Re: Replacement of Cardinality

Am Thu, 01 Aug 2024 12:27:27 +0000 schrieb WM:

Le 01/08/2024 à 02:09, Richard Damon a écrit :

On 7/31/24 10:27 AM, WM wrote:

Le 31/07/2024 à 03:28, Richard Damon a écrit :

On 7/30/24 1:37 PM, WM wrote:

Le 30/07/2024 à 03:18, Richard Damon a écrit :

On 7/29/24 9:11 AM, WM wrote:

>

But what number became ω when doubled?

ω/2

And where is that in {1, 2, 3, ... w} ?

In the midst, far beyond all definable numbers, far beyond ω/10^10.

In other words, outside the Natural Nubmer, all of which are defined
and definable.

That is simply nonsense. Do you know what an accumalation point is?
Every eps interval around 0 contains unit fractions which cannot be
separated from 0 by any eps. Therefore your claim is wrong.

No. There is ALWAYS an epsilon.

The input set was the Natural Numbers and w,

ω/10^10 and ω/10 are dark natural numbers.

They may be "dark" but they are not Natural Numbers.

They are natural numbers.

How are they defined?

Natural numbers, by their definition, are reachable by a finite number
of successor operations from 0.

That is the opinion of Peano and his disciples. It holds only for
potetial infinity, i.e., definable numbers.

I assume completness.

I guess you definition of "completeness" is incorrect.
If I take the set of all cats, and the set of all doges, can there not
be a gap between them?

What is the reason for the gap before omega? How large is it? Are these
questions a blasphemy?

A "gap" implies some sort of space that is not filled. There is no such
space (it would be filled with infinitely many natural numbers).
We just condense the whole of N into one concept and call that omega,
or add it on the next level of infinity.
Your questions are only a display of your unwillingness to understand
infinity, even though you would like to imagine yourself as some sort
of martyr.

∀n ∈ ℕ: 1/n - 1/(n+1) > 0. Note the universal quantifier.

Right, so we can say that ∀n ∈ ℕ: 1/n > 1/(n+1), so that for every
unit fraction 1/n, there exists another unit fraction smaller than
itself.

No. My formula says ∀n ∈ ℕ.

Right, for ALL n in ℕ, there exist another number in ℕ that is n+1,

That does ny formula not say. It says for all n which have successors,
there is  distance between 1/n and 1/(n+1).

If k did not have a successor, what would k+1 be?

Remember, one property of Natural numbers that ∀n ∈ ℕ: n+1 exists.

Not for all dark numbers.

Maybe not for dark numbers, but it does for all Natural Numbers, as
that is part of their DEFINITION.

It is the definition of definable numbers. Study the accumulation point.
Define (separate by an eps from 0) all unit fractions. Fail.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.