Re: how

On 4/27/24 10:36 AM, WM wrote:

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:

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The set is bounded by ω.

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

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

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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.

They do if you want to use them to step into the set, like you did.

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

 Therefore this set ends before.

Only if it HAS an "end"
But if it did, then that value would be the least upper bound, but no such value exists in the set, so we can sho

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

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

 Then something else must be used.

Like a LIE I guess.
The fact that you logic needs for there to be a value there is what breaks it.

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

 In actual infinity something is below ω.

Nope, at least not just one step below.
What is "below" ω is the whole set of Natural Numbers.

 Regards, WM