Re: KISS-decorations

["J. Wolff von Gudenberg" <wolff@xxxxxxxxxxxxxxxxxxxxxxxxxxx>]

Motion 27: the example with the floor function has been changed
version 1.4 is on the website
Juergen

Am 03.07.2011 22:32, schrieb Ralph Baker Kearfott:
> OK;  that makes sense to me at this point.
>
> Baker
>
> On 07/03/2011 01:46 AM, Jürgen Wolff von Gudenberg wrote:
> > Yes, I agree
> > floor[1,1.2] is continuous, so is floor[1,1],
> > our view was too cautious.
> > We will changee our proposal
> > Juergen
> >
> > Am 02.07.2011 16:24, schrieb Kreinovich, Vladik:
> > > Oops, yes, I goofed, it is continuous on this interval :-(
> > >
> > > ________________________________________
> > > From: Corliss, George [george.corliss@xxxxxxxxxxxxx]
> > > Sent: Saturday, July 02, 2011 8:21 AM
> > > To: Kreinovich, Vladik
> > > Cc: Corliss, George; Nate Hayes; Jürgen Wolff von Gudenberg; John
> > > Pryce; Dominique Lohez; stds-1788@xxxxxxxx
> > > Subject: Re: KISS-decorations
> > >
> > > Vladik,
> > >
> > > Did you look at the interval argument before you made your comment?
> > > For floor([1, 3/2]) = 1, which IS continuous in the calculus
> > > epsilon-delta sense.
> > >
> > > Or am I missing something?
> > >
> > > George
> > >
> > > On Jul 2, 2011, at 9:07 AM, Kreinovich, Vladik wrote:
> > >
> > > > Of course, the notion of contnuity depends on topology, but when
> > > > people talk about continuity without explaining what this means, they
> > > > mean continuity in the normal calculus epsilon-delta sense in which
> > > > the floor function is clearly NOT continuous.
> > > >
> > > > ________________________________________
> > > > From: stds-1788@xxxxxxxx [stds-1788@xxxxxxxx] On Behalf Of Nate Hayes
> > > > [nh@xxxxxxxxxxxxxxxxx]
> > > > Sent: Friday, July 01, 2011 3:37 PM
> > > > To: Jürgen Wolff von Gudenberg
> > > > Cc: John Pryce; Dominique Lohez; stds-1788@xxxxxxxx
> > > > Subject: Re: KISS-decorations
> > > >
> > > > 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
> > >
> > > Dr. George F. Corliss
> > > Electrical and Computer Engineering
> > > Marquette University
> > > P.O. Box 1881
> > > 1515 W. Wisconsin Ave
> > > Milwaukee WI 53201-1881 USA
> > > 414-288-6599; GasDay: 288-4400; Fax 288-5579
> > > George.Corliss@xxxxxxxxxxxxx
> > > www.eng.mu.edu/corlissg

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