is impossible. Hence y is a limit point of a cluster, i.e., a limit point of a sequence of intervals with rational endpoints.

>
> > But that makes y an endpoint of two clusters, contrary to the assumption.
>
> It certainly does not - which two clusters exactly have y as an
> endpoint?

Is that really important? Who forces us to stick with clusters at all? (They are useful in the forgoing proof but they are simply arbitrary definitions.) Return to intervals. In any case y is the limit of a sequence of intervals. Remember that every interval has finite measure. And remember that if y is not a limit of a sequence of intervals, then it has a distance from this limit and from every interval (on the right-hand side). This distance is a subset of the complement. That is forbidden.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 05:58:42 UTC+1 schrieb George Greene:


> In fact it is easy to envision a cluster where the lowest-numbered and therefore widest interval completely covers all the others, which never overlap either of its endpoints.

No. The endpoints are rational numbers and therefore are covered by other intervals.

> There is nothing requiring the sequence of right-endpoints-of-a-cluster to have a limit at all, even though the cluster must have a lub.

Wrong. If the endpoint is rational, then it is covered by an interval with rational endpoint. And so on. There must be a limit and this limit must be irrational.
>
> He likes to say that every point of the complement separates 2 clusters, but it
> obviously separates more.

A single point splits the real line into two disconnected parts.

> Between any 2 disjoint clusters, there are uncountably infinitely many points of the complement, so assigning the role of separator to any one of them seems strange.

A point disconects two clusters or it has a neigbourhood containing irrational points of the complement.

Regards, WM

[jvr]

On Tuesday, January 1, 2019 at 10:24:21 PM UTC+1, WM wrote:
> Am Dienstag, 1. Januar 2019 20:12:03 UTC+1 schrieb jvr:
> > On Tuesday, January 1, 2019 at 7:27:56 PM UTC+1, WM wrote:
> > > Am Dienstag, 1. Januar 2019 19:04:42 UTC+1 schrieb jvr:
> > > > On Tuesday, January 1, 2019 at 5:02:02 PM UTC+1, WM wrote:
> > >
> > >
> > > > > > You are claiming that C is countable,
> > >
> > > I have proved it.
> > >
> >
> > You have proved no such thing.
>
> I have proved the following:
>

> If every cluster has two endpoints in C ...

It does. The red herring that you call "cluster" is an open interval.

> ... and if every point in C is separated by at least one cluster from every other point in C, ...

It is, because between two distinct reals there is a rational.

> then there are countably many points in C.

This is false and, of course, doesn't follow from preceding statements.

[jvr]

On Wednesday, January 2, 2019 at 8:40:44 AM UTC+1, WM wrote:
> Am Mittwoch, 2. Januar 2019 00:26:14 UTC+1 schrieb jvr:
> > On Tuesday, January 1, 2019 at 10:24:21 PM UTC+1, WM wrote:
>
>
> > > If every cluster has two endpoints in C and if every point in C is separated by at least one cluster from every other point in C, then there are countably many points in C.
> >
> > Your "proof" is incorrect, the assertion is false,
>
> The assertion is the non-existence of irrational numbers without clusters between them.

That assertion is correct; however, it doesn't imply the countability of C.

>
> > and you have been given counter-examples.
>
> The "counter example", the Cantor set,

That isn't the only counter-example. Another one is the implication that the real line is the union of intervals of total length less than 3.

>... is is based on the same fallacies

Please be more specific. The Cantor set construction can be formulated entirely without reference to sets and topological properties, using only the properties of real numbers.

> but not as easy to disprove because

.. because it is valid and you don't understand it.

> it does not have irrational numbers without rationals between them and vice versa.

It can be formulated entirely in terms of the ternary expansion of the reals.

[WM]

Am Mittwoch, 2. Januar 2019 13:17:10 UTC+1 schrieb jvr:
> On Tuesday, January 1, 2019 at 10:24:21 PM UTC+1, WM wrote:

> > if every point in C is separated by at least one cluster from every other point in C, ...
>
> It is, because between two distinct reals there is a rational.

Fine.

>
> > then there are countably many points in C.
>
> This is false and, of course, doesn't follow from preceding statements.

It follows from the countability of clusters. But you cannot see it. The reason is explained here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 344.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 13:28:25 UTC+1 schrieb jvr:
> On Wednesday, January 2, 2019 at 8:40:44 AM UTC+1, WM wrote:


> > The "counter example", the Cantor set,
>
> That isn't the only counter-example. Another one is the implication that the real line is the union of intervals of total length less than 3.

It is not! It *would be* in case all rational numbers could be put into a sequence. That is not even possible for the natural numbers.

>
> >... is is based on the same fallacies
>
> Please be more specific. The Cantor set construction can be formulated entirely without reference to sets and topological properties, using only the properties of real numbers.

>

> It can be formulated entirely in terms of the ternary expansion of the reals.

The fundamental fallacy here as well as in all paradoxical cases is assuming actual infinity, i.e., assuming that infinite digit sequences (= series of fractions) could approach their limit, i.e., could define a real number.

The infinite sequence 0.333333333333333333333333-and-so-on will never define 1/3. 1/3 is defined by finite expressions like "0.333..." or simply "1/3".

Therefore it is impossible to remove all thirds from the unit interval or to enumerate all rational numbers and to put them into a sequence. But all that becomes most clear when we assume it in cases like the present one or, equally lucid, taking the intersection of all endsegments of natural numbers with limit { } or removing all finite initial segments F_k of natural numbers from the set F to be unioned or considering the Binary Tree.

Regards, WM

[jvr]

You are hopelessly confused by all the dead red herrings that you yourself have introduced.

"The time has come," the Walrus said,
"To talk of many things:
Of shoes--and ships--and sealing-wax--
Of cabbages--and kings--
And why the sea is boiling hot--
And whether pigs have wings."
(Lewis Carroll)

[jvr]

You do understand that your "proof" implies that the measure of the real is finite?

And you do understand that you are claiming to disprove by elementary means mathematical theorems that for the last 100 years every mathematician has studied and understood and applied?

There is no logical reason why an amateur like you, who does not have a command of basic logic and doesn't know enough math to pass Analysis I, might not stumble upon a fact that every mathematician since 1900 has overlooked. But it is rather unlikely, don't you think?

So wouldn't it be a good idea to remove all the undefined terms and all the unnecessary obfuscations and reduce your insight to whatever you think the essential problem is?

[WM]

Am Mittwoch, 2. Januar 2019 14:45:33 UTC+1 schrieb jvr:
> On Wednesday, January 2, 2019 at 1:36:44 PM UTC+1, WM wrote:


> You do understand that your "proof" implies that the measure of the real is finite?

There is no measure of the reals. The measure of a distance is defined by its endpoints. What are points of a real line? See here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 363: MatheRealism in geometry.

>
> And you do understand that you are claiming to disprove by elementary means mathematical theorems that for the last 100 years every mathematician has studied and understood and applied?

First this is not quite correct. See chapter V of the source book with over 300 sceptical quotes, in particular Weyl, Bernays, Schütte. (I have collected this chapter because initially I had the same scruples, when faced with giant bookshelves full of set theory in the libraries.) Secondly, I have not been indcotrinated by a complete study of mathmatics but have only studied parts and have maintained my independ thoughts. This is very important as I can learn from my students and other persons, for instance in Volkshochschulen: Nobody has problems to understand that the infinite cannot be finished. And it is easy to see that finished infinity is the basis of transfinite set theory.
>
> There is no logical reason why you, who does not have a command of basic logic and doesn't know enough math to pass Analysis I,

That is wrong. I have passed Analysis III and taught Analysis II.

> might not stumble upon a fact that every mathematician since 1900 has overlooked. But it is rather unlikely, don't you think?

For an explanation see above. Further many have expressed their doubts. No intelligent person would give a dime for set theory if not being forced by the official doctrine or mislead by the denial of the difference between potential and actual infinity. See in particular Wildberger's essay but also Zeilberger's and Petry's in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, pp. 346-350.

