Covering all rational points by irrational intervals

[transf...@gmail.com]

Consider the covering of all rational numbers q_n, of the positive axis by intervals

I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)] .

Then all intervals cover (2/9)sqrt(2) or less of the positive axis. The complement has unknown measure. The structure of the complement is not known and is not of interest. There are only two important facts:

_ There is no rational number in the complement because every rational is in an interval.
_ All endpoints of the complement are irrational.

Since there cannot exist two irrational numbers without a rational number between them, there is no irrational number in the complement either. The remainder of the positive axis of infinite length is empty.

For more of this Kind see https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

Regards, WM

[Jim Burns]

On 11/22/2018 5:59 AM, transf...@gmail.com wrote:

> Consider the covering of all rational numbers q_n, of
> the positive axis by intervals
> I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)] .
>
> Then all intervals cover (2/9)sqrt(2) or less of the positive
> axis. The complement has unknown measure. The structure of
> the complement is not known and is not of interest.
> There are only two important facts:
> _ There is no rational number in the complement because
> every rational is in an interval.
> _ All endpoints of the complement are irrational.
>
> Since there cannot exist two irrational numbers without a
> rational number between them,

... and then a miracle occurs, and ...

> there is no irrational number
> in the complement either. The remainder of the positive
> axis of infinite length is empty.

You put all of the rationals in intervals, and they obligingly
cease to exist. For your next trick, use the open interval
(0,1) to prove that there are no real numbers between 0 and 1.

I suspect that WMaths should not be ingested without a
prescription.

> For more of this Kind see https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
>

http://www.sciencecartoonsplus.com/pages/gallery.php
...then a miracle occurs...

"I think you should be more explicit here in step two."

[transf...@gmail.com]

Am Freitag, 23. November 2018 02:20:52 UTC+1 schrieb Jim Burns:
> On 11/22/2018 5:59 AM, transf...@gmail.com wrote:
>
> > Consider the covering of all rational numbers q_n, of
> > the positive axis by intervals
> > I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)] .
> >
> > Then all intervals cover (2/9)sqrt(2) or less of the positive
> > axis. The complement has unknown measure. The structure of
> > the complement is not known and is not of interest.
> > There are only two important facts:
> > _ There is no rational number in the complement because
> > every rational is in an interval.
> > _ All endpoints of the complement are irrational.
> >
> > Since there cannot exist two irrational numbers without a
> > rational number between them,
>
> ... and then a miracle occurs, and ...
>
> > there is no irrational number
> > in the complement either. The remainder of the positive
> > axis of infinite length is empty.
>
> You put all of the rationals in intervals, and they obligingly
> cease to exist.

No. They exist but only within the intervals. All the intervals have irrational endpoints. Therefore between intervals, there cannot exist irrationals (and also not rationals, of Course, because they are in the intervals).

> For your next trick, use the open interval
> (0,1) to prove that there are no real numbers between 0 and 1.

This sentence Shows that you have not understood at all.

Regards, WM

[Jim Burns]

On 11/23/2018 5:31 AM, transf...@gmail.com wrote:
> Am Freitag, 23. November 2018 02:20:52 UTC+1
> schrieb Jim Burns:
>> On 11/22/2018 5:59 AM, transf...@gmail.com wrote:

>>> Consider the covering of all rational numbers q_n, of
>>> the positive axis by intervals
>>> I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)] .
>>>
>>> Then all intervals cover (2/9)sqrt(2) or less of the
>>> positive axis. The complement has unknown measure. The
>>> structure of the complement is not known and is not
>>> of interest.
>>> There are only two important facts:
>>> _ There is no rational number in the complement because
>>> every rational is in an interval.
>>> _ All endpoints of the complement are irrational.
>>>
>>> Since there cannot exist two irrational numbers without a
>>> rational number between them,
>>
>> ... and then a miracle occurs, and ...
>>
>>> there is no irrational number
>>> in the complement either. The remainder of the positive
>>> axis of infinite length is empty.
>>
>> You put all of the rationals in intervals, and they
>> obligingly cease to exist.
>
> No. They exist but only within the intervals.

Then there are still rationals between irrationals.
Within the intervals.

> All the intervals have irrational endpoints. Therefore
> between intervals, there cannot exist irrationals (and
> also not rationals, of Course, because they are in the
> intervals).

Between intervals, there can only be isolated irrational
points.

This has nothing to do with your open intervals. Look at the
real number line. The set of irrationals is a set of isolated
points, with gaps == missing rational points, surrounding every
irrational point at every distance down to zero.

Or, we could think about Cantor's diagonal. Suppose we have a
list of all the _rationals_ alone. You've been given such lists,
I think. From that list, we can define an antidiagonal number x
that is at least 1/10 from the first listed rational q[1], at
least 1/100 from the second listed rational q[2], ... , at least
1/10^k from the k_th listed rational q[k], ... , and so on, for
every listed rational, which is to say, all of the rationals.

For every rational q[k], there is a positive distance
d(x,q[k]) > 1/10^k > 0 between it and the antidiagonal.
_And yet_ , there _is no_ positive distance eps > 0
between x and every rational q[k]. It's important to get
the quantifiers right here.

"No eps > 0" is what I mean by "isolated point". And that
is the reason that all of the rationals being in your
intervals is not a problem.

[WM]

Am Samstag, 24. November 2018 02:08:31 UTC+1 schrieb Jim Burns:
> On 11/23/2018 5:31 AM, transf...@gmail.com wrote:


> >> You put all of the rationals in intervals, and they
> >> obligingly cease to exist.
> >
> > No. They exist but only within the intervals.
>
> Then there are still rationals between irrationals.
> Within the intervals.

Yes, but that is only within the intervals covering not more than (2/9)sqrt(2) of the unit. More than 2/3 of the unit are complementary intervals free of rationals and having irrational endpoints. Therefore they cannot contain irrational numbers either.

>
> > All the intervals have irrational endpoints. Therefore
> > between intervals, there cannot exist irrationals (and
> > also not rationals, of Course, because they are in the
> > intervals).
>
> Between intervals, there can only be isolated irrational
> points.

No. An isolated irrational point would lie between two intervals with irrational endpoints. That is impossible since there is always a rational number between two irrational numbers.

>
> This has nothing to do with your open intervals.

The intervals are not open but closed (including their irrational endpoints).

> Look at the
> real number line. The set of irrationals is a set of isolated
> points, with gaps == missing rational points, surrounding every
> irrational point at every distance down to zero.

That should be the case but cannot be if the rational numbers are countable and have measure zero.

>
> Or, we could think about Cantor's diagonal.

No, let us stay with the topic. A countable set of rationals can be included into aleph_0 closed intervals of measure 0. The intervals gave irrational endpoints. Therefore between the intervals there cannot exist anything.

> "No eps > 0" is what I mean by "isolated point". And that
> is the reason that all of the rationals being in your
> intervals is not a problem.

I do not talk about isolated points but about an arbitrarily large part of the unit interval that cannot contain any number. The distance between intervals is certainly less than any eps. But the free space is say 2/3 or if you like 99/100 of the unit interval.

Regards, WM

[Jim Burns]

On 11/24/2018 6:07 AM, WM wrote:
> Am Samstag, 24. November 2018 02:08:31 UTC+1
> schrieb Jim Burns:
>> On 11/23/2018 5:31 AM, transf...@gmail.com wrote:

>>>> You put all of the rationals in intervals, and they
>>>> obligingly cease to exist.
>>>
>>> No. They exist but only within the intervals.
>>
>> Then there are still rationals between irrationals.
>> Within the intervals.
>
> Yes, but that is only within the intervals covering not more
> than (2/9)sqrt(2) of the unit. More than 2/3 of the unit are

> complementary *intervals* free of rationals and having

> irrational endpoints. Therefore they cannot contain
> irrational numbers either.

*emphasis added*

There are no complementary _intervals_ -- unless you mean
single-point intervals. Each irrational point in the
complement is an isolated point, a single-point interval,
if you'd like to call it that. But a _single-point_ interval
creates no contradiction.

The argument for single points is basically your own argument:
between any two irrationals in the complement there is a
rational, which will not be in the complement. Therefore,
there is no closed interval subset [x,y] _in the complement_
with x < y.

But that does not rule out "closed intervals" of the form
[x,x], which is to say {x}. So, without looking any closer,
we have narrowed the possibilities down to one of these two:
(i) Every point in the complement is isolated.
(ii) There are no points in the complement, and therefore
no real numbers at all.

How do you eliminate possibility (i)?
It looks to me as though you write on the chalkboard
"and then a miracle occurs". This is what I mean by
http://www.sciencecartoonsplus.com/pages/gallery.php

But you can expand on your reasoning, can't you?
Just answer the question: How do you know that there
are no isolated points in the complement?
Thanks in advance.

----
Trying to imagine the complement of the sequence of intervals
is a mess. This is why I think it's good to look at the
anti-diagonal generated by Cantor's list-of-reals argument.
We can use that machinery to produce one specific real
number in the complement, for one list, one antidiagonal-
-digit rule, and one factor (1/10) by which the intervals
around the rationals shrink.

>>> All the intervals have irrational endpoints. Therefore
>>> between intervals, there cannot exist irrationals (and
>>> also not rationals, of Course, because they are in the
>>> intervals).
>>
>> Between intervals, there can only be isolated irrational
>> points.
>
> No. An isolated irrational point would lie between two
> intervals with irrational endpoints. That is impossible
> since there is always a rational number between two
> irrational numbers.

An isolated point would lie between _infinitely many_
closed intervals one one side and on the other. _None_
of those intervals is closest to the isolated point,
so there is no impossible rational point to worry about.

Consider the union of the closed intervals
[-1,-0.1], [0.1,1],
[-1,-0.01], [0.01,1],
[-1,-0.001], [0.001,1],
[-1,-0.0001], [0.0001,1],
[-1,-0.00001], [0.00001,1],
...

Every point in [-1,1] is in at least one of those intervals,
and therefore in their union, _except_ for 0.
The point 0 is an isolated point in the union of countably
many closed intervals.

Note that _none_ of the intervals is closest to 0, even though
each individual interval is some positive distance from 0.
If there were only finitely many intervals, that would not
be possible. However, there are infinitely many intervals.

I'm not sure if it matters to you that all of the endpoints
and the isolated point are rational. But, if it matters,
add sqrt(2) to everything, and the endpoints and the isolated
point will be irrational.

>> This has nothing to do with your open intervals.
>
> The intervals are not open but closed (including their
> irrational endpoints).
>
>> Look at the
>> real number line. The set of irrationals is a set of isolated
>> points, with gaps == missing rational points, surrounding every
>> irrational point at every distance down to zero.
>
> That should be the case but cannot be if the rational
> numbers are countable and have measure zero.
>
>> Or, we could think about Cantor's diagonal.
>
> No, let us stay with the topic.

Perhaps you see now why Cantor's diagonal is staying with

the topic.

> A countable set of rationals can be included into aleph_0
> closed intervals of measure 0. The intervals gave

> irrational endpoints. *Therefore* between the intervals
> there cannot exist anything.
*emphasis added*

...and then a miracle occurs...

>> "No eps > 0" is what I mean by "isolated point". And that
>> is the reason that all of the rationals being in your
>> intervals is not a problem.
>
> I do not talk about isolated points but about an arbitrarily
> large part of the unit interval that cannot contain any number.

... cannot contain any number that is not isolated.

The existence of non-isolated numbers is all that your argument
addresses. This makes your not talking about isolated points
a problem for you.

> The distance between intervals is certainly less than any eps.
> But the free space is say 2/3 or if you like 99/100 of the
> unit interval.

Sure, or 999/1000 of the unit interval, or 999999/1000000 of
the unit interval. I can see how this might be counter-intuitive.
Counter-intuitive is not the same as wrong.

Perhaps it would help to look at the rational numbers as
the bones of the number line, its skeleton, and all of the
rest, the Dedekind-complete numbers, as the muscles etc
of the number line. If you remove the bones, there is still
quite a lot left. If we're talking about turkeys or something
like that, "a lot" is measurably less than what we had with
the bones in. But this is mathematics, so we can have our
bones weigh in at zero. When we _perfectly_ de-bone our
unit interval, we are left with a unit interval's worth
of meat and skin and so on.

Huh. Suddenly, I am in the mood for transcendental pumpkin pi.

[Mike Terry]

We had a similar thread some time ago, and WM was making exactly the
same mistake - he claimed that every point in the complement must be the
endpoint of one of the removed intervals. [I think the removed
intervals on that occasion might have been open intervals, and this time
around they are closed, hence his claim that the complement must be
empty this time.]

https://groups.google.com/d/msg/sci.math/wFlm4liBfmE/aQ7SSSJQDpQJ

I (and others) pointed out that a point of the complement could have on
each side an infinite *sequence* of removed intervals approaching it,
rather than having a directly adjacent interval, but WM couldn't see
this. Also he couldn't prove his claim that there MUST be a directly
adjacent interval, instead claiming no proof was necessary as it was
self-evident (to him).

I think that was the point at which I realised that regardless of any
possible merits or arguments in favour of a potential infinity vs actual
infinity distinction, when it comes down to basic undergrad real
analysis, and "actually doing mathematics", WM is a complete duffer!

So to literally answer your question: "No, WM CAN'T expand his reasoning
(in a reasonable fashion)."

