Re: how

On 4/17/24 2:49 PM, WM wrote:

Le 17/04/2024 à 01:27, Richard Damon a écrit :

On 4/16/24 10:59 AM, WM wrote:

 

Note, no "Natural Number" is actually infinite, that distinction falls on omega. But the SET of the Natural Numbers is Actually Infinite in Size, having a size of Aleph_0.

 The size of ℕ is |ω|. That means ℕ extends on the ordinal line from 0 to ω. By multiplying every natural number the extendion is doubled. That is mathematics. Every contrary opinion is foolish.

Nope, the size of ℕ, that is |ℕ| is aleph_0. ω is an ORDINAL number (showing order), not a cardinal number (showing size). This means that the first transfinite ordinal above the set ℕ would be the value ω. So yes, as you say below, there are no "finite" numbers between the set ℕ and the value ω, but since ℕ has no "highest" member there is no "predecessor" to ω, just as there is no predecessor to 0 in the Natural Numbers.
Note, there are some extended transfinite systems that do put transfinite numbers in that gap. (just like the Rationals put numbers between 0 and 1 which are consecutive in the Natural Numbers).

>
Does that mean the Natural Numbers themselves are individually only "Potentially Infinite" but the set of them is "Actually Infinite" by your definitions?

 The visible natural numbers are potentially infinite. The set ℕ is assumed to be actualy infinite. This cannot be known let alone be proven. But we can assume it and draw conclusions. One of them is that nothing fits between all natural numbers and ω.

There is no defined set of "visible natural numbers" except in your own broken logic. ALL Natural numbers fit your definition of "visible", but your logic can not handle that this set is actually infinite in size.
The set ℕ IS ACTUALLY infinite in size, as BY DEFINITION, there is no finite number that can count all its members, there is no "highest" member.
Refusing to beleive the definition, just proves that you logic system can not handle infinite sets.

 Regards, WM