It is always shocking how far the pure denial goes to accept or understand the important difference between infinities. This unqualified rejection and devotion to the doctrine calls to mind the behaviour of Sect members.

> So wouldn't it be a good idea to remove all the undefined terms and all the unnecessary obfuscations and reduce your insight to whatever you think the essential problem is?

There is nothing undefined. If Cantor is right, then Clusters have two endpoints and endpoints have two clusters around them and that's all and Cantor is wrong. It is so simple. And it is also obvious because no-one can give an explanation. See https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf: Arguing over "scattered space". There is not the least argument besides pure asserting like you do. And that is what makes me confident to be right.

Regards, WM

[j4n bur53]

Do you mean by nobody such that ..:

~exists x (x in Person /\ ...) ?

Yes exactly "Nobody understands that the infinite
cannot be finished". Why should this be an issue?
How do you define finished?

Last element of omega? No omega has no last element.
Has omega all natural numbers. Yes.

[George Greene]

On Tuesday, January 1, 2019 at 1:18:02 PM UTC-5, Mike Terry wrote:
> - When WM says "the limits of a cluster" he just means one
> of the two end-points of the cluster interval.

Oh, come ON!! You CAN'T concede that! He does NOT GET to NOT KNOW what
"limit" means! I am not *sure* you are glossing him correctly here! I think
he thinks that what you have said here (which is a thing -- or 2 things, one on each end - THAT MUST exist) MUST BE THE SAME as THE ACTUAL limit (of the l&r sequences of endpoints), WHICH MAY NOT exist! Or, rather, those 2 sequences may both wind up having bunches and bunches of limit points.

Here is a thing one can do, in his paradigm: one can pick ONE interval --
I_1 = [p_1,q_1], say -- and note that since its endpoints are rational, IT MUST BE in a cluster WITH I_n(p_1) and I_n(q_1) -- yes, we now find ourselves needing notation for INVERTING the mapping/function q_n, for finding n(q) for an arbitrary rational -- and note further that this process is iterable at both ends ad inf. THAT gets you ONE limit at each end, FROM ANY starting I_n.
But THOSE limits are NOT the ones that matter because 1)THEY actually COULD be rational - you would just have an infinite sequence of rationals approaching a rational limit -- and 2) you would get a different left-endpoint-limit and right-endpoint-limit FOR EACH starting interval, even when two different starting intervals were "connected" and therefore in THE SAME "cluster"! What the clusters have are a glb and lub, or, as you said, a left and a right endpoint that they do not contain.

[George Greene]

On Tuesday, January 1, 2019 at 1:18:02 PM UTC-5, Mike Terry wrote:

> George - I think you've misunderstood how the term "cluster" is being
> used.

"Used"? Oh, please. That implies a level of coherence that was not present.
And is not even normally present in WM's presentations. Does it not strike you
that this is perhaps NOT ORIGINAL content with him here? The mere fact that
you know a rational treatment of it implies that somebody WITH some sense (UNlike WM) has written this up ALREADY and that he is just parroting it.

Clusters are not simply arbitrary unions of arbitrary intervals,
> or even arbitrary unions of WM's countable set of intervals I_n centred
> on the rationals.
>
> Take I_n as the interval centred on q_n as defined elsewhere, then lets
> define:
>
> D = Union {I_n: n in |N}
> so C = D' (complement of D)

>
>
> a cluster is a maximal path-connected component of D.

Thank you for telling me what he meant. Thank you for getting it right
when I got it wrong. Thank you for explaining it to me when I was just
ranting making a fool of myself. But I do not need to apologize for not
READING HIS MIND. He *began* by talking about things that were "the only"
way to define the irrational numbers in C. That is how I got misled and that
was HIS fault.

>
> [Well, strictly this is my interpretation, but I'm pretty sure it's what
> WM would say if he could express himself better mathematically speaking.]

Indeed, but given that topics will come and go, while WM will unfortunately be staying, I would personally almost rather concentrate on the expression part
than on your correct explanation of the topic, because the expression part IS WHAT GETS HIM INTO all his contradictions.

[George Greene]

On Tuesday, January 1, 2019 at 2:12:17 PM UTC-5, WM wrote:
> Am Dienstag, 1. Januar 2019 19:18:02 UTC+1 schrieb Mike Terry:

> > a cluster is a maximal path-connected component of D.
> >

> > [Well, strictly this is my interpretation, but I'm pretty sure it's what
> > WM would say if he could express himself better mathematically speaking.]
> >

> Perhaps you have overlooked the definition in the OP:
> "Such an infinite set of overlapping intervals is called a cluster."

He didn't overlook shit, and neither did I. Perhaps YOU overlooked YOUR use of the word "Such". Such AS WHAT?? YOU claimed about the overlapping intervals that were supposed to be forming this cluster that "the particles of Cantor dust can only be limits of infinite sequences of endpoints of overlapping intervals I_n." This is NOT TRUE but because you claimed you thought it, I thought clusters as you were presenting them had to be something that MIGHT actually HAVE this property, i.e., that MIGHT contain intervals in a way that would cause their endpoints to approach all reals. Correctly understood, of course, THEY CAN'T, since there are only countably many of them. But that is obviously enough for them to approach all reals arbitrarily closely, JUST AS THE RATIONALS do.

> > Some properties of the clusters:
> >
>
> > Clusters are never degenerate (single point) intervals as
> > each cluster contains some I_n.
>
> Infinitely many I_n, in order to get a limit.

Not IN ORDER to get a limit. You in fact canNOT even prove that they will always even HAVE a limit!! The reason why there are infnitely many of them is that the endpoints ARE RATIONAL, so each endpoint of each newly-arriving I_n (at both ends) will ITself be a NEW rational endpoint FURTHER OUT, with a new interval AROUND IT (which will in turn have a new rational endpoint further out). THAT process necessarily goes on ad inf. at both ends of the starting/seed interval. The problem is, while all those intervals will be in the cluster, they may fail to exhaust it -- starting at a different starting interval that was also in the cluster will produce a different pair of endpoint-sequences. The cluster is ultimately going to be bigger than all of that and it is even possible that some of these individual endpoint-sequences (from some starting-points) may even have RATIONAL limits (even though the cluster as a whole cannot), so after that process has reached its rational limits, the I_n around the rational limit will also have to be added to the cluster, RE-triggering an infinite process. Aleph_ONE starts to get relevant to THAT.

[George Greene]

On Monday, December 31, 2018 at 5:39:32 AM UTC-5, WM wrote:

> Let all rational numbers q_n of the interval (0, oo) be covered by intervals I_n of measure |I_n| = 2^−n, such that q_n is the centre of I:n. Then the endpoints are rational numbers. The irrational numbers x of the complement of infinite measure, not covered by the intervals I_n, form particles of a totally disconnected space, so-called "Cantor dust" C.


It really does help TO NAME things.
It would really help to stick to the convention that if a big important
set is named X, the generic element of it is named x.

> Every particle x of C must be separated from every particle y of C by at least one rational number q_n and hence by at least one interval I_n covering q_n. Since the end points of the I_n are rational numbers too, also being covered by their own intervals,

Everything is correct up to this point, except maybe the "Since".

> the particles of Cantor dust can only be limits
> of infinite sequences of endpoints of overlapping intervals I_n.

Mike Terry has already gone into why this is wrong. In the first place, you haven't even proved that the ends of clusters HAVE "limits". They have endpoints but it's really not the same thing. The iteration of adding new intervals around the endpoints of the previous ones is NOT guaranteed to produce an irrational limit (although adding overlapping intervals in general IS so guaranteed).

[WM]

Am Mittwoch, 2. Januar 2019 16:28:10 UTC+1 schrieb j4n bur53:


> Last element of omega? No omega has no last element.

It is the last element before we continue to count omega + 1.

> Has omega all natural numbers. Yes.

All = completely exhausted = exhaustion finished.
Otherwise a one-to-one correspondence could not be claimed.

McDuck is rich for every step that is not the last one. How does he get bankrupt?

The intersection of endsegments E_n {n, n+1, n+2, ...} is infinite for every n. How does it get empty?