BTW, your use of the term "isolated point" is not standard - an isolated
point (of a subset of a topological space) is one which has a
neighborhood containing no other point of the set. In the referenced
thread above, we were calling them "singleton" points, meaning points
belonging to a singleton set, which can't be extended to any larger
interval. (Such point aren't isolated in the topologial sense.)

Mike.

[Jim Burns]

On 11/24/2018 3:09 PM, Mike Terry wrote:

> BTW, your use of the term "isolated point" is not standard
> - an isolated point (of a subset of a topological space) is
> one which has a neighborhood containing no other point of
> the set. In the referenced thread above, we were calling
> them "singleton" points, meaning points belonging to a
> singleton set, which can't be extended to any larger interval.
> (Such point aren't isolated in the topologial sense.)

You make a good point, but I need to call such points something.

Calling them "singleton" points is roughly just as
clear/unclear. _Every_ point is a member of some singleton
set, in standard usage. I don't mean that as a criticism,
you need to call them something, too.

What if I call them _path-isolated_ points?[1] Is this also
a term that already has a use? A little googling doesn't
show it to me.

The reference to paths is an improvement, I think, since WM's
argument is that there can't be any non-trivial paths in
the complement of his rational-number closed cover. My
answer is that there doesn't need to be any paths. This is
a discussion of paths in WM's complement, so it's nice to
tag the points in an apparently relevant way.

[1]
Let a point x in topological space Y be _path-isolated_
iff there is no point z in Y such that there is a path
in Y from x to z.

Let a _path_ f in Y from x to z be a continuous function
f: [0,1] -> Y with f(0) = x and f(1) = z.

Every point in [0,1]/Q is path-isolated.

[Jim Burns]

D'Oh!

... no _distinct_ point z, z ~= x ...

[Mike Terry]

On 25/11/2018 00:42, Jim Burns wrote:
> On 11/24/2018 7:38 PM, Jim Burns wrote:
>> On 11/24/2018 3:09 PM, Mike Terry wrote:
>
>>> BTW, your use of the term "isolated point" is not standard
>>> - an isolated point (of a subset of a topological space) is
>>> one which has a neighborhood containing no other point of
>>> the set. In the referenced thread above, we were calling
>>> them "singleton" points, meaning points belonging to a
>>> singleton set, which can't be extended to any larger interval.
>>> (Such point aren't isolated in the topologial sense.)
>>
>> You make a good point, but I need to call such points something.
>>
>> Calling them "singleton" points is roughly just as
>> clear/unclear. _Every_ point is a member of some singleton
>> set, in standard usage. I don't mean that as a criticism,
>> you need to call them something, too.

Yes, perhaps "singleton interval" was more the intention. (The term
wasn't my invention.) The rationale behind it would be in the context
that any subset of R can be uniquely represented as a disjoint union of
maximal intervals, and when this is done for the complement set above,
the resulting intervals are all singleton sets. This is tying things in
to the specific order properties of R, but then we /are/ talking
specifically about the set R, with its interval structure, as that's how
the set was defined.

We can also say that the complement set is "nowhere dense", but that
might not capture what you are looking for?

>>
>> What if I call them _path-isolated_ points?[1] Is this also
>> a term that already has a use? A little googling doesn't
>> show it to me.

Sounds OK to me - and it could apply more generally in other topological
spaces. OTOH it is maybe overkill. (That's just a feeling I have. I
suppose my reservation is that the complexity of introducing continuous
functions from [0,1] to R is unneeded and hence should be inappropriate
in some way. E.g. there are uncountably many paths from 0 to 1 in R,
but in our context none of that is needed or important in any meaningful
way - all such paths amount to the same thing, which can be said in
terms of simple order properties without this redundancy. Hmm, that's
not very convincing I guess, but as I say it's only a feeling...)

>>
>> The reference to paths is an improvement, I think, since WM's
>> argument is that there can't be any non-trivial paths in
>> the complement of his rational-number closed cover. My
>> answer is that there doesn't need to be any paths. This is
>> a discussion of paths in WM's complement, so it's nice to
>> tag the points in an apparently relevant way.

Sure no problem there.

That the complement *is* nonempty follows from a standard measure theory
argument, viz if a countable union of intervals {I_j} covers an interval
I, then the sum of lengths of I_j must be at least the length of I.
(This result holds for R, but fails for Q, so the result is more subtle
than it first appears - somewhere in its proof the completeness of R
will need to be invoked. The result certainly isn't just "obvious".)

Of course, here it is WM who is making the opposite claim: that the
complement IS empty, so it is his obligation to prove that, not
something vice versa.

His arguments so far all have "silly" errors:

WM: _ There is no rational number in the complement because
WM: every rational is in an interval.
[Correct]
WM: _ All endpoints of the complement are irrational.
[Correct, with a suitable interpretation for "endpoints of the
complement", e.g. endpoints of the maximal intervals comprising the
complement]
WM:
WM: Since there cannot exist two irrational numbers without
WM: a rational number between them, there is no irrational
WM: number in the complement either. The remainder of the
WM: positive axis of infinite length is empty.

[As you pointed out, the "since" paragraph simply doesn't follow.]


WM: No. They [rationals] exist but only within the intervals.
WM: All the intervals have irrational endpoints.
WM: Therefore between intervals, there cannot exist irrationals (and

also not rationals, of Course, because they are in the intervals).

[The argument here seems to amount to:
- all (removed) intervals have irrational endpoints which
are not in the complement set [Correct]
- all irrationals are either internal to some removed
interval, or the endpoint of some removed interval
[FALSE]
- and so in either case the irrational is not in the
complement set [FALSE due to previous mistake]
]


WM: Yes, but that is only within the intervals covering
WM: not more than (2/9)sqrt(2) of the unit. More than 2/3
WM: of the unit are complementary intervals free of rationals
WM: and having irrational endpoints.
[Correct, allowing degenerate intervals]
WM: Therefore they cannot contain irrational numbers either.
[FALSE - as you said, they can be "path-isolated"/"singleton"
(degenerate) intervals consisting of a single irrational number which is
not part of any removed interval, and also not "adjacent" to any interval.]

I think WM can't get his head around the "not adjacent" aspect, since
that was definitely his problem 6 years ago...


Interestingly (at least I think so) if we denote the complement set by
C, then although there is no non-degenerate interval I satisfying:

I ⋂ C = I [equivalently: I ⊆ C]

(or put another way, C is "nowhere dense"), it IS the case that there
are intervals I satisfying:

measure(I ⋂ C) > 0.99 * measure(I)

I.e. such that we could say "most" of I is within C (in terms of the
measures of the sets involved).


Regards,
Mike.

[Mike Terry]

Um, sorry that's not right - it turns out C IS nowhere dense, but
nowhere dense is not equivalent to containing no open set, so the "put
another way" is wrong :(.

[WM]

Am Samstag, 24. November 2018 19:32:23 UTC+1 schrieb Jim Burns:
> On 11/24/2018 6:07 AM, WM wrote:
> > Am Samstag, 24. November 2018 02:08:31 UTC+1
> > schrieb Jim Burns:
> >> On 11/23/2018 5:31 AM, transf...@gmail.com wrote:
>
> >>>> You put all of the rationals in intervals, and they
> >>>> obligingly cease to exist.
> >>>
> >>> No. They exist but only within the intervals.
> >>
> >> Then there are still rationals between irrationals.
> >> Within the intervals.
> >
> > Yes, but that is only within the intervals covering not more
> > than (2/9)sqrt(2) of the unit. More than 2/3 of the unit are
> > complementary *intervals* free of rationals and having
> > irrational endpoints. Therefore they cannot contain
> > irrational numbers either.
> *emphasis added*
>
> There are no complementary _intervals

If we use intervals

[q_n - eps*sqrt(2)*10^(-n), q_n + eps*sqrt(2)*10^(-n)]

with eps small enough, then 99% of the unit interval is empty space.

_ -- unless you mean
> single-point intervals. Each irrational point in the
> complement is an isolated point, a single-point interval,
> if you'd like to call it that. But a _single-point_ interval
> creates no contradiction.

In the empty space there can no points exist - no rational numbers and no irrational numbers.

>
> The argument for single points is basically your own argument:
> between any two irrationals in the complement

There are no irrationals in the complement. They would lack a rational between the themselves and the irrational ends of intervals. Whatever the complement may be, it lies between irrational numbers, contains no rational points and hence contains no irrational points either.

> there is a
> rational, which will not be in the complement.

And it will be shielded or isolated from the complement by the irrational end of its interval.

Therefore,
> there is no closed interval subset [x,y] _in the complement_
> with x < y.

There is nothing in the complement. But the complement can consists of measure 1.

>
> But that does not rule out "closed intervals" of the form
> [x,x], which is to say {x}. So, without looking any closer,
> we have narrowed the possibilities down to one of these two:
> (i) Every point in the complement is isolated.

No, There cannot be points.

> (ii) There are no points in the complement, and therefore
> no real numbers at all.

So it is - iff the rationals are countable.

>
> How do you eliminate possibility (i)?

See above. The complement is disconnected from any rational point by the rational ends of the intervals.

> But you can expand on your reasoning, can't you?
> Just answer the question: How do you know that there
> are no isolated points in the complement?

The rationals are covered by intervals by definition.
An irrational cannot exist with another irrational next to it.

> Thanks in advance.

You are welcome.

>
> ----
> Trying to imagine the complement of the sequence of intervals
> is a mess.

We need not imagine. We can prove that there is space of measure 1 outside of intervals and shielded from rational numbers.

Regards, WM

[WM]

Am Sonntag, 25. November 2018 05:20:38 UTC+1 schrieb Mike Terry:


> His arguments so far all have "silly" errors:
>
> WM: _ There is no rational number in the complement because
> WM: every rational is in an interval.
> [Correct]
> WM: _ All endpoints of the complement are irrational.
> [Correct, with a suitable interpretation for "endpoints of the
> complement", e.g. endpoints of the maximal intervals comprising the
> complement]
> WM:
> WM: Since there cannot exist two irrational numbers without
> WM: a rational number between them, there is no irrational
> WM: number in the complement either. The remainder of the
> WM: positive axis of infinite length is empty.
>
> [As you pointed out, the "since" paragraph simply doesn't follow.]

You are wrong. There are no irrational numbers without a rational between them. That should be basic mathematics.

But we can easily prove it. Assume two different irrational numbers. They have decimal representations. Take the first digit where they differ. Cut the larger one behind that digit. It is a rational number between them.

Regards, WM

[WM]

Am Samstag, 24. November 2018 19:32:23 UTC+1 schrieb Jim Burns:
> On 11/24/2018 6:07 AM, WM wrote:


> >>> All the intervals have irrational endpoints. Therefore
> >>> between intervals, there cannot exist irrationals (and
> >>> also not rationals, of Course, because they are in the
> >>> intervals).
> >>
> >> Between intervals, there can only be isolated irrational
> >> points.
> >
> > No. An isolated irrational point would lie between two
> > intervals with irrational endpoints. That is impossible
> > since there is always a rational number between two
> > irrational numbers.
>
> An isolated point would lie between _infinitely many_
> closed intervals one one side and on the other. _None_
> of those intervals is closest to the isolated point,

But there is no rational number between the isolated point and the border of the complement.

> so there is no impossible rational point to worry about.
>
> Consider the union of the closed intervals
> [-1,-0.1], [0.1,1],
> [-1,-0.01], [0.01,1],
> [-1,-0.001], [0.001,1],
> [-1,-0.0001], [0.0001,1],
> [-1,-0.00001], [0.00001,1],
> ...
>
> Every point in [-1,1] is in at least one of those intervals,
> and therefore in their union, _except_ for 0.
> The point 0 is an isolated point in the union of countably
> many closed intervals.

Of course this example shows the same as my example, namely that finished infinity leads to contradictions. (Here the existence of an interval that touches 0.) But this example does not contradict my example.

>
> Note that _none_ of the intervals is closest to 0, even though
> each individual interval is some positive distance from 0.

Of course. That is potential infinity and does not cover all rational numbers. But set theory has "all" intervals including those not leaving any number less and larger zero.

> If there were only finitely many intervals, that would not
> be possible. However, there are infinitely many intervals.


>
> I'm not sure if it matters to you that all of the endpoints
> and the isolated point are rational.

It yields a contradiction because between every two rational numbers there is an irrational number.

> But, if it matters,
> add sqrt(2) to everything, and the endpoints and the isolated
> point will be irrational.

As I said, it is the same contradiction to mathematics.

> Perhaps you see now why Cantor's diagonal is staying with
> the topic.

Of course. The diagonal does not exist because ot must have a finished infinity of digits. But there is none. So there are always infinitely many digits missing. So there is no numerical value obtaimable. But I would prefer to stay with the more obvious proof: There is no irrational number x without a rational between x and another irrational number y.

>

> > I do not talk about isolated points but about an arbitrarily
> > large part of the unit interval that cannot contain any number.
>
> ... cannot contain any number that is not isolated.
>
> The existence of non-isolated numbers is all that your argument
> addresses. This makes your not talking about isolated points
> a problem for you.

Isolated irrational points cannot exist in a space with irrational endpoints.

> Sure, or 999/1000 of the unit interval, or 999999/1000000 of
> the unit interval. I can see how this might be counter-intuitive.
> Counter-intuitive is not the same as wrong.

Isolated irrational points in a space with only irrational endpoints are wrong!

Regards, WM

[Mike Terry]

So you think I said that there are distinct irrationals with no
rationals between them? You can't even read and understand what people
are saying when it comes to maths! This is confirming you're a
"duffer"! (Don't worry, there are plenty of other things in life beyond
mathematics...)

