So I'm having a hard time grasping how a relation can be both antisymmetric and symmetric, or neither.

Are my examples correct? symmetric & antisymmetric

R ={(1,1),(2,2),(3,3)}

not symmetric & not antisymmetric

R = { (1,2),(2,1),(3,4) }
  • $\begingroup$ By "synthetic" do you mean "symmetric"? $\endgroup$ – wj32 Oct 31 '12 at 4:12
  • $\begingroup$ @wj32 yeah! sorry bout that will edit $\endgroup$ – Hoodlum Oct 31 '12 at 4:13

Here’s a way to think about symmetry and antisymmetry that some people find helpful. A relation $R$ on a set $A$ has a directed graph (or digraph) $G_R$: the vertices of $G_R$ are the elements of $A$, and for any $a,b\in A$ there is an edge in $G_R$ from $a$ to $b$ if and only if $\langle a,b\rangle\in R$. Think of the edges of $G_R$ as streets. The properties of symmetry, antisymmetry, and reflexivity have very simple interpretations in these terms:

  • $R$ is reflexive if and only if there is a loop at every vertex. (A loop is an edge from some vertex to itself.)

  • $R$ is symmetric if and only if every edge in $G_R$ is a two-way street or a loop. Equivalently, $G_R$ has no one-way streets between distinct vertices.

  • $R$ is antisymmetric if and only every edge of $G_R$ is either a one-way street or a loop. Equivalently, $G_R$ has no two-way streets between distinct vertices.

This makes it clear that if $G_R$ has only loops, $R$ is both symmetric and antisymmetric: $R$ is symmetric because $G_R$ has no one-way streets between distinct vertices, and $R$ is antisymmetric because $G_R$ has no two-way streets between distinct vertices.

To make a relation that is neither symmetric nor antisymmetric, just find a digraph that has both a one-way street and a two-way street, like this one:

$$0\longrightarrow 1\longleftrightarrow 2$$

It corresponds to the relation $R=\{\langle 0,1\rangle,\langle 1,2\rangle,\langle 2,1\rangle\}$. on $A=\{0,1,2\}$.


Symmetric means if $(a,b)$ is there then so is $(b,a)$. Antisymmetric means if $(a,b)$ is there then $(b,a)$ isn't there.

Look at your second example. $(3,4)$ is there. Is $(4,3)$ there? Is the relation symmetric? $(1,2)$ and $(2,1)$ are both there. Is the relation antisymmetric?

  • $\begingroup$ Yey! So I'm correct in both examples? $\endgroup$ – Hoodlum Oct 31 '12 at 4:42
  • 2
    $\begingroup$ The intuition you provided is really good, however, not precise enough and confusing in the case $a = b$. The way to think about it is that anti symmetric, if you draw a diagonal line, then the opposite side to (a,b) (i.e. b,a) is not there, but the diagonal is ok to be included. i.e. (a,a) is fine to be included in an antisymmetric relation. i.e. if (b,a) is in R and (b,a) is in R then a=b. $\endgroup$ – Charlie Parker Oct 9 '14 at 21:40

One definition of antisymmetry is $$ [(a,b) \wedge (b,a)] \Rightarrow a = b. $$ That is, if $a$ is related to $b$ and vice-versa, then $a$ and $b$ are actually the same element.

In your first example, the lefthand side of the definition above is never invoked for distinct $a$ and $b$. For this reason, you might say the relation is vacuously antisymmetric. The argument for its symmetry is similar. In fact, the only way a relation can be both symmetric and antisymmetric is if all its members are of the form $(x,x)$, like in the example you give.

To violate symmetry or antisymmetry, all you need is a single example of its failure, which Gerry Myerson points out in his answer.


protected by Zev Chonoles Feb 23 '16 at 22:09

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