A digit sequence does not define a real number or point at the real line as long as another digit can follow. What makes the antidiagonal a real number?

Regards, WM



Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 17:00:08 UTC+1 schrieb George Greene:


> > the particles of Cantor dust can only be limits
> > of infinite sequences of endpoints of overlapping intervals I_n.
>
> Mike Terry has already gone into why this is wrong. In the first place, you haven't even proved that the ends of clusters HAVE "limits".

They are irrational limits. Otherwise they were not ends.

> They have endpoints but it's really not the same thing.

They are irrational limits. Otherwise they were not ends.

> The iteration of adding new intervals around the endpoints of the previous ones is NOT guaranteed to produce an irrational limit (although adding overlapping intervals in general IS so guaranteed).

They are irrational limits. Otherwise they were not ends.

Regards, WM

[George Greene]

On Monday, December 31, 2018 at 5:39:32 AM UTC-5, WM wrote:
> Let all rational numbers q_n of the interval (0, oo) be covered by intervals I_n of measure |I_n| = 2^−n, such that q_n is the centre of I:n. Then the endpoints are rational numbers. The irrational numbers x of the complement of infinite measure, not covered by the intervals I_n, form particles of a totally disconnected space, so-called "Cantor dust" C.

> Since the end points 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.

This is not true.
It's also not the definition of "cluster".

> (If intervals don't overlap, then a limit must lie between them, but in any

> case infinitely many endpoints are required to establish a limit.)
>

> Such an infinite set of overlapping intervals is called a cluster.

This is NOT a definition and it is an appalling use of the weasel-word "such".
There are MANY THINGS THAT CAME BEFORE this use of the word. It is NOT clear WHICH of them "such" is referring to in trying to describe clusters.
Mike Terry correctly defined a cluster as the union of any set of I_n's that are connected. That IS NOT THE SAME THING AS what you get by starting with 1 I_n and just growing it at both ends by adding in the intervals around the rational endpoints. Even if that sequence were to approach 2 limits at both ends, it might still be the case that NEITHER of those limits was relevant to the endpoints of the cluster, because there might berational q_n BEYOND those limits with a low enough n's (and wide enough widths) that their intervals overlapped the limits.

You have TO DEFINE "cluster".
Or maybe you don't, now, SINCE MIKE did. But you started down the road toward defining it as the closure of a starting interval under the operation of adding I_n's at its endpoints. THAT'S TOO SMALL. That IS NOT what you EVEN MEANT, let alone enough to have all reals (or even all members of the complement) as limits. It is, however, enough to have a countable set that approaches all members of the complement arbitrarily closely (just as the rationals do).

[George Greene]

On Monday, December 31, 2018 at 5:39:32 AM UTC-5, WM wrote:

> Since

Which was in and of itself a red flag for bullshit.
It might as well have read, "Alert -- WM is about to draw an unsound
inference." The word MEANS that what follows 2nd after this is a logical
consequence of what follows immediately afterward, but my point, here, is that, this being WM, it doesn't ACTUALLY follow -- as usual.

> the end points of the I_n are rational numbers too,
> also being covered by their own intervals,

Right. They are. This means you can start a pair of infinite processes
generating a pair of infinite sequences of endpoints, ONE AT EACH END, for
ANY given start-interval. Since the right endpoint is rational, you can
add the interval around the right endpoint to "a cluster" that now has 2 intervals in it, the original one and the one you added. You can similarly do the same thing at the left endpoint (now giving you a cluster of 3 intervals). The cluster now has a union that is wider than the original start interval, since the added interval on the right has a greater right endpoint and the added interval on the left has a lesser left endpoint. PLEASE NOTE THAT NOTHING WAS SAID about the left end of the interval added on the right, or the right end of the interval added on the left. PLEASE FURTHER NOTE that if by chance (or by design of q_n), the endpoint at the center of an end-added interval had a much lower n than the original I_n, the other (inner, non-extreme) end of the added interval COULD COMPLETELY OVERLAP the whole original interval. The limit on that side would very suddenly have NOTHING WHATSOEVER TO DO WITH the bound on that end of the cluster.

[j4n bur53]

Well you error of McDuck is well known. Its again
this inference rule:

/* WMs usual fallacy */

forall n in omega P(n) => P(omega)

But the above inference rule doesn't exist. Neither
for McDuck and some wealth at omega.

[jvr]

On Wednesday, January 2, 2019 at 4:22:25 PM UTC+1, WM wrote:
> Am Mittwoch, 2. Januar 2019 14:45:33 UTC+1 schrieb jvr:
> > On Wednesday, January 2, 2019 at 1:36:44 PM UTC+1, WM wrote:
>
>
> > You do understand that your "proof" implies that the measure of the real is finite?
>
> There is no measure of the reals. The measure of a distance is defined by its endpoints. What are points of a real line? See here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 363: MatheRealism in geometry.

Evidently you don't know what "measure" means in mathematics. Then why do you keep talking about measure?

> >
> > And you do understand that you are claiming to disprove by elementary means mathematical theorems that for the last 100 years every mathematician has studied and understood and applied?
>
> First this is not quite correct.

No reputable mathematician has claimed to have found a single one of the contradictions that you claim to have demonstrated. There is no professional mathematician who agrees with your "proof" that the rationals are not countable, nor with your "proof" that the reals are countable.

> See chapter V of the source book ...

No, Mücke, I have seen enough of this collection of quackery and illogic, of quotes that you have misunderstood and taken out of context. I am not downloading it again.

> with over 300 sceptical quotes, in particular Weyl, Bernays, Schütte.
(I have collected this chapter because initially I had the same scruples, when faced with giant bookshelves full of set theory in the libraries.) Secondly, I have not been indcotrinated by a complete study of mathmatics but have only studied parts and have maintained my independ thoughts. This is very important as I can learn from my students and other persons, for instance in Volkshochschulen: Nobody has problems to understand that the infinite cannot be finished. And it is easy to see that finished infinity is the basis of transfinite set theory.

Your main problem isn't even that you know nearly nothing about math; it is that you can't think logically.

> >
> > There is no logical reason why you, who does not have a command of basic logic and doesn't know enough math to pass Analysis I,
>
> That is wrong. I have passed Analysis III and taught Analysis II.
>

Among many errors in your textbook the most embarrassing one - the pseudo-proof of the Fourier inversion formula and your inability to correct it - proves that you should not have passed.

> > might not stumble upon a fact that every mathematician since 1900 has overlooked. But it is rather unlikely, don't you think?
>
> For an explanation see above. Further many have expressed their doubts.

And nobody has found an inconsistency.

> No intelligent person would give a dime for set theory if not being forced by the official doctrine or mislead by the denial of the difference between potential and actual infinity. See in particular Wildberger's essay but also Zeilberger's and Petry's in https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, pp. 346-350.

These 3 people, though they are also nuts, do not agree with your line of quackery. They have their own.

etc etc etc

[Mike Terry]

Yes, that's all true, and I'm not sure whether WM understands that
point. It seems to me perhaps WM thinks he has constructed a
well-defined sequence of intervals, which when we take the limit of
their LH endpoints, these constitute the LH limit of the cluster. As
you say, that's not the end of it by any means! Sure, working to the
left of the initial interval we can construct a chain of intervals each
centred on the LH endpoint of the previous, and that sequence will have
a limit (yeah, possibly rational I guess), but that point might also be
in various other I_n intervals and we could begin a new chain, and so
on. I suspect the full order structure of the countable ordinals could
be represented in there somewhere!

This is why I produced my own definition of what a cluster *should* be.
It enables us to clearly talk about the endpoints of the cluster without
getting bogged down with useless sequences of intervals which have
limits inside the cluster. I sort of new what was intended from a
previous thread from a few years back...

Here are a couple of other rigorous ways we could define clusters: given
a point x in D (D = union of all I_n), we could define the cluster
containing x to be the interval:

Cluster(x) := ( sup {y ∈ C : y < x} , inf {y in C : y > x} )
or
Cluster(X) := Union { interval [y,z] : x ∈ [y,z] and [y,z] ⊂ D }

Upon reflection, I'd go with the latter as the simplest so far,
reflecting the intuition that the cluster extends as far as it can while
remaining wholly in D. (Of course, it amounts to the same as my or
Jim's previous definitions.)

