Greg Cunt schrieb am Dienstag, 24. August 2021 um 23:17:13 UTC+2:
> On Tuesday, August 24, 2021 at 10:25:29 PM UTC+2, WM wrote:
> > FredJeffries schrieb am Dienstag, 24. August 2021 um 18:25:07 UTC+2:
> > > On Monday, August 23, 2021 at 12:33:01 PM UTC-7, WM wrote:
> > > >
> > > > In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
> > >
> > > Prove it
> > >
> > ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .
> This is not a proof, but a spoof! :-)

Why? If q is any rational number, then q + n is another rational number in the distance n. Simple mathematics, Volksschule 1. Klasse.

And if not, are you ready foolish enough to doubt this clear fact in order to continue believing in Cantor's nonsense?

Regards, WM

Greg Cunt schrieb am Dienstag, 24. August 2021 um 23:20:11 UTC+2:
> On Tuesday, August 24, 2021 at 10:43:27 PM UTC+2, WM wrote:
>
> > There are the first thre steps: 1, 2, 3. There are even more. If they cease then <bla bla>
>
> THE DON'T CEASE,
>
> There is no largest natural number.

Simple minds may think so. But when considering the unit fractions, then you should be able to see that somewhe zero is reached. Then the smallest unit fraction has been passed.

Regards, WM

On Wednesday, August 25, 2021 at 9:15:57 AM UTC-3, WM wrote:

> ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .

This shows that the cardinality of the set of rationals in (0,1] is the same as the cardinality of the rationals in (n,n+1]. Don't stop the presses.
However cardinality is not density. "Density" is nonsense when dealing with infinite cardinalities.

--
William Hughes

William schrieb am Mittwoch, 25. August 2021 um 01:56:46 UTC+2:
> On Tuesday, August 24, 2021 at 4:51:46 PM UTC-4, WM wrote:
> > William schrieb am Dienstag, 24. August 2021 um 17:20:26 UTC+2:
> > > On Monday, August 23, 2021 at 2:36:31 PM UTC-4, WM wrote:
> >
> > > > Here is an incomplete mapping: In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
> > > Nope. Let the mapping be f from |N_F to the rationals. Then if r is a rational there exists an element of |N_F, x(r), such that f(x(r))= r. It is nonsense to say the mapping is "incomplete". There is no rational that is not mapped to.
> > Mathematics tells the contrary.
> Nope, mathematics says there is no rational that is not mapped to.

That is not mathematics but matheology.
>
> >Half...
>
> "Half" has no meaning when applied to infinite cardinalities .

Here it is applied to mathematics. https://math.stackexchange.com/questions/3708845/relative-abundance-of-rationals-in-cantors-bijection
https://hsm.stackexchange.com/questions/11938/has-cantors-irregular-enumeration-of-rationals-ever-been-discussed

> The cardinality of a set is not the number of elements in the set.

Neveretheless Matheologians claim that Cantor indiexes all fractions. It is provable however that he indexes less fractions in every intervall than in the first unit interval. Of course what he indexes can be put in bijection, but it is clearly not complete.

Can you see that half of all fractions are 1 or less than 1 here?
1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...

> "Number of elements" has no meaning for a set with infinite cardinality like the cardinality of the rationals in an interval, or the cardinality of any Peano set like the set of FISONs.

It has meaning in mathematics:
∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1]

> >There is no *definable* rational that is not mapped to.
> Correct there is no rational that you can write down that is not mapped to.

Right.

> And there is no rational that you cannot write down that is not mapped to..

What about the 99.9 % missing in interval (1000, 1001]?
What about the first unit fraction when starting from 0?

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 25. August 2021 um 07:07:51 UTC+2:
> tisdag 24 augusti 2021 kl. 22:57:38 UTC+2 skrev WM:

> ALL rational are indiced. It is trivial to show there is a bijection betwee N and Q+

It is trivial to show the contrary:
∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1]
1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...

Regards, WM

William schrieb am Mittwoch, 25. August 2021 um 14:25:34 UTC+2:
> On Wednesday, August 25, 2021 at 9:15:57 AM UTC-3, WM wrote:
>
> > ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .
> This shows that the cardinality of the set of rationals in (0,1] is the same as the cardinality of the rationals in (n,n+1]. Don't stop the presses.

