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Partially ordered set

  1. Nov 19, 2007 #1


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    1. The problem statement, all variables and given/known data


    3. The attempt at a solution

    I am quite confused as to how to define the function. Any help would be appreciated.
  2. jcsd
  3. Nov 19, 2007 #2
    The new set, [tex] f(a) [/tex], is a set consisting of all [tex] x \epsilon A [/tex] s.t. [tex] x \leq a [/tex].

    For example, consider [tex] A = \mathbb{N} [/tex]. Look at [tex] f(5) [/tex]. This is simply [tex] \{1,2,3,4,5\} [/tex].

    So, for the first part, you need to show that

    [tex] f(a)=f(b) \Rightarrow a=b [/tex]

    I'll stop here, just in case you need more assistance.
    Last edited: Nov 19, 2007
  4. Nov 20, 2007 #3


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    In other words, f(a) is the set of all members of A that are less than or equal to a.

    If A were the set of real numbers with the usual order (which is, or course, not "partial"), then f(3)= ([itex]-\infty[/itex],3]. If A were a collection of sets, with "<" meaning set inclusion (that is a "partial" order) then f(a) is the collection of all subsets of a.
  5. Nov 20, 2007 #4


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    What's wrong with [itex]f(a) = \{ \, x \mid x \in A \wedge x \leq a \, \}[/itex]?
  6. Nov 20, 2007 #5


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    Yea I was confused last night and reading too much into the question. I got it (i think) now.
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