One definite difference from WM's use of the word, is that my definition
makes the cluster an interval, whereas for WM the cluster is an infinite
set of I_n intervals, and what I call the cluster is the union of those
intervals. (No big deal as we can easily go from my cluster to the
contained I_n intervals and vice versa.)

BTW, of course I wasn't criticising you in any way for failing to read
WM's mind, or for taking what he wrote literally, or for expecting that
he would write rigorously what he meant etc., and of course no
appologies are expected of anyone who does so! Just thought I might
save some time. (I did not really intend to start conversing with WM,
as I'd gone through all this a few years ago and it was a dead end, but
we'll see if this time he produces any proof of his claims, and if so
I'll respond further.)

Mike.

[WM]

Am Mittwoch, 2. Januar 2019 18:28:36 UTC+1 schrieb j4n bur53:
> Well you error of McDuck is well known. Its again
> this inference rule:
>

> forall n in omega P(n) => P(omega)
>
> But the above inference rule doesn't exist. Neither
> for McDuck and some wealth at omega

nor for the bijection of any set with omega.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 19:09:35 UTC+1 schrieb jvr:
> On Wednesday, January 2, 2019 at 4:22:25 PM UTC+1, WM wrote:
> > Am Mittwoch, 2. Januar 2019 14:45:33 UTC+1 schrieb jvr:
> > > On Wednesday, January 2, 2019 at 1:36:44 PM UTC+1, WM wrote:
> >
> >
> > > You do understand that your "proof" implies that the measure of the real is finite?
> >
> > There is no measure of the reals. The measure of a distance is defined by its endpoints. What are points of a real line? See here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 363: MatheRealism in geometry.
>
> Evidently you don't know what "measure" means in mathematics.

You are confused by the fact that I know in addition (and sometimes use that meaning) what measure means in physics and what it meant in mathematics before the invention of set theory? That is required to apply measure after set theory has been exocized.

> Then why do you keep talking about measure?

I see no resaon to restrict myself to what you know.

>
> > >
> > > And you do understand that you are claiming to disprove by elementary means mathematical theorems that for the last 100 years every mathematician has studied and understood and applied?
> >
> > First this is not quite correct.
>
> No reputable mathematician has claimed to have found a single one of the contradictions that you claim to have demonstrated.

You are lying. Many mathematicians have claimed that definable numbers are restricted to a countable set. An undefinable number is a contradiction. Brouwer has claimed that uncountable sets do not exist. That is a contradiction with Cantor's theory.

> There is no professional mathematician who agrees with your "proof" that the rationals are not countable, nor with your "proof" that the reals are countable.

My proof proves that nothing is countable because countable is a meanimgless notion.

>
> > See chapter V of the source book ...
>

> I have seen enough of this collection of quackery and illogic, of quotes that you have misunderstood and taken out of context.

Chuckle. The typical reaction of a politician, not of a mathematician.

Luitzen E.J. Brouwer: "Cantor's second number class does not exist." What context could reverse the meaning of that statement?

> And nobody has found an inconsistency.

Undefinable real number is an inconsistency for every not brain-damaged person.

Regards, WM

[jvr]

[...]

>
> Luitzen E.J. Brouwer: "Cantor's second number class does not exist." What context could reverse the meaning of that statement?
>

“A stupid man's report of what a clever man says can never be accurate, because he unconsciously translates what he hears into something he can understand.”
― Bertrand Russell, A History of Western Philosophy

[Mike Terry]

On 02/01/2019 10:07, WM wrote:
> Am Mittwoch, 2. Januar 2019 02:47:20 UTC+1 schrieb Mike Terry:
>
>

>> Suppose there exists a point of C which is not an endpoint
>> of any cluster. Let y be such a point.
>>
>> y does not belong to any cluster, since every cluster
>> is a subset of D, and y is in C (the complement of D).
>>
>> Let x be an endpoint of some cluster F. So x != y. WLOG we
>> can consider the case x > y.
>> (Just to aid stating comparisons below)
>>
>> Between x and y there is some rational q_n. Its
>> corresponding interval I_n belongs to a unique cluster which
>> we'll label
>>
>> F_1 = (f_1, g_1). (F_1 being an open interval...)
>>
>> Since x is not in F_1, and y < q_n, we have
>>
>> y ≤ f_1 < g_1 ≤ x.
>>
>> (since otherwise f1 < y < q_n, which would imply y ∈ F_1)
>>
>> No problem so far - you've basically deduced that there is a cluster F_1
>> to the right of y. You have not shown that y is an endpoint of F_1 or
>> reached any contradiction.
>>
>> So what next?
>
> If y is not an endpoint of a cluster (on the right-hand side), then there is a neighbourhood of y (on the right-hand side) that does not contain points of D. These points are irrational.

No, that's not right.

First of all, every neighbourhood of y contains y, which is a point of D.

Also, your phrase "neighbourhood (on the right-hand side)" is a little
odd. Anyway, here is something equivalent to what you meant to say

∃e>0 [y, y+e) ⊂ C

I think this capture what you intended, but it does not follow from "y
is not an endpoint of a cluster".

The fact that no cluster has y as an endpoint doesn't mean there aren't
clusters "arbitrarily close" to y. That's the bit of your claim that
requires the proof - it's no good just rewording the claim slightly and
calling that a proof.

Anyway in case it saves time, I think we all agree on the following:

∀e>0 ∃x ∈ [y, y+e) : x is lhs endpoint of some cluster

your challenge is to go beyond that to a proof that y itself is the lhs
endpoint of a cluster...

Regards,
Mike.

[Mike Terry]

On 02/01/2019 10:30, WM wrote:
> Am Mittwoch, 2. Januar 2019 02:47:20 UTC+1 schrieb Mike Terry:
>> On 01/01/2019 19:12, WM wrote:
>
>>> But if you claim another limit, say y, of a sequence of clusters or of any arbitrary sequence of points (can you tell me what should be the merits of a sequence of clusters here?), then y cannot float in free space.If it is not an endpoint x of a cluster, then it must be disconnected
>> from every endpoint x by at least one cluster. In every neighbourhood
>> there must be points of a cluster at the right-hand side and of a
>> clusters at the left-hand side.
>>
>> Yes of course - in the following sense: If N is any neighbourhood of y,
>> then there exists a point x ∈ N, with x > y and x is the endpoint of
>> some cluster. [Similarly with x < y].
>>
>> NOTE: neither of the clusters identified above necessarily have y as an
>> endpoint.

>
> If y is not an endpoint of any cluster, then y has a neighbourhood not containing points of D.

No, that's wrong. You need to expand this bit of the proof!

> This is impossible. Hence y is a limit point of a cluster, i.e., a
limit point of a sequence of intervals with rational endpoints.

It's true that every neighbourhood of y contains a point of D, but
there's no contradiction as you have not proved there exists any
contradicting neighbourhood - you've just claimed it.

>>
>>> But that makes y an endpoint of two clusters, contrary to the assumption.
>>
>> It certainly does not - which two clusters exactly have y as an
>> endpoint?
>
> Is that really important?

You claimed the existence of two clusters having y as an endpoint, so
you have to prove said clusters exist.

> Who forces us to stick with clusters at all? (They are useful in the forgoing proof but they are simply arbitrary definitions.)

Well, you were claiming that every point in C was the endpoint of a
cluster, and the clusters are countable, therefore C is countable.

If you want to drop that claim, that's fine, otherwise you need to deal
with clusters!

Are you dropping that argument? (E.g. in favour of an argument based on
intervals... That would be another thing to consider in its own right.)

> Return to intervals. In any case y is the limit of a sequence of
intervals.

OK, intervals do not in fact converge to points, but I guess we can take
the lhs endpoints of the intervals, or the central q_i points of the
intervals, and then we get a sequence of real (or even rational) numbers
converging to y.

No problem...

