Inner and outer enclosures

[Nate Hayes <nh@xxxxxxxxxxxxxxxxx>]

Folks,

It seems Arnold lost focus on the point I was making in the last thread. So
I just want to finish making my point here and summarize.

For classical interval arithmetic, there are inner and outer enclosures.
Look at the example 1/[1,2]=[1/2,1]. There is definition for inner inclosure

    Y** \subseteq [1/2,1] <=> (for all y in Y**)(there exists x in [1,2])
y=1/x

and for outer enclosure

    [1/2,1] \subseteq Y* <=> (for all x in [1,2])(there exists y in Y*)
y=1/x.

In exact arithmetic, we have

    Y* = [1/2,1] = Y**.

However if we change our example to 1/[0,2], then inner enclosure is

    Y** \subseteq [1/2,Inf) <=> (for all y in Y**)(there exists x in [0,2])
y=1/x

and outer enclsoure

    NaI \subseteq Y* <=> (for all x in [0,2])(there exists y in Y*) y=1/x,
i.e.,

it is undefined. So even in exact arithmetic, we have Y* != Y** (the
operation is undefined).

My understanding of Sunaga's definition is it should give the outer
enclosure when implemented inside a computer in floating-point arithmetic.
But even in exact arithmetic, Y* != Y** for the operation 1/[0,2]. This is
why in his definition he says "provided that zero does not fall in Y."

Note that he does not say "provided that zero is dropped from Y."

I also want to emphaise that just because I use "for all" and "there exists" 
does *not* mean I'm talking about modal intervals. Please!!

Nate