# Understanding the definition of congruences over a lattice

Let $$(L, \land, \lor)$$ a lattice and $$\theta$$ a binary relation over $$L$$. We say $$\theta$$ is a congruence iff

$$x_0\theta x_1, y_0 \theta y_1 \Rightarrow (x_0 \lor y_0)\theta(x_1 \lor y_1)$$

(and the same for $$\land$$).

I am confused by the required cardinality of $$\theta$$. Any equivalence relation is isomorphic to a partition over its set. The definition does not require (explicitly) that $$x_0, x_1$$ and $$y_0, y_1)$$ be distinct. So does that mean we could have a congruence where an element is only equivalent to itself?

To take a simple example, let $$(\{1,2,3,4\}, \max, \min)$$ be a lattice. Let $$\theta$$ be the binary relation induced by the partition $$\{\{1\}, \{2\}, \{3\}, \{4\}\}$$. How could we apply the definition of congruence to test if $$\theta$$ is one? Can we say that it is a congruence because

$$1\theta 1, 2\theta 2, \text{ and } (1\max 2)\theta(1 \max 2)$$

and so to for all other elements?

• Yes, a congruence can have singleton classes. Commented Jul 14 at 5:11
• Equality is a congruence with only singleton classes. This has nothing to do with lattices. Commented Jul 14 at 8:04
• Sorry, there was a typo, the statement was: "any equivalence relation is isomorphic to a partition over a set". What I mean by this is that there is a perfect correspondance (i.e. a bijection) between the set of possible equivalence relations and the set of possible partitions of a set; any equivalence relation can be represented through a partition and vice-versa. If this is not the correct use of the term "isomorphic", I apologize - I'm not a mathematician. Commented Jul 14 at 17:38
• Thanks @NoahSchweber for answering my question. Commented Jul 14 at 17:38