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Ordered Sum of Sets

  1. Nov 27, 2007 #1

    Office_Shredder

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    If [tex]M_1[/tex] and [tex]M_2[/tex] are ordered sets, the ordered sum [tex]M_1+M_2[/tex] is the set [tex]M_1\cupM_2[/tex] with the ordering defined as:

    If [tex]a,b \epsilon M_1[/tex] or [tex]a,b \epsilon M_2[/tex] then order them as they would be in the original orderings. If [tex]a \epsilon M_1[/tex] and [tex]b \epsilon M_2[/tex] then [tex]a<b[/tex]

    The question then is if [tex]a \epsilon M_1[/tex] and [tex]a \epsilon M_2[/tex], then we get [tex]a< a[/tex] which is impossible. In general, it seems you'll get a is less than and greater than some elements, which means [tex]M_1+M_2[/tex] isn't really ordered at all

    (I use epsilon as the 'element of' symbol as I couldn't find a more appropriate one in the latex pdfs)
     
  2. jcsd
  3. Nov 28, 2007 #2
    It seems like there's a loophole here since if [tex]a \epsilon M_1[/tex] and [tex]a \epsilon M_2[/tex], then you should be interpreting the ordering there under the first clause, not the second clause (since [tex]a,a \epsilon M_1[/tex]).

    Alternately how is "[tex]M_1_2[/tex]" defined? If this is a union, then shouldn't the a = a case never come up since something cannot be a member of a set "more than once"?

    Otherwise maybe whoever you're getting this from just made a mistake in their wording...

    Well, this again comes back to the wording being kind of confusing. Let's say you have a, b where [tex]a,b \epsilon M_1[/tex] and also [tex]a,b \epsilon M_2[/tex]. And let's say by [tex]M_1[/tex]'s ordering a < b, and by [tex]M_2[/tex]'s ordering b < a. What do you do here?

    However if you can somehow resolve this case, for example if the original wording gives one some excuse to declare that [tex]M_1[/tex]'s ordering takes precedence, then I think (a < b ?) will always be unambiguous.
     
  4. Nov 28, 2007 #3

    Office_Shredder

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    Right, I see what you're talking about with the case [tex]a,a \epsilon M_1[/tex] cutting you off from 'seeing' the [tex]a \epsilon M_1[/tex] and [tex]a \epsilon m_2[/tex] case. But you still have trouble if [tex]b \epsilon M_1[/tex], [tex]c \epsilon M_2[/tex] and a<b in [tex]M_1[/tex] c<a in [tex]M_2[/tex] then a<b<c<a which means it's not transitive.

    I have to apologize, when I wrote [tex]M_{12}[/tex] it was just poorly writing [tex]M_1 \cup M_2[/tex] so it didn't come out right. But even though a can only be an element of that set once, the way the ordering is defined it still works out fishily, unless it's modified to be if a is in [tex]M_2[/tex] and not [tex]M_1[/tex] then it gets ordered as if it was in [tex]M_2[/tex].

    EDIT: This isn't well defined, as if [tex]M_1=M_2=N[/tex] then the order type of [tex]\omega + \omega = \omega[/tex] which certainly isn't true if [tex]M_1=N M_2=Z_-[/tex] where the negative integers are ordered by their absolute value
     
    Last edited: Nov 28, 2007
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