Re: "still unclear on infinite intervals"
On May 25 2011, Michel Hack wrote:
I think Nick intended "that approach" to refer to an arithmetic where
intervals are defined over the extended reals, so 1/0 can be a member
of an interval. Motion 3 specifically excluded that, so dividing by
an interval spanning zero is simply unbounded.
Not really. Even if intervals are defined over the extended reals,
the only contiguous inteval that 1/0 is part of is (-infinity,+infinity),
because it doesn't have a sign.
My first point is that 0*any_interval = 0 makes sense only if the
intervals are defined over the UNEXTENDED reals, and hence are not
closed. Otherwise, it is saying that 0*infinity is 0. My second is
related, and is that 1/(-1,+1) is not a subset of ANY interval if
that approach is adopted, because it includes 1/0.
If 0*any_interval = 0 and 1/0 is part of an interval, then we can
easily prove that 1 = 2.
Regards,
Nick Maclaren.