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Re: overflow question

Alexandre Goldsztejn wrote:

So, what is the point? Why introducing families of bounded intervals
to model one unbounded interval? Only for being able to justify at
Level 1a a definition of the midpoint? Anyway, the definition of the
midpoint of an unbounded interval is arbitrary, so your proposal tries
to justify something arbitrary, which is dangerous with many regards.


The proposal does not define the midpoint of an unbounded interval; it
allows the midpoint of an overflow family at Level 1a to be defined in the
same manner and for the same reasons that P1788 has already discussed
defining midpoint of an unbounded interval at Level 2.

If such an idea is really so dangerous, P1788 should not define midpoint for
unbounded intervals at Level 2 in the current model, either.





To be more homogeneous, I think Omega should return a set containing
only one interval in case there is no overflow, so that the process
above applies to both bounded and unbounded intervals, otherwise two
cases have to be handled.



Yes. This is exactly what Omega(X) does (see definition (6) in the paper
in
the case X=[a,b] is a subset of the interval H=[-h,h]).


No: Your definition in the paper explicitly says that Omega(X) gives
"a nonempty, closed and bounded interval [u,v]" if [u,v] subset
[-h,h]. In this case, Omega([u,v])=[u,v] is an interval. What I am
saying is that you should return a family that contains only one
interval, i.e. Omega([u,v])={ [u,v] }, so as to be homogeneous with
the other cases. Although, you would probably want to define
Omega(emptyset)={ emptyset }, i.e. a family that contains only the
emptyset.


Now I see your point. Yes, I agree.

Nate