> Remember that every interval has finite measure. And remember that if
y is not a limit of a sequence of intervals, then it has a distance from
this limit and from every interval (on the right-hand side).

Yes, *IF* y were not a limit [...] *THEN* it would have some strictly
positive distance from [...], which would certainly be a problem!

But of course nobody claims that y is not such a limit - it IS, so no
problem.


What now? Also, as you're switching to intervals, it's not clear any
more what you're actually trying to prove? Is it that every point of C
is the endpoint of some interval I_n? [Rather than the endpoint of a
cluster? I think you know that's false..., so what is the claim here?]

Regards,
Mike.

[j4n bur53]

You are crazy, for a bijection you only need
a proof certificate of the form:

forall n in omega P(n)

You don't need a proof certificate of the form

P(omega)

You can check yourself:

- Injectivity: means:
forall x,y in omega (f(x)=f(y) => x=y)

- Surjectivity: means
forall y in omega exists x in omega (f(x)=y)

Its only forall y in omega and exists y in omega.
The value omega itself is not used except for
the bounded quantifiers.

Whats wrong with you? Quantifier dyslexia?

[WM]

Am Mittwoch, 2. Januar 2019 21:38:47 UTC+1 schrieb Mike Terry:
> On 02/01/2019 10:07, WM wrote:


> > If y is not an endpoint of a cluster (on the right-hand side), then there is a neighbourhood of y (on the right-hand side) that does not contain points of D. These points are irrational.
>
> No, that's not right.
>
> First of all, every neighbourhood of y contains y, which is a point of D.

No, you are confused. There are three possible locations of a point y: Either it is in the interior of a cluster, then y ∈ D, or it is an endpoint of a cluster or it is outside of any cluster. Here we consider the third case. Then there is a neighbourhood of y that does not contain points of D.

>
> Also, your phrase "neighbourhood (on the right-hand side)" is a little
> odd. Anyway, here is something equivalent to what you meant to say
>
> ∃e>0 [y, y+e) ⊂ C
>
> I think this capture what you intended,

y ∈ C means y ∉ C (C = complement).

> but it does not follow from "y

> is not an endpoint of a cluster".#

It follows from y is outside of any cluster.

>
> The fact that no cluster has y as an endpoint doesn't mean there aren't
> clusters "arbitrarily close" to y.

You are wrong.

> That's the bit of your claim that
> requires the proof - it's no good just rewording the claim slightly and
> calling that a proof.

There are three possible situations: Inside cluster, in the limit, outside. The latter means that there is a distance from every point of D.

>
> Anyway in case it saves time, I think we all agree on the following:
>
> ∀e>0 ∃x ∈ [y, y+e) : x is lhs endpoint of some cluster
>
> your challenge is to go beyond that to a proof that y itself is the lhs
> endpoint of a cluster...

If not, then the lhs endpoint of every cluster must be larger than y and we have for the irrational numbers y < x with X the set of irrationals and Q the set of rationals:

~∀x ∈ X ∃q ∈ Q: |q - y| < |x - y| .

In correct mathematics however we have

∀x ∈ X ∃q ∈ Q: |q - y| < |x - y|
and
∀q ∈ Q ∃x ∈ X: |x - y| < |q - y| .

Regards, WM

[j4n bur53]

A lot of gibberish.

Example: WM wrote:
> y ∈ C means y ∉ C (C = complement).

Whats wrong with you? Please try again.
Here are some rules:

1) In set theory there is no general complement as a set,
since there is no universal set. If there were
a general complement comp, then the universal
set would be comp({}).

2) So all that one can hope for is relative complement,
also known as set difference. So if you want a
relative complement of A subset X, you can write:

X \ A

3) In set theory you can nevertheless NOT prove,
this would be a translation of the verb "means":

/* WMs fallacy, Nr. 479579345 */

y ∈ A <=> y ∉ X \ A

You can only prove:

y ∈ A => y ∉ X \ A

y ∉ X v y ∈ A <=> y ∉ X \ A

4) If you have classes, you can define a general
complement, for example comp(A) = V \ A. Works
also when A is proper class. You then have indeed:

y ∈ A <=> y ∉ comp(A)

[Python]

Doktor Frankenheim wrote:
...

> y ∈ C means y ∉ C

This is quite a adequate way for a crank like you, Herr Mueckenheim,
to begin the year 2019 with.

[Mike Terry]

On 02/01/2019 22:14, WM wrote:
> Am Mittwoch, 2. Januar 2019 21:38:47 UTC+1 schrieb Mike Terry:
>> On 02/01/2019 10:07, WM wrote:
>
>
>>> If y is not an endpoint of a cluster (on the right-hand side), then there is a neighbourhood of y (on the right-hand side) that does not contain points of D. These points are irrational.
>>
>> No, that's not right.
>>
>> First of all, every neighbourhood of y contains y, which is a point of D.
>
> No, you are confused.

hehe you're right! y of course is a point of C, not D.

So ignore the "first of all" bit :)

> There are three possible locations of a point y: Either it is in the interior of a cluster, then y ∈ D, or it is an endpoint of a cluster or it is outside of any cluster. Here we consider the third case.

Agreed, y is a point in C, not D, and y is not the endpoint of any
cluster. I'd understood that, but somehow still got confused regarding
C and D above. (Not below, though...)

BTW, since the endpoints of a cluster are not in the cluster (clusters
are open) they are also outside of any cluster, so I don't like that
phrase, but I guess we must refer to the "in C but not the endpoint of
any cluster" points somehow...

>
Then there is a neighbourhood of y that does not contain points of D.
>>
>> Also, your phrase "neighbourhood (on the right-hand side)" is a little
>> odd. Anyway, here is something equivalent to what you meant to say
>>
>> ∃e>0 [y, y+e) ⊂ C
>>
>> I think this capture what you intended,
>
> y ∈ C means y ∉ C (C = complement).

Eh? Typo?? Guess you meant y ∈ C means y ∉ D, which I agree with.

When I wrote ∃e>0 [y, y+e) ⊂ C, my understanding of the context for y
was actually the same as yours despite my earlier confusion: y ∈ C,
and y is not the endpoint of any cluster. (In your words, I think, "y
is outside of any cluster").

With that cleared up, I'm guessing you would agree that what you claimed
was indeed equivalent to what I stated:

∃e>0 [y, y+e) ⊂ C

The way I'm putting it, [y, y+e) is what I think you mean by
"neighbourhood of y (on the right-hand side)", and I'm saying this
interval is wholly contained in C. I.e. it does not contain any points
of D.

I'm pretty sure that's also what you're intending. Sorry for any confusion.

>
>> but it does not follow from "y
>> is not an endpoint of a cluster".#
>
> It follows from y is outside of any cluster.

Ah, I see - you're thinking I had spotted a technicality where it
doesn't follow because you'd perhaps overlooked that y could be in D?

No no no, I correctly understood that y was in C, in addition to not
being the endpoint of any cluster (or that "y is outside any cluster")
so I was saying that your claim does not follow from "y is outside of
any cluster".

So... we have not advanced here at all. Your last sentence above is
(still) wrong.

>>
>> The fact that no cluster has y as an endpoint doesn't mean there aren't
>> clusters "arbitrarily close" to y.
>
> You are wrong.

Well, fair enough - we'll see in the end. If I'm wrong you need to
prove your claim rather than just saying I'm wrong! Hopefully that will
follow...

>
>
>> That's the bit of your claim that
>> requires the proof - it's no good just rewording the claim slightly and
>> calling that a proof.
>
> There are three possible situations: Inside cluster, in the limit, outside.

Hmm, you've changed your wording from what you said above, and I don't
know if that's deliberate or even significant.

Above you were saying (and I agreed) there are three possible situations:

a) inside cluster
b) cluster endpoint
c) "outside cluster"

Your wording made it clear that those 3 were mutually exclusive
scenarios for y, so I read "outside cluster" as "y is in C, and y is not
a cluster endpoint". (And that makes sense, because it is the scenarion
we've actually been busy discussing in the recent posts. Also it's
the literal negation of (a) and (b), making the claim (nearly) a
tautology nobody would dispute!)

