Is the congruence relations lattice of a lattice a sublattice of all equivalence relations on it?

Question

By this Wikipedia link, it seem the set of all congruence relaions on a lattice $(X,\le)$ is a complete lattice with inclusion.

Is this lattice a (complete) sublattice of the lattice of all equivalence relations on $X$?

Answer 1

Let $\phi,\psi$ be equivalence relations on a set $X$. It is easy enough to show that the meet of $\phi$ and $\psi$ is $\phi\cap\psi$. Let $\phi\circ\psi=\{(a,b)\in X^2:\text{ there is }c\in X\text{ such that }(a,c)\in\phi \text{ and }(c,b)\in\psi\}.$ It would be nice if $\phi\circ\psi$ were $\psi\vee\psi$, but unfortunately $\phi\circ\psi$ is not transitive (show that it contains $\phi$ and $\psi$, is reflexive, and is symmetric though). However, the transitive closure is the join: $\phi\vee\psi=\phi\circ\psi\cup\phi\circ\psi\circ\phi\cup\phi\circ\psi\circ\phi\circ\psi\cup\dots$

Now assume $\phi$ and $\psi$ are congruences and show that the meet and join I defined above are also congruences. Also, see if you can guess what the join and meet of an arbitrary collection of equivalence relations is.