Why is 'Antisymmetry' named so? 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?
 A: 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.
A: 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:


*

*Every pair of distinct elements of $X$ are $R$-comparable.

*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.
