Join of two preorders and of two equivalence relations

Question

I'm sorry for the silly doubt. What is the join of two preorders? And of two equivalence relations? The meet is given by intersections. But in general the union of two preorders (resp. equivalence relations) fails to be a preorder (resp. equivalence relations). In the case of equivalence relation, is it the transitive closure of their union?

Answer 1

In general, any poset which has all meets also has all joins. Proof: the join of the set $S$ is the meet of the set $\{x | \forall s \in S, s \leq x\}$.

So in this case, the join of two preorders $a, b$ is just the intersection of all pre-orders containing both $a$ and $b$.

Similarly, the join of two equivalence relations $a, b$ is just the intersection of all pre-orders containing both $a$ and $b$.

It turns out that in both of these cases, the transitive closure of the union of $a$ and $b$ is the join.

For if $a, b$ are reflexive, then the transitive closure of $a \cup b$ is also reflexive.

And if $a, b$ are symmetric, then $a \cup b$ is symmetric, and thus the transitive closure of $a \cup b$ is symmetric.