If $\theta_1$ and $\theta_2$ are two equivalence relations, how do I show that
$$\theta_1 \cup (\theta_1 \circ \theta_2) \cup (\theta_1 \circ \theta_2 \circ \theta_1) \: \cup ...$$
is it an equivalence relation too?
My problem is I am not sure what the $\cup$ does with the parentheses. I would certainly use that $\theta_1$ and $\theta_2$ are already equivalence relations, but not sure, how to proceed with the $\circ$ and $\cup$.
Thank you for any help!
The "$\circ$" means relational product:
the relational product $r\circ s$ of two binary relations $r,s$ on $A$ is given by: $\langle a,b \rangle \in r\circ s$ iff for some $c,\langle a,c \rangle \in r,\langle c,b\rangle \in s$
Some thoughts:
Call this union $R$. If $x \in X$ (the set on which the relations "live") then $(x,x) \in \theta_1 \subseteq R$. So $R$ is trivially reflexive.
If $(x,y) \in R$ then either $(x,y)\in \theta_1$ so $(y,x) \in R$ or
$(x,y) \in \theta_1 \circ \theta_2 \circ \ldots \circ \theta_i$ for $i\in \{1,2\}$ and some finite alternating composition. All such finite compositions are sub-equivalence relations of $R$ so again symmetric.
For transitivity we have to consider $(x,y)$ in some finite such chain (say length $n$) and $(y,z)$ in another such chain (length $m$) and perhaps we can reason that $(x,z)$ is in some $n+m$ length "composition chain" (or one shorter or longer), depending on the last $\theta$ in the chain perhaps?
I'd say it's not about how unions commute with compositions but a case by case proof of the different cases provided by the union..