Motion P1788/M0034.01: a conditional NO
P1788
I vote NO on this motion. I will change my vote to YES if Ulrich confirms that the motion is consistent with the "I" in IR, \overline{IR}, etc., being an operator. Ideally I would have liked the motion changed to say this, but as we're now in the voting period that would entail withdrawing it, and I don't want to hold things up. So I just seek clarification.
My reasons, stated several times recently, are that at Level 2 it will be very convenient to write stuff like
\overline{I}(binary64)
to mean "the inf-sup type derived from IEEE754 binary64", i.e. all real intervals whose bounds are binary64 numbers.
But the conventions of the motion seem to require introducing a symbol to refer to the format:
"\overline{IF} where F is the format binary64."
Then, what if we have 3 such types to compare? Do we call them
"\overline{IF_1}, \overline{IF_2}, \overline{IF_3}, where F_1, F_2, F_3 are such & such formats?"
Or are we allowed
"\overline{IF}, \overline{IG}, \overline{IH}, where F, G, H are such & such formats?"
(Excuse the bastard LaTeX.)
However done, this will be clumsy.
Getting a notation that is self-consistent and pleases everyone is likely impossible. Ulrich thinks of \overline as denoting a completion in the lattice sense for ordered sets. Nate seeks a completion that converts a cancellative monoid to a group. I seek concise denotations for what we'll need to express in the Level 2 text.
I made my points in detail in an email to P1788 of 7 May 2012, at 21:52, so I won't repeat them here, but I append two off-list emails between me and Ulrich.
John Pryce
On 14 May 2012, at 09:53, Ulrich Kulisch wrote:
> John,
>
> I agree that we are very close. But I have a problem with your bar upon the I. It seems to me not being consistent with conventional use of the bar.
>
> R and IR are both conditionally completely ordered sets (every subset has an infimum and a supremum). Such sets can be completed by joining a least and a greatest element which leads to a complete lattice. The completion is indicated by overlining the notation for the set. So we get \overline{R} and \overline{IR}. So putting a bar upon the symbol of a set means adding some elements to the set.
>
> Let now F denote the set of floating-point numbers and \overline{F} := F U {-oo, +oo}. Then the set I\overline{F} contains the elements
> (1) [-oo, -oo], [-oo, a], [a, +oo], [-oo, +oo], with a \in F
> the bounds always included in the set.
> If putting a bar upon the I just means adding the empty set, then \overline{I}\overlineF} still cantains the elements (1) which are not in \overline{IF} in my notation.
> If you require that \overline{I}\overlineF} should be the same as \overline{IF} then putting a bar upon the I means taking the lements (1) out of I\overline{F} which is a totally unusual behavior of the overlining.
>
> Let me make another remark: I would prefer calling the elements of the set \overline{IR} "closed and connected sets of real numbers" instead of "closed real intervals". In colloquial English a real interval is a closed and bounded set of real numbers, i.e., an element of IR.
>
> Best wishes
> Ulrich
>
>
> Am 12.05.2012 11:00, schrieb John Pryce:
>> Ulrich
>>
>> On 11 May 2012, at 18:04, Ulrich Kulisch wrote:
>>> ...a preamble that states about the following:
>>>
>>> A real interval (or interval for short) is defined as a closed and connected set of real numbers (a set of real numbers is called closed, if its complement is open). A real interval can be bounded or unbounded. If it is unbounded the bounds -oo and +oo are not elements of the interval.
>> Indeed that is what is meant. It is said clearly in the standard text. And also in my email
>>> \overline{IR} the set of closed real intervals, including unbounded intervals and the empty set.
>> The term "real interval" means a subset of the reals, not of the extended reals.
>>
>> So I think we agree. If so, do you think my notation idea is OK?
>>
>> John
>