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Math Help - A partial order set with a non-unique maximal element

  1. #1
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    A partial order set with a non-unique maximal element

    Find an example of a partial order set with a maximal element that is not unique.

    My question here is, since a maximal element is define as:

    m is a max of a set X if  m \leq x , then x = m.

    Not unique means that there are at least two, but how can I find two such elements? If we have two different max, say m and n, then either m \leq n or  n \leq m since  m \neq n , but then by definition of a maximal element, m is n...

    Thank you!
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  2. #2
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    let x\le y if x^2\le y^2. And to this on the interval [-a,a] for a>0.
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  3. #3
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    Quote Originally Posted by tttcomrader View Post
    Find an example of a partial order set with a maximal element that is not unique.
    Remember this is a partial order not a total ordering.
    Consider \{1,2,3,4,5,6,7,8,9\} ordered by divides, <br />
x \prec y \Leftrightarrow x|y.
    Does that answer the question?
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  4. #4
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    Or just make up something like {a, b, c, d} with a< b< d, a< c< d with b and c "non-comparable". ("Non-comparable meaning none of b< c, c< b, or b= c are true. d is maximal (in fact, the maximum) because every other member is less than it. That would not be possible in a linearly ordered set but is in a partially ordered set.)

    If you want something a little more concrete use sets. Supose A= {a}, B= {a, b}, C= {a, c}, D= {a, b, c} with " \le" defined by "inclusion": [tex]X\le Y[tex] if and only if X is a subset of Y. That is partially ordered because neither B nor C is a subset of the other. B\le C is false because B is not a subset of C. C\le B is not true because C is not a subset of B. Clearly B= C is not true. So this is a partially ordered set in which D is the maximum (and so "maximal").

    Yet another example, where D is maximal but not "the" maximum, is {A, B, C, D} where A= {a}, B= {a,b}, C= {c}, D= {c, d}. Now both B and D are maximal.
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