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Can a partial order by symmetric in addition to being reflexive, antisymmetric, and transitive?

Also, can an equivalence relation be antisymmetric aside from being reflexive, symmetric, and transitive?

All of the definitions I see only state that a relation has to be those things in order for it to be considered a partial order or an equivalence relation. The definitions do not state that it has to be NOT antisymmetric or NOT symmetric.

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    Please don't rely on the title for content. The body of the post should be self-contained. It's rather jarring to start reading a post that starts with "Also..."2011-04-05

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The only reflexive, symmetric, and antisymmetric relation on a set $X$ is $\{(x,x):x\in X\}$. Reason: If $(x,y)$ is in the relation, then by symmetry so is $(y,x)$, then by antisymmetry $x=y$. This shows that the relation is contained in $\{(x,x):x\in X\}$, and the other containment is the definition of reflexivity.

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Yes, although a poset whose partial ordering is symmetric will have a 'trivial' partial ordering, namely $a \leq b \Rightarrow a=b$

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The only way for a partially ordered set to be symmetric is if no two distinct elements $x$ and $y$ are comparable. Otherwise, if there is some $x, then by symmetry $y. Transitivity then implies $x which contradicts the fact that partial orders are irreflexive.

Edit: I was using a slightly different definition of partial order. (Irreflexive and transitive.)

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    @joriki: Thanks, I didn't notice that, although I should have!2011-04-04