aa strictLess bb <==> a1 < b1 && a2 < b2
I shall use the analogue of "lessEqual":
for all b in bb, there exists a in aa such that a < b
and
for all a in aa, there exists b in bb such that a < b.
This agrees with the level 3 definition for bounded intervals, I
think. But not for unbounded ones, e.g. ([-oo,0] strictLess
[-oo,1]), and (Entire strictLess Entire) are true in my definition
but not in the level 3 one. Comments please.
On 30 Apr 2010, at 15:14, Vincent Lefevre wrote:
The motion says:
The greatest lower bound and the least upper bound of an interval
with the empty set are both the empty set.
So, this would mean that X <= Empty and Empty <= X for any X, but
with the consequence that <= would no longer be an order relation
(since no antisymmetric).
I think that X <= Empty and Empty <= X should be false for any
non-empty X. If defined as operations of P1788, glb and lub
should return NaI on (X,Empty).