Re: KISS-decorations

["Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>]

I'm thinking along the lines of Definition 2.2.3 on pp. 55-56 in:
   http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF
By that definition, wouldn't
   floor([1,3/2])
be continuous?

Nate


----- Original Message ----- 
From: "Jürgen Wolff von Gudenberg" <wolff@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
To: "Nate Hayes" <nh@xxxxxxxxxxxxxxxxx>
Cc: "John Pryce" <j.d.pryce@xxxxxxxxxxxx>; "Dominique Lohez"
<dominique.lohez@xxxxxxx>; "stds-1788@xxxxxxxx" <stds-1788@xxxxxxxx>
Sent: Friday, July 01, 2011 2:59 PM
Subject: Re: KISS-decorations


> Nate
>
> Am 01.07.2011 17:09, schrieb Nate Hayes:
> > One other comment:
> >
> > What about
> > floor([1,3/2])?
> > I would think it should be defined and continuous, i.e., "safe", since
> > when
> > restricted to the domain [1,3/2] the floor function at 1 is defined and
> > continuous when approached from the right and cannot be approached from
> > the
> > left.
> >
> > But it seems this would imply
> > floor([1,1])
> > should also be "safe"; but motion 27 gives "defined" for this example?
> >
> > Nate
> I think 1 is a discontinuity point of the floor function, hence both cases
> are defined only
> Juergen