# Lecture 19: equivalence classes

## Equivalence classes

### Definitions

Definition: If $$R$$ is an equivalence relation on $$A$$ and $$x \in A$$, then the equivalence class of $$x$$, denoted $$[x]_R$$, is the set of all elements of $$A$$ that are related to $$x$$, i.e. $$[x]_R = \{y \in A \mid x R y\}$$. If $$R$$ is clear from context, we leave it out.

In the example above, $$[a] = [b] = [e] = [f] = \{a,b,e,f\}$$, while $$[c] = [d] = \{c,d\}$$ and $$[g] = [h] = \{g,h\}$$. The equivalence classes are easy to see in the diagram:

Definition: The set of all equivalence classes of $$A$$ is denoted $$A / R$$ (pronounced "$$A$$ modulo $$R$$" or "$$A$$ mod $$R$$"). Notationally, $$A/R = \{[x] \mid x \in A\}$$.

In the example above, $$A/R = \{[a], [c], [g]\}$$.

Definition: If $$c \in A/R$$ and $$x \in c$$, then $$x$$ is called a representative of $$c$$.

In the example above, $$a$$ is a representative of $$[b]$$, and $$d$$ is a representative of $$\{c,d\}$$.

### Examples

Equivalence classes let us think of groups of related objects as objects in themselves. For example

• if $$A$$ is the set of people, and $$R$$ is the "is a relative of" relation, then $$A/R$$ is the set of families

• if $$A$$ is the set of hash tables, and $$R$$ is the "has the same entries as" relation, then $$A/R$$ is the set of functions with a finite domain.

We'll see equivalence classes in several places in the remainder of the course:

• (cardinality) if $$A$$ is the set of all sets (this is actually problematic, but we'll pretend it makes sense), and $$R$$ is the "has a same cardinality as" relation, then $$A/R$$ is the set of cardinal numbers; which is how you would define $$|X|$$.

• (combinatorics) we'll see later that if $$A$$ is the set of sequences of length $$n$$, and $$R$$ is the "can be rearranged to" relation, then $$A/R$$ is the set of subsets of size $$n$$.

• (automata) we'll take $$A$$ to be the set of states of a machine, and $$R$$ to be the "behaves the same as" relation, and then $$A/R$$ will be the states of an optimized machine.

• (number theory) if $$A$$ is the set of integers, and $$R$$ is the "has the same remainder when divided by $$n$$ as" relation, then $$A/R$$ will be the modular numbers.

• (graphs) if $$A$$ is the set of vertices, and $$R$$ is the "is reachable from" relation, then $$A/R$$ is the set of connected components of the graph.

### Properties

Claim: if $$R$$ is an equivalence relation on $$A$$, then the equivalence classes of $$R$$ form a partition of $$A$$. That is, every element of $$x$$ is in some equivalence class, and no two different equivalence classes overlap.

Proof sketch: (you could fill in the details as an exercise)

• first part: every $$x$$ is in $$[x]$$ because $$R$$ is reflexive.

• second part: if $$[a]$$ and $$[b]$$ overlap, then there is some $$c$$ in the intersection. Then we can use symmetry and transitivity to show that every element of $$[a]$$ is related to $$d$$, and thus to $$b$$, and is thus in $$[b]$$; likewise, every element of $$[b]$$ is in $$[a]$$, so $$[a]$$ and $$[b]$$ are the same.

### Functions

We ended lecture with the following question. Let $$A$$ be a set of people, and let $$R$$ be the "is related to" relation (where everyone is assumed to be related to themselves).

Suppose I wrote down the following rule:

Let $$f : A/R → A$$ be defined by letting $$f([a])$$ be $$a$$'s oldest living relative

Is $$f$$ a function?

To see why it might or might not be, compare it with the following rule:

Let $$g : A/R → \mathbb{N}$$ be defined by letting $$g([a])$$ be $$a$$'s age.

One of these is a function, and the other is not. $$f$$ is a function, but not obviously so. $$g$$ is not a function. To see why, suppose that $$a$$'s age is 13 and $$b$$'s age is 25. Then $$g([a]) = 13$$ and $$g([b]) = 25$$. But $$[a] = [b]$$, so we have a single input giving multiple outputs, depending on how we write it down.

$$f$$ is a function, because if $$[a] = [b]$$ and if $$a$$'s oldest living relative is $$c$$, then $$b$$'s oldest living relative must also by $$c$$. So choosing different representatives of the input leads to the same value; the function is well-defined.