But NOW, you're changing to say the three scenarios are:

a) inside cluster
b') "in the limit"
c) "outside cluster"

Just to be clear, perhaps you could clarify the definitions of both the
following?

- Point y is "in the limit"

- Point y is "outside (of any) cluster"

To save you unnecessary typing, here is how I'm interpreting the two
phrases at the moment:

- Point y is "in the limit" iff y is the endpoint of some cluster

- Point y is "outside (of any) cluster" iff (y is in C, and y is not
the endpoint of any cluster).

So hopefully you can just agree that's what you mean.

No point going further until this is clarified.


Mike.

[WM]

Am Mittwoch, 2. Januar 2019 22:23:54 UTC+1 schrieb Mike Terry:
> On 02/01/2019 10:30, WM wrote:


> > If y is not an endpoint of any cluster, then y has a neighbourhood not containing points of D.
>
> No, that's wrong. You need to expand this bit of the proof!

If you have two sets on the real line which are separated by a point x and if another point y is different from x, then y has a neighbourhood within its set.

>
> > This is impossible. Hence y is a limit point of a cluster, i.e., a
> limit point of a sequence of intervals with rational endpoints.
>
> It's true that every neighbourhood of y contains a point of D

Chuckle. Do you want to cheat? Of course the rational numbers are dense on the real line. But you forget that we make the assumption that all rationals can be included in clusters of measure 1. Then an infinite complement C remains uncovered.

Assume, for sake of definiteness, that the measures of C and D are 100000 : 1. That is a very conservative estimate. Then the average distance between clusters is 100000 times the average distance of the endpoints of clusters. That means that the complement is as empty as the matter surrounding us. And this estimation is (1) very conservative and (2) independent of the real distances. It cannot be refuted by smaller and smaller clusters.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 23:08:17 UTC+1 schrieb j4n bur53:
> for a bijection you only need
> a proof certificate of the form:
>
> forall n in omega P(n)

That does not biject sets.

For McDuck you only need a proof forall n in omega: McDuck's wealth increases.

>
> You don't need a proof certificate of the form
>
> P(omega)
>
> You can check yourself:
>
> - Injectivity: means:
> forall x,y in omega (f(x)=f(y) => x=y)
>
> - Surjectivity: means
> forall y in omega exists x in omega (f(x)=y)

And McDuck's case means: for all n he is not bankrupt.


Regards, WM

[WM]

Am Donnerstag, 3. Januar 2019 04:47:52 UTC+1 schrieb Mike Terry:

> On 02/01/2019 22:14, WM wrote:

> > There are three possible locations of a point y: Either it is in the interior of a cluster, then y ∈ D, or it is an endpoint of a cluster or it is outside of any cluster. Here we consider the third case.
>
> Agreed, y is a point in C, not D, and y is not the endpoint of any
> cluster. I'd understood that, but somehow still got confused regarding
> C and D above. (Not below, though...)
>
> BTW, since the endpoints of a cluster are not in the cluster (clusters
> are open) they are also outside of any cluster, so I don't like that
> phrase, but I guess we must refer to the "in C but not the endpoint of
> any cluster" points somehow...

Agreed.

>
> >
> Then there is a neighbourhood of y that does not contain points of D.
> >>
> >> Also, your phrase "neighbourhood (on the right-hand side)" is a little
> >> odd. Anyway, here is something equivalent to what you meant to say
> >>
> >> ∃e>0 [y, y+e) ⊂ C
> >>
> >> I think this capture what you intended,
> >
> > y ∈ C means y ∉ C (C = complement).
>
> Eh? Typo?? Guess you meant y ∈ C means y ∉ D, which I agree with.

Yes thank you.

>
> When I wrote ∃e>0 [y, y+e) ⊂ C, my understanding of the context for y
> was actually the same as yours despite my earlier confusion: y ∈ C,
> and y is not the endpoint of any cluster. (In your words, I think, "y
> is outside of any cluster").
>
> With that cleared up, I'm guessing you would agree that what you claimed
> was indeed equivalent to what I stated:
>
> ∃e>0 [y, y+e) ⊂ C

Yes, that is correct.

>
> > It follows from y is outside of any cluster.
>
> Ah, I see - you're thinking I had spotted a technicality where it
> doesn't follow because you'd perhaps overlooked that y could be in D?
>
> No no no, I correctly understood that y was in C, in addition to not
> being the endpoint of any cluster (or that "y is outside any cluster")
> so I was saying that your claim does not follow from "y is outside of
> any cluster".
>
> So... we have not advanced here at all. Your last sentence above is
> (still) wrong.
>
> >>
> >> The fact that no cluster has y as an endpoint doesn't mean there aren't
> >> clusters "arbitrarily close" to y.
> >
> > You are wrong.
>
> Well, fair enough - we'll see in the end. If I'm wrong you need to
> prove your claim rather than just saying I'm wrong! Hopefully that will
> follow...
>

See the previous post.

> >
> >> That's the bit of your claim that
> >> requires the proof - it's no good just rewording the claim slightly and
> >> calling that a proof.
> >
> > There are three possible situations: Inside cluster, in the limit, outside.
>
> Hmm, you've changed your wording from what you said above, and I don't
> know if that's deliberate or even significant.
>
> Above you were saying (and I agreed) there are three possible situations:
>
> a) inside cluster
> b) cluster endpoint
> c) "outside cluster"
>
> Your wording made it clear that those 3 were mutually exclusive
> scenarios for y, so I read "outside cluster" as "y is in C, and y is not
> a cluster endpoint". (And that makes sense, because it is the scenarion
> we've actually been busy discussing in the recent posts. Also it's
> the literal negation of (a) and (b), making the claim (nearly) a
> tautology nobody would dispute!)
>
> But NOW, you're changing to say the three scenarios are:
>
> a) inside cluster
> b') "in the limit"
> c) "outside cluster"
>
> Just to be clear, perhaps you could clarify the definitions of both the
> following?

The limit is the same as the cluster endpoint.

>
> - Point y is "in the limit"
>
> - Point y is "outside (of any) cluster"

Here I mean not inside and not in the limit.

>
> To save you unnecessary typing, here is how I'm interpreting the two
> phrases at the moment:
>
> - Point y is "in the limit" iff y is the endpoint of some cluster
>
> - Point y is "outside (of any) cluster" iff (y is in C, and y is not
> the endpoint of any cluster).
>
> So hopefully you can just agree that's what you mean.
>

> Agreed.


Now if y is outside of any cluster, then it has neighbourhood containg only points of the complement C and no rational number of D. Then we have for X = set of irrationals and Q = set of rationals:
> >
> > ~∀x ∈ X ∃q ∈ Q: |q - y| < |x - y| . (*)

> >
> > In correct mathematics however we have
> >
> > ∀x ∈ X ∃q ∈ Q: |q - y| < |x - y|
> > and
> > ∀q ∈ Q ∃x ∈ X: |x - y| < |q - y| .

Of course we cannot find any neighbourhood of y satisfying (*). But IF we could confine all rational numbers in intervals of total measure 1 and had an infinite complement, THEN (*) would necessarily happen, because, in the average, the distance between clusters was infinitely larger than the, always finite, clusters.

Regards, WM

[Alan Smaill]

WM <wolfgang.m...@hs-augsburg.de> writes:

> Am Mittwoch, 2. Januar 2019 22:23:54 UTC+1 schrieb Mike Terry:
>> On 02/01/2019 10:30, WM wrote:
>
>
>> > If y is not an endpoint of any cluster, then y has a neighbourhood
>> > not containing points of D.
>>
>> No, that's wrong. You need to expand this bit of the proof!
>
> If you have two sets on the real line which are separated by a point x
> and if another point y is different from x, then y has a neighbourhood
> within its set.

Let's see: set A, all a in A a < x
set B = {y}, y > x.

y has no neighbourhood within B.

Try again.


--
Alan Smaill

[WM]

Don't confuse quantifiers. Every y of B has a neighbourhood in B. (There is no neighbourhood in B for all y.)

