1.2.4: Equivalence Relations
The identity relation on a set is reflexive, symmetric, and transitive. Relations \(R\) that have all three of these properties are very common.
A relation \(R \subseteq A^2\) that is reflexive, symmetric, and transitive is called an equivalence relation . Elements \(x\) and \(y\) of \(A\) are said to be \(R\) -equivalent if \(Rxy\) .
Equivalence relations give rise to the notion of an equivalence class . An equivalence relation “chunks up” the domain into different partitions. Within each partition, all the objects are related to one another; and no objects from different partitions relate to one another. Sometimes, it’s helpful just to talk about these partitions directly . To that end, we introduce a definition:
Let \(R \subseteq A^2\) be an equivalence relation. For each \(x \in A\) , the equivalence class of \(x\) in \(A\) is the set \(\equivrep{x}{R} = \Setabs{y \in A}{Rxy}\) . The quotient of \(A\) under \(R\) is \(\equivclass{A}{R} = \Setabs{\equivrep{x}{R}}{x \in A}\) , i.e., the set of these equivalence classes.
The next result vindicates the definition of an equivalence class, in proving that the equivalence classes are indeed the partitions of \(A\) :
If \(R \subseteq A^2\) is an equivalence relation, then \(Rxy\) iff \(\equivrep{x}{R} = \equivrep{y}{R}\) .
Proof. For the left-to-right direction, suppose \(Rxy\) , and let \(z \in \equivrep{x}{R}\) . By definition, then, \(Rxz\) . Since \(R\) is an equivalence relation, \(Ryz\) . (Spelling this out: as \(Rxy\) and \(R\) is symmetric we have \(Ryx\) , and as \(Rxz\) and \(R\) is transitive we have \(Ryz\) .) So \(z \in \equivrep{y}{R}\) . Generalising, \(\equivrep{x}{R} \subseteq \equivrep{y}{R}\) . But exactly similarly, \(\equivrep{y}{R} \subseteq \equivrep{x}{R}\) . So \(\equivrep{x}{R} = \equivrep{y}{R}\) , by extensionality.
For the right-to-left direction, suppose \(\equivrep{x}{R} = \equivrep{y}{R}\) . Since \(R\) is reflexive, \(Ryy\) , so \(y \in \equivrep{y}{R}\) . Thus also \(y \in \equivrep{x}{R}\) by the assumption that \(\equivrep{x}{R} = \equivrep{y}{R}\) . So \(Rxy\) . ◻
A nice example of equivalence relations comes from modular arithmetic. For any \(a\) , \(b\) , and \(n \in \Nat\) , say that \(a \equiv_n b\) iff dividing \(a\) by \(n\) gives remainder \(b\) . (Somewhat more symbolically: \(a \equiv_n b\) iff \((\exists k \in \Nat)a - b = kn\) .) Now, \(\equiv_n\) is an equivalence relation, for any \(n\) . And there are exactly \(n\) distinct equivalence classes generated by \(\equiv_n\) ; that is, \(\equivclass{\Nat}{\equiv_n}\) has \(n\) elements. These are: the set of numbers divisible by \(n\) without remainder, i.e., \(\equivrep{0}{\equiv_n}\) ; the set of numbers divisible by \(n\) with remainder \(1\) , i.e., \(\equivrep{1}{\equiv_n}\) ; …; and the set of numbers divisible by \(n\) with remainder \(n-1\) , i.e., \(\equivrep{n-1}{\equiv_n}\) .
Show that \(\equiv_n\) is an equivalence relation, for any \(n \in \Nat\) , and that \(\equivclass{\Nat}{\equiv_n}\) has exactly \(n\) members.