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I was looking at particular examples and I observed that they were always reflective, antisymmetric and transitive.

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And so are $\leq$, $=$. –  copper.hat Dec 7 '12 at 17:09
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@copper.hat I would think that equality "$=$" is not antisymmetric ;-) –  dtldarek Dec 7 '12 at 17:26
    
I was trying to be discrete... –  copper.hat Dec 7 '12 at 17:37

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up vote 2 down vote accepted

Indeed, it is (if you're using the $\ge$ relation that I think you are). In fact, it's a total order, since comparability holds, as well.

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so is it a partial order or what? they're suggesting that it's not. –  CuriousJoe Dec 8 '12 at 22:22
    
If the relation is defined in the way I suspect it is, then it is a partial order. Just in case I'm thinking of the wrong one, how are you defining the relation? –  Cameron Buie Dec 8 '12 at 22:49
    
In particular, I was assuming the natural order of the real numbers induced the relation. That gives a (non-strict) total order. It also totally orders every subset of the reals, and so partially orders them. –  Cameron Buie Dec 8 '12 at 23:18

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