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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
@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
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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