RE: Motion 52: final "Expressions" text for vote

["Kreinovich, Vladik" <vladik@xxxxxxxx>]

Yes, this is exactly what I descrube (and this is what forms the basis of our standard).

 

We have discussed, from the very beginning, whether we want to consider only real numbers as elements of our intervals, or we also want to consider two infinite values. Both approaches are consistent, both have their advantages and disadvantages.

 

The majority decided (it was long ago, one of the first motions) that we should NOT include infinite values in intervals, infinite values can only serve to indicate bounds of unbounded intervals, but all their elements are finite.

 

In this case, 1/0 is definitely an empty set, since we start with arithmetic of real numbers (NOT with arithmetic of extended real numbers) in which 1/0 is not defined.

 

Alternatively, we could base our formalism on extended real numbers, but as nof now, it is too late to change -- and if I remember there were convincing pragmatic arguments for considering only real numbers (in the mathematical sense( and not symbols of plus minus infinity.

 

In short, there is no contradiction:

 

* if we consider only real numbers, then 1/0 is not defined in real numbers, and not defined in intervals

 

* if we consider extended real numbers, then 1/0 is defined and in the corresponding interval arithmetic (which is different from what we consider now) infinity would be an element of what we consider too.

 

I apologize for being somewhat sloppy in my notations, and thanks for the clarification: since we only consider real numbers, the range means the set of all values f(x1,...,xn) when x_i are real numbers (and for which f(x1,...,xn) is defined).

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From: Bill Walster [billwalster@xxxxxxxxx] on behalf of G. William (Bill) Walster [bill@xxxxxxxxxxx]
Sent: Saturday, November 23, 2013 4:04 PM
To: Kreinovich, Vladik; Michel Hack; stds-1788
Subject: Re: Motion 52: final "Expressions" text for vote

On 11/23/13 1:07 PM, Kreinovich, Vladik wrote:

We have gone through these questions when we started the standard. 

The range of a function over the interval is defined as the set of all possible values of the function when its arguments are in the range. If for some values within the range, the function is not defined, there are no values to add to the set.  

Do you mean instead of the above:

The range of a function over the interval is defined as the set of all possible values of the function when its arguments are in the function's domain of definition. 

That is:

	f(X) = { f(x) | x \in X \cap D_f };

where X is an interval and D_f is f's domain of definition.

Therefore, the interval extension of any function returns the empty set { } for any argument outside the function's domain of definition. 

In this case, the containment set of 1 / [0, 0] is { }, the empty set.

Is this what you propose?

If so, then I claim this is a containment failure because in the projective real system 1/0 = oo, which is not the empty set.  Therefore there is an inconsistency in the proposed theory.

Cheers,

Bill