Mike.

[George Greene]

> Since there cannot exist two irrational
> numbers without a rational number between them,

True.

> there is no irrational number in the complement either.

False. For every *Defined* bijeciton q_n, you can find many irrational
numbers in the complement. You can find lots of them, in fact, but
how you go about it depends on the definition of q_n.

Obviously, for any given irrational number -- say 2/(1+sqrt(5))= 0.618....
you can find rationals on either side of it as close as you like but you can't
prove that the associated interval with either rational endpoint is big
enough to actually reach the rational between them. Or certainly you can, if
you define the *correct* q_n. But for any defined one, you have also (whether
you like it or not) defined many irrational numbers as not being covered.

[George Greene]

On Friday, November 23, 2018 at 5:31:23 AM UTC-5, transf...@gmail.com wrote:
> No. They exist but only within the intervals.

Right.

> All the intervals have irrational endpoints.

Right.

> Therefore

Wrong.
You have never known how to PROVE *shit*.

> between intervals, there cannot exist irrationals

Yes, there can, and there do, exist irrational numbers not in any of
these intervals. As soon as you present anyone with a definition of
q_n, they can define one for you.

> (and also not rationals, of Course, because they are in the intervals).

Yes, all the rationals are in one of these intervals, but many of the irrationals are not. Since the intervals overlap, the length of what they
cover is LESS THAN 2/9sqrt(2).
THE VAST majority of irrational numbers are not in any interval.
Yes I know cardinality is weird there.

[George Greene]

On Friday, November 23, 2018 at 8:08:31 PM UTC-5, Jim Burns wrote:
> This has nothing to do with your open intervals. Look at the
> real number line. The set of irrationals is a set of isolated
> points, with gaps == missing rational points, surrounding every
> irrational point at every distance down to zero.

The rationals don't "surround" the irrationals. It's really very much the other way around.

> Or, we could think about Cantor's diagonal. Suppose we have a
> list of all the _rationals_ alone.

He's presuming one. He calls it q_n. The nth element of this list is a
rational number, and every rational is on it somewhere.

> You've been given such lists,

HE'S GIVING YOU ONE, in this problem. JEEzus!!

> I think. From that list, we can define an antidiagonal number x
> that is at least 1/10 from the first listed rational q[1], at
> least 1/100 from the second listed rational q[2], ... , at least
> 1/10^k from the k_th listed rational q[k], ... , and so on, for
> every listed rational, which is to say, all of the rationals.
>
> For every rational q[k],

Do you HAVE to call it q[k] when he called it q_n ? Isn't
that just being petty??

> there is a positive distance
> d(x,q[k]) > 1/10^k > 0 between it and the antidiagonal.

It's not necessarily strictly greater than 1/10^k.
It very probably is but there
could be a number where the the kth digit of the rational is a 4 and the
anti-diagonal digit in that kth slot is a 5, and all the other digits
on the the kth row MATCH the anti-diagonal. Probably not but it could
theoretically be greater-than-or-equal-to 1/10^k. The problem in any
case is that 1/10^k ISN'T BIG ENOUGH!! The width of the interval
is greater than 2.82/10^k !

[WM]

Am Sonntag, 25. November 2018 16:22:39 UTC+1 schrieb Mike Terry:
> On 25/11/2018 11:41, WM wrote:
> > Am Sonntag, 25. November 2018 05:20:38 UTC+1 schrieb Mike Terry:


> >> WM: Since there cannot exist two irrational numbers without
> >> WM: a rational number between them, there is no irrational
> >> WM: number in the complement either. The remainder of the
> >> WM: positive axis of infinite length is empty.
> >>
> >> [As you pointed out, the "since" paragraph simply doesn't follow.]
> >
> > You are wrong. There are no irrational numbers without a rational between them. That should be basic mathematics.
> >
> > But we can easily prove it. Assume two different irrational numbers. They have decimal representations. Take the first digit where they differ. Cut the larger one behind that digit. It is a rational number between them.

> So you think I said that there are distinct irrationals with no
> rationals between them?

I wrote "there is no irrational number in the complement either" in the "since paragraph" and you said "the "since" paragraph simply doesn't follow."

> You can't even read and understand what people
> are saying when it comes to maths!

Try to express yourself clearer if you don't like to blurr things.

Fact is: There is no irrational number in the complement of measure 1.

Regards, WM

[WM]

Am Montag, 26. November 2018 05:33:27 UTC+1 schrieb George Greene:
> > Since there cannot exist two irrational
> > numbers without a rational number between them,
>
> True.
>
> > there is no irrational number in the complement either.
>
> False. For every *Defined* bijeciton q_n

Every definied bijection in ludes all rationals in a measure less than eps. So the remainder of measure 1 is empty and has irrational endpoints. No irrational number can inhabit it.

> you can find many irrational
> numbers in the complement.

In fact you can find irrationals everywhere because it is impossible to include all rational numbers in intervals of total measure eps. But that is another topic.

If the measure 1 is void of rationals but has irrational endpoints everywhere, then you cannot find irrational there.

> But for any defined one, you have also (whether
> you like it or not) defined many irrational numbers as not being covered.

Not in mathematics. In a space with only irrational endpoints there would nothing exist in between.

Regards, WM

[WM]

Am Montag, 26. November 2018 05:40:37 UTC+1 schrieb George Greene:
> On Friday, November 23, 2018 at 5:31:23 AM UTC-5, transf...@gmail.com wrote:
> > No. They exist but only within the intervals.
>
> Right.
>
> > All the intervals have irrational endpoints.
>
> Right.
>
> > Therefore

> > between intervals, there cannot exist irrationals
>
> Yes, there can, and there do, exist irrational numbers not in any of
> these intervals.

Of course, since the rational numbers are not countable.

> As soon as you present anyone with a definition of
> q_n, they can define one for you.

Not necessary. If the rational numbers are assumed to be countable, then there is a complementary set of measure 1 with countably many irrational endpoints. This set is empty.

>
> > (and also not rationals, of Course, because they are in the intervals).
>
> Yes, all the rationals are in one of these intervals, but many of the irrationals are not.

So you dismiss mathematical proof in order to adhere to matheology.

> THE VAST majority of irrational numbers are not in any interval.
> Yes I know cardinality is weird there.

Cardinality is here obviously in contradiction with mathematics.

Regards, WM

[George Greene]

On Monday, November 26, 2018 at 5:00:32 AM UTC-5, WM wrote:
> I wrote "there is no irrational number
> in the complement either" in the "since paragraph"
> and you said "the "since" paragraph simply doesn't follow."

Because indeed it doesn't. Just to be able to give a size argument,
let's pretend we're talking about rational and irrational reals in the
intervale [0,1]. In that case, "most" irrational numbers are in the complement.
And I don't see why you felt you needed to use sqrt(2) as the multiplier.
Any number, including rational numbers or integers, smaller than 4.4 would have
sufficed. The interval around q_n is of width 2*10^-n, so if you start at n=1
(which you shouldn't; you should start at n=0) then the interval around q_1 is .2*some_constant. If you sum this all the way up you get .22222222....*some_constant, which is 2/9 of the constant, so as long as the constant is less than 9/2, the total length has to be less than 1, so some irrational numbers have to be missing. Forget sqrt(2)! Even at sqrt(22), which is almost 4.7 -- you STILL cannot prove that every point is covered because
even though the interval around q_1 takes up almost 94% of the unit interval
(if q_1 is near the middle), and the interval around q_2 takes up almost 9.4% (which would sum to over 103%), it will usually happen (since the second
interval is 10 times smaller than the first) that the
interval around q_2 winds up WHOLLY INSIDE the interval around q_1 and
thereby CONTRIBUTES NOTHING to the length covered.

It would be a good exercise for you to ACTUALLY FIND AN ENUMERATION
q_n and come up with the smallest constant k FOR THAT enumeration,
for which you can actually prove that the intervals [q_n-k*10^-n,q_n+k*10^-n] ACTUALLY COVER THE WHOLE unit interval.
Bets about whether k is bigger than 5 ?
There probably are provably enumerations for which k is arbitrarily close to 4.5 but It's a safe bet YOU'LL never find one!

[George Greene]

On Monday, November 26, 2018 at 5:00:32 AM UTC-5, WM wrote:

> Fact is: There is no irrational number in the complement of measure 1.

Fact is, you don't know what a fact is, since you can't prove that
(and since a very simple length argument proves that there are in fact
infinitely many).

But it is a nice challenge to construct one from a given enumeration.
It is a provable fact that the denial of what you just said is a fact.

Are you afraid that if you just published it as a challenge problem,
nobody would care?
I repeat, the irrationality of your constant has nothing whatsoever to
do with your argument.
This is another reason why everybody knows you are fucktarded.
You could have MUCH BIGGER intervals -- you could replace sqrt(2)
by * 4 * -- and still be WRONG for the SAME reason even with intervals
MORE THAN DOUBLE THE WIDTHS of the ones you are using.

[George Greene]

On Sunday, November 25, 2018 at 6:31:40 AM UTC-5, WM wrote:
> If we use intervals
>
> [q_n - eps*sqrt(2)*10^(-n), q_n + eps*sqrt(2)*10^(-n)]
>
> with eps small enough, then 99% of the unit interval is empty space.

That constitutes that proof that the complement is non-empty and that there
are irrational numbers in it. We already know there are no rational numbers in it. If there were no irrational numbers in it either then it would be empty.
If it were empty then the length of the all the intervals summed would have to be at least 1. IT ISN'T. That IS A PROOF that the complement is non-empty.

Nobody can help it if you can't recognize a proof when you see one.
Proof-recognition IS PRIMITIVE-recursive -- modulo the LENGTH of the proof,
IT'S EASY.
If it's not easy FOR YOU then that's YOUR PROBLEM.

[George Greene]

On Sunday, November 25, 2018 at 6:31:40 AM UTC-5, WM wrote:

> If we use intervals
>
> [q_n - eps*sqrt(2)*10^(-n), q_n + eps*sqrt(2)*10^(-n)]
>
> with eps small enough, then 99% of the unit interval is empty space.

You continue to fail to understand that sqrt(2) is playing no role
whatsoever in this problem. You could replace it with 1. You could replace
it with 4. You could replace it with any sufficiently small rational up to and
inculding 4.4. IT DOESN'T MATTER.
Given that sqrt(2) can't matter eps *CAN'T*MATTER*EITHER*.

as long as what you are multipliying by 10^-n IS LESS THAN 9/2,
the total length has to be less than 1 so THERE HAVE TO BE points
in the complement. Given an actual enumeration, YOU CAN CONSTRUCT ONE.

[WM]

Am Mittwoch, 28. November 2018 05:41:42 UTC+1 schrieb George Greene:
> On Monday, November 26, 2018 at 5:00:32 AM UTC-5, WM wrote:
> > Fact is: There is no irrational number in the complement of measure 1.
>
> Fact is, you don't know what a fact is, since you can't prove that

Theorem: Between two irrational numbers x and y there is always a rational number q.

Proof: Assume any two irrational numbers x < y. They have decimal representations. Take the first digit where they differ. Cut the decimal representation of y beyond that digit. It is a rational number q between them:
x < q < y.

In the space between the closed intervals with irrational endpoints there cannot exist any real number.

> (and since a very simple length argument proves that there are in fact
> infinitely many).

Of course. That is fact because the countability of the rationals is nonsense.

>
> But it is a nice challenge to construct one from a given enumeration.
> It is a provable fact that the denial of what you just said is a fact.

What I said is correct, but you may have misunderstood. I say:

If the rational numbers can be included in intervals of measure 0, then the remainder of measure 1 does not contain any real number. Note the implication. Note further the contraposition:

If there are real numbers everywhere, then the rational numbers cannot be included in intervals of measure 0.

>
> Are you afraid that if you just published it as a challenge problem,
> nobody would care?
> I repeat, the irrationality of your constant has nothing whatsoever to
> do with your argument.

It is only a simplified case. Of course my proof holds for all kinds of endpoints.

Regards, WM

[WM]

Am Mittwoch, 28. November 2018 06:01:56 UTC+1 schrieb George Greene:
> On Sunday, November 25, 2018 at 6:31:40 AM UTC-5, WM wrote:
> > If we use intervals
> >
> > [q_n - eps*sqrt(2)*10^(-n), q_n + eps*sqrt(2)*10^(-n)]
> >
> > with eps small enough, then 99% of the unit interval is empty space.
>
> That constitutes that proof that the complement is non-empty and that there
> are irrational numbers in it. We already know there are no rational numbers in it. If there were no irrational numbers in it either then it would be empty.
> If it were empty then the length of the all the intervals summed would have to be at least 1. IT ISN'T. That IS A PROOF that the complement is non-empty.

Of course. Note the implication: If the rational numbers can be included in intervals of measure 0, then the remainder of measure 1 does not contain any real number.

Note further the contraposition: If there are real numbers everywhere, then the rational numbers cannot be included in intervals of measure 0.

Regards, WM

[Jim Burns]

On 11/25/2018 11:50 PM, George Greene wrote:
> On Friday, November 23, 2018 at 8:08:31 PM UTC-5,
> Jim Burns wrote:

>> This has nothing to do with your open intervals. Look at the
>> real number line. The set of irrationals is a set of isolated
>> points, with gaps == missing rational points, surrounding
>> every irrational point at every distance down to zero.
>
> The rationals don't "surround" the irrationals.
> It's really very much the other way around.

