On Thursday, August 26, 2021 at 3:21:03 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. August 2021 um 17:26:38 UTC+2:
> > On Thursday, August 26, 2021 at 9:50:32 AM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 26. August 2021 um 01:12:57 UTC+2:
> > >
> > > > "Number of points" is nonsense when you try to apply it to sets with infinite cardinality.
> > > Cardinality is nonsense.
> > Nope, As you note there is a bijection between the rationals in (0,1] and the rationals in (1000,1001]. So the two sets have the same cardinality. It is nonsense to say that the range of the bijection has "more points] in (0,1] than in (1000,1001].
> My bijection f(q) = q + 1000 shows same number

"same number" is nonsense when talking about infinite cardinalities.

--
William Hughes

On Thursday, August 26, 2021 at 3:23:10 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. August 2021 um 20:18:53 UTC+2:
> > On Thursday, August 26, 2021 at 3:12:41 PM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 26. August 2021 um 17:34:12 UTC+2:
> > > > On Thursday, August 26, 2021 at 10:02:15 AM UTC-3, WM wrote:
> > > > > William schrieb am Mittwoch, 25. August 2021 um 14:54:01 UTC+2:
> > > > s
> > > > > > > What about the first unit fraction when starting from 0?
> > > > > > >
> > > > > > This does not exist.
> > > > > >
> > > > > What about the first existing unit fraction? Or are there no unit fractions existing?
> > > > There are an infinite number of existing elements of U. However, no element of U has the property that is it first.
> > > How is that possible?
> > Simple consequence of the fact that no element of the set |N_F has the property that it is last.
> For the definable elements this is true.

It is also true for the elements that you can't write down.

--
William Hughes

On Thursday, August 26, 2021 at 3:25:38 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. August 2021 um 17:17:13 UTC+2:
> > On Thursday, August 26, 2021 at 9:46:43 AM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 26. August 2021 um 01:20:41 UTC+2:
> > > > On Wednesday, August 25, 2021 at 5:42:20 PM UTC-3, WM wrote:
> > >
> > > > > 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].
> > > >
> > > Yes, there is a bijection between the [rationals] enumerated in (0, 1] and in (1000, 1001].
> >
> > Indeed, there is a bijection, that is a mapping that is both injective and surjective.
> Alas it is not between all fractions of these intervals.

Piffle. You just showed it and called it a bijection.

--
William Hughes

Jim Burns schrieb am Mittwoch, 25. August 2021 um 20:30:42 UTC+2:
> On 8/25/2021 8:35 AM, WM wrote:

> > 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).

There is no reason to claim "no more". It is sufficient to know that every second will be in (0, 1].
>
> Your unstated assumption is that, at some point,
> _there are no more_ to be indexed.

Not at all! Look at the matrix https://en.wikipedia.org/wiki/File:Diagonal_argument.svg It is infinite. But below the diagonal there re as many fractions as above.

> That can't happen.

Every zig or zag has precisely as many fractions below and above the diagonal. This can never chang. No "no more" required gto see it.

> 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, ...

Of course it has. This is proven for every finite initial segment.
>
> 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, ...

Binaries are not required. Simply multiply 1, 2, 3, 4, 5, 6, 7, ... by 2.
>
> 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.

That's not wrong but owed to undefinable numbers. It does not entitle you to believe that half of the infinite set of rationals is in (0, 1] when we use
1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...
or in (1000, 1001], when we use
1/1 + 1000, 1/2 + 1000, 2/1 + 1000, 1/3 + 1000, 3/1 + 1000, 1/4 + 1000, ...

Regards, WM

William schrieb am Donnerstag, 26. August 2021 um 20:41:14 UTC+2:
> On Thursday, August 26, 2021 at 3:25:38 PM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 26. August 2021 um 17:17:13 UTC+2:
> > > On Thursday, August 26, 2021 at 9:46:43 AM UTC-3, WM wrote:
> > > > William schrieb am Donnerstag, 26. August 2021 um 01:20:41 UTC+2:
> > > > > On Wednesday, August 25, 2021 at 5:42:20 PM UTC-3, WM wrote:
> > > >
> > > > > > 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].
> > > > >
> > > > Yes, there is a bijection between the [rationals] enumerated in (0, 1] and in (1000, 1001].
> > >
> > > Indeed, there is a bijection, that is a mapping that is both injective and surjective.
> > Alas it is not between all fractions of these intervals.
> Piffle. You just showed it and called it a bijection.
>
I showed it between all rationals of these intervals. Cantor uses less that 1/1000 of all rationals of the interval (1000, 1001]. https://mathoverflow.net/questions/362791/what-fraction-of-fractions-does-cantors-famous-sequence-enumerate

