Re: Motion P1788/M0013.04 - Comparisons - Overflow / Infinity

["Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>]

John Pryce wrote:
> > On 20 Sep 2010, at 23:22, Ian McIntosh wrote:
> > > You're right, as I said Overflow is an infinite set just as Infinity is 
> > > an infinite set, so [a, Overflow] represents an infinite family of 
> > > intervals just as [a, Infinity] represents an infinite family of 
> > > intervals. How does one evaluate
> > > [1, Infinity] \subseteq [1, Infinity] ? Which Infinity is larger? It's 
> > > the same problem, one which I did not claim to solve.
> >
> > NO! You are falling into two traps at once.
> >
> > First, our interval model says intervals are subsets of the reals R; our 
> > notation says [1, Infinity] is just shorthand for {x in R | 1 <= x < oo}. 
> > Hence it is ONE set, not an infinite family of sets like [1, Overflow]. 
> > And standard set theory unambiguously says [1, Infinity] \subseteq [1, 
> > Infinity] is true.

Here is where I stick my neck out a little and, I think, differ with Ian's 
interpretation, too.

As John mentions, I also see [1,Infinity] as shorthand for
   { x in R | 1 <= x < oo }.
However, unlike Ian, I see [1,Overflow] as shorthand for
   { x in R | 1 <= x <= u },
where MAXREAL < u is some unknown real number.

Hence, [1,Infinity] is a closed, unbounded interval, but [1,Overflow] is a 
compact interval with a "really big" (but finite) upper bound.

I don't claim to know (at least at this point) what [1,Overflow] \subseteq 
[1,Overflow] means in this case. A bool_set result might be one possible 
answer.

Nate