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    • CommentAuthorRoy Maclean
    • CommentTimeDec 2nd 2010 edited
     
    The question http://mathoverflow.net/questions/48117/multi-inequality was closed.

    As Ryan Budney pointed out, multiple inequalities occur in the Simplex algorithm and in describing polyhedra. Thanks Ryan for pointing that out to me.

    To clarify, I was asking about single expressions like 2+3i(> <)1+4i in which each side is related by a multi-inequality symbol. I have edited the question to make this clearer.

    I thought it would be interesting to consider taking an inequality expression and replacing the inequality symbol with a multi-inequality symbol, and let the variables be members of non-real sets.

    Perhaps those closing knew what I meant and closed anyway, in which case I'll have to accept that. Could this question please be considered for reopening ?
  1.  

    I see that you've revised the question, but it doesn't look like you've made any effort to reconcile the notation you've invented with whatever is in place among people who study polyhedra. Among the reasons you might introduce a new concept or notation are: (a) it is sufficiently unambiguous that people don't just get confused, (b) it makes a problem easier to solve, or (c) it makes complicated objects easier to conceive or manipulate. While this may not be a comprehensive list, I think it covers most of the important parts, and as it stands, you haven't provided any evidence that your multi-inequality does any of these.

    It's not clear from the question whether you are really looking for an answer, or are trying to advertise this peculiar concept. I expect the answer to your first question is "no, no one has used that notation," and the answer to your second question is "inequalities are useful, and multi-inequalities don't offer substantial advantages."

  2.  

    The question has the form "Has anyone seen this mathematical concept before?" But I don't see a new mathematical concept: the concept of finite systems of inequalities has been around for centuries. I just see a proposed new notation for the concept. It seems reasonable to close questions of the form "Has this notation been used before?" because 1) they are quite localized in nature, perhaps not of interest outside of the user of the notation, 2) in most cases the answer is probably going to be "no", but how do you conclusively establish that a certain notation has never before been used?, and 3) such questions could be endlessly generated.

    It sounds like you wish to ask about something beyond notation, but honestly whatever else you may have in mind is far from clear. That is why you were asked to edit the question before asking it to be reopened. As far as I can see you have barely edited the question at all. Remember that the burden is always on the questioner to make himself/herself clear. Please put a little more effort into this.

    • CommentAuthorRoy Maclean
    • CommentTimeDec 2nd 2010 edited
     
    "no, no one has used that notation"

    The question wasn't about the notation but about the concept: considering a suitable collection of partial orders as a multi-order in a single expression. I did want to know if that had been investigated before. Knowing if something has been studied before helps you to avoid reinventing the wheel.

    "doesn't look like you've made any effort to reconcile the notation you've invented with whatever is in place among people who study polyhedra."

    If several different polynomials are required to describe a polyhedron then I didn't think this was the same thing as the multi-inequality concept I was asking about. A multi-inequality in this question is not about starting with arbitrary collections of multiple inequalities. It's about taking expressions in partially-ordered variables and relating the expressions.
  3.  
    I think the answer to your question is that multi-inequality notation brings no added value. Multi-inequalities are just a bunch of inequalities with "and" statements stuck between them. There's no compelling reason to have a general-purpose notation based on multi-inequalities as they're so closely related to things we already use.
  4.  

    @Roy Maclean: I'm not trying to be argumentative, but I honestly don't understand why the "concept: considering a suitable collection of partial orders as a multi-order in a single expression" is not a notational concept. As Ryan explains, this is mathematically equivalent to considering a certain Boolean combination of inequalities, which is very well studied.

    Again, if you intend more than this, you'll have to say more, at least if you want others on this site to understand and consider your question, which I presume is why you asked it.

    Added a few minutes later: "A multi-inequality in this question is not about starting with arbitrary collections of multiple inequalities. It's about taking expressions in partially-ordered variables and relating the expressions." Yes, I think I agree with this. It looks you are restricting to a special kind of system of inequalities. What I (and I assume the five others who closed the question) don't understand is why you're doing this and exactly what you hope to gain, what problems you're trying to solve, and so forth. All you've given us is the notation.