Regards, WM

William schrieb am Donnerstag, 26. August 2021 um 20:38:23 UTC+2:
> On Thursday, August 26, 2021 at 3:23:10 PM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 26. August 2021 um 20:18:53 UTC+2:
> > > On Thursday, August 26, 2021 at 3:12:41 PM UTC-3, WM wrote:
> > > > William schrieb am Donnerstag, 26. August 2021 um 17:34:12 UTC+2:
> > > > > On Thursday, August 26, 2021 at 10:02:15 AM UTC-3, WM wrote:
> > > > > > William schrieb am Mittwoch, 25. August 2021 um 14:54:01 UTC+2:
> > > > > s
> > > > > > > > What about the first unit fraction when starting from 0?
> > > > > > > >
> > > > > > > This does not exist.
> > > > > > >
> > > > > > What about the first existing unit fraction? Or are there no unit fractions existing?
> > > > > There are an infinite number of existing elements of U. However, no element of U has the property that is it first.
> > > > How is that possible?
> > > Simple consequence of the fact that no element of the set |N_F has the property that it is last.
> > For the definable elements this is true.
> It is also true for the elements that you can't write down.

Maybe. But then there is nothing next to zero. Then ther is a gap between zero and the smallest unit fraction that can be written down. Never this gap can be closed.

Regards, WM

William schrieb am Donnerstag, 26. August 2021 um 20:31:13 UTC+2:
> On Thursday, August 26, 2021 at 3:21:03 PM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 26. August 2021 um 17:26:38 UTC+2:
> > > On Thursday, August 26, 2021 at 9:50:32 AM UTC-3, WM wrote:
> > > > William schrieb am Donnerstag, 26. August 2021 um 01:12:57 UTC+2:
> > > >
> > > > > "Number of points" is nonsense when you try to apply it to sets with infinite cardinality.
> > > > Cardinality is nonsense.
> > > Nope, As you note there is a bijection between the rationals in (0,1] and the rationals in (1000,1001]. So the two sets have the same cardinality. It is nonsense to say that the range of the bijection has "more points] in (0,1] than in (1000,1001].
> > My bijection f(q) = q + 1000 shows same number
> "same number" is nonsense when talking about infinite cardinalities.

This is not cardinality but identity of number. It is simple to see by
∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1]
that identity of number bis justified. Also the matrix
https://en.wikipedia.org/wiki/File:Diagonal_argument.svg
proves identity of number.

By the way this is also claimed by Cantor but obviously he cannot satify this claim.

Regards, WM

On Thursday, August 26, 2021 at 3:46:09 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. August 2021 um 20:41:14 UTC+2:
> > On Thursday, August 26, 2021 at 3:25:38 PM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 26. August 2021 um 17:17:13 UTC+2:
> > > > On Thursday, August 26, 2021 at 9:46:43 AM UTC-3, WM wrote:
> > > > > William schrieb am Donnerstag, 26. August 2021 um 01:20:41 UTC+2:
> > > > > > On Wednesday, August 25, 2021 at 5:42:20 PM UTC-3, WM wrote:
> > > > >
> > > > > > > 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].
> > > > > >
> > > > > Yes, there is a bijection between the [rationals] enumerated in (0, 1] and in (1000, 1001].
> > > >
> > > > Indeed, there is a bijection, that is a mapping that is both injective and surjective.
> > > Alas it is not between all fractions of these intervals.
> > Piffle. You just showed it and called it a bijection.
> >
> I showed [a bijection] between all rationals of these intervals.

Correct. The bijecction is injective and surjective. Thus it is "between all fractions of these intevals"

--
William Hughes

On Thursday, August 26, 2021 at 3:52:44 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. August 2021 um 20:31:13 UTC+2:
> > On Thursday, August 26, 2021 at 3:21:03 PM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 26. August 2021 um 17:26:38 UTC+2:
> > > > On Thursday, August 26, 2021 at 9:50:32 AM UTC-3, WM wrote:
> > > > > William schrieb am Donnerstag, 26. August 2021 um 01:12:57 UTC+2:
> > > > >
> > > > > > "Number of points" is nonsense when you try to apply it to sets with infinite cardinality.
> > > > > Cardinality is nonsense.
> > > > Nope, As you note there is a bijection between the rationals in (0,1] and the rationals in (1000,1001]. So the two sets have the same cardinality. It is nonsense to say that the range of the bijection has "more points] in (0,1] than in (1000,1001].
> > > My bijection f(q) = q + 1000 shows same number
> > "same number" is nonsense when talking about infinite cardinalities.
> This is not cardinality but identity of number.

