Re: Motion P1788/M0013.04 - Comparisons
<Resending because of a typo: in the Proof, an "A" should be "a". Discard previous version>
P1788
On 18 Sep 2010, at 17:45, Arnold Neumaier wrote:
>> Someone please say which of these is right:
>> (1) It is false, because the endpoints of the "inside" set _must_ be different from those of the "outside" one.
>> (2) It is true, because what we want is "topologically interior to".
>
> Fronm a mathematical point of view (since motion 3 says intervals are sets), version (2) is the right explanation.
>
>> (3) I couldn't care less, because this relation is only useful for bounded intervals.
>> Or, of course, something else.
>> Answers from those with experience in writing interval software especially welcome.
>
> From usage in algorithms, one needs the interiorness check for some existence tests. Usually these also require boundedness, and the above would not matter.
>
> But in view of Hadamard's theorem,
> ...In this case, the topological definition is a necessity.
On 18 Sep 2010, at 15:06, Nate Hayes wrote:
> The purist in me might say (3)... the pragmatist in me might say (1).
Nate, I'm sorry that whether purist or pragmatist you avoided (2). Arnold, thanks for a possible use for (2) in actual algorithms.
Reason (IMO) to choose "topologically interior", i.e. (2):
- Motion 3's definition of intervals is topological:
they are closed connected subsets of the reals R.
- It automatically gives "A is interior to B" a meaning
for ANY intervals A,B (empty, bounded or unbounded)
-- indeed for arbitrary subsets of R.
Namely it means
B is a neighbourhood of each a in A. (eqn1)
By contrast a definition based on endpoints is totally ad-hoc for Empty, and somewhat ad-hoc for unbounded intervals.
Theorem. If B is a connected subset of R (it needn't be closed) and A is any subset of R, then A is topologically interior to B iff, using Marco & Juergen's notation,
\forall_a \exist_{b,b'} b < a < b'. (eqn2)
In particular this holds when A, B are arbitrary P1788 intervals.
Proof. Suppose (eqn2) holds. Let a be any point of A, and let b, b' be as in (eqn2). Since B is assumed connected and contains b,b', it must contain all points between b and b'. That is, it contains the open interval (b,b') which contains a. Hence by definition, B is a neighbourhood of a. Since a is arbitrary, this proves (eqn1) holds. The converse is similar. QED
Proposal. Whatever set of comparisons P1788 decides to support, if (as seems very likely) "is interior to" is among them, then (eqn2) shall be its definition.
Anyone wish to speak against this? If no one, I shall assume agreement and include it in the upcoming revised draft standard text. If anyone, I will propose it as a motion.
Note it implies the following.
Empty is interior to anything, including itself.
Anything, including Entire, is interior to Entire.
[1,oo] is interior to [0,oo].
John Pryce