RE: KISS-decorations
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.
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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
>