# Thread: Prove or disprove that the complement of a relation is symmetric

1. ## Prove or disprove that the complement of a relation is symmetric

Consider a relation $\displaystyle R$ on a set $\displaystyle A$. Suppose that you are told that $\displaystyle R$ is symmetric. Prove or disprove that $\displaystyle \overline{R}$ is also symmetric.

I only really know this much:
$\displaystyle R\subset A\times A$ and $\displaystyle \overline{R}\subset A\times A-R$

2. In fact, $\displaystyle \bar{R}$ is not only a subset, but is equal to $\displaystyle (A\times A)-R$ (by definition).

To get a better intuition, consider an example, such as $\displaystyle xRy$ iff $\displaystyle x-y$ is even ($\displaystyle x,y$ are integers), or $\displaystyle A\,R\,B$ iff $\displaystyle A\cap B\ne\emptyset$ ($\displaystyle A,B$ are sets). Take a pair $\displaystyle (x,y)$ from $\displaystyle \bar{R}$ and see if $\displaystyle (y,x)\in\bar{R}$. Do this several times and then try to have a different outcome for whether $\displaystyle (y,x)\in\bar{R}$.

3. Originally Posted by emakarov
In fact, $\displaystyle \bar{R}$ is not only a subset, but is equal to $\displaystyle (A\times A)-R$ (by definition).

To get a better intuition, consider an example, such as $\displaystyle xRy$ iff $\displaystyle x-y$ is even ($\displaystyle x,y$ are integers), or $\displaystyle A\,R\,B$ iff $\displaystyle A\cap B\ne\emptyset$ ($\displaystyle A,B$ are sets). Take a pair $\displaystyle (x,y)$ from $\displaystyle \bar{R}$ and see if $\displaystyle (y,x)\in\bar{R}$. Do this several times and then try to have a different outcome for whether $\displaystyle (y,x)\in\bar{R}$.
I'm afraid that didn't really help me. Would it be possible for you to show me?

4. If $\displaystyle (x,y)\in \overline{R}$ then $\displaystyle (x,y)\notin R$.
But $\displaystyle R$ is symmetric so $\displaystyle (y,x)\notin R$.
That means that $\displaystyle (y,x)\in \overline{R}$. DONE.