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Re: Intersection example (was Re: Motion 26: NO)

Arnolde Neumaier wrote:

On 07/31/2011 10:21 PM, Nate Hayes wrote:

Arnold Neumaier wrote:

On 07/31/2011 09:16 PM, Nate Hayes wrote:

Arnold Neumaier wrote:

On 07/28/2011 10:30 PM, Nate Hayes wrote:

Arnold Neumaier wrote:

I had given an example where Motion 27 gives erroneous results for
decorated intersections:

The expression f(x)= x/((x+1) intersect x^2) is undefined for any
x in
[1,3], but Definition 7 claims a safe answer for f([1,3]).


Arnold, this is not any example of erroneous results.


For example, the expression is undefined
for
x=1, for x=2, and for x=3, although all these are in [1,3].


u(1) = (1+1) = 2
v(1) = 1^2 = 1

f(1) = 1/(u(1) intersect v(1))
= 1/(2 intersect 1)
= 1/empty
= empty

The set-theoretic intersection of u(1) and v(1) is empty and is not
undefined (it is safe).


Never before I was told that the set theoretic intersection of two
numbers
should be empty.


There are three main traditions for defining numbers.

In Zermelo-Fraenkel set theory, the intersection of two cardinal numbers
is the smaller one.

In Dedekind's definition of real numbers as , the intersection of two
real numbers is also the smaller one.


[...]


But P1788 so far (that I'm aware of) only considers the intersection of
singleton sets, so I assume in your example
1 intersect 2
is meant to be shorthand for
[1,1] intersect [2,2].


If you argue that the meaning of an expression for real input is the
special case of the meaning of the corresponding interval expression,
you'd also conclude that
     1/0=[1,1]/[0,0]=Empty,    0/0=[0,0]/[0,0]=Entire,
If this argumentation applies, division is always defined, and the
decoration should be at least def. (Though this is useless for the
applications.)



Motion 5 says:
   "when evaluating a function... points outside its domain are simply
ignored"




But so far in P1788, the argumentation was _always_ the reverse: the
interval expression gets its semantics from the corresponding real
expression. (This is the mode needed in the applications.)
If this argumentation applies, intersection is not always defined,
and the decoration should be at most con.

Your scheme violates the conclusion of both cases.



Motion 5 says:
   "the result of an operation is a set"
So the intersection operation in the function does not operate on real
numbers.

Nate