Re: Revised interval flavors motion (36.02)
John Pryce wrote:
I asked, e.g., if "operation compatibility", as in (2), is to hold for
A = [1,2]/[-1,1] ?
B = sqrt([-1,1]) ?
C = sign([0,7]) ?
In the language introduced above, which of these operation instances are
internal?
Instance A can't be internal, because its set-based value cannot be a
common interval (whether we take the exact range, or its hull).
Agreed.
However, when 1/[0,1] is restricted to the intersection of the natural
domain of the reciprocal operation and the input [0,1] (so that the
operation is effectivly restricted to the semi-open interval (0,1]), then
(all x in (0,1])(exists y in [1,+oo]) : y = 1/x // *-theorem
(all y in [1,+oo])(exists x in (0,1]) : y = 1/x // **-theorem
are both true and this is proof the solution [1,+oo] is unique. Although it
involves unbounded intervals, it is also a common solution between the
set-based and modal flavors.
B has the set-based value [0,1], a common interval. Here, the function is
continuous wherever defined, but isn't defined everywhere on the input. I
don't know the modal/Kaucher value, or if it has one.
For classical intervals,
sqrt([-1,1]) = { sqrt(x) : x in [-1,1] } = somewhereUndefined (1)
isn't everywhere defined, either, unless one restricts the sqrt operatation
to the intersection of its natural domain Df and the input, i.e.,
sqrt([-1,1]) = { sqrt(x) : x in [-1,1] and x in Df } = [0,1] (2)
This is how P1788 gets the decorated result
sqrt([-1,1]) = ([0,1],somewhereUndefined),
which gives both interpretations (1) and (2).
IMO, it ok to have both interpreations for modal intervals, too.
C also has the set-based value [0,1], a common interval. Here, the
function is defined everywhere, but isn't continuous everywhere on the
input. Again, I don't know the modal/Kaucher value, or if it has one.
Strictly speaking, classical intervals don't have a value, either, since
sign([0,7]) = { sign(x) : x in [0,7] ) = { 0, 1 }
is not a classical interval. One can take the hull to obatain the interval
[0,1], and this satisfies the semantic
(all x in [0,7])(exists y in [0,1]) : y = sign(x) // *-theorem
but the semantic
(all y in [0,1])(exists x in [0,7]) : y = sign(x) // **-theorem
is false, because the image of sign([0,7]) is not actually an interval.
Question to modal folk: Is it necessary to make the quite restricted
definition:
An operation instance is internal iff the operation is
everywhere defined and continuous on its inputs
or can modal theory go beyond that in some circumstances?
IMO it can go beyond that in some circumstances, e.g., when the operation is
still defined and continuous on the intersection of the input interval and
the natural domain of the real function, as in the case of B or 1/[0,1].
Note. Since this is about "exceptional cases", it is also about how we
make
decorations interoperate with flavors.
I notice that if decorations are defined as what is mathematically true
about the restriction of a real function f to interval input X, then the
decorations are the same for modal and set-based flavors. For example, the
decoration of 1/[-1,1] and sqrt([-1,1]) is always somewhereUndefined,
regardless what value is returned for the interval portion.
Nate