I have tried without luck for a few hours now...
given $A=\{1,2,3\}$ find $R \subseteq A \times A$ such that $R \cup R^2$ is not transitive.
I have tried without luck for a few hours now...
given $A=\{1,2,3\}$ find $R \subseteq A \times A$ such that $R \cup R^2$ is not transitive.
What about $R=\{(1,2),(2,3),(3,1)\}$.
Then $R^2=\{(1,3),(2,1),(3,1)\}$, which implies $S=R\cup R^2= \{(1,2),(1,3),(2,1),(2,3),(3,1),(3,2)\}.$
The relation $S$ is not transitive since $(1,2),(2,1)\in S$ but $(1,1)\notin S$.
Let me explain how it is possible to come up with such examples.
You can visualize relations as arrow between elements of the sets. A pair $(x,y)$ is in $R^2$ if there is path of length 2 between $x$ and $y$. Thus $S=R\cup R^2$ contains all pairs joined by a path of length at most 2. Every transitive relation containing $S$ must contain all pairs for paths of lengths 3 and 4. So I was trying to construct a relation such that I have 2 elements joined by a path of length 3, but there is no path of length 2 between them.
The same thing explained a bit differently.
Lemma. A relation $S$ is transitive if and only if $S\circ S\subseteq S$.
Maybe you had this result on your lectures. You can find the proof e.g. at proofwiki.
If we use it for your case, we have $S\circ S=(R\cup R^2)\circ (R\cup R^2) = R^2 \cup R^3\cup R^4.$ This implies $S\circ S\setminus S= R^3\cup R^4 \setminus (R^2\cup R)$. So if I am able to find a relation such that $R^3\setminus (R^2\cup R)$ is non-empty, this will give a counterexample. (And this is basically looking for two elements such that the shortest path has length 3, as I explained above.)
If $R$ is transitive, then $R^2$ is a subset of $R$, so the union is transitive.
Therefore, $R$ cannot be chosen transitive.
If you don't have this condition, you can take a three-circle.