  5.  
    There is a distinction here between a partially ordered set and a multi-ordered set. As far as I know a tree can have a partial order, and so can the complex numbers, but the complex numbers being component-based with each of the components having a total order means that the complex numbers have a natural multi-order i.e. a collection of relations that cover the whole set, so that any two complex numbers are related by one of the relations (<<), (<>), (<=),( ><),(>>),(>=),(=<),(=>),(==). The real inequality a<b is transformed into -a>-b. The question is if you have a multi-order then what transform do you need to turn say w(<<)z into Tw(>>)Tz, or Sw(<>)Sz, Rw(><)Rz etc. There probably aren't any such transforms for the complex numbers, but maybe there are in other situations, so the question becomes are there any component-based objects where the inequalities between the components change in a predictable way when the object as a whole is transformed.

    My motivation for this is the same as for my earlier questions about cyclic order relations http://mathoverflow.net/questions/25505/cyclic-order-relation-in-zn and about set inclusions at http://www.google.com/buzz/114134834346472219368/FAYeqjLsKYK/Much-of-hard-analysis-relies-extensively-on: namely to look for ungrazed pastures - there is a huge literature on inequalities, so if the concept of inequality is widened all the questions about inequalities can be asked again and perhaps give rise to non-real constants http://mathoverflow.net/questions/24740/non-real-constants
    • CommentAuthorRoy Maclean
    • CommentTimeDec 3rd 2010 edited
     
    I've just googled "multi-ordered set" and found page 471 of "Applied geometry and discrete mathematics:
    the Victor Klee festschrift" by Peter Gritzmann, Bernd Sturmfels
    http://books.google.com/books?id=Fe1M5i6SvooC&pg=PA471&lpg=PA471

    I did not have the phrase "multi-order" in mind when I initially asked the question. It was only on this meta thread that I came to that term and now I've found a book chapter about multi-ordered sets (mosets ?)
  6.  

    @Roy Maclean: In fact I just looked up multi-order on mathscinet. I got some hits, most of which refer to the order of a differential equation and one in mathematical logic. Sieving these out, I found the same paper you reference above. In fact this is the unique hit for "multi-ordered set" on mathscinet.

    The fact that you're interested in this particular kind of structure does clarify your question a bit.

    But I still don't understand the part about the complex numbers. The multi-order you define is not compatible with the product of complex numbers in any reasonable way. So what do you mean by a "transform"? One way to transform (<,<) into (>,>) is simply to send (x,y) to (-x,-y). But this seems trivial.

    If you can clarify this point, you might have a "real question" and I would consider voting to reopen.

    • CommentAuthorRoy Maclean
    • CommentTimeDec 3rd 2010 edited
     
    I used the complex numbers because they are a simple example of vectors over the reals with vector multiplication, i.e. an algebra.

    I know that multiplying complex numbers does not change the order type in a consistent way. It is a bad example. I don't know if there is a good example. That's my question.

    Even in a good example I expect that going from all less thans (<,<,<,<,<) to all more thans (>,>,>,>,>) will be relatively trivial compared to going between any old signatures like from (<,>,<,>>) to (>,>,<,>,<).

    Since multiplying complex numbers doesn't work in this scenario, then I wondered if maybe there is some other operation that can be performed on both sides of the inequation that would result in a modified inequation with the order type changed and I just used the word "transform" to describe such an operation.
  7.  

    An example would be the product algebra R x R x ... R with coordinatewise multiplication. Then as I said above, you can track the change on multi-orders just by looking at the signs on the components.

    Again, this is not a very interesting example. I'm trying to show you that your question is under-determined. It's some kind of question, but not the sort of precise, specific question that is appropriate on MO. "Is there an algebra over R with a multi-order [you will have to precisely define this] satisfying the following properties...?" would be roughly the right format for a MO question. To me, it seems that you're throwing out rather vague ideas and hoping that someone else can make them precise. That's what friends are for, not MO.

    • CommentAuthorHarry Gindi
    • CommentTimeDec 3rd 2010 edited
     

    That's what friends are for, not MO.

    I consider MO one of my dearest friends.