Regards, WM

[WM]

Russel was wrong in many other respects too. See https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf p. 340:

Comments on a paper by Bertrand Russell. [B. Russell: "Mathematics and the metaphysicians" in "Mysticism and logic", Allen & Unwin, London (1918) pp. 57-74]

1) His {{Boole's}} book was in fact concerned with formal logic, and this is the same thing as mathematics. [p. 57]
"Kant already has taught [...] that mathematics includes a solid contents that is independent of logic and therefore cannot be substantiated by logic." [D. Hilbert: "Über das Unendliche", Math. Annalen 95 (1925)]
"Gradually the pendulum swung in the direction of pure logic and abstraction, actually so much that a dangerous separation between "pure" mathematics and essential realms of application emerged. [...] But it seems and it is to hope that this period of isolation has ended." [R. Courant, H. Robbins: "Was ist Mathematik?", Springer, Berlin (1962)]
"To create mathematics from pure logic has not yet been managed." [Gerhard Hessenberg: "Grundbegriffe der Mengenlehre", Vandenhoeck & Ruprecht, Göttingen (1906)]

2) Pure mathematics consists entirely of assertions to the effect that, if such and such proposition is true of anything, then such and such another proposition is true of that thing. [p. 57]
Of course every statement can be formulated as an implication – but need not. Pure mathematics consists of statement like "1 + 1 = 2", "1 < 2", "2 is a prime number", "2m2 = n2 has no solution in integers", "pi and e are transcendental", "a point is what cannot be separated into parts", "the cube, one of five Platonic solids, has 8 vertices, 12 edges, and 6 sides" – without ifs and buts!

3) These rules of inference constitute the major part of the principle of formal logic. [p. 58]
True but irrelevant. In mathematics formal logic is merely applied as an auxiliary tool, in particular if the mental capacity of mathematicians otherwise would be stretched too far, and crutches are required to pass through difficult landscape.

4) Thus mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we are saying is true. [p. 58]
That may be true for Russell's vision of mathematics but not for real mathematics.

...

9) The proofs favourable to infinity, on the other hand, involved no principle that had evil consequences. [p. 66]
How that? Russell himself later mentions Tristram Shandy [p. 69f]. And before that he explicitly says The fundamental numbers are not ordinal but are what is called cardinal [p. 67]. Nevertheless he does not become aware of the fact that in the Tristram Shandy case only the sequence of cardinal numbers has mathematical meaning?

"This is an instance of the amazing power of desire in blinding even very able men to fallacies which would otherwise be obvious at once." [Bertrand Russell: "What I believe" in "Why I am not a Christian and other essays on religion and related subjects", Paul Edwards (ed.), Allen & Unwin, London (1957)]

... and so on.

Do something for your education and read the whole article.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 22:23:54 UTC+1 schrieb Mike Terry:

> >> It certainly does not - which two clusters exactly have y as an
> >> endpoint?
> >
> > Is that really important?
>
> You claimed the existence of two clusters having y as an endpoint, so
> you have to prove said clusters exist.

That is proven by simple logic: if y is not an endpoint (= limit) of any cluster then it is surrounded by points of the complement on both sides. Then it has a neighbourhood of only irrational numbers.

> > Who forces us to stick with clusters at all? (They are useful in the forgoing proof but they are simply arbitrary definitions.)
>
> Well, you were claiming that every point in C was the endpoint of a
> cluster, and the clusters are countable, therefore C is countable.

Nevertheless when you claim that y is the limit of some imagined sequence that comes arbitrary close, then intervals, the constituents of clusters, must come arbitrary close.

> Are you dropping that argument?

No. I said simply that clusters coming arbitrarily close imply intervals coming arbitrarily close But that is irrelevant.

> > Remember that every interval has finite measure. And remember that if
> y is not a limit of a sequence of intervals, then it has a distance from
> this limit and from every interval (on the right-hand side).
>
> Yes, *IF* y were not a limit [...] *THEN* it would have some strictly
> positive distance from [...], which would certainly be a problem!

If y has no strictly positive distance from all clusters, then it is in a cluster or is its limit.

>
> But of course nobody claims that y is not such a limit - it IS, so no
> problem.
>
>
> What now?

Either we have countably many separated points = limits of the complement. Then their measure is zero. Or there are uncountably many separated points of the complement, then there must be something separating them. But there is nothing. It is hard to believe how rather intelligent mathematicians can claim uncountably many points of the complement without asking what separates them.

Regards, WM

[Alan Smaill]

WM <wolfgang.m...@hs-augsburg.de> writes:

> Am Donnerstag, 3. Januar 2019 09:00:02 UTC+1 schrieb Alan Smaill:
>> WM <wolfgang.m...@hs-augsburg.de> writes:
>>
>> > Am Mittwoch, 2. Januar 2019 22:23:54 UTC+1 schrieb Mike Terry:
>> >> On 02/01/2019 10:30, WM wrote:
>> >
>> >
>> >> > If y is not an endpoint of any cluster, then y has a neighbourhood
>> >> > not containing points of D.
>> >>
>> >> No, that's wrong. You need to expand this bit of the proof!
>> >
>> > If you have two sets on the real line which are separated by a point x
>> > and if another point y is different from x, then y has a neighbourhood
>> > within its set.
>>
>> Let's see: set A, all a in A a < x
>> set B = {y}, y > x.
>>
>> y has no neighbourhood within B.
>>
> Don't confuse quantifiers.

I don't. I explicitly included the quantifier where needed.

> Every y of B has a neighbourhood in
> B. (There is no neighbourhood in B for all y.)

y is a particular real in this example.
As is x, of course.

Look above: the set B in this case is {y}.
A neighbourhood of y, by definition, must be a superset of an *open*
subset of |R of which y is a member. There is no such open set.

Try again.


> Regards, WM
>

--
Alan Smaill

[jvr]

An excellent demonstration of your confusion. You quote texts that you don't understand, culled from popular science and philosophical essays, that often mean the exact opposite of what you are claiming. I have given examples of this kind of fraud before and it isn't worthwhile to repeat them. You are a muddlehead.

So even if it isn't a universal truth, in your case it is applicable:

[jvr]

> >
> > > This is impossible. Hence y is a limit point of a cluster, i.e., a
> > limit point of a sequence of intervals with rational endpoints.
> >
> > It's true that every neighbourhood of y contains a point of D
>
> Chuckle. Do you want to cheat? Of course the rational numbers are dense on the real line. But you forget that we make the assumption that all rationals can be included in clusters of measure 1. Then an infinite complement C remains uncovered.
>
> Assume, for sake of definiteness, that the measures of C and D are 100000 : 1. That is a very conservative estimate. Then the average distance between clusters is 100000 times the average distance of the endpoints of clusters. That means that the complement is as empty as the matter surrounding us. And this estimation is (1) very conservative and (2) independent of the real distances. It cannot be refuted by smaller and smaller clusters.

Kindly tell us what you mean by "average distance between clusters" and "average distance of the endpoints of clusters". Neither makes any sense as it stands.

[WM]

Am Mittwoch, 2. Januar 2019 18:07:06 UTC+1 schrieb George Greene:
> PLEASE NOTE THAT NOTHING WAS SAID about the left end of the interval added on the right, or the right end of the interval added on the left.

They are irrelevant.

> PLEASE FURTHER NOTE that if by chance (or by design of q_n), the endpoint at the center of an end-added interval had a much lower n than the original I_n, the other (inner, non-extreme) end of the added interval COULD COMPLETELY OVERLAP the whole original interval. The limit on that side would very suddenly have NOTHING WHATSOEVER TO DO WITH the bound on that end of the cluster.

Nevertheless the bound will always be an irrationl limit (if the whole construction is sensible - of course it is not - but we have to assume it in order to contradict it). Every rational point, wherever and whenever it is introduced is the centre of an interval. Therefore only irrational limits can exist. Countably many.

If further points exist in the complement, then they are not limits but interior points and have a neighbourhood of only irrational numbers. Contradiction.

If no further points exist, then the complement has measure zero. Contradiction.