Maybe you should say what you mean by "surround".
It doesn't seem to be what I'm calling "surround" above.

In the above sense, the rationals surround the irrationals
AND the irrationals surround the rationals.

Maybe a better image to use would be "thoroughly mixed",
like chocolate stirred into cake batter.

>> Or, we could think about Cantor's diagonal. Suppose we
>> have a list of all the _rationals_ alone.
>
> He's presuming one. He calls it q_n. The nth element of
> this list is a rational number, and every rational is on
> it somewhere.

q_n slipped my mind.

It would have made a better example to use the *WM*
list to generate a number not covered by *his* collection
of closed intervals.

>> You've been given such lists,
> HE'S GIVING YOU ONE, in this problem. JEEzus!!
>
>> I think. From that list, we can define an antidiagonal
>> number x that is at least 1/10 from the first listed
>> rational q[1], at least 1/100 from the second listed
>> rational q[2], ... , at least 1/10^k from the k_th listed
>> rational q[k], ... , and so on, for every listed rational,
>> which is to say, all of the rationals.
>>
>> For every rational q[k],
>
> Do you HAVE to call it q[k] when he called it q_n ?
> Isn't that just being petty??

So, this is why I'm bothering you (George Greene):
It's not that I don't want to insult WM, I'm fine with that.
However, I would rather insult him (when I do) intentionally,
not by accident.

I don't understand what you mean by my using q[k] instead
of q_n being petty. Is it your opinion that I am insulting
WM in that way? Could you explain what you see me as saying,
if that's not too much trouble?

>> there is a positive distance
>> d(x,q[k]) > 1/10^k > 0 between it and the antidiagonal.
>
> It's not necessarily strictly greater than 1/10^k.

For the strictly greater claim, I need to be careful in
my choice of rules for the anti-diagonal digit. But all
we need is for the diagonal digit and anti-diagonal digit
to be at least 2 apart. There are 10 options. We can do this.

I think the following is a good rule, in large part because
it's simple to write down:

For a list of reals f(k) and anti-diagonal x, define the
k_th digit of x as
x[k] = ( f(k)[k] + 5 ) mod 10

We get 0's and 9's in the anti-diagonal this way, but
never when there's a 9 or a 0 in the diagonal, so there
isn't a problem created by some real number being
represented one way in the list and the other way as an
anti-diagonal.

----
This is a good rule to use (as a first step) to show that
there are uncountably many anti-diagonals. Because it's
simple, we can get away with complicating it a bit more.

Let g: N -> {-1,1}
Then, for that particular g, there is an anti-diagonal x_g
defined, digit-by-digit, as
x_g[k] = ( f(k)[k] + 5 + g(k) ) mod 10

x_g is distinct from f(k) for every k, and
x_g is distinct from any x_h, also an anti-diagonal
defined similarly, for a distinct h: N -> {-1,1}

So, we can show there isn't a list of all _anti-diagonals_
to the list f(k) in a manner analogous to showing that
f(k) is not a list of all real numbers, by producing
an anti-diagonal anti-diagonal (I suppose we could call it).

----
Maybe the anti-diagonal anti-diagonal would be helpful to
anyone who wants to show that we can list all of
the reals by doing some Hilbert Hotel hocus-pocus on
"the" anti-diagonal. There is no "the" anti-diagonal.

[transf...@gmail.com]

Am Mittwoch, 28. November 2018 17:57:02 UTC+1 schrieb Jim Burns:
> On 11/25/2018 11:50 PM, George Greene wrote:

> We get 0's and 9's in the anti-diagonal this way, but
> never when there's a 9 or a 0 in the diagonal, so there
> isn't a problem created by some real number being
> represented one way in the list and the other way as an
> anti-diagonal.

There is no real number represented by an infinite string of digits! A representation allowing to identify a number must be complete but an infinite string is incomplete. Even in actual infinity of set theory there is no last digit and no last one before the last one and so on in infinity.

It is possible to distinguish two "given" real numbers by their digits but it is impossible to distinguish a real number from all other real numbers by its string of digits. For that sake you need a finite formula like "0.999...".

The main topic however is the empty space of measure 1 between closed intervals with irrational endpoints. Of course there is no possibility to maintain the three features:

- Cantor's axiom: Every real number determines a point on the real line.
- Mathematics: There cannot exist irrational numbers in a space without rational numbers and with irrational endpoints.
- Set thdeory: All rational numbers can be indexed by natural numbers.

But I would be interested to see attempts to try it.

Regards, WM

[George Greene]

On Wednesday, November 28, 2018 at 5:13:35 AM UTC-5, WM wrote:
> Theorem: Between two irrational
> numbers x and y there is always a rational number q.

Bitch, PLEASE, EVERYBODY ALREADY KNOWS THIS.
>
> Proof:

FUCK you, "proof". EVERYBODY ALREADY KNOWS.
More to the point, between any two RATIONAL numhers there is a rational number,
namely p+q/2. To get an irrational number also between, JUST ADD A SMALL
irrational number. JEEzus!!

> Assume any two irrational numbers x < y. They have decimal representations.

This is irrelevant.
How can you defend ADDING complexity to something SIMPLE??
One rational number guaranteed to be between rational p and q is (p+q)/2. QED.

> In the space between the
> closed intervals with irrational endpoints there cannot exist any real number.

There IS NO "the space between the closed irrational intervals with irrational endpoints" -- there are A DIFFERENT SET OF spaces for EACH DIFFERENT ENUMERATION OF the rationals, for each DIFFERENT q_n. And it DOES NOT MATTER whether the endpoints are rational or irrational. THE ONLY thing that matters IS THEIR LENGTH. YOU DO NOT USE the irrationality of the endpoints ANYWHERE in your proof.

>
> > (and since a very simple length argument proves that there are in fact
> > infinitely many).
>
> Of course. That is fact because the countability of the rationals is nonsense.

You're nonsense. If you had sense you would know that THIS IS MATH, which means that BEFORE you dismiss something as nonsense, YOU HAVE TO PROVE it so.

> > But it is a nice challenge to construct one from a given enumeration.
> > It is a provable fact that the denial of what you just said is a fact.
>
> What I said is correct,

No, it isn't.

> but you may have misunderstood.

That is irrelevant, since YOU misunderstand YOUR OWN ravings.

> I say:
>
> If the rational numbers can be included in intervals of measure 0,

Well, they can't, so what's your point? EVERY INTERVAL HAS POSITIVE measure!!
DO YOU *EVEN*KNOW* what AN INTERVAL *IS*???

[George Greene]

On Wednesday, November 28, 2018 at 5:13:35 AM UTC-5, WM wrote:

> If there are real numbers everywhere,
> then the rational numbers cannot be included in intervals of measure 0.

No individual interval has measure 0, so what do you THINK you (mis)understand by "interval*S*", plural, "of measure 0" ?

> > I repeat, the irrationality of your constant has nothing whatsoever to
> > do with your argument.
>
> It is only a simplified case.

IT IS NOT a simplified case, DIPSHIT!!
THE SIMPLE case would be *4* (which would put a ceiling of 8/9 on the
fraction of the interval covered)! Choosing an irrational constant
IS ADDING complexity AND REDUCING simplicity! *4* is A SIMPLER number than sqrt(2)!!

[George Greene]

On Sunday, November 25, 2018 at 6:31:40 AM UTC-5, WM wrote:

> There are no irrationals in the complement.

Of course there are.

> They would lack a rational between the themselves

THEY WOULD NOT, dumbass.

> and the irrational ends of intervals.

The irationality of the ends of the intervals has nothing to do with
anything.

> Whatever the complement may be, it lies between irrational numbers,

Nothing "lies between" ANYTHING here. You can't say a distributed set of points "lies between" something. The fact that you cannot speak English is always
relevant.

[George Greene]

On Sunday, November 25, 2018 at 6:31:40 AM UTC-5, WM wrote:

> If we use intervals
>
> [q_n - eps*sqrt(2)*10^(-n), q_n + eps*sqrt(2)*10^(-n)]
>
> with eps small enough, then 99% of the unit interval is empty space.

You don't NEED eps and you don't NEED sqrt(2). You could replace sqrt(2) by *4*
and eps by *1* AND STILL come up with 11.11111% of the unit interval as NOT COVERED BY THESE INTERVALS (which is NOT THE SAME AS "being empty space", but you're illiterate anyhow, so fuck you).

[George Greene]

On Monday, November 26, 2018 at 5:09:11 AM UTC-5, WM wrote:
> Every definied bijection includes all rationals in a measure

WRONG.
DAMN, you're stupid.
SPEAK ENGLISH or SHUT *THE*FUCK* UP.
> less than eps.
No,
It includes them in *A*SET* of measure less than eps.
THERE IS NO SUCH THING as being "in a measure less than eps".
These rationals are included in A SET OF measure less than eps.
They ARE a set of measure less than eps.
If you don't put intervals around them then they are a set of measure 0.

They are also "included" -- another completely non-mathematical verb
that you have no business being allowed to use here -- if you
surround them by intervals on either side shrinking by powers
of 10 at each step with that power being
multiplied by some constant k, in a collection of intervals whose
SUMMED measures add up to less than 2k/9. So if we are talking about
the rationals in/on the open unit interval (0,1) here,

> So the remainder of measure 1 is empty

The remainder IS NOT empty, you idiot! The empty set is not of measure 1!
DAMN, you're stupid! You might as well be saying that there are no irrational
numbers!

> and has irrational endpoints

The endpoints don't matter.

You don't even have to use intervals. You can just use the set of points
ITSELF. *THAT*SET* has measure 0. Its complement has measure 1 and is not
empty. Once you add the intervals, the measure of <the set that is the union of all the intervals> is LESS THAN 2k/9 but that is all you can say about it until you have an enumeration specified.
Given any eps in advance, you can set k=eps and make this measure less than eps,
but THAT HAS NOTHING TO DO WITH its being or possibly being 0. IT IS ALWAYS POSITIVE once you add the intervals. The measure of the complement is NOT 1
but it is certainly greater than (9-2k)/9, i.e. greater than 1/9 if k<4,
greater than 7/9 if k<1, and as close to 1 as you like if k is small enough.

[George Greene]

On Monday, November 26, 2018 at 5:00:32 AM UTC-5, WM wrote:
> Fact is:

Oh, shut up.
Math DOESN'T DO "fact".
Facts are facts about what ACTUALLY EXISTS.
Math is about ABSTRACTIONS.

> There is no irrational number in the complement of measure 1.

There are gazillions. If you ever actually PRESENT a bijection q_n,
there will be a countable infinity of them DEFINABLE and PROVABLY extant.

I'm still kind of amazed that nobody has done that yet.

I *dangit*HAD* an explicit bijection available a while back but all the
ones I can find now are for the whole line as opposed to (0,1).

[Me]

On Thursday, November 29, 2018 at 11:02:15 AM UTC+1, transf...@gmail.com wrote:

> A [decimal] representation [identifying] a number must be complete

Sure.

> but an infinite string is incomplete.

Huh? Where did you get that nonsense from, man?

What exactly is MISSING in that infinite string? (Just curious.)

Hint: That is does not have an "end" (i.e. a last character) does not mean that it is "incomplete".

[WM]

Am Montag, 10. Dezember 2018 05:41:50 UTC+1 schrieb Me:


> What exactly is MISSING in that infinite string? (Just curious.)

In order to understand, simply try to write an infinite string - no, not the finite formula defining it!

Regards, WM

[WM]

Am Samstag, 8. Dezember 2018 16:29:28 UTC+1 schrieb George Greene:
> On Wednesday, November 28, 2018 at 5:13:35 AM UTC-5, WM wrote:
> > Theorem: Between two irrational
> > numbers x and y there is always a rational number q.
>

> > Assume any two irrational numbers x < y. They have decimal representations.
>
> This is irrelevant.
> How can you defend ADDING complexity to something SIMPLE??
> One rational number guaranteed to be between rational p and q is (p+q)/2. QED.

Here we have to find a rational between two *irrationals*.

>
> > In the space between the
> > closed intervals with irrational endpoints there cannot exist any real number.
>

> And it DOES NOT MATTER whether the endpoints are rational or irrational. THE ONLY thing that matters IS THEIR LENGTH. YOU DO NOT USE the irrationality of the endpoints ANYWHERE in your proof.

I use it here: In correct mathematics of real numbers we have with X = R\Q and with x =/= y:

∀x ∈ X ∀q ∈ Q ∃y ∈ X: |x - y| < |x - q| (1)
∀x ∈ X ∀y ∈ X ∃q ∈ Q: |x - q| < |x - y| (2)

But if we accept
- Cantor's axiom: Every real number determines a point of the real line.
- All rational numbers can be included in intervals of arbitrary small total measure.

Then we violate (2).

Regards, WM

[burs...@gmail.com]

You cannot write x =/= y outside of the Quantifiers.

The correct writing would be:
∀x ∈ X ∀q ∈ Q (x =/= y => ∃y ∈ X: |x - y| < |x - q|) (1)
∀x ∈ X ∀y ∈ X (x =/= y => ∃q ∈ Q: |x - q| < |x - y|) (2)

[j4n bur53]

Or maybe:

∀x ∈ X ∀q ∈ Q (x =/= q => ∃y ∈ X: |x - y| < |x - q|) (1)

∀x ∈ X ∀y ∈ X (x =/= y => ∃q ∈ Q: |x - q| < |x - y|) (2)

Who knows what you mean?

burs...@gmail.com schrieb:

[George Greene]

On Thursday, November 22, 2018 at 5:59:22 AM UTC-5, transf...@gmail.com wrote:
> For more of this Kind see https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf

Are you somehow not the only one?

https://njwildberger.com/2016/01/01/uncomputable-decimals-and-measure-theory-is-it-nonsense

Or is this also you?
The re-occurrence of "nonsense" (as well as nonsense) invites the identification.

