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So when we talk about order relations for the familiar number systems, we are always introduced to the antisymmetry property which is $x \le y, x \ge y \implies x=y$.

When I think of the word 'antisymmetry', I think of something being the opposite of symmetry but not asymmetry. Is there any meaningful way to interpret it?

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Maybe you can think of it this way (which I just made up). The antisymmetry of a relation $R$ is equivalent to the condition that for any two $x\ne y$, if $x R y$ then not $y R x$. So symmetry between $x R y$ and $y R x$ is forbidden for $x\ne y$. – Rahul Aug 16 '12 at 5:25
I like that. Thanks – StuartHa Aug 16 '12 at 5:28
I've added tags relations, elementary-set-theory and order-theory - to me these tags seem suitable for questions about partial orders and similar relations. Feel free to change them, if you think a different set of tags would be more appropriate. – Martin Sleziak Aug 16 '12 at 6:07
@RahulNarain: Good explanation. The $\le$ fuzzes things up a bit. But works beautifully as an explanation for $\lt$, which is where the intuition comes from anyway. – André Nicolas Aug 16 '12 at 6:18

A relation $R$ on a set $A$ is symmetric if $a\,R\,b$ implies that $b\,R\,a$ for all $a,b\in A$. In other words, the implication $a\,R\,b\implies b\,R\,a$ holds in every possible case. It would be natural to use the term antisymmetric for a relation that was at the opposite extreme.

A relation $R$ for which the implication $a\,R\,b\implies b\,R\,a$ never holds would be at the opposite extreme. Of course there is no such relation, because the implication must always hold when $a=b$, so at best the opposite extreme is a relation $R$ such that if $a\,R\,b\implies b\,R\,a$, then $a=b$. Unfortunately, the implication $a\,R\,b\implies b\,R\,a$ is vacuously true if $a\,\not R\,b$, so the only relations with the property that $a\,R\,b\implies b\,R\,a$ implies that $a=b$ are the two relations on a one-element set. It hardly seems worthwhile to waste a nice name like antisymmetric on such an uninteresting property.

It seems that the property $(a\,R\,b\implies b\,R\,a)\implies a=b$ is really only interesting in those cases in which $a\,R\,b$. In other words, barring the uninteresting case noted above, the real opposite extreme from symmetry is a relation $R$ such that if $a\,R\,b\implies b\,R\,a$ and $a\,R\,b$, then $a=b$. But $$(a\,R\,b\implies b\,R\,a)\quad\mathbf{and}\quad a\,R\,b$$ is logically equivalent to the simpler statement $$a\,R\,b\quad\mathbf{and}\quad b\,R\,a\;,$$ so we’re now talking about a relation $R$ such that if $a\,R\,b$ and $b\,R\,a$, then $a=b$. In other words, we’re talking about what we do in fact call an antisymmetric relation.

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I want to like this interpretation, but the condition "if $a R b \implies b R a$ then $a = b$" is violated by any relation that has two distinct $a$ and $b$ such that $a R b$ does not hold. So it can only be satisfied by a relation on a set with at most one element. – Rahul Aug 16 '12 at 5:38
@Rahul: You’re absolutely right; that was a silly oversight. But I still like the basic idea, so I’ve modified the discussion to avoid the problem. It isn’t quite so straightforward as I’d hoped, but I think that it still works. – Brian M. Scott Aug 16 '12 at 5:57

There is probably no historical truth in what I am about to write, but one may consider it.

Given a symmetric relation $R$ on a set $X$, for every $x,y \in X$ there are exactly two possibilities: $$x \mathrel{R} y \wedge y \mathrel{R} x; \quad\text{or}\quad x \not\mathrel{R} y \wedge y \not\mathrel{R} x.$$ The natural way to define anti-symmetry would then be to only allow the other two possibilities, i.e., for all $x,y \in X$ exactly one of the following holds: $$x \mathrel{R} y \wedge y \not\mathrel{R} x; \quad\text{or}\quad x \not\mathrel{R} y \wedge y \mathrel{R} x. \tag{1}$$ Of course, this is inconsistent, as taking $x = y$ will easily show.

Let us now consider a relation $R$ such that (1) holds for all distinct $x , y \in X$. We get the following simple facts:

  1. Every pair of distinct elements of $X$ are $R$-comparable.
  2. If $R$ is also transitive, then $R$ is essentially a linear order (the only problem being that it may be neither reflexive nor irreflexive).

This first fact is quite strong, and the second fact basically says that it gives no new natural class of relations in the presence of transitivity. We would then like to weaken the condition, and the natural weakening would seem to be to allow distinct elements to be $R$-incomparable, meaning that for distinct $x, y$ one of the following three conditions hold: $$x \not\mathrel{R} y \wedge x \not\mathrel{R} y; \quad\text{or}\quad x \mathrel{R} y \wedge y \not\mathrel{R} x; \quad\text{or}\quad x \not\mathrel{R} y \wedge y \mathrel{R} x.$$ Of course, this is the same as demanding $$x \neq y \rightarrow ( x \not\mathrel{R} y \vee y \not\mathrel{R} x )$$ which is exactly what we mean by anti-symmetry.

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The disjunction of your three conditions does not contain $x\mathrel Rx$. – Rahul Aug 17 '12 at 1:18
@Rahul: I'm not certain what you are asking, but anti-symmetric relations are not necessarily reflexive. All we require is that whenever $x \mathrel{R} y$ and $y \mathrel{R} x$ both hold, then $x = y$. (My "triple disjunction" is only meant to apply to distinct $x,y$, as stated in the line above.) – arjafi Aug 17 '12 at 2:13
Sorry, I missed the part where it said "for distinct $x, y$". – Rahul Aug 17 '12 at 2:26

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