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.