[George Greene]

On Monday, December 10, 2018 at 9:03:34 AM UTC-5, WM wrote:
> Here we have to find a rational between two *irrationals*.

You are ONE IGNORANT LYING DIPSHIT!!!

THERE ARE *TRIVIALLY* *COUNTABLY* *I*N*F*I*N*I*T*E*L*Y*
many rationals between *A*N*Y* two *R*E*A*L* numbers!!!

IT DOESN'T EVEN MATTER whether the reals are rational or irrational!!

[George Greene]

On Monday, December 10, 2018 at 8:57:22 AM UTC-5, WM wrote:
> In order to understand, simply try
> to write an infinite string - no, not the finite formula defining it!

Humans are finite -- they can't physically write anything infintite.
That IS NOT relevant to any discussion of real numbers, MOST of which are
unwriteable. The fact that you think that something is undefinable means it
can't exist is laughable. For any regime of definability, ONCE YOU'VE DEFINED IT,
you CAN THEN define, IN TERMS OF IT, something bigger and badder than it that IT DID NOT COVER.

For somebody who claims TO BELIEVE in potential infinity, your inability to understand this is maddening.

[George Greene]

On Sunday, November 25, 2018 at 7:33:59 AM UTC-5, WM wrote:
> Isolated irrational points cannot exist in a space with irrational endpoints.

Isolated points can't exist on the real line period.
All reals have uncountably many other reals arbitrarily close to them.
NO point on the real line is "isolated" IN ANY meaningful sense.

[George Greene]

On Sunday, November 25, 2018 at 7:33:59 AM UTC-5, WM wrote:

> Isolated irrational points in a space
> with only irrational endpoints are wrong!

Isolated anything is wrong.
I don't know how you came up with the concept.
There are countably infinitely many rationals arbitrarily close TO ANY real
number -- rational or irrational DOESN'T EVEN MATTER.

There are similarly UNcountably infinitely many IRrational real numbers arbitrarily close TO ANY real number -- again, rational or irrational DOESN'T EVEN MATTER.

[George Greene]

On Monday, November 26, 2018 at 5:17:22 AM UTC-5, WM wrote:
> Am Montag, 26. November 2018 05:40:37 UTC+1 schrieb George Greene:
> > Yes, there can, and there do, exist irrational numbers not in any of
> > these intervals.
>
> Of course, since the rational numbers are not countable.

No, the rational numbers ARE countably infinite, which is what allows
us to compute the upper bound on the SUMMED lengths of the intervals.

> > As soon as you present anyone with a definition of
> > q_n, they can define one for you.
>
> Not necessary.

It IS SO TOO necessary to convince YOUR DUMB ASS, since you DON'T
believe it YET!! If it were not necessary to present a q_n, YOU WOULD
ALREADY BELIEVE us!!

> If the rational numbers are assumed to be countable,
> then there is a complementary set of measure 1 with
> countably many irrational endpoints.

Of course.

> This set is empty.

It IS NOT, DUMBASS! It is UNcountably INFINITE!
You certainly haven't presented anything vaguely resembling a proof
that it is empty. You haven't clarified why you need closed vs. open
intervals, and if your endpoints are all in terms of sqrt(2), that is
certainly no reason to presume that they would include irrationals based
on sqrt(3) or e or pi. The endpoints don't matter anyway (and neither does
their irrationality) but your inability to recognize a simplifying assumption
forces us to have arguments about ever more trivial things than the question
you are actually asking.

> > Yes, all the rationals are in on

> > of these intervals, but many of the irrationals are not.
>
> So you dismiss mathematical proof in order to adhere to matheology.

NO, you STUPID-ASS m#f#r, YOU do that.
WE PROVABLY CONSTRUCT an irrational number not in your intervals
FROM ANY AND EVERY q_n that YOU COULD POSSIBLY present.
YOU dismiss THAT.
While claiming you have a proof that some set is empty WHEN YOU DON'T.

Seriously, at the top you said "the structure of the complement is unknown"
and NOW you are claiming the complement IS EMPTY??
WHO'S got THE ACTUAL contradiction here??
Empty is a VERY known and well-understood structure!

[j4n bur53]

Typical N J Wildberger nonsense, who has is
totally clueless. Here is a non-computable real:

"In computability theory, a Specker sequence is
a computable, monotonically increasing, bounded
sequence of rational numbers whose supremum
is not a computable real number. The first
example of such a sequence was constructed
by Ernst Specker (1949)."
https://en.wikipedia.org/wiki/Specker_sequence

George Greene schrieb:

[WM]

Am Montag, 10. Dezember 2018 15:45:27 UTC+1 schrieb George Greene:
> On Monday, December 10, 2018 at 8:57:22 AM UTC-5, WM wrote:
> > In order to understand, simply try
> > to write an infinite string - no, not the finite formula defining it!
>
> Humans are finite -- they can't physically write anything infintite.

Neither can Gods.

> That IS NOT relevant to any discussion of real numbers

It is relevant, since infinite mans without end. But without end there is nothing finally defined. The next bit of information could change the preceding information. The last bit does not even exist in matheology.

> The fact that you think that something is undefinable means it
> can't exist is laughable.

It is fact for humans who can think logically and for Gods.

Regards, WM

[WM]

Am Montag, 10. Dezember 2018 15:19:28 UTC+1 schrieb j4n bur53:
> Or maybe:
>
> ∀x ∈ X ∀q ∈ Q (x =/= q => ∃y ∈ X: |x - y| < |x - q|) (1)
> ∀x ∈ X ∀y ∈ X (x =/= y => ∃q ∈ Q: |x - q| < |x - y|) (2)

No. x =/= q is obvious from y ∈ X = R\Q.

REgards, WM

[WM]

Am Montag, 10. Dezember 2018 15:18:13 UTC+1 schrieb burs...@gmail.com:
> You cannot write x =/= y outside of the Quantifiers.

I can:

> > I use it here: In correct mathematics of real numbers we have with X = R\Q and with x =/= y:
>

> The correct writing would be:
> ∀x ∈ X ∀q ∈ Q (x =/= y => ∃y ∈ X: |x - y| < |x - q|) (1)
> ∀x ∈ X ∀y ∈ X (x =/= y => ∃q ∈ Q: |x - q| < |x - y|) (2)

Not wrong. Go on! What is your conclusion?

Regards, WM

[WM]

Am Montag, 10. Dezember 2018 15:31:16 UTC+1 schrieb George Greene:
> On Monday, December 10, 2018 at 9:03:34 AM UTC-5, WM wrote:
> > Here we have to find a rational between two *irrationals*.
>

> THERE ARE *TRIVIALLY* *COUNTABLY* *I*N*F*I*N*I*T*E*L*Y*
> many rationals between *A*N*Y* two *R*E*A*L* numbers!!!

No. If the intervals have only measure 1/3 then they have endpoints where they touch the complement. The complement cointains no rational points. Then for two different irrationa points x and y, we have

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

Regards, WM

[Me]

On Monday, December 10, 2018 at 2:57:22 PM UTC+1, WM wrote:
> Am Montag, 10. Dezember 2018 05:41:50 UTC+1 schrieb Me:
> >
> > What exactly is MISSING in that infinite string? (Just curious.)
> >
> In order to understand, simply try to write an infinite string

Huh?! Relevance?

We didn't talk about the ability or inability of a human being to write down an infinite string - which OF COURSE is impossible.

Your claim WAS:

> > an infinite string is incomplete.

And I ASKED:

> What exactly is MISSING in that infinite string? (Just curious.)
>

> [ Hint: That is does not have an "end" (i.e. a last character) does not
> mean that it is "incomplete". ]

So no answer? Well, then don't claim such nonsense, man!

[WM]

Am Montag, 10. Dezember 2018 20:46:45 UTC+1 schrieb Me:
> On Monday, December 10, 2018 at 2:57:22 PM UTC+1, WM wrote:
> > Am Montag, 10. Dezember 2018 05:41:50 UTC+1 schrieb Me:
> > >
> > > What exactly is MISSING in that infinite string? (Just curious.)
> > >
> > In order to understand, simply try to write an infinite string
>
> Huh?! Relevance?

To show you that it is impossible write or read or access an infinite string.

>
> We didn't talk about the ability or inability of a human being to write down an infinite string - which OF COURSE is impossible.

That is not a matter of ability but of existence. There does not exist and end signal. Therefore every information can be reversed.

>
> Your claim WAS:
>
> > > an infinite string is incomplete.

An infinite string is infinite. Incomplete is a synonym. Complete: accessible as a whole, all parts being simultaneously present and receivable.

>
> And I ASKED:
>
> > What exactly is MISSING in that infinite string? (Just curious.)
> >
> > [ Hint: That is does not have an "end" (i.e. a last character) does not
> > mean that it is "incomplete". ]
>
> So no answer?

No answer that you could understand.

Regards, WM

[George Greene]

On Monday, December 10, 2018 at 1:23:14 PM UTC-5, WM wrote:
> Am Montag, 10. Dezember 2018 15:31:16 UTC+1 schrieb George Greene:
> > On Monday, December 10, 2018 at 9:03:34 AM UTC-5, WM wrote:
> > > Here we have to find a rational between two *irrationals*.
> >
> > THERE ARE *TRIVIALLY* *COUNTABLY* *I*N*F*I*N*I*T*E*L*Y*
> > many rationals between *A*N*Y* two *R*E*A*L* numbers!!!
>
> No. If the intervals have only measure 1/3 then they have endpoints where they touch the complement.

THERE IS NO SUCH THING as "touching" the complement!! WHAT IN THE HELL is wrong with you?? How do you keep INVENTING THESE VERBS that have NO defined mathematical meaning???

We are talking about the interval (0,1) here. I have written it as open.
Do you want it closed?
The intervals you are talking about do not need to include their endpoints.
I would write them as open. If you are talking about all these intervals
being closed then they all CONTAIN their irrational endpoints so NONE OF THEIR ENDPOINTS *IS*IN* the complement! An irrational endpoint that is NOT IN the
complement *CAN*NOT* be said to "touch" the complement *IN*ANY* way!!

[George Greene]

On Monday, December 10, 2018 at 1:23:14 PM UTC-5, WM wrote:

> No. If the intervals have only measure 1/3 then they have endpoints where they touch the complement.

Dipshit:
The measure of the intervals has NOTHING WHAT*SO*FUCKING*EVER to do WITH ANYthing!!
"The measure of the intervals" DOES NOT EXIST,
you IGNORANT-ass ILLITERATE dipshit!

The SUM OF the measureS, PLURAL, of the intervals, WHICH IS WHAT YOU MEANT
to say, can be ANY POSITIVE REAL NUMBER YOU LIKE, and whether it is or isn't 1/3 has nothing whatsoever to do WITH ANYthing.
Neither does the rationality or irrationality of the endpoints.
Irrational endpoints DO NOT "touch the complement" -- your dumb ass canNOT even DEFINE "touch" in such a way as to make it coherent here, and even it were coherent, it still wouldn't be relevant. If the intervals are closed then the endpoints are not in the complement. NOTHING "TOUCHES" anything around here.
No real number "touches" another because "touch" connotes adjacency
but there are INFINITELY MANY OTHER (also non-touching) real numbers between ANY two real numbers, REGARDLESS of rationality or irrationality. Numbers with an infinite number of numbers SEPARATING them are NOT *touching*, YOU *I*D*I*O*T*.

[George Greene]

On Monday, December 10, 2018 at 1:23:14 PM UTC-5, WM wrote:

> ~∀x ∈ X ∀y ∈ X ∃q ∈ Q: |x - q| < |x - y|

Just because you think you know how to write that does not make it true.
Do you just pick theorems at random and assert their denials to troll?
There is no reason for you to use X without saying what it might be.
You should go ahead and choose R for X, or (),1), AND SAY that.
The only way your
~∀xy∈R[∃q∈Q[|x - q| < |x - y|]]
can be true is if x and y are equal -- in that case the rightmost
absolute value is 0 so obviously the left one can't be less than it.
In all other cases, it is trivial to increase the lesser of x and y
by some very small amount to get to a rational that does not get up
to y, and there's your q. To make this non-trivial, the reals in question
have to be different, so one of them has to be less, so call that one x.
A CORRECT statement of the relevant theorem would be
∀xy∈R[ x<y -> ∃q∈Q[ x<q /\ q<y] ]
You DON'T NEED any 0's and you DON'T NEED any absolute values.
Your capacity for introducing irrelevant bullshit remains flabbergasting.

This a trivial theorem.
Informally, here is a trivial proof.
Real numbers have decimal representations.
If x<y and they are in (0,1) then there is a first(least/lowest/leftmost) decimal place number m where x's mth digit is less than y's.
Find the next digit-place n, after m -- n>m -- where x's nth digit is not a 9.
There has to be one because if the decimal expansion of x ended in all 9s then
x would be rational;or, just as well, it could be rewritten EQUIVALENTLY with the mth digit incremented (x's mth digit can't possibly be 9 because it has to be less than y's) followed by all 0's, so the first of these 0's would be the
mth place. Increment x's nth digit and throw away (or equivalently, set to all 0's) the rest of its decimal expansion. The resulting real number is rational because its decimal expansion terminates (or ends in all 0's) and it is less than y because it is less than y in the nth
place (and also less in the mth, which is the first place they differ -- unless x's mth got incremented to match y's, but in that case, x is less because it has nothing after this matching digit while y has something).

