Re: KISS-decorations
Yes, in this context, one is looking either at the topology over the
reals, or that topology restricted to, say [1,3/2]. Whichever way
we go with the floor function, I think we can explain it in such
terms. What is the practical impact of one way over the other?
Baker
On 07/02/2011 09:07 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
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