Re: how

On 4/19/24 11:20 AM, WM wrote:

Le 19/04/2024 à 00:35, Richard Damon a écrit :

On 4/18/24 11:05 AM, WM wrote:

ω is amidst the interval (0, ω2) because in the image there are as many ordinals in (ω, ω2) as in (0, ω).

>
But all those ordinals are transfinite ordinals, and none are the value of double a finite Natural Number.

 You are in error. Counting goes like this: 1, 2, 3, ..., ω, ω+1, ω+2, .. You simply pass ω although no known natural number k+1 will reach ω. But by multiplication, which goes faster, you cannot pass ω?
 Regards, WM

Nope, "counting individual numbers" NEVER gets to ω.
ω is what you get to when you go BEYOND just "counting", which is why your bounded logic can't handle it, as it needs to count to things.
ω isn't just some "big" number, but a step above what you get by counting and moving into dealing with infinite sets.