[George Greene]

On Monday, December 10, 2018 at 4:04:18 PM UTC-5, WM wrote:
> Am Montag, 10. Dezember 2018 20:46:45 UTC+1 schrieb Me:
> > On Monday, December 10, 2018 at 2:57:22 PM UTC+1, WM wrote:
> > > Am Montag, 10. Dezember 2018 05:41:50 UTC+1 schrieb Me:
> > > >
> > > > What exactly is MISSING in that infinite string? (Just curious.)
> > > >
> > > In order to understand, simply try to write an infinite string
> >
> > Huh?! Relevance?
>
> To show you that it is impossible write or read or access an infinite string.

The fact that finite human beings cannot read or write these infinite numbers in finite lifespans with finite brains DOES NOT STOP THEM FROM EXISTING, MATHEMATICALLY, *D*I*P*S*H*I*T*.
"WM's finite brain can write this down" IS NOT a necessary condition for this being a number.

It's a shame Wildberger couldn't ever get put the heck out of the academy.
I guess he could have made some contributions in other arenas while people
ignored his philosophy as harmless. I don't think any platonists actually
feel threatened by him.

You, on the other hand, have destroyed what used to be a relevant forum.
Of course, you had help
(Peter Olcott, Graham Cooper, Nam Nguyen, AD NAUSEAM).

[George Greene]

On Monday, December 10, 2018 at 2:46:45 PM UTC-5, Me wrote:
> > [ Hint: That is does not have an "end" (i.e. a last character) does not
> > mean that it is "incomplete". ]

Well, you know this, and I know this, but WM....
One thing he can truthfully say is that if it has an order,
at every POINT IN it, it is incomplete -- No element in it is its greatest/last
element. If it has a least upper bound, then that bound is also not in it.
There are of course also ordered sets with no least or first element,
e.g., (0,1) as an open interval, or "the positive rationals" or "the positive reals".

[WM]

Am Dienstag, 11. Dezember 2018 04:51:50 UTC+1 schrieb George Greene:


> > No. If the intervals have only measure 1/3 then they have endpoints where they touch the complement.
>
> THERE IS NO SUCH THING as "touching" the complement!!

If there is an interval in a nonempty complement, then there are two points where interval and complement touch.

>
> We are talking about the interval (0,1) here.

We are talking about the real axis and aleph_0 intervals with total measure less than 1/3.

> The intervals you are talking about do not need to include their endpoints.

They are defined as closed intervals. Your statement is wrong.

> I would write them as open.

Then make up your own thread. Here we discuss the intervals

I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)] with n ∈ ℕ.

> If you are talking about all these intervals
> being closed then they all CONTAIN their irrational endpoints so NONE OF THEIR ENDPOINTS *IS*IN* the complement!

Correct. But the complement is open and end at these endpoints.

> An irrational endpoint that is NOT IN the
> complement *CAN*NOT* be said to "touch" the complement *IN*ANY* way!!

The complement C can be said to touch the complement, because every point between x ∈ C and and endpoint belongs to the complement. All such points are irrational without a rational between them. Therefore the statement

∀x ∈ X ∀y ∈ (X \ {x}) ∃q ∈ Q: |x - q| < |x - y|

fails.

Regards, WM

[WM]

Am Dienstag, 11. Dezember 2018 05:25:53 UTC+1 schrieb George Greene:
> On Monday, December 10, 2018 at 1:23:14 PM UTC-5, WM wrote:
> > ~∀x ∈ X ∀y ∈ X ∃q ∈ Q: |x - q| < |x - y|
>
> Just because you think you know how to write that does not make it true.

No, it is true because there is an irrational number in the complement.

> There is no reason for you to use X without saying what it might be.

X = R\Q.

> You should go ahead and choose R for X, or (),1), AND SAY that.
> The only way your
> ~∀xy∈R[∃q∈Q[|x - q| < |x - y|]]
> can be true is if x and y are equal

Right, but wrong in set theory. The complement consists only of irrationals.

Regards, WM

[WM]

Am Dienstag, 11. Dezember 2018 05:28:52 UTC+1 schrieb George Greene:
> On Monday, December 10, 2018 at 4:04:18 PM UTC-5, WM wrote:


> > To show you that it is impossible write or read or access an infinite string.
>
> The fact that finite human beings cannot read or write these infinite numbers in finite lifespans with finite brains DOES NOT STOP THEM FROM EXISTING, MATHEMATICALLY,

It has nothing to do with your inability to write, but with your inability to think: There is no final bit of the string. Therefore it can change its meaning on and on and doe snever fix it. It is a matter of existence!

Regards, WM

[me]

Am Dienstag, 11. Dezember 2018 12:41:10 UTC+1 schrieb WM:

> Therefore it can change its meaning on and on and does never fix it.

Mathematical objects don't "change", man. The decimal expansion of each and every real number is FIXED: it does not "change". (Otherwise the number pi of today might not be the number pi of tomorrow. :-)

Hint: (Pure) Mathematics does not deal with physical objects or living beings (which can change).

[George Greene]

On Tuesday, December 11, 2018 at 5:56:14 AM UTC-5, WM wrote:
> If there is an interval in a nonempty complement,

There obviously ARE NO intervals in the non-empty complement.
JEEzus.

[WM]

Obviously. But if the rationals were countable and included in ℵ_0 intervals of measure 1/3, then at most ℵ_0 intervals of infinite measure would be the complement.

Regards, WM

[WM]

Am Dienstag, 11. Dezember 2018 13:19:41 UTC+1 schrieb me:
> Am Dienstag, 11. Dezember 2018 12:41:10 UTC+1 schrieb WM:
>
> > Therefore it can change its meaning on and on and does never fix it.
>
> Mathematical objects don't "change", man.

Therefore infinite strings do not exist in mathematics.

> The decimal expansion of each and every real number is FIXED: it does not "change".

Therefore it is must be given by a finite formula.

> (Otherwise the number pi of today might not be the number pi of tomorrow.

Hint: pi is a limit, not a string of digits and it cannot be represented by a string of digits but merely approximated.

> Hint: (Pure) Mathematics does not deal with physical objects or living beings (which can change).

Therefore we use limits but not infinite strings. https://www.quora.com/Why-isnt-0-999-1-common-knowledge/answer/Wolfgang-Mueckenheim-1

Regards, WM

[Me]

Am Mittwoch, 12. Dezember 2018 12:21:39 UTC+1 schrieb WM:
> Am Dienstag, 11. Dezember 2018 13:19:41 UTC+1 schrieb me:
> > Am Dienstag, 11. Dezember 2018 12:41:10 UTC+1 schrieb WM:
> >
> > > Therefore it can change its meaning on and on and does never fix it.
> >
> > Mathematical objects don't "change", man.
> >
> Therefore infinite strings do not exist in mathematics.

We can represent [or encode] these "strings" in set theory. :-)

Usually this is done by using (infinit) sequences. See: https://en.wikipedia.org/wiki/Sequence

> > The decimal expansion of each and every real number is FIXED: it does not
> > "change".
> >
> Therefore it is must be given by a finite formula.

Nonsens.

> > (Otherwise the number pi of today might not be the number pi of tomorrow.
> >
> Hint: pi is a limit,

pi is a real number in my book. :-)

> not a string of digits

We can introduce/define real numbers as "string of digits" (i. e. infinite sequences). THEN pi actually _is_ a "string of digits".

> and it cannot be represented by a string of digits [...]

It can.

See: Wolfgang Rautenberg, Elementare Grundlagen der Analysis. Mannheim / Leipzig / Wien / Zürich : BI-Wiss.-Verl., 1993

> > Hint: (Pure) Mathematics does not deal with physical objects or living
> > beings (which can change).
> >

> Therefore <bla>

Shut up, man.

[WM]

Am Mittwoch, 12. Dezember 2018 14:24:14 UTC+1 schrieb Me:
> Am Mittwoch, 12. Dezember 2018 12:21:39 UTC+1 schrieb WM:


> > Therefore infinite strings do not exist in mathematics.
>
> We can represent [or encode] these "strings" in set theory. :-)

They can be encoded in mathematics too. But they do not exist without a finite formula or encoding.

>
> Usually this is done by using (infinit) sequences. See: https://en.wikipedia.org/wiki/Sequence

Infinite sequences cannot be used without finite encoding.

>
> > Hint: pi is a limit,
>
> pi is a real number in my book. :-)

Many real numbers are limits without decimal representation.

>
> > and it cannot be represented by a string of digits [...]
>
> It can.

Try it. Fail. Recognize that you cannot because you are a finite human. Claim that God could if he only would. Thereby proving that you are a fool of matheology.

Regards, WM

[Me]

Am Mittwoch, 12. Dezember 2018 18:05:18 UTC+1 schrieb WM:

> Many real numbers are limits without decimal representation.

Not in mathematics.

See: https://en.wikipedia.org/wiki/Decimal_representation

[George Greene]

On Wednesday, December 12, 2018 at 6:21:30 AM UTC-5, WM wrote:
> Am Dienstag, 11. Dezember 2018 22:08:16 UTC+1 schrieb George Greene:
> > On Tuesday, December 11, 2018 at 5:56:14 AM UTC-5, WM wrote:
> > > If there is an interval in a nonempty complement,
> >
> > There obviously ARE NO intervals in the non-empty complement.
>
> Obviously.

If it's obvious, then why do you continue?

> But if the rationals were countable and included
> in ℵ_0 intervals of measure 1/3,

Which they are, and of measure as small as you like -- there is nothing
special about 1/3. There is no reason for you to even put intervals around
them at all. You could just let them be isolated points. There is a function f(x) defined as 0 when x is rational and 1 when x is irrational. Its graph
looks like two parallel horizontal lines BUT NEITHER OF THEM has ANY INTERVALS at all. The bottom one has measure 0 and the top one has measure 1 but NO
intervals are relevant FOR EITHER line.

> then at most ℵ_0 intervals of infinite measure

Let's not do the whole line. Let's just do (0,1), please.
NO intervals of infinite meaure are relevant.
And since the complement has NO INTERVALS AT ALL, i.e. *0* intervals,
that CERTAINLY AGREES with your "at most ℵ_0 intervals".
But since there aren't any at all, there can't be any "of infinite measure".
The complement (if you go back to the whole line) may have "infinite measure"
but infinity is not a real number so I don't think you want to attribute that result to measure theory.
Maybe YOU SHOULD LEARN SOME before pontificating.

[George Greene]

On Wednesday, December 12, 2018 at 12:05:18 PM UTC-5, WM wrote:
> They can be encoded in mathematics too.
> But they do not exist without a finite formula or encoding.

You CAN'T *just*say* that somethign doesn't exist.
You have to derive a contradiction from the assumption of its existence
*BEFORE* you can say that. If you can't say that then we can just have
an axiom POSTULATING the thing's existence and THERE WON'T BE SHIT
that YOU can do or say about it.

More to the point, Everybody trivially CAN derive a contradiction from your
claim that only definable things exist, since we can just diagonalize your
class of definitions. By your own requirement your class of definitions has
tbe definable, but based on it, anybody can always define a more complex and
comprehensive class, so your class of all definitions canNOT be definable,
so non-definable things MUST exist.

[Julio Di Egidio]

On Thursday, 13 December 2018 04:36:18 UTC+1, George Greene wrote:

> More to the point, Everybody trivially CAN derive a contradiction from your
> claim that only definable things exist, since we can just diagonalize your
> class of definitions. By your own requirement your class of definitions has
> tbe definable, but based on it, anybody can always define a more complex and
> comprehensive class, so your class of all definitions canNOT be definable,
> so non-definable things MUST exist.

That only holds in the standard, classical paradigm. Which I assume you know,
but most just don't, so you better be explicit on the scope of your assertions.
Because, now that that is clear, it's also fucking clear that there is nothing granted about the truth of your assertions: nobody can guarantees that that
paradigm isn't just badly broken, as in fact it is and not just according to
few delusional cranks around here. Read Wittgenstein just to begin with.

Julio

[transf...@gmail.com]

Am Donnerstag, 13. Dezember 2018 04:33:02 UTC+1 schrieb George Greene:
> On Wednesday, December 12, 2018 at 6:21:30 AM UTC-5, WM wrote:
> There is no reason for you to even put intervals around
> them at all. You could just let them be isolated points.

But I don't. I use intervals, each one of of finite measure, and each one embedded either in other intervals (they are irrelevant) or bordering on points of the complement.

> Let's not do the whole line. Let's just do (0,1), please.

> NO intervals of infinite meaure are relevant.
> And since the complement has NO INTERVALS AT ALL

That is nonsense.

Take a ring with measure 1:
If three intervals are embedded into the continuum, then the complement is split into at most three intervals too. If ℵ_0 intervals are embedded into the continuum, then the complement is split into at most ℵ_0 intervals too.

As Long as we have finite intervals, they split the continuum into intervals too.

> Maybe YOU SHOULD LEARN SOME before pontificating.

Then try to explain how a measure of more than 2/3 can be accumulated in the complement.

Regards, WM

[transf...@gmail.com]

Am Mittwoch, 12. Dezember 2018 18:20:33 UTC+1 schrieb Me:
> Am Mittwoch, 12. Dezember 2018 18:05:18 UTC+1 schrieb WM:
>
> > Many real numbers are limits without decimal representation.
>
> Not in mathematics.
>

Then give the decimal representation of pi.

