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Re: Motion 3

Dominique Lohez schrieb:

I agree with the principles introduced in this motion
However I find the formulation of exclusion of infinity too abstract to be useful For example it does not provide a way to evaluate a expresssion such that f( (-\infty, a] )
As a tentative solution to this difficulty, I propose the following approach 

The number  set under consideration is R
The basic intervals have the general form [a,b] where a and b are reals numbers such that a<= b At least two order relations can be defined over basic intervals 

The usual inclusion

and rthe <=_I relation defined by
<=_i ={([a,b], [c,d] )   | a <= c and b <= d }

Only this last relation structure the basic interval set as a lattice
This lattice is incomplete. To insure the completeness one has to had to add the following intervals (-infty,a] = inf_x {[x,a] } 
           [b,+infty] = sup_y { [b,y] }
(-infty, +infty) = sup_y{ (-infty, y]) = inf_x ({[x, +infty)}) (-infty,-infty) is the infinum of the whole interval set ( basic intervals + extensions) (+infty,+infty) is the supremum of the whole interval set ( basic intervals + extensions)
The notations (-infty, -infty) and (+infty, +infty) are very confusing. Let (IR) denote the set of (bounded and unbounded) real intervals. Then {(IR), <=} is a lattice. It is not complete. It shares this property with the real numbers. Both sets are only conditionally complete, i.e., every bounded subset has an infimum and a supremem. If there is a need for a particular application to join a least and a greatest element to the set {(IR), <=} I would not mind denoting these elements by [-oo,-oo] and [+oo, +oo]. It should be cleaar, however, that these elements are not real intervals and they do not obey many rules which otherwise hold for interval arithmetic. For instance for a point interval A we usually have A - A = 0, or A/A = 1. 

Best whishes, Ulrich Kulisch
In such a context the value of f( (-infty,a] ) is defined only as the limit for /any /sequence of f([x,a]) when [x,a] goes to (-infty, a] 
It is undefined if such a limit does not exist.

As an illustration we consider the sine fonction
Of course the calculation of sin( (-infty,1] is problematic due to the difficulties of evaluating corerectly sin( (x,1]) However we ca note that clearly range (f, (-infty, 1]) = [-1, 1] despite the fact that sin(-infty) cannot be defined . 

Conversely on has  clearly 1/[0,1] =  [1,+infty)



I hope this can help

Best regards

Dominique LOHEZ








. Wolff v. Gudenberg a écrit :

Dear members of the working group.
Please note the following motion concerning the semantics of intervals
So it is not about structure and procedure, not to fiX the final formulation, but about cootents 
Comments welcome
Juergen



===Motion P1788/M0003.01_Set_of_reals===
Proposer: Juergen Wolff von Gudenberg
Seconder: required

===Motion text===
The P1788 Interval arithmetic standard defines intervals as closed and connected sets of real numbers. That means that +/- Infinity may be used to denote an unbounded interval but are never considered as members of an interval. 

===Rationale===
The Vienna Proposal section 1.2
Kulisch's position paper "Complete Interval Arithmetic"
The extensive discussion in the mailing list


--
Dr Dominique LOHEZ
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Email: Dominique.Lohez@xxxxxxx