1) You are right, I have missed "!" .
Duh! You never read carefully what others write!.
2) So you actually agree that no matter what pair of different dims are used, it is always based on ≠ (non-locality)\some dim (locality) linkage.
The term linkage is a forced way so you can have something to say about something that "standard" Math takes for granted.
So it is more verbalisation of something that is implicit in "standard" Math.
3) (2) is an example of a generalization.
Hey! I agree! But it does not answer the question!
4) Covering dim by itself is trivial, we are talking about fully cover n+1-dim by n-dim, that cannot be done.
So, prove that there is one point that is not covered on that n+1-dim by the n-dim!
The ≠ is not a proof. It is a statement in the vein "I decree it so", and so it means absolutely nothing.
Show me the starting point, then work up to it as a proof.
We wasted a goodly amount of posts on this, and you still have shown bupkiss.
To explain my question in your own lingo:
Again:
• is 0-dim element.
_____ is 1-dim element.
It is shown that for any arbitrary • on _____ there is • < • < • , where < is possible for any amount of • on _____ exactly because no collection of • elements can fully cover _____
Yes? Ok, now the question is:
Using your quoted and modified logic:
• is 0-dim element.
_____ is 1-dim element.
It is shown that for any arbitrary ≠ on _____ there is • ≠ • ≠ • because no collection of elements can fully uncover _____
So, according to you, any 1-dim element is at once covered and uncovered by • and ≠ simultaneously?
It all exists and does not exist.
I understood that much already from pages ago. But somehow I did not get that through.
Now, my question is:
Can you, in OM, point to any specific point on that 1-dim and state that it is uncovered?
If so, how?
In "standard" Math this is easy. All points are covered.
There is a succinct need for that equation in OM, because otherwise the information that a 1-dim holds is of no use.