Saying "identity of number" instead of "same number" is still nonsense.

--
William Hughes

On Thursday, August 26, 2021 at 8:31:13 PM UTC+2, William wrote:
> On Thursday, August 26, 2021 at 3:21:03 PM UTC-3, WM wrote:
> >
> > My bijection f(q) = q + 1000 shows same number
> >
> "same number" is nonsense when talking about infinite cardinalities.

Well, not so.

Hint: Same CARDINALnumber.

For all A,B: If A ~ B, then card(A) = card(B) .

On Thursday, August 26, 2021 at 8:56:56 PM UTC+2, William wrote:
> On Thursday, August 26, 2021 at 3:52:44 PM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 26. August 2021 um 20:31:13 UTC+2:
> > > On Thursday, August 26, 2021 at 3:21:03 PM UTC-3, WM wrote:
> > > >
> > > > My bijection f(q) = q + 1000 shows same [cardinal --GC] number

Right!

> > > "same number" is nonsense when talking about infinite cardinalities.

Not so.

> > This is not cardinality but identity of number.

Oh, he just got that wrong.

Should read: This is about the identity of the cardinality numbers.

> Saying "identity of number" instead of "same number" is still nonsense.

Not so! :-)

Hint: A ~ B implies card(A) = card(B) für any sets A, B.

WM is making progress!

On 8/26/2021 8:54 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 25. August 2021 um 23:51:47 UTC+2:

>> Cantor succeeds.
>
> There is a bijection between the fractions enumerated in (0, 1]
> and in (1000, 1001].

What that means is that
a collection F of pairs (p,q) exists such that

for each rational p in interval (0,1],
one and only one rational q is in interval (1000,1001]
such that (p,q) is in F, and

for each rational q in interval (1000,1001],
one and only one p is in interval (0,1]
such that (p,q) is in F.

That's true.
There is the collection of pairs of the form (p, q)
such that p is a rational in interval (0, 1],
q is a rational in interval (1000, 1001],
and q = p+1000.

We can prove that's true from the properties of addition of
rationals.

(Other collections exist which meet the same conditions.)

> But Cantor does not enumerates all fractions.

Cantor enumerates all rationals.

What that means is that
a collection G of pairs (k,p) exists such that

for each natural k in N+,
one and only one rational p is in Q+
such that (k,p) is in G, and

for each rational p in Q+,
one and only one natural k is in N+
such that (k,p) is in G.

There is the collection of pairs of the form (k,p)
such that k is a natural in N+,
p is a rational in Q+,
and exists r,s in N+, r/s = p, gcd(r,s) = 1,
k = s*s*r*r/rad(r)

rad(r) == "the radical of r"
https://en.wikipedia.org/wiki/Radical_of_an_integer

You are rejecting a claim which, allowing for different
infinite sets being paired and different infinite sets of
pairs, means the same as the claim you accept.

> There are less enumerated in (1000, 1001] than in (0, 1]

Define the set Indexed[k] of the first k indexed rationals.
Indexed[k] = { p e Q+ | j e {1,...,k} & (j,p) e G }

Assume, for the sake of argument, that, for all k e N+,
there are 1000 times as many elements of Indexed[k] in (0,1]
as there are in (1000,1001].

That doesn't imply that there are fewer enumerated in
(1000,1001] than are enumerated in (0,1].
None of these Indexed[k] are Q+.

Compare the even naturals to all naturals the same way
Even[k] = { j e N+ | j e {1,...,k} & exists m, 2*m = j }

Up to any k, there are only half as many even numbers.
But _all_ the naturals can be mapped to a unique even.
This is equivalent to the inverse being surjective.

It's not even wrong to talk about "half of aleph_0".

> but there are not less in (1000, 1001].
> This excludes a bijection between |N and Q+
> by lack of surjectivity.

k = q*q*p*p/rad(p) is surjective onto Q+.

For each p'/q' in Q+,
there is a unique p/q = p'/q' in lowest terms.
For p/q, there is unique q*q*p*p/rad(p).
Define k = q*q*p*p/rad(p).
For each p'/q' in Q+, there is a unique index k.