No, this is a true bijection, not a fake bijection a la Cantor. It shows precisely the same number. Identity.

> However cardinality is not density. "Density" is nonsense when dealing with infinite cardinalities.

Density is indicating the number of rationals. You shoul have learnt this in basic school: If q is a rational number, then q + 1 is a rational number to. And vice versa. It holds even for real numbers and complex numbers too.

Regards, WM

Jim Burns schrieb am Mittwoch, 25. August 2021 um 07:08:59 UTC+2:

> Another claim pretending infinite sets are finite:
> Half of all indexes are issued to the first
> unit interval.

Provable by 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... for the infinite sequence. Or see the areas in Cantor's first diagonal argument.

Regards, WM

On Wednesday, August 25, 2021 at 9:32:40 AM UTC-3, WM wrote:
> William schrieb am Mittwoch, 25. August 2021 um 14:25:34 UTC+2:
> > On Wednesday, August 25, 2021 at 9:15:57 AM UTC-3, WM wrote:
> >
> > > ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .
> > This shows that the cardinality of the set of rationals in (0,1] is the same as the cardinality of the rationals in (n,n+1]. Don't stop the presses.
> No, this is a true bijection,

Yes, you do not use a step by step method and you do not get nonsense. You show that the two sets have the same cardinality.

>not a fake bijection a la Cantor. It shows precisely the same number. Identity.
> > However cardinality is not density. "Density" is nonsense when dealing with infinite cardinalities.
> Density is indicating the number of rationals.

Cardinality is not "the number of". If you pretend it is you get nonsense.

--
WilliamHughes

On Wednesday, August 25, 2021 at 9:26:53 AM UTC-3, WM wrote:
> William schrieb am Mittwoch, 25. August 2021 um 01:56:46 UTC+2:

> > ... there is no rational that you cannot write down that is not mapped to.
>
> What about the 99.9 % missing in interval (1000, 1001]?

Percentages applied to infinite cardinalities give nonsense. There is no rational in (1000,1001] that is not mapped to.

> What about the first unit fraction when starting from 0?
>

This does not exist.

--
William Hughes

On 8/25/2021 7:18 AM, WM wrote:
> Greg Cunt schrieb am Dienstag, 24. August 2021 um 23:20:11 UTC+2:
>> On Tuesday, August 24, 2021 at 10:43:27 PM UTC+2, WM wrote:
>>
>>> There are the first thre steps: 1, 2, 3. There are even more. If they cease then <bla bla>
>>
>> THE DON'T CEASE,
>>
>> There is no largest natural number.
>
> Simple minds may think so. But when considering the unit fractions, then you should be able to see that somewhe zero is reached. Then the smallest unit fraction has been passed.
>
> Regards, WM
>

Wrong again. you failed the test, *you get an F*. Is WM always wrong ?

If you want to pass, study the following, from MIT, UIC, and Berkeley

The Limit of a Sequence
http://www-math.mit.edu/~apm/ch03.pdf

Limits of Sequences
http://homepages.math.uic.edu/~saunders/MATH313/INRA/INRA_Chapter2.pdf

Limits of Sequences
https://math.berkeley.edu/~willij/1b/limit-of-a-sequence.pdf

On 8/25/2021 7:26 AM, WM wrote:
> William schrieb am Mittwoch, 25. August 2021 um 01:56:46 UTC+2:
>> On Tuesday, August 24, 2021 at 4:51:46 PM UTC-4, WM wrote:
>>> William schrieb am Dienstag, 24. August 2021 um 17:20:26 UTC+2:
>>>> On Monday, August 23, 2021 at 2:36:31 PM UTC-4, WM wrote:
>>>
>>>>> Here is an incomplete mapping: In the interval (1000, 1001] the density of rationals is precisely the same as in the interval (0, 1].
>>>> Nope. Let the mapping be f from |N_F to the rationals. Then if r is a rational there exists an element of |N_F, x(r), such that f(x(r))= r. It is nonsense to say the mapping is "incomplete". There is no rational that is not mapped to.
>>> Mathematics tells the contrary.
>> Nope, mathematics says there is no rational that is not mapped to.
>
> That is not mathematics but matheology.

