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



> From: "Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>
> To: "John Pryce" <j.d.pryce@xxxxxxxxxxxx>,
>  "stds-1788" <stds-1788@xxxxxxxxxxxxxxxxx>
> Subject: Re: Motion P1788/M0013.04 - Comparisons - Overflow / Infinity
> Date: Tue, 21 Sep 2010 08:19:45 -0500
> 
> John Pryce wrote:
> >> . . .
> >>
> >> 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.

	John is right.  There ARE conceptual traps here.

> 
> 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.

	Hmm.  I think John's approach is better here in that
	there is only one unbounded or infinite endpoint &
	that intervals are open at that endpoint.

	So I suspect you cannot make that distinction
	consistent mathematically.

	But even supposing you could, we cannot make that
	distinction computationally.

	We have (& should have) but one infinite endpoint
	to represent intervals whose bound cannot be
	computed.  It make no difference whether we arrive
	at that endpoint through overflow or some other
	means.

	And since we have only one such endpoint we can
	have only one interpretation for an interval
	possessing that endpoint.  That is that we know
	no bound for that interval.

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

	Well, first, I think of [1,infinity] as a semi-open
	interval.

	And, second, whether we know no bound because there
	is no bound or we know no bound because we were unable
	to compute & represent that bound amounts to the
	same thing: we know no bound.

	For us these are the same.

> 
> 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

	I have no comment on this last as it is outside
	what I believe we must consider as our conceptual
	framework.

	John is also right historically: it has always
	been hard to deal with the infinite.


			   Dan