I will point out for you that N+ is a proper subset of Q+,
that N+ at least _looks_ much smaller than Q+.
( There are finitely-many elements of N+ between two
( elements of N+
( There are infinitely-many elements of Q+ between two
( elements of Q+

*This is why* it's not even wrong to talk about
"half of aleph_0".

> Why is this so? Dark numbers are the reason.

Dark numbers have nothing to do with this.

For each element k of N+,
steppable 1,...,k with adjacent i, j = i+1 exists.

For each element p of Q+,
there are elements r,s in N+ such that r/s = p.

William schrieb am Donnerstag, 26. August 2021 um 20:56:56 UTC+2:
> On Thursday, August 26, 2021 at 3:52:44 PM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 26. August 2021 um 20:31:13 UTC+2:
> > > On Thursday, August 26, 2021 at 3:21:03 PM UTC-3, WM wrote:
> > > > William schrieb am Donnerstag, 26. August 2021 um 17:26:38 UTC+2:
> > > > > On Thursday, August 26, 2021 at 9:50:32 AM UTC-3, WM wrote:
> > > > > > William schrieb am Donnerstag, 26. August 2021 um 01:12:57 UTC+2:
> > > > > >
> > > > > > > "Number of points" is nonsense when you try to apply it to sets with infinite cardinality.
> > > > > > Cardinality is nonsense.
> > > > > Nope, As you note there is a bijection between the rationals in (0,1] and the rationals in (1000,1001]. So the two sets have the same cardinality. It is nonsense to say that the range of the bijection has "more points] in (0,1] than in (1000,1001].
> > > > My bijection f(q) = q + 1000 shows same number
> > > "same number" is nonsense when talking about infinite cardinalities.
> > This is not cardinality but identity of number.
> Saying "identity of number" instead of "same number" is still nonsense.

No, we can prove it using mathematics. For an intuitive argument I frequently pointed you to https://en.wikipedia.org/wiki/File:Diagonal_argument.svg an extended version of
1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...
You can find a strictly mathematical argument in https://math.stackexchange..com/questions/3708845/relative-abundance-of-rationals-in-cantors-bijection . On the other hand we have
∀ q ∈ (0, 1] ∃ q+n ∈ (n, n+1] und ∀ q+n ∈ (n, n+1] ∃ q ∈ (0, 1]

Only denying will not help. Of course for you personally it may help to avoid the insight that set theory is nonsense.

Regards, WM

Jim Burns schrieb am Donnerstag, 26. August 2021 um 21:34:24 UTC+2:
> On 8/26/2021 8:54 AM, WM wrote:

> There is the collection of pairs of the form (p, q)
> such that p is a rational in interval (0, 1],
> q is a rational in interval (1000, 1001],
> and q = p+1000.
>
> We can prove that's true from the properties of addition of
> rationals.
>
> > But Cantor does not enumerates all fractions.
> Cantor enumerates all rationals.

In (1000, 1001] he misses about 99.9 %. That is fact. Conjuration of a contradicted
> collection G of pairs (k,p)
cannot avoid this fact.

Regards, WM

On Thursday, August 26, 2021 at 4:37:47 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. August 2021 um 20:56:56 UTC+2:
> > On Thursday, August 26, 2021 at 3:52:44 PM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 26. August 2021 um 20:31:13 UTC+2:
> > > > On Thursday, August 26, 2021 at 3:21:03 PM UTC-3, WM wrote:
> > > > > William schrieb am Donnerstag, 26. August 2021 um 17:26:38 UTC+2:
> > > > > > On Thursday, August 26, 2021 at 9:50:32 AM UTC-3, WM wrote:
> > > > > > > William schrieb am Donnerstag, 26. August 2021 um 01:12:57 UTC+2:
> > > > > > >
> > > > > > > > "Number of points" is nonsense when you try to apply it to sets with infinite cardinality.
> > > > > > > Cardinality is nonsense.
> > > > > > Nope, As you note there is a bijection between the rationals in (0,1] and the rationals in (1000,1001]. So the two sets have the same cardinality. It is nonsense to say that the range of the bijection has "more points] in (0,1] than in (1000,1001].
> > > > > My bijection f(q) = q + 1000 shows same number
> > > > "same number" is nonsense when talking about infinite cardinalities..
> > > This is not cardinality but identity of number.
> > Saying "identity of number" instead of "same number" is still nonsense.
> No, we can prove it

Indeed you can prove that both "identity of number" and "same number" are nonsense