Regards, WM

[transf...@gmail.com]

Am Donnerstag, 13. Dezember 2018 04:36:18 UTC+1 schrieb George Greene:
> On Wednesday, December 12, 2018 at 12:05:18 PM UTC-5, WM wrote:
> > They can be encoded in mathematics too.
> > But they do not exist without a finite formula or encoding.
> You CAN'T *just*say* that somethign doesn't exist.
> You have to derive a contradiction from the assumption of its existence
> *BEFORE* you can say that.

There is no infinite decimal representation of a real number x (without a finite formula) that allows to distinguish x from all other real numbers.

Proof: Up to every digit there are infinitely many real numbers possibe.

>
> More to the point, Everybody trivially CAN derive a contradiction from your
> claim that only definable things exist, since we can just diagonalize your
> class of definitions.

That is wrong. Diagonalization requires that the anti-diagonal number differs from all numbers of the list. The list however is as incomplete a sequence as every infinite string of digits. Unless the list has a finite construction formula. Then the list and its antidiagonals belong to a countabe set.

> By your own requirement your class of definitions has
> tbe definable, but based on it, anybody can always define a more complex and
> comprehensive class, so your class of all definitions canNOT be definable,
> so non-definable things MUST exist.

Of course in potential infinity there are many undefined objects. But each one is definable. There is nothing uncountable.

Regards, WM

[George Greene]

On Tuesday, December 11, 2018 at 5:56:14 AM UTC-5, WM wrote:

> Am Dienstag, 11. Dezember 2018 04:51:50 UTC+1 schrieb George Greene:
> If there is an interval in a nonempty complement,
> then there are two points where interval and complement touch.

There ISN'T an interval in the compelement, so you're irrelevant and stupid,
but even if there were, your attempt to define "touch" here STILL FAILS:
The interval IS ALREADY IN the complement, so you were NOT even TRYING to say that the INTERVAL and the COMPLEMENT touch!!

[Me]

Huh?! Your claim was: "Many real numbers are limits without decimal representation."

In mathematics each and every real number (hence especially pi too) does have a decmal representation.

See: https://en.wikipedia.org/wiki/Decimal_representation

Hint: Mathematics is not reduced to your moronic "pocket calculator muckenmatics".

[George Greene]

On Tuesday, December 11, 2018 at 6:16:57 AM UTC-5, WM wrote:
> Am Dienstag, 11. Dezember 2018 05:25:53 UTC+1 schrieb George Greene:
> > On Monday, December 10, 2018 at 1:23:14 PM UTC-5, WM wrote:
> > > ~∀x ∈ X ∀y ∈ X ∃q ∈ Q: |x - q| < |x - y|
> >
> > Just because you think you know how to write that does not make it true.
>
> No, it is true because there is an irrational number in the complement.

ALL the numbers in the complement are irrational, YOU IDIOT!! THAT DOES NOT
make what you said true! THERE IS STILL a rational number between any two
real numbers! YOU DIDN'T SAY WHAT X was! Not that it matters.

> > There is no reason for you to use X without saying what it might be.
>
> X = R\Q.

I thought you yourself were claiming TO PROVE that there was a rational
number between any two irrational numbers! So why are you here
asserting the denial of that??

> > You should go ahead and choose R for X, or (),1), AND SAY that.
> > The only way your
> > ~∀xy∈R[∃q∈Q[|x - q| < |x - y|]]
> > can be true is if x and y are equal
>
> Right, but wrong in set theory.

> The complement consists only of irrationals.

Set theory doesn't disagree with that. The complement does consist only
of irrationals. That does not make what you said any less false.
You seem to think that if x and y are in the complement, then everything
between them has to be too. This is not the case. There are no intervals
in the complement.
This would be equally true EVEN IF YOU DIDN'T put intervals around the rationals.
I repeat, you should consider the graph of the function f(x)= 0 if x is rational, = 1 if x is irrational. That graph looks like 2 straight lines but there are NO intervals IN EITHER of those lines.

[George Greene]

On Wednesday, December 12, 2018 at 12:05:18 PM UTC-5, WM wrote:
> Many real numbers are limits without decimal representation.

You are LYING. The fact that a decimal representation is countably infinitely wide (they all are) DOES NOT STOP it from existing.

[George Greene]

On Wednesday, December 12, 2018 at 12:05:18 PM UTC-5, WM wrote:
> Many real numbers are limits without decimal representation.

This is a lie. All real numbers by definition have a decimal representation.
If you are going to doubt the existence of the representation, how can you be any LESS doubtful about the existence of THE REAL NUMBER *ITSELF*?? All the undesirable properties you wanted to ascribe to the decimal repreesentation -- containing an infinitude of information, countably infinitely many BITS of information, basically -- ARE EQUALLY problematic in THE NUMBER ITSELF.

[WM]

Am Freitag, 14. Dezember 2018 07:09:04 UTC+1 schrieb George Greene:
> On Wednesday, December 12, 2018 at 12:05:18 PM UTC-5, WM wrote:
> > Many real numbers are limits without decimal representation.
>
> This is a lie. All real numbers by definition have a decimal representation.

No. And if one exists, you cannot write or read or else use it. It has the same status as the devil. It is religion.

> If you are going to doubt the existence of the representation, how can you be any LESS doubtful about the existence of THE REAL NUMBER *ITSELF*??

Easy, to answer. Every mathematician should know that. There is the limit, defined by the sequence or series like SUM 1/n!.

> All the undesirable properties you wanted to ascribe to the decimal repreesentation -- containing an infinitude of information, countably infinitely many BITS of information, basically -- ARE EQUALLY problematic in THE NUMBER ITSELF.

No. LIM SUM 1/n! = e. It allows you to find for every rational number q whether or not e > q. But contrary to every decimal representation that you can read or write SUM1/n! is unique, i.e., it differs from all real numbers except itself.

Every decimal representation that you can apply allows only to distinguish e from every other *given* real number.

Regards, WM

[WM]

Am Donnerstag, 13. Dezember 2018 13:09:43 UTC+1 schrieb George Greene:
> On Tuesday, December 11, 2018 at 5:56:14 AM UTC-5, WM wrote:
> > Am Dienstag, 11. Dezember 2018 04:51:50 UTC+1 schrieb George Greene:
> > If there is an interval in a nonempty complement,
> > then there are two points where interval and complement touch.
>
> There ISN'T an interval in the compelement,

There are at most aleph_0 intervals. At least one of them has measure > 0. Otherwise the measure of 2/3 cannot be realized.

> but even if there were, your attempt to define "touch" here STILL FAILS:
> The interval IS ALREADY IN the complement,

Every interval has irrational endpoints. Between two intervals there is at least one irrational number. Otherwise the measure of 2/3 which is larger than that of the intervals could not be realized.

> so you were NOT even TRYING to say that the INTERVAL and the COMPLEMENT touch!!

Every interval has irrational endpoints. An endpoint is defined as a point beyond which the interval does not exist.

Regards, WM

[WM]

Am Donnerstag, 13. Dezember 2018 14:25:35 UTC+1 schrieb Me:
> Am Donnerstag, 13. Dezember 2018 10:23:55 UTC+1 schrieb transf...@gmail.com:
> > Am Mittwoch, 12. Dezember 2018 18:20:33 UTC+1 schrieb Me:
> > > Am Mittwoch, 12. Dezember 2018 18:05:18 UTC+1 schrieb WM:
> > > >
> > > > Many real numbers are limits without decimal representation.
> > > >
> > > Not in mathematics.
> > >
> > Then give the decimal representation of pi.
>
> Huh?! Your claim was: "Many real numbers are limits without decimal representation."
>
> In mathematics each and every real number (hence especially pi too) does have a decmal representation.

No. And if one exists, you cannot write or read or else use it. It has the same status as the devil. It is Matheology.

>
> See: https://en.wikipedia.org/wiki/Decimal_representation
>
> Hint: Mathematics is not reduced to your moronic "pocket calculator muckenmatics".

You cannot give the decimal representation of pi. You prefer to believe in inapplicable nonsense that has the same status of reality as the devil of theology. Your choice.

Regards, WM

[WM]

Am Freitag, 14. Dezember 2018 07:06:15 UTC+1 schrieb George Greene:
> On Tuesday, December 11, 2018 at 6:16:57 AM UTC-5, WM wrote:
> > Am Dienstag, 11. Dezember 2018 05:25:53 UTC+1 schrieb George Greene:
> > > On Monday, December 10, 2018 at 1:23:14 PM UTC-5, WM wrote:
> > > > ~∀x ∈ X ∀y ∈ X ∃q ∈ Q: |x - q| < |x - y|
> > >
> > > Just because you think you know how to write that does not make it true.
> >
> > No, it is true because there is an irrational number in the complement.
>
> ALL the numbers in the complement are irrational

Of course.

> THERE IS STILL a rational number between any two
> real numbers!

That is not possible, because there are aleph_0 intervals with rationals. By aleph_0 endpoints aleph_0 intervals of the complement are created. Each one lies between two intervals with irrational endpoints. At least one of them has finite measure in order to get 2/3.

> > > There is no reason for you to use X without saying what it might be.
> >
> > X = R\Q.
>
> I thought you yourself were claiming TO PROVE that there was a rational
> number between any two irrational numbers!

I mathematics that is true. In the topology of set theory it is impossible.

> So why are you here
> asserting the denial of that??

Because it is impossible to collect a Lebesgue measure of 2/3 when observing this restriction.

>
>
> > > You should go ahead and choose R for X, or (),1), AND SAY that.
> > > The only way your
> > > ~∀xy∈R[∃q∈Q[|x - q| < |x - y|]]
> > > can be true is if x and y are equal
> >
> > Right, but wrong in set theory.
>
> > The complement consists only of irrationals.
>
> Set theory doesn't disagree with that. The complement does consist only
> of irrationals.

And it has measure 2/3. That is impossible by aleph_0 degenerate intervals. More intervals however do not exists, because there are only aleph_0 endpoints of intervals.

> You seem to think that if x and y are in the complement, then everything
> between them has to be too.

If x is in the complement, then the next rational number is shielded by the endpoint of an interval.

> This is not the case. There are no intervals
> in the complement.

You are wrong. There are aleph_0 endpoints. Therefore there are aleph_0 intervals with rationals and aleph_0 intervals without rationals.

> I repeat, you should consider the graph of the function f(x)= 0 if x is rational, = 1 if x is irrational. That graph looks like 2 straight lines but there are NO intervals IN EITHER of those lines.

I recommend to use intervals. All of them are finite. That disallows to shut the eyes in the manner you try to do.

Regards, WM

[jrenne...@googlemail.com]

On Friday, December 14, 2018 at 12:15:14 PM UTC+1, WM wrote:
[...]
> ... there are aleph_0 intervals with rationals. By aleph_0 endpoints >aleph_0 intervals of the complement are created.

This is an error. The complement does not contain any intervals; nevertheless it is uncountable.
Since you don't see how that is possible, try to understand the construction of Cantor's "middle thirds" set.

> Each one lies between two intervals with irrational endpoints. At least >one of them has finite measure in order to get 2/3.

This is also an error. To understand why this statement is false google "nowhere dense perfect sets with nonzero Lebesgue measure".

It is rather astonishing that you have fallen into such an elementary trap: The complement of the rationals already contains no intervals - how can you then believe that a covering of the rationals has a complement containing intervals?

>
> Because it is impossible to collect a Lebesgue measure of 2/3 when >observing this restriction.

Wrong. Obviously you haven't understood what Lebesgue measure is.

>
> And it has measure 2/3. That is impossible by aleph_0 degenerate >intervals. More intervals however do not exists, because there are only >aleph_0 endpoints of intervals.

This statement is more or less correct: Every countable set is Lebesgue-measurable and has measure 0.

>
> If x is in the complement, then the next rational number is shielded by >the endpoint of an interval.

There is no "next rational number".

>
> You are wrong. There are aleph_0 endpoints. Therefore there are aleph_0 >intervals with rationals and aleph_0 intervals without rationals.

Again: This is a beginner's error.

[WM]

Am Samstag, 15. Dezember 2018 18:49:04 UTC+1 schrieb jrenne...@googlemail.com:
> On Friday, December 14, 2018 at 12:15:14 PM UTC+1, WM wrote:
> [...]
> > ... there are aleph_0 intervals with rationals. By aleph_0 endpoints >aleph_0 intervals of the complement are created.
>
> This is an error.

It is mathematics. Consider a ring. N distinct intervals leave N distinct intervals in the complement. This holds for every finite N. Should it change in case of aleph_0?

> Since you don't see how that is possible, try to understand the construction of Cantor's "middle thirds" set.

We should stay with mathematics and remember that all intervals are finite. There is no limit that shrinks all intervals to points.

>
> > Each one lies between two intervals with irrational endpoints. At least >one of them has finite measure in order to get 2/3.
>
> This is also an error. To understand why this statement is false google "nowhere dense perfect sets with nonzero Lebesgue measure".

We should stay with mathematics, not switch to nonsense.

>
> It is rather astonishing that you have fallen into such an elementary trap: The complement of the rationals already contains no intervals

It does not. Without intervals there is no measure 2/3 or even infinity possible.

> > Because it is impossible to collect a Lebesgue measure of 2/3 when >observing this restriction.
>
> Wrong. Obviously you haven't understood what Lebesgue measure is.

