Transitive closure of $p=\{(1,3),(2,1),(3,2),(4,1)\}$

Question

What is the transitive closure of the relation $p$? I thought it would just be $t=p \cup p^2$. But in the solution I have, there is also $p^3$. Why is this so? What I showed is already the smallest transitive set!

Answer 1

In general: if $R$ is a relation then $S:=\bigcup_{n=1}^{\infty}R^n$ is its transitive closure.

If $T$ is a transitive relation with $R\subseteq T$ then with induction it can be shown that $R^n\subseteq T$ for each $n\in\{1,2,\dots\}$, so that $S\subseteq T$.

Conversely it can be shown that $S$ is a transitive relation with $R\subseteq S$.