Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 11.01.2025 10:41, joes wrote:

Am Sat, 11 Jan 2025 09:50:02 +0100 schrieb WM:

On 10.01.2025 22:51, joes wrote:

Am Fri, 10 Jan 2025 22:38:51 +0100 schrieb WM:

On 10.01.2025 21:06, joes wrote:

Am Fri, 10 Jan 2025 20:17:37 +0100 schrieb WM:

On 10.01.2025 19:28, joes wrote:

Am Fri, 10 Jan 2025 18:21:43 +0100 schrieb WM:

>

I have no expectations about cardinality. I know that for every
finite initial segment the even numbers are about half of the
natural numbers.
This does not change anywhere. It is true up to every natural
number.

You wrongly expect this to hold in the infinite.

No, I expect it is true for all natural numbers, none of which is
infinite.

But it is true for every natural

Of course. Otherwise you would have to find a counterexample.

Good. It is not true for the infinite sets.

The natural numbers are an infinite set. For all of them it is true,

But not for omega, which is not a natural.

Therefore it is irrelevant. No bijection from ℕ contains it.

 

(if you formalise it correctly)!

Irrelevant.

Mathematics is all about formalising.

No, that is only a habit of the last century.

Informal reasoning gets you nowhere, see the centuries before that.

There mathematics has flourished. Now mainly nonsense is produced.

 

∀n ∈ ℕ: |{1, 2, 3, ..., 2n}|/|{2, 4, 6, ..., 2n}| = 2.

Those are not N and E.

Find an element of N or E that is not covered by the equation.

Not what I said. Every natural is finite, and so are the
starting segments of N and E.

And which are not?

The whole sets (which can be seen
as the limits) are not finite.

My claim holds for all numbers only. That is mathematics.

 

That doesn't make it true for N and G.

I am not interested in these letters but only in all natural numbers.
All natural numbers are twice as many as all even natural numbers. If
your N and G denote all natural numbers and all even numbers, then 2
is true also for them.

No. For n->oo,

Every n is finite.

The *set* of all of them isn't.

Irrelevant. My claim holds for all natnumbers only.

 

G is both the set {2, 4, ..., 2n} and {2, 4, ..., 42n};
indeed, {2, 4, ..., 2kn} for every k e N.

And all of them can be denoted by n.

All what?

All natnumbers which Cantor uses in bijections: "such that every element of the set stands at a definite position of this sequence". If this has been accomplished, and then more numbers are created, the bijection fails. This must not happen.

"thus we get the epitome (ω) of all real algebraic numbers [...] and
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]
Afterwards no extension by 42 is allowed.

There is no "after" an infinity.

Cantor maps all natural numbers to a set. Afterwards these natural numbers can be multiplied by 2. Not all remain those which Cantor has applied.
Regards, WM