Re: how

Le 26/04/2024 à 20:13, Richard Damon a écrit :

On 4/26/24 1:02 PM, WM wrote:

Le 26/04/2024 à 16:49, Richard Damon a écrit :

On 4/26/24 10:41 AM, WM wrote:

 

>
The set is bounded by ω.

>
But ω isn't in the set, so it can't be the upper bound of the set that is below ω

 Learn the meaning of upper bound. ω*2 is an upper bound too.

 But since it isn't in the set, you can't use it as the value to "step back" to.

The question was whether it is an upper bound. Bounds need not belong to a set.

 Note, that value isn't "of the set that is below ω".

Therefore this set ends before.

 

Fact is that below ω there are natnumbers and above there are none. Hence they cease below the bound ω.

 Yes, but it can't be the "Upper Bound" used to step back to from ω.

Then something else must be used.

 There is not "Upper Bound" for the Natural Numbers *IN* the Natural Numbers to be the "last" value to step back to.
 Thus, your logic about ω-1 breaks. The value does not exist,

In actual infinity something is below ω.
Regards, WM