Re: min / max and empty intervals (was: Friendly amendment to Motion 25)
> Date: Tue, 7 Jun 2011 12:18:19 +0200
> From: Vincent Lefevre <vincent@xxxxxxxxxx>
> To: "stds-1788@xxxxxxxx" <stds-1788@xxxxxxxx>
> Subject: Re: min / max and empty intervals (was: Friendly amendment to Motion 25)
>
> On 2011-06-06 21:06:32 +0100, John Pryce wrote:
> > . . .
> >
>
> . . .
>
> Moreover if I is a subset of J, then f(I) should be a subset of f(J),
> even in the case where I is empty. This would no longer be true with
> (2). For instance, min(sqrt(X),[3,4]) with:
> * X = [-2,-1] -> result = [3,4]
> * X = [-2,0] -> result = [0,0]
>
> --
> Vincent Lefèvre <vincent@xxxxxxxxxx> - Web: <http://www.vinc17.net/>
> 100% accessible validated (X)HTML - Blog: <http://www.vinc17.net/blog/>
> Work: CR INRIA - computer arithmetic / Arénaire project (LIP, ENS-Lyon)
OOoo, Vincent. I find your example both illuminating
& disturbing.
If we have a min that returns the non-empty interval
in the case where the other is empty then, as you say:
xx = [-2,-1] \subset yy = [-2,0] but for
f(tt) = min(sqrt(tt),[3,4]) we have that
f(xx) = [3,4] NOT \subset of f(yy) = [0,0].
But if we use a min that returns the empty we have:
f(xx) = empty \subset of f(yy) = [0,0].
Is this a fluke of this example or the nose in the tent
of a more general principle?
Actually, it is sufficient to simplify your example to
f(tt) = min(tt,[3,4]). Then the interval returning min
has:
f(empty) = [3,4] NOT \subset f([0,1]) = [0,1]
and with the empty returning min we have
f(empty) = empty \subset f([0,1]) = [0,1].
This suggests the more general principle to me.
If this is convincing to all of you I entirely withdraw
my argument for the other approach.
Wow, I didn't see that one coming.
Dan