Note that here https://mathoverflow.net/questions/88267/covering-the-rationals-a-paradox?rq=1 is an answer by Gerald Edgar: "There are, in fact, uncountably many gaps between the intervals of your construction."

It is hard to believe that no intelligent person has asked *what* separates these gaps from each other. But the question has been closed by some leading matheologians who obviously are afraid that too difficult questions could damage their sect.

Regards, WM

[WM]

Am Mittwoch, 2. Januar 2019 17:13:43 UTC+1 schrieb George Greene:

> But you started down the road toward defining it as the closure of a starting interval under the operation of adding I_n's at its endpoints. THAT'S TOO SMALL ... let alone enough to have all reals (or even all members of the complement) as limits.

Of course not.

> It is, however, enough to have a countable set that approaches all members of the complement arbitrarily closely (just as the rationals do).

No, here the sitiation is different. All rationals are confined in intervals and the intervals in clusters. The remaining irrational points of the complement, if not being limits of clusters, must be separated from the countably many limits. What separates them?

Regards, WM

[WM]

Try to apply physics. There are clusters of galaxies. We measure their sizes and their distances and can find interesting results. Same applies to clusters in the present case where we have not enough information and computing capacity to calculate the complete configuration. Therefore we apply statictical methods. The complement between 0 and 100001 has 100000 meters measure compared to 1 meter of the complete set of clusters. Then this vacuum must be filled by something.

Either there are uncountably many scattered atoms, then we ask what scatters (= separates) them? Or there are intervals, then we get a contradiction.

Note: It is always helpful to use physical methods. They can open the mind. Some centuries after the physicists also the mathematicians can learn the truth about their "matter".

Regards, WM

[WM]

Am Donnerstag, 3. Januar 2019 12:00:02 UTC+1 schrieb Alan Smaill:
> WM <wolfgang.m...@hs-augsburg.de> writes:
>
> > Am Donnerstag, 3. Januar 2019 09:00:02 UTC+1 schrieb Alan Smaill:
> >> WM <wolfgang.m...@hs-augsburg.de> writes:

> > Every y of B has a neighbourhood in
> > B. (There is no neighbourhood in B for all y.)
>
> y is a particular real in this example.
> As is x, of course.
>
> Look above: the set B in this case is {y}.

We are interested in elements y, i.e., in points y of the complement. A set of only irrational points cannot be assumed at all.

> A neighbourhood of y, by definition, must be a superset of an *open*
> subset of |R of which y is a member. There is no such open set.

That is irrelevant.

Regards, WM

[jvr]

On Wednesday, January 2, 2019 at 8:33:03 PM UTC+1, WM wrote:
> Am Mittwoch, 2. Januar 2019 19:09:35 UTC+1 schrieb jvr:
> > On Wednesday, January 2, 2019 at 4:22:25 PM UTC+1, WM wrote:
> > > Am Mittwoch, 2. Januar 2019 14:45:33 UTC+1 schrieb jvr:
> > > > On Wednesday, January 2, 2019 at 1:36:44 PM UTC+1, WM wrote:
> > >
> > >
> > > > You do understand that your "proof" implies that the measure of the real is finite?
> > >
> > > There is no measure of the reals. The measure of a distance is defined by its endpoints. What are points of a real line? See here: https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf, p. 363: MatheRealism in geometry.
> >
> > Evidently you don't know what "measure" means in mathematics.
>
> You are confused by the fact that I know in addition (and sometimes use that meaning) what measure means in physics and what it meant in mathematics before the invention of set theory? That is required to apply measure after set theory has been exocized.

Nobody will understand what you mean if you don't define your terms. The fact that you don't know what "measure" means in mathematics isn't mitigated by your claim to know other meanings of "measure".

>
> > Then why do you keep talking about measure?
>
> I see no resaon to restrict myself to what you know.
> >
> > > >
> > > > And you do understand that you are claiming to disprove by elementary means mathematical theorems that for the last 100 years every mathematician has studied and understood and applied?
> > >
> > > First this is not quite correct.
> >
> > No reputable mathematician has claimed to have found a single one of the contradictions that you claim to have demonstrated.
>
> You are lying. Many mathematicians have claimed that definable numbers are restricted to a countable set. An undefinable number is a contradiction. Brouwer has claimed that uncountable sets do not exist. That is a contradiction with Cantor's theory.

Brouwer's statements refer to different basic axioms for the foundations and for logic than is customary. His are more restrictive.

>
> > There is no professional mathematician who agrees with your "proof" that the rationals are not countable, nor with your "proof" that the reals are countable.
>
> My proof proves that nothing is countable because countable is a meanimgless notion.

You *prove* no such thing; but you do keep repeating it.

"Just the place for a Snark! I have said it twice:
That alone should encourage the crew.
Just the place for a Snark! I have said it thrice:
What I tell you three times is true."

> >
> > > See chapter V of the source book ...
> >
> > I have seen enough of this collection of quackery and illogic, of quotes that you have misunderstood and taken out of context.
>
> Chuckle. The typical reaction of a politician, not of a mathematician.
>
> Luitzen E.J. Brouwer: "Cantor's second number class does not exist." What context could reverse the meaning of that statement?

Standard foundations of mathematics. Brouwer's logical system is more restrictive.

>
> > And nobody has found an inconsistency.
>
> Undefinable real number is an inconsistency for every not brain-damaged person.

That depends upon what you mean by "undefinable".

[j4n bur53]

Well if countable leads to a contradiction, then
all that is left is uncountable.

G, ~A |- f
----------
G |- A

Reductio ad absurdum. What is your problem?

[j4n bur53]

If you don't want this result, you need to use
another theory.

[Alan Smaill]

WM <wolfgang.m...@hs-augsburg.de> writes:

Oh look, WM has snipped his own claim,
all the better to muddy the waters.

WM claimed upthread:

> If you have two sets on the real line which are separated by a point x
> and if another point y is different from x, then y has a neighbourhood
> within its set.

A counterexample is below.

> Am Donnerstag, 3. Januar 2019 12:00:02 UTC+1 schrieb Alan Smaill:
>> WM <wolfgang.m...@hs-augsburg.de> writes:
>>
>> > Am Donnerstag, 3. Januar 2019 09:00:02 UTC+1 schrieb Alan Smaill:
>> >> WM <wolfgang.m...@hs-augsburg.de> writes:
>
>> > Every y of B has a neighbourhood in
>> > B. (There is no neighbourhood in B for all y.)
>>
>> y is a particular real in this example.
>> As is x, of course.
>>
>> Look above: the set B in this case is {y}.
>
> We are interested in elements y, i.e., in points y of the
> complement. A set of only irrational points cannot be assumed at all.

rational or irrational does not matter here,
your claim is wrong either way.

Take your claim above:

A and B are two sets separated by a point x.
Assume A is to the left of x, no need for further specification.
y is a particular real greater than x. B is the set {y};
y may be rational or irrational, it does not matter.

Under these conditions, there is *no* neighourhood of y
within its set, ie there is *no* neighbourhood of y inside
the set {y}.

Your claim fails.

>> A neighbourhood of y, by definition, must be a superset of an *open*
>> subset of |R of which y is a member. There is no such open set.
>
> That is irrelevant.

It is entirely relevant to your claim, and its refutation.

[jvr]

There are no "physical methods" applicable to mathematics. There are physical analogies that are often helpful in analysis.

Here the "physical methods" served your purpose of muddling the issue and avoiding the question.

Again: Kindly tell us what you mean by "average distance between clusters" and "average distance of the endpoints of clusters". Neither makes any sense as it stands.

[Me]

On Wednesday, January 2, 2019 at 9:23:31 PM UTC+1, jvr wrote:

> “A stupid man's report of what a clever man says can never be accurate, because
> he unconsciously translates what he hears into something he can understand.”
> ― Bertrand Russell, A History of Western Philosophy

Right.

What do you expect from a man who does not understand that there's a difference between

ES({} e S & (x e S => x u {x} e S))

and

ES({} e S & Ax(x e S => x u {x} e S)) .