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Math Help - Partial Orders

  1. #1
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    Partial Orders

    Hi there, I have a question relating to partial orders, I can prove only part of the question but can't seem to finish it off, any help would be greatly appreciated!

    Let T denote the set of real-valued functions that are differentiable at the point x=0 we define a relation R on T by 'f R g iff f'(0) \le g'(0)'. Does R give rise to a partial ordering of T? If not, then which of reflexive, antisymmetry, transitive fails?

    So for R to be a partial order on T then R must be reflexive, antisymmetric and transitive.

    I can show R is reflexive, for example, let h be any real valued function that is differentiable at the point x=0, ie, h \in T

    Thus h'(0) \le h'(0) so (h,h) \in R and thus R is reflexive.

    I am unsure of how to go about showing R is antisymmetric and transitive (or how to go about showing if one of these fails)

    Thanks again!

    -----------------------------------------------------------------------------------------

    Actually after thinking more, I can show R is a partial order but the answer says it's not?

    I showed R is antisymmetric like this:

    If (f, g) \in R and (g,f) \in R then it is clear that f'(0)=g'(0)

    I also showed R is transitive like this:

    If f'(0) \le g'(0) and g'(0) \le z'(0) then f'(0) \le z'(0) and thus (f, z) \in R

    So where in my working have I gone wrong...? :S
    Last edited by usagi_killer; September 6th 2010 at 09:18 AM. Reason: Posting what I worked out
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  2. #2
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    Consider the functions f(x)=x^2~\&~g(x)=cos(x).
    Is the relation antisymmetric?
    i.e. Does f=g?
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  3. #3
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    Why don't you start by writing what it means for any relation R on T to be antisymmetric and then substituting the actual definition for the letter R in that expression? You will get a proposition of the form, "For every two functions f and g in T, ...". Then we can decide whether it is true or false.
    Last edited by emakarov; September 6th 2010 at 09:21 AM. Reason: I am always three minues behind Plato :-)
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  4. #4
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    Oh right, sigh, what a careless mistake by me, thank you Plato (you're always helping me ^_^) and emakarov.

    I assume my reasoning behind reflexivity and transitivity are valid...?
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  5. #5
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    I assume my reasoning behind reflexivity and transitivity are valid...?
    Yes.
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