Wrong.

>>
>>> Half...
>>
>> "Half" has no meaning when applied to infinite cardinalities .
>
> Here it is applied to mathematics. https://math.stackexchange.com/questions/3708845/relative-abundance-of-rationals-in-cantors-bijection
> https://hsm.stackexchange.com/questions/11938/has-cantors-irregular-enumeration-of-rationals-ever-been-discussed

wrong and misleading

>
>> The cardinality of a set is not the number of elements in the set.
>
> Neveretheless Matheologians claim that Cantor indiexes all fractions. It is provable however that he indexes less fractions in every intervall than in the first unit interval. Of course what he indexes can be put in bijection, but it is clearly not complete.

two wrongs.

>
> Can you see that half of all fractions are 1 or less than 1 here?
> 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...

wrong. that is not proof, its is spoof.

>
>> "Number of elements" has no meaning for a set with infinite cardinality like the cardinality of the rationals in an interval, or the cardinality of any Peano set like the set of FISONs.
>

>
>>> There is no *definable* rational that is not mapped to.
>> Correct there is no rational that you can write down that is not mapped to.
>
> Right.
>
>> And there is no rational that you cannot write down that is not mapped to.
>
> What about the 99.9 % missing in interval (1000, 1001]?

wrong there are no missing

> What about the first unit fraction when starting from 0?

wrong, there are no missing

>
> Regards, WM
>
>

Seven wrong in one post. WM is always wrong.

On 8/25/2021 7:35 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 25. August 2021 um 07:08:59 UTC+2:
>
>> Another claim pretending infinite sets are finite:
>> Half of all indexes are issued to the first
>> unit interval.
>
> Provable by 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... for the infinite sequence. Or see the areas in Cantor's first diagonal argument.
>
> Regards, WM
>

Obviously Wrong. You like to fool little students. No wonder you suck at infinite sequences.

On 8/25/2021 8:35 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 25. August 2021 um 07:08:59 UTC+2:

>> Another claim pretending infinite sets are finite:
>> Half of all indexes are issued to the first
>> unit interval.
>
> Provable by
> 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1,
> 1/5, 5/1, 1/6, ... for the infinite sequence.
> Or see the areas in Cantor's first diagonal argument.

How do you count the sheep in a pasture?

Match an unmatched sheep to the next unmatched natural
number. Halt when there are _no more unmatched sheep_

What if, for each sheep, there is another sheep
still out in the pasture? Then the process does not halt.
Then counting the sheep this way fails.

We cannot count through the natural numbers until
there are no more natural numbers.
For each steppable 0,...,k, there is k+1 not in it.
There is always another sheep still out in the pasture.

This has nothing to do with dark numbers.
The k with steppable 0,...,k is "definable" and
k+1 also has steppable 0,...,l+1 and is definable.
There is always another _definable_ sheep still
out in the pasture.

> Provable by
> 1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1,
> 1/5, 5/1, 1/6, ... for the infinite sequence.

Your unstated argument is that, all along the infinite
sequence, half of the indexed fractions are in (0,1)
Therefore _when there are no more_ to be indexed,
half of them will be in (0,1).

Your unstated assumption is that, at some point,
_there are no more_ to be indexed. That can't happen.
There is always another _definable_ sheep still
out in the pasture.

What happens is that the match-until-no-unmatched
process fails in certain situations.

However, we don't need to match-until-no-unmatched
if all we do is match sheep with ribbons, for example.
Suppose you and your friends go through a pasture
tying pretty satin ribbons around the necks of
sheep. At some point in the future, each sheep has
one ribbon around its neck, and only one.
There is no last sheep required in this procedure.
You and your friends might not know how many sheep
got ribbons, in the match-until-no-unmatched sense.
But you know the sheep and the ribbons match.

If things worked the way you would like them to,
this sequence
1, 2, 3, 4, 5, 6, 7, ...
has twice as many elements as this sequence
2, 4, 6, 8, 10, 12, 14, ...

