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.

  • $\begingroup$ Consider a very tiny set, or a very very tiny set. $\endgroup$ Apr 4, 2011 at 23:28
  • 2
    $\begingroup$ To be precise, it's the partial order, not the partially ordered set, that is reflexive, antisymmetric and transitive. $\endgroup$
    – joriki
    Apr 4, 2011 at 23:30
  • $\begingroup$ 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..." $\endgroup$ Apr 5, 2011 at 2:08

3 Answers 3


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.


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


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 $x<y$, then by symmetry $y<x$. Transitivity then implies $x<x$ which contradicts the fact that partial orders are irreflexive.

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

  • $\begingroup$ There are different definitions of partial orders, corresponding to $<$ and $\le$. The question implied that the one where a partial order is reflexive ($\le$) is being used. $\endgroup$
    – joriki
    Apr 4, 2011 at 23:28
  • $\begingroup$ @joriki: Thanks, I didn't notice that, although I should have! $\endgroup$ Apr 4, 2011 at 23:34

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