I have understood that you cannot explain but ttry to cheat.

> > And it has measure 2/3. That is impossible by aleph_0 degenerate >intervals. More intervals however do not exists, because there are only >aleph_0 endpoints of intervals.
>
> This statement is more or less correct: Every countable set is Lebesgue-measurable and has measure 0.

aleph_0 endpoints imply aleph_0 intervals. By symmetry they are also in the complement.

> >
> > If x is in the complement, then the next rational number is shielded by >the endpoint of an interval.
>
> There is no "next rational number".

Not in mathematics. But if all rational numbers could be included into intervals with irrational endpoints, then an irrational number in the complement would be shielded from every rational number by at least one irrational endpoint. Magic will not help.

> >
> > You are wrong. There are aleph_0 endpoints. Therefore there are aleph_0 >intervals with rationals and aleph_0 intervals without rationals.
>
> Again: This is a beginner's error.

Your statemenzt is that of a charlatan, a deliberate defrauder or a fool diving into matheology and unable to think.

At least one finite interval is in the complement of measure 2/3 or infinity.

Regards, WM

[jrenne...@googlemail.com]

Stupidity is a lack of intelligence, understanding, reason, wit, or common sense. Stupidity may be innate, assumed or reactive – a defence against grief or trauma. [Wikipedia!!]

[Tucso...@me.com]

How can there be any interval in the complement? This even especially stupid for you. It is a fact that every interval of non-zero measure must contain a rational number. Now, you’re construction has excluded each and every rational number from the complement. Hence, the complement only consist of isolated points, all of which are irrational most obviously.

You are most likely confused by the fact that a set of isolated points can have positive measure, but that’s how it works. Consider, for instance, if you take the set of irrational points between 0 and 1; that is, X = [0,1] \ Q; then X has measure 1. If you going to use measure theory in your fruitless pseudo-refutations of set theory, then you must stick with facts. And, the fact is you are clueless.

> Regards, WM

ZG

[WM]

Am Sonntag, 16. Dezember 2018 01:50:27 UTC+1 schrieb Tucso...@me.com:

> > At least one finite interval is in the complement of measure 2/3 or infinity.
>

> How can there be any interval in the complement? It is a fact that every interval of non-zero measure must contain a rational number.

Of course. But that is not possible if all rational numbers can be covered by intervals of total measure les than 1/3.

> Hence, the complement only consist of isolated points, all of which are irrational most obviously.

There cannot be more isolated points than endpoints isolating them. There are aleph_0 such endpoints. What else should isolate the points? Further the endpoints are irrational themselves, such that for an isolated irrational point x ∈ X = R\Q we have

∀x ∈ X ∀q ∈ Q ∃y ∈ X\{x}: |x - y| < |x - q| (1)

but

~∀x ∈ X ∀y ∈ X\{x} ∃q ∈ Q: |x - q| < |x - y| (2)

> you must stick with facts.

Fact is: the failure of (2) shows a contradiction of set theory and mathematics.

> You are most likely confused by the fact that a set of isolated points can have positive measure,

A set of aleph_0 isolated points cannot have positive measure.

What is isolating more points in your opinion?
There are only aleph_0 endpoints.

And these are irrational.

> but that’s how it works.

Try to answer the questions and find out how it works.

Regards, WM

[WM]

Am Sonntag, 16. Dezember 2018 00:35:05 UTC+1 schrieb jrenne...@googlemail.com:


>
> Stupidity is a lack of intelligence, understanding, reason, wit, or common sense.

Obviously you have first-hand information from your own brain. But instead of trying to comprehend my arguments you can simply apply logic. A machine could hepl you.

If the real axis contains ℵ_0 finite closed intervals

I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)]

with irrational endpoints and of total measure < 1/3 then the complement contains at most ℵ_0 finite intervals of infinite total measure. This complement contains at least one irrational number x such that for X = R\Q

∀x ∈ X ∀q ∈ Q ∃y ∈ X\{x}: |x - y| < |x - q| (1)

but

~∀x ∈ X ∀y ∈ X\{x} ∃q ∈ Q: |x - q| < |x - y| (2)

The failure of (2) shows a contradiction of set theory and mathematics.

Regards, WM

[jrenne...@googlemail.com]

On Sunday, December 16, 2018 at 11:30:14 AM UTC+1, WM wrote:
> Am Sonntag, 16. Dezember 2018 00:35:05 UTC+1 schrieb jrenne...@googlemail.com:
>
>
> >
> > Stupidity is a lack of intelligence, understanding, reason, wit, or common sense.
>
> Obviously you have first-hand information from your own brain. But instead of trying to comprehend my arguments you can simply apply logic. A machine could hepl you.
>
> If the real axis contains ℵ_0 finite closed intervals
>
> I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)]
>
> with irrational endpoints and of total measure < 1/3 then the complement contains at most ℵ_0 finite intervals of infinite total measure.

One more time: The complement contains no non-degenerate interval and
Lebesgue measure is countably additive; therefore the measure of the complement is infinite.

> This complement contains at least one irrational number x such that for X = R\Q
>
> ∀x ∈ X ∀q ∈ Q ∃y ∈ X\{x}: |x - y| < |x - q| (1)

"It contains at least one x such that for all x in X ..."? That makes no sense. But yes, between any two distinct irrationals there is a rational number. Nobody has doubts it.

>
> but
>
> ~∀x ∈ X ∀y ∈ X\{x} ∃q ∈ Q: |x - q| < |x - y| (2)

Can't you write an unambiguous formula? What's the scope of the negation?
Apparently you would like to claim that there are distinct irrationals x and y in the complement of the covering that are not separated by a rational q.
The only argument that you have given to support this claim is your erroneous belief that the complement of the covering must contain intervals. It doesn't. There are no intervals in the complement of the rationals.

And, obviously, the Lebesgue measure of [0,1]/Q is 1.

>
> The failure of (2) shows a contradiction of set theory and mathematics.
>

Poppycock.

[jrenne...@googlemail.com]

You had better look up what "isolated" means in this context. The misuse of technical terms not only exposes your ignorance but also makes the gibberish that you write even harder to understand than it would otherwise be.

[WM]

Am Sonntag, 16. Dezember 2018 13:56:29 UTC+1 schrieb jrenne...@googlemail.com:


> You had better look up what "isolated" means in this context.

It is completely irrelevant what "isolated" means in this or other contexts. Relevant is only that

∀x ∈ X ∀y ∈ X\{x} ∃q ∈ Q: |x - q| < |x - y|

is violated in this context - if all rational numbers can be covered by intervals of masure < 1/3.

Regards, WM

[WM]

Am Sonntag, 16. Dezember 2018 13:51:19 UTC+1 schrieb jrenne...@googlemail.com:
> On Sunday, December 16, 2018 at 11:30:14 AM UTC+1, WM wrote:


> > I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)]
> >
> > with irrational endpoints and of total measure < 1/3 then the complement contains at most ℵ_0 finite intervals of infinite total measure.
>
> One more time: The complement contains no non-degenerate interval and
> Lebesgue measure is countably additive; therefore the measure of the complement is infinite.

One more time: The complement cannot even contain degenerate intervals since then

∀x ∈ X ∀y ∈ X\{x} ∃q ∈ Q: |x - q| < |x - y|

would be violated.

>
> > This complement contains at least one irrational number x such that for X = R\Q
> >
> > ∀x ∈ X ∀q ∈ Q ∃y ∈ X\{x}: |x - y| < |x - q| (1)
>
> "It contains at least one x such that for all x in X ..."?

∀x ∈ X ∀y ∈ X\{x} ∃q ∈ Q: |x - q| < |x - y|

> That makes no sense.

X = R\Q is too hard to understand fpor you?

> But yes, between any two distinct irrationals there is a rational number. Nobody has doubts it.

Not in the complement.

> >
> > but
> >
> > ~∀x ∈ X ∀y ∈ X\{x} ∃q ∈ Q: |x - q| < |x - y| (2)
>
> Can't you write an unambiguous formula?

(2) is unambiguous.

> Apparently you would like to claim that there are distinct irrationals x and y in the complement of the covering that are not separated by a rational q.
> The only argument that you have given to support this claim is your erroneous belief that the complement of the covering must contain intervals.

No. One irrational x in the complement is enough - it is one too much.

Regards, WM

[Ben Bacarisse]

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

> X = R\Q is too hard to understand fpor you?

Remember: it's too hard for /you/. At least you were never able to
define set difference. I asked (repeatedly) for your definition of \
because you asserted that in WMaths there is a set S with e ∈ S but with
S \ {e} = S. You were not able to give a definition of \ to explain
this "fact" of WMaths. How can anything you say be taken seriously if
you can't even define set difference?

Reference:
Message-ID: <672edadc-6c24-44b8...@googlegroups.com>
Date: Wed, 23 May 2018 08:02:46 -0700 (PDT)

<snip>
--
Ben.

[jrenne...@googlemail.com]

Are you really unable to understand the construction of Cantor's ternary set and its properties?
Or the proof that Cantor gave for the fact that a nowhere dense perfect set must be uncountable?
Or Volterra's function?
Or Weierstrass' example of a continuous nowhere differentiable function?
Or this standard example of a nowhere dense set with nonzero measure: https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set?

Then why don't you develop a hobby more in line with your intellectual capacity?

[jrenne...@googlemail.com]

On Sunday, December 16, 2018 at 3:10:27 PM UTC+1, WM wrote:
> Am Sonntag, 16. Dezember 2018 13:56:29 UTC+1 schrieb jrenne...@googlemail.com:
>
>
> > You had better look up what "isolated" means in this context.
>
> It is completely irrelevant what "isolated" means in this or other contexts.

You used a technical term without understanding what it means. And now you tell us that the meaning is irrelevant. Apparently that serves your purpose.
Would you tell us, please, what this purpose is?

[WM]

Am Sonntag, 16. Dezember 2018 16:28:32 UTC+1 schrieb Ben Bacarisse:

> > X = R\Q is too hard to understand for you?

>
> I asked (repeatedly) for your definition of \

If you are unable to understand it, simply learn that X is the set of irrational numbers.

Further off-topic diversion deleted.

Regards, WM

[j4n bur53]

Fantastic gibberish.

Am Donnerstag, 22. November 2018 11:59:22 UTC+1 schrieb transf...@gmail.com:
> Consider the covering of all rational numbers q_n, of the positive axis by intervals
>
> I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)] .
>
> Then all intervals cover (2/9)sqrt(2) or less of the positive axis. The complement has unknown measure. The structure of the complement is not known and is not of interest. There are only two important facts:
>
> _ There is no rational number in the complement because every rational is in an interval.
> _ All endpoints of the complement are irrational.
>
> Since there cannot exist two irrational numbers without a rational number between them, there is no irrational number in the complement either. The remainder of the positive axis of infinite length is empty.
>
> For more of this Kind see https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf
>
> Regards, WM

[WM]

Am Sonntag, 16. Dezember 2018 16:57:47 UTC+1 schrieb jrenne...@googlemail.com:
> On Sunday, December 16, 2018 at 3:16:56 PM UTC+1, WM wrote:


> > > > ~∀x ∈ X ∀y ∈ X\{x} ∃q ∈ Q: |x - q| < |x - y| (2)
> > >

> > > Can't you write an unambiguous formula? (*)

> >
> > (2) is unambiguous.
> >
> > > Apparently you would like to claim that there are distinct irrationals x and y in the complement of the covering that are not separated by a rational q.
> > > The only argument that you have given to support this claim is your erroneous belief that the complement of the covering must contain intervals.

(**) You have not understood. Isolated points are excluded too.

> >
> > No. One irrational x in the complement is enough - it is one too much.

> Are you really unable to understand the construction of Cantor's ternary set and its properties?

Here we show that Cantor's theory incompatible with mathematics. Therefore his set and its properties are irrelevant. But it would fail as a counter example anyhow because the intervals used here

I_n = [q_n - sqrt(2)*10^(-n), q_n + sqrt(2)*10^(-n)]

are not limits but are finite and have irrational endpoints. (***)

> Or the proof that Cantor gave for the fact that a nowhere dense perfect set must be uncountable?
> Or Volterra's function?
> Or Weierstrass' example of a continuous nowhere differentiable function?

Stop name-dropping. Here we see by (*), (**), (***) that you have no clue.

Regards, WM

[Ben Bacarisse]

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

> Am Sonntag, 16. Dezember 2018 16:28:32 UTC+1 schrieb Ben Bacarisse:
>
>> > X = R\Q is too hard to understand for you?
>>
>> I asked (repeatedly) for your definition of \
>
> If you are unable to understand it, simply learn that X is the set of
> irrational numbers.

So you still won't define what you mean by set difference? I am not
surprised.

> Further off-topic diversion deleted.

The fact that your symbol \ is as yet undefined and has the property
that e ∈ S and S \ {e} = S is hardly off-topic if you challenge readers
about it. Yes, it's embarrassing for you, but not off-topic. If you
write \ the reader is entitled to know what you mean, or as here, that
you don't know what you mean.

(A bit of background: WM has written a textbook that contains the usual
definition of \, but that definition does not give the magic result that
S \ {e} = S despite e ∈ S. Thus he had to pretend that that definition
does not apply in all cases but he would or could not say what \ means
in general.)

If you are in any doubt, here's the post in question:

Message-ID: <672edadc-6c24-44b8...@googlegroups.com>
Date: Wed, 23 May 2018 08:02:46 -0700 (PDT)

--
Ben.