If we wrote the first sequence in binary
1, 10, 11, 100, 101, 110, 111, ...
and appended 0 on the right,
we have the second sequence in binary
10, 100, 110, 1000, 1010, 1100, 1110, ...

According to the way you would like things to work,
appending 0 on the right removes half the elements.

That's wrong. What's wrong about it is the use of
"half the elements" when we do not count until
there are no more.

On 8/23/2021 8:08 AM, WM wrote:
> Jim Burns schrieb
> am Montag, 23. August 2021 um 06:55:33 UTC+2:

>> Perhaps you accept different assumptions.
>> | A finite index is a finite index.
>> | A positive rational is a positive rational.
>> | Preserving truth preserves truth.
>>
>> Do you (WM) accept these assumptions?
>
> Of course. Including the results.
> Do you understand and accept quantifier logic?

Yes.

x is a _variable_ name. It refers to an _individual_

In a given context, there are individuals we regard as
_possible_ referents of x. These individuals are
contained in the _domain_ which x ranges over.
Any individual which we regard as NOT a possible
referent is NOT in the domain. And vice versa.

For an individual x in the domain , a _predicate_ P(x)
has one unchanging truth value true xor false. However,
for different individuals, P(x) may have different
truth values.

Consider some collection of claims about the individual
to which x refers. I'll call these claims P(x).

Consider these four statements:

| It's possible for x to refer to an individual
| such that P(x) is true.

| It's NOT possible for x to refer to an individual
| such that P(x) is true.

| It's possible for x to refer to an individual
| such that ~P(x) is true.

| It's NOT possible for x to refer to an individual
| such that ~P(x) is true.

We don't need to know what individuals x possibly
refers to and we don't need to know what P(x) claims
in order to know that two of them are true and
two of them are false.

If it's possible for x to refer to an individual
such that P(x) is true, then some individual *exists*
in the domain for which, if x refers to it,
P(x) is true. We call this existential quantification,
and we write it exists x, P(x) or Ex, P(x).

If it's NOT possible for x to refer to an individual
such that P(x) is true, then there is *no* individual
in the domain such that, if x refers to it,
P(x) is true. We can write this not exists x, P(x)
or ~Ex, P(x).

For each individual in the domain, if x refers to it,
P(x) is true xor false. So, if there are no
individuals x in the domain such that P(x), then
for all the individuals in the domain, ~P(x).
We call this universal quantification, and we
write it forall x, ~P(x) or Ax, P(x).

Therefore,
not exists x, P(x) iff forall x, ~P(x)

as you note.

> ~∃n ∈ ℕ: P(n) <==> ∀n ∈ ℕ: ~P(n)
>
> Please apply this to
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

∀n ∈ ℕ_def: | ℕ_def\{1,2,3,...,n} | = ℵo.

because
∀n ∈ ℕ_def: n+1 ∈ ℕ_def\{1,2,3,...,n}

The natural numbers can be stepped through one-by-one
either forward or backward *from any natural number*
*to any other natural number*

If there were also a first and a last natural number,
then there would be finitely-many natural numbers.
However, a last natural number would be a bug, a
potential cause of the sheep-counting process to
halt incorrectly. There is no last natural number.

> In the interval (1000, 1001] the density of rationals
> is precisely the same as in the interval (0, 1].
> Alas Cantor enumerates less than one promille.

Cantor enumerates them all.

> When going from 1 to 0, then there is no unit fraction
> more later. Either, forced by the linearity of the
> sequence 1, 1/2, 1/3, ..., 0, a last unit fraction
> has been passed. Or there are dark numbers.

If there were a last unit fraction,
passing all the unit fractions would entail
passing the last unit fraction.

It does not work the other way around.
Passing all the unit fractions *does not entail*
passing a last unit fraction.

That is a fallacy, affirming the consequent.
https://en.wikipedia.org/wiki/Affirming_the_consequent

For example
| If an animal is a dog, then it has four legs.
| My cat has four legs.
| Therefore, my cat is a dog.

Apart form that being a fallacy, we know, separately,
that there is no last unit fraction.
For each 1/k, 1/k is not last, there is 1/(k+1) after it.