--
William Hughes

torsdag 26. august 2021 kl. 20:47:56 UTC+2 skrev WM:
> William schrieb am Donnerstag, 26. August 2021 um 20:38:23 UTC+2:
> > On Thursday, August 26, 2021 at 3:23:10 PM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 26. August 2021 um 20:18:53 UTC+2:
> > > > On Thursday, August 26, 2021 at 3:12:41 PM UTC-3, WM wrote:
> > > > > William schrieb am Donnerstag, 26. August 2021 um 17:34:12 UTC+2:
> > > > > > On Thursday, August 26, 2021 at 10:02:15 AM UTC-3, WM wrote:
> > > > > > > William schrieb am Mittwoch, 25. August 2021 um 14:54:01 UTC+2:
> > > > > > s
> > > > > > > > > What about the first unit fraction when starting from 0?
> > > > > > > > >
> > > > > > > > This does not exist.
> > > > > > > >
> > > > > > > What about the first existing unit fraction? Or are there no unit fractions existing?
> > > > > > There are an infinite number of existing elements of U. However, no element of U has the property that is it first.
> > > > > How is that possible?
> > > > Simple consequence of the fact that no element of the set |N_F has the property that it is last.
> > > For the definable elements this is true.
> > It is also true for the elements that you can't write down.
> Maybe. But then there is nothing next to zero. Then ther is a gap between zero and the smallest unit fraction that can be written down. Never this gap can be closed.
>
> Regards, WM

Is there a largest natural number (n)? No there is not. I think you agree with that.
Then there is no smallest inverse (1/n) which would be nearest to 0. Get it?
KON

William schrieb am Donnerstag, 26. August 2021 um 21:46:18 UTC+2:
> On Thursday, August 26, 2021 at 4:37:47 PM UTC-3, WM wrote:

> > No, we can prove it
> Indeed you can prove that both "identity of number" and "same number" are nonsense

I prove that there are more rationals in the interval (1000, 1001] than Cantor enumerates. Why do you call this proof is nonsense?

Regards, WM

konyberg schrieb am Donnerstag, 26. August 2021 um 21:54:15 UTC+2:
> torsdag 26. august 2021 kl. 20:47:56 UTC+2 skrev WM:

> > > > > Simple consequence of the fact that no element of the set |N_F has the property that it is last.
> > > > For the definable elements this is true.
> > > It is also true for the elements that you can't write down.
> > Maybe. But then there is nothing next to zero. Then ther is a gap between zero and the smallest unit fraction that can be written down. Never this gap can be closed.
> >
> Is there a largest natural number (n)? No there is not.

If dark numbers are existing, then there is no gap between omega and these numbers. But it is useless to call one of them the largest because dark numbers cannot be distinguished. Same with unit fractions and zero.

Regards, WM

On Thursday, August 26, 2021 at 3:47:56 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. August 2021 um 20:38:23 UTC+2:
> > On Thursday, August 26, 2021 at 3:23:10 PM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 26. August 2021 um 20:18:53 UTC+2:
> > > > On Thursday, August 26, 2021 at 3:12:41 PM UTC-3, WM wrote:
> > > > > William schrieb am Donnerstag, 26. August 2021 um 17:34:12 UTC+2:
> > > > > > On Thursday, August 26, 2021 at 10:02:15 AM UTC-3, WM wrote:
> > > > > > > William schrieb am Mittwoch, 25. August 2021 um 14:54:01 UTC+2:
> > > > > > s
> > > > > > > > > What about the first unit fraction when starting from 0?
> > > > > > > > >
> > > > > > > > This does not exist.
> > > > > > > >
> > > > > > > What about the first existing unit fraction? Or are there no unit fractions existing?
> > > > > > There are an infinite number of existing elements of U. However, no element of U has the property that is it first.
> > > > > How is that possible?
> > > > Simple consequence of the fact that no element of the set |N_F has the property that it is last.
> > > For the definable elements this is true.
> > It is also true for the elements that you can't write down.
> Maybe. But then there is nothing next to zero

Correct. There is no element of U closest to zero. There is a gap between any *element of the set U* and zero. There is no gap between the *set U* and 0.

--
William Hughes

On Thursday, August 26, 2021 at 4:59:55 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 26. August 2021 um 21:46:18 UTC+2:

>
> > ... both "identity of number" and "same number" are nonsense
> I prove that there are more rationals in the interval (1000, 1001] than Cantor enumerates.

Nope. Your "proof" is nonsense

>Why do you call this proof is nonsense?

