I was told to give an example of one of each of these kinds, and this is what I came up with:
(these are both relations on the set of all positive integers)
R = { (a,b) | a = b} is an example of a relation of a set that is both symmetric and antisymmetric. It is both symmetric because if (a,b) ∈ R, then (b,a) ∈ R (if a = b). Since (a,b) ∈ R and (b,a) ∈ R if and only if a = b, then it is anti-symmetric.
R = { (a,b) | a <= b }
It is not symmetric because a < b and b < a can never both be true. The antisymmetric part kinda confuses me..but I guess (a,b) ∈ R and (b,a) ∈ R ONLY when a = b, which I believe is true in this case.
Did I do this correctly?
Best Answer
Your first answer is correct for the reason that you give; your second is not. The relation $\le$ on $\Bbb Z^+$ is not symmetric, but it is antisymmetric: if $m\le n$ and $n\le m$, then $m=n$.
The easiest way to find a relation $R$ that is neither symmetric nor antisymmetric is to build one from scratch. To ensure that $R$ is not symmetric, we must put two distinct elements, say $0$ and $1$, into the underlying set $A$ and put exactly one of the ordered pairs $\langle 0,1\rangle$ and $\langle 1,0\rangle$ into $R$; I’ll put $\langle 0,1\rangle$ into $R$ and leave $\langle 1,0\rangle$ out. So far, then, we have $0,1\in A$ and $\langle 0,1\rangle\in R$.
To ensure that $R$ is not antisymmetric, we must have two elements of $A$ — call them $a$ and $b$ for a moment — such that $a\ne b$, but both of the ordered pairs $\langle a,b\rangle$ and $\langle b,a\rangle$ belong to $R$. We can’t use $0$ and $1$ for $a$ and $b$, since we’ve already required that $\langle 1,0\rangle\notin R$, but I can add $2$ to $A$ and use $0$ and $2$ for $a$ and $b$. That is, I’ll set $A=\{0,1,2\}$ and $R=\{\langle 0,1\rangle,\langle 0,2\rangle,\langle 2,0\rangle\}$; then