Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ordered Sum of Sets

  1. Nov 27, 2007 #1


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Ordered Sum of Sets
  1. Well-ordered set (Replies: 9)