William schrieb am Mittwoch, 25. August 2021 um 14:38:50 UTC+2:
> On Wednesday, August 25, 2021 at 9:32:40 AM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 25. August 2021 um 14:25:34 UTC+2:
> > > On Wednesday, August 25, 2021 at 9:15:57 AM UTC-3, WM wrote:
> > >
> > > > ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .
> > > This shows that the cardinality of the set of rationals in (0,1] is the same as the cardinality of the rationals in (n,n+1]. Don't stop the presses.
> > No, this is a true bijection,
> Yes, you do not use a step by step method and you do not get nonsense. You show that the two sets have the same cardinality.

More. Same cardinality is also true for very different sets. Here I show exatly the same number of rational points by translation invariance.

> >not a fake bijection a la Cantor. It shows precisely the same number. Identity.
> > > However cardinality is not density. "Density" is nonsense when dealing with infinite cardinalities.
> > Density is indicating the number of rationals.
> Cardinality is not "the number of".

Correct!

> If you pretend it is you get nonsense.

I do not. But what I can say is that the number of rational points in every interval is exactly the same:

∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1]
Can you see this? Cantor however gets a different result by

1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ... .

About half of all rationals are indexed in (0, 1]. This does not contradict his cardinality. Of course the cardinality of rationals in every interval is aleph_0. But it contradicts the claimed completeness.

Regards, WM

William schrieb am Mittwoch, 25. August 2021 um 14:54:01 UTC+2:
> On Wednesday, August 25, 2021 at 9:26:53 AM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 25. August 2021 um 01:56:46 UTC+2:
> > > ... there is no rational that you cannot write down that is not mapped to.
> >
> > What about the 99.9 % missing in interval (1000, 1001]?
> Percentages applied to infinite cardinalities give nonsense.

Percetage applied to density gives mathematical result: https://mathoverflow.net/questions/362791/what-fraction-of-fractions-does-cantors-famous-sequence-enumerate

> There is no rational in (1000,1001] that is not mapped to.

But there are less mapped in (1000,1001] than in (0,1]. That is not doubted. What went wrong?

> > What about the first unit fraction when starting from 0?
> >
> This does not exist.

What about the first that exists?

Regards, WM

Jim Burns schrieb am Mittwoch, 25. August 2021 um 20:30:42 UTC+2:

> How do you count the sheep in a pasture?

I count so: ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1].
Cantor fails.

Regards, WM

On Wednesday, August 25, 2021 at 10:39:21 PM UTC+2, WM wrote:

> ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1]
>
> Can you see this?

Yeah, idiotic nonsense.

It's getting worse with you, Mückenheim.

On Wednesday, August 25, 2021 at 10:42:20 PM UTC+2, WM wrote:
> William schrieb am Mittwoch, 25. August 2021 um 14:54:01 UTC+2:
> > On Wednesday, August 25, 2021 at 9:26:53 AM UTC-3, WM wrote:
> > >
> > > What about the first unit fraction when starting from 0?
> > >
> > This does not exist.
> >
> What about the first that exists?

Holy shit! You are beyond help, man.

On 8/25/2021 4:44 PM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 25. August 2021 um 20:30:42 UTC+2:

>> How do you count the sheep in a pasture?
>
> I count so:
> ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und
> ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1].
> Cantor fails.

∀ q ∈ Q, ∃! n ∈ N: n = index(q) und
∀ n ∈ N, ∃! q ∈ Q: index(q) = n.

In particular, for r/s = q in lowest terms, let
index(r/s) = s*s*r*r/rad(r)

Cantor succeeds.

Crank Wolfgang Mueckenheim, aka WM wrote:
> Jim Burns schrieb am Mittwoch, 25. August 2021 um 20:30:42 UTC+2:
>
>> How do you count the sheep in a pasture?
>
> I count so: ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und [sic] ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1].

This does make much sense, Crank Wolfgang Mueckenheim, from Hochschule
Augsburg. Especially the "∀ q+n" and "∃ q+n" are logically unsound. What
would make some sense is:

∀ q ∈ (0, 1] ∃ q' ∈ (n, n+1] and ∀ q ∈ (n, n+1] ∃ q'' ∈ (0, 1]
by considering q' = q+n and q'' = q-n.