Because it is based on trying to compare the "number of elements" of sets with infinite cardinality.
Terms like "number of elements", "same number", "identity of number", "more", "half", ad nauseum are nonsense for sets with infinite cardinality.

--
William Hughes

William schrieb am Donnerstag, 26. August 2021 um 22:17:49 UTC+2:
> On Thursday, August 26, 2021 at 4:59:55 PM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 26. August 2021 um 21:46:18 UTC+2:
>
> >
> > > ... both "identity of number" and "same number" are nonsense
> > I prove that there are more rationals in the interval (1000, 1001] than Cantor enumerates.
> Nope. Your "proof" is nonsense

It is proven by the number of rationals in the first unit interval.

> >Why do you call this proof is nonsense?
> Because it is based on trying to compare the "number of elements" of sets with infinite cardinality.

That is usual in mathematics of infinite series.

> Terms like "number of elements", "same number", "identity of number", "more", "half", ad nauseum are nonsense for sets with infinite cardinality.

That is wrong. Crdinality is nonsense. It leads to McDuck's bankruptcy, to the empty intersection of infinite endsegments, and to more silly results which are irrelevant for any application of mathematics.

You should ponder about that for some days and nights. Perhaps you will be able to understand that it is nonsense to find half of all rationals in the first unit interval. In particular because the existence of more rationals than Cantor finds in the interval (1000, 1001] can be proved by
1/1 + 1000, 1/2 + 1000, 2/1 + 1000, 1/3 + 1000, 3/1 + 1000, 1/4 + 1000, ...

Regards, WM

William schrieb am Donnerstag, 26. August 2021 um 22:07:23 UTC+2:

> Correct. There is no element of U closest to zero. There is a gap between any *element of the set U* and zero. There is no gap between the *set U* and 0.

The points of the real line are immovable. No quantifier tricks are possible. If there is no gap between "all points" and zero, then there are points next to zero.

Regards, WM

On 8/26/2021 3:43 PM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 26. August 2021 um 21:34:24 UTC+2:
>> On 8/26/2021 8:54 AM, WM wrote:

>> There is the collection of pairs of the form (p, q)
>> such that p is a rational in interval (0, 1],
>> q is a rational in interval (1000, 1001],
>> and q = p+1000.
>>
>> We can prove that's true from the properties of addition of
>> rationals.
>>
>>> But Cantor does not enumerates all fractions.
>> Cantor enumerates all rationals.
>
> In (1000, 1001] he misses about 99.9 %. That is fact.

For each p'/q' in (1000,1001] there is a unique p/q = p'/q'
in lowest terms. gcd(p,q) = 1.
For each p/q in lowest terms, there is a unique q*q*p*p/rad(p).
Define the index of p'/q' in (1000,1001] to be q*q*p*p/rad(p)

For each rational p'/q' in (1000,1001], there is
a unique index. None are missed.

> Conjuration of

There is no conjuration.
I can prove it exists, that thing I say exists.
It's not difficult to do.

> Conjuration of a contradicted
>> collection G of pairs (k,p)
> cannot avoid this fact.

I think that you (WM) think that we (matheologians) are just
assuming into existence the things we need. _Conjuring_

No. _It doesn't work like that_
We describe the naturals in N+.
We describe the rationals in Q+.
These descriptions are intended to find very wide agreement
that what are described are naturals, are rationals.

We reason truth-preserving-ly from the descriptions.
Because truth-preserving, the further claims coming after
a finite sequence of truth-preserving inferences are
true of the same individuals as the widely-agreed-upon
descriptions are true of.

Therefore (assuming everything has been done correctly),
there will be wide agreement on the further claims,
even if those further claims fly in the face of what
everyone expected.

The reason for that wide agreement is NOT some decision of
some Vast Conspiracy assembled to stop you from getting your
deserved recognition. The reason is _truth-preservation_

WM laid this down on his screen :
> William schrieb am Donnerstag, 26. August 2021 um 22:07:23 UTC+2:
>
>> Correct. There is no element of U closest to zero. There is a gap between
>> any *element of the set U* and zero. There is no gap between the *set U* and
>> 0.
>
> The points of the real line are immovable.

The whole point of points is that they have a location.

Jim Burns schrieb am Donnerstag, 26. August 2021 um 22:43:41 UTC+2:

> I can prove it exists, that thing I say exists.

In an inconsistent theory everything can be proved.

I can prove that Cantor enumerates half of all rationals in (0, 1]. Call their number X. And I can prove that there are also X rationals in every other interval.

Regards, WM