Suppose that $(X,\wedge,\vee)$ is a complete lattice and that $\equiv$ is a congruence relation on $(X,\wedge,\vee)$, i.e, for all $x_1$, $x_2$ , $y_1$ and $y_2$ in $X$
- $x_1\equiv y_1$ and $x_2\equiv y_2$ implies $x_1\wedge x_2\equiv y_1\wedge y_2$, and
- $x_1\equiv y_1$ and $x_2\equiv y_2$ implies $x_1\vee x_2\equiv y_1\vee y_2$.
The congruence relation is an equivalence relation that preserves the lattice structure of $(X,\wedge,\vee)$.
My question is: does the congruence relation also preserve the completeness property? That is, if $(x_i)_{i\in I}$ and $(y_i)_{i\in I}$ are two families of elements in $X$ such that $x_i\equiv y_i$ for each $i$ in $I$, then do we have $\bigwedge_{i\in I}x_i\equiv\bigwedge_{i\in I}y_i$ and $\bigvee_{i\in I}x_i\equiv\bigvee_{i\in I}y_i$?
Given any infinite set $A$, define the relation $S \equiv T$ on $P(A)$ (the power set of $A$) to mean that the symmetric difference $S \triangle T$ is finite.
One can then use the fact that the finite subsets of $A$ form an ideal of $P(A)$ to show that $\equiv$ is in fact a lattice congruence.
However, $\equiv$ is not a complete lattice congruence. For example, let $I=A$, and $S_i=\{i\}$ and $T_i=\emptyset$ for each $i \in I$. Then, $S_i \equiv T_i$ for any $i \in I$, but $\bigcup_{i \in I} S_i \not\equiv \bigcup_{i \in I} T_i$.