Anyway it does not at all implies that:

> Cantor fails.

But that you (again) failed, Crank Wolfgang Mueckenheim, from Hochschule
Augsburg.

On Wednesday, August 25, 2021 at 5:39:21 PM UTC-3, WM wrote:
> William schrieb am Mittwoch, 25. August 2021 um 14:38:50 UTC+2:
> > On Wednesday, August 25, 2021 at 9:32:40 AM UTC-3, WM wrote:
> > > William schrieb am Mittwoch, 25. August 2021 um 14:25:34 UTC+2:
> > > > On Wednesday, August 25, 2021 at 9:15:57 AM UTC-3, WM wrote:
> > > >
> > > > > ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] and ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1] .
> > > > This shows that the cardinality of the set of rationals in (0,1] is the same as the cardinality of the rationals in (n,n+1]. Don't stop the presses.
> > > No, this is a true bijection,
> > Yes, you do not use a step by step method and you do not get nonsense. You show that the two sets have the same cardinality.
> More. Same cardinality is also true for very different sets. Here I show exatly the same number of rational points by translation invariance.

"Number of points" is nonsense when you try to apply it to sets with infinite cardinality.

> > >not a fake bijection a la Cantor. It shows precisely the same number. Identity.
> > > > However cardinality is not density. "Density" is nonsense when dealing with infinite cardinalities.
> > > Density is indicating the number of rationals.
> > Cardinality is not "the number of".
> Correct!
> > If you pretend it is you get nonsense.
> I do not.

He said and then said "the number of rational points" which is of course nonsense.

On Wednesday, August 25, 2021 at 5:42:20 PM UTC-3, WM wrote:
> William schrieb am Mittwoch, 25. August 2021 um 14:54:01 UTC+2:
> > On Wednesday, August 25, 2021 at 9:26:53 AM UTC-3, WM wrote:
> > > William schrieb am Mittwoch, 25. August 2021 um 01:56:46 UTC+2:
> > > > ... there is no rational that you cannot write down that is not mapped to.
> > >
> > > What about the 99.9 % missing in interval (1000, 1001]?
> > Percentages applied to infinite cardinalities give nonsense.
> Percetage applied to density gives mathematical result: https://mathoverflow.net/questions/362791/what-fraction-of-fractions-does-cantors-famous-sequence-enumerate
> > There is no rational in (1000,1001] that is not mapped to.
> But there are less mapped in (1000,1001] than in (0,1].

Nope; there are aleph_0 mapped in (0,1] and aleph_0 mapped in (1000,1001].

--
William Hughes

On Wednesday, August 25, 2021 at 11:51:47 PM UTC+2, Jim Burns wrote:

> Cantor succeeds.

WM fails (as usual).

Metatheorem: For any statement A: if WM claims A, then A is either false or "not even wrong".

A rather safe bet.

On Thursday, August 26, 2021 at 12:08:57 AM UTC+2, Python wrote:
> Crank Wolfgang Mueckenheim, aka WM wrote:
> >
> > [bla bla:] ∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1].
> >
> This does make much sense, Crank Wolfgang Mueckenheim, from Hochschule
> Augsburg. Especially the "∀ q+n" and "∃ q+n" are logically unsound.

Right ("unsyntactical").
> ∀ q ∈ (0, 1] ∃ q' ∈ (n, n+1] and ∀ q ∈ (n, n+1] ∃ q'' ∈ (0, 1]
> by considering q' = q + n and q'' = q - n.

Actually, what WM TRIED to claim is even (syntactically) simpler:

∀q ∈ (0, 1]: q+n ∈ (n, n+1]

and

∀q ∈ (n, n+1]: q-n ∈ (0, 1].

Ha ha ha!

Incredible insights!

@psychotic asshole full of shit, Mückenheim:

Yeah, the function f: (0, 1] --> (n, n+1] defined with f(x) = x + n (for all x e (0, 1]) is a bijection from (0, 1] onto (n, n+1].

In other words, (0, 1] ~ (n, n+1]. HENCE card((0, 1]) = card((n, n+1]) // since card(A) = card(B) iff A ~ B (for any sets A, B).

And right, f^-1(x) = x - n, in this case.