Prove that if a relation $R$ on $X$ is not symmetric, but transitive, the collection of pseudoequivalence classes does not partition $X$.

Question

I'm trying to work through this problem for the class I'm teaching, but am getting stuck. I think the key is that there exists some $(x,y)\in R$ such that $(y,x) \notin R$, so $x$ won't be in a partition, but I'm not sure how to combine this with transitivity. Any help is appreciated.

Answer 1

Suppose $R$ is a relation on $X$ which is transitive but not symmetric. Because $R$ is not symmetric, there are two elements $x,y\in X$ such that $x\mathrel R y$ but not $y \mathrel R x$.

Now either there is a $z\in X$ such that $y\mathrel R z$ or there isn't.

If there isn't such a $z$ then $[y]=\varnothing$, which is not allowed in a partition.

Otherwise $z$ does exist. Now either there is a $w\in X$ such that $w \mathrel R x$ or there isn't.

If $w$ doesn't exist, then $x$ is in no pseudoequivalence class.

Otherwise both $w$ and $z$ exist and we have $w\mathrel R x \mathrel R y \mathrel R z $. Because $R$ is transitive we have $z\in [w]$ and $z\in[y]$. However, $x\in [w]$ but $x\notin [y]$ by our original choice of $x$ and $y$. So $z$ is in two different classes, which means that the collection is not a partition.