RE: Decorations and Motion 22
Dear Friends,
Many thanks for the explanations, they help me (and I hope others) to understand the issues better. Arnold's example is good, but I would like to be able to reach a level where I can run similar examples on my own.
Please check whether my understanding is correct.
Bounded is clear, that we have finite bounds.
Continuous, if I remember correctly, means that this interval came as a result of continuous operations, so it may be an enclosure of the actual range, but at least we know that it did not come from computing the interval hull of a discrete set such as sign([-1,1]).
About everywhere defined: if I understand correctly, this means that in all previous steps, we did not have possible un-defined values. For example, sqrt([0,4]) is [0,2] and everywhere defined while sqrt([-1.4]) is still [0,2] but this time NOT everywhere defined, only possible everywhere defined -- since it could be that [-1,4] is only an enclosure for the actual interval [0,4] for which the previous operation is everywhere defined.
Is this correct?
Re different between nowhere defined and not valid I am not so sure.
Vladik
P.S. This was probably explained in previous motions and position papers, but I am probably not the only one still somewhat confused.
-----Original Message-----
From: stds-1788@xxxxxxxx [mailto:stds-1788@xxxxxxxx] On Behalf Of Arnold Neumaier
Let me write the presence of the decorations x.dec=0,...,4 as
d0: safe = everywhere defined, continuous, and bounded)
d1: everywhere defined
d2: possibly everywhere defined
d3: nowhere defined
d4: not valid
The decorations are partially ordered according to information content
as d0>d1>d2<d3<d4. Thus d2 is the least informative decoration, where
nothing is claimed anymore; it is the bare decoration equivalent of
the bare interval Entire. d4 is sticky in all arithmetic operations.