I have this problem I'm trying to solve.
Given is a set $X$ and a relation $r$ on $X$. For an equivalence relation $c$ on $X$ we say that $c$ is produced by r, if $r \subseteq c$ and for all equivalence relations $d$ on $X$ with $r \subseteq d$ it follows that that $c \subseteq d$.
(1) Let $s$ be the transitive-reflexive closure of $r$. Show that: If $r$ is symmetric, then $s$ is symmetric. (2) Show that there is at most one equivalence relation that is produced by $r$.
My solution attempts:
(1) Let $x,y \in X$ and $(x,y) \in s$. If $(x,y) \in r$, then also $(y,x) \in r$, because $r$ is symmetric and because $r \subseteq s$, we have $(y,x) \in s$. If $(x,y) \not \in r$ and $x=y$, then clearly $(y,x) \in s$, because $s$ is reflexive. If $(x,y) \not \in r$ and $x \not=y$, then there is $z \in X$ such that $(x,z) \in r$ and $(z,y) \in r$. But since $r$ is symmetric, also $(z,x) \in r$ and $(y,z) \in r$ and because $s$ is transitive, also $(y,x) \in s$.
(2) Let $c,e$ be equivalence relations produced by $r$. It follows that $r \subseteq c$ and $r \subseteq e$. Since for all equivalence relations $d$ on $X$ with $r \subseteq d$, we have $c \subseteq d$ and $e \subseteq d$, we must also have $c \subseteq e$ and $e \subseteq c$. Therefore, we must have $c=e$.
I'm sure there are some mistakes. So I'd be very glad if you could point them out. I'd also appreciate if you could propose an alternative proof.