Lecture 7: Equivalence classes

Drawing binary relations

We can draw a binary relation \(A\) on \(R\) as a graph, with a vertex for each element of \(A\) and an arrow for each pair in \(R\).

For example, the following diagram represents the relation \(\{(a,b), (b,e), (b,f), (c,d), (g,h), (h,g), (g,g)\}\):

diagram of R (click for LaTeX source)

diagram of R (click for LaTeX source)

Using these diagrams, we can describe the three equivalence relation properties visually:

If we have a relation that we know is an equivalence relation, we can leave out the directions of the arrows (since we know it is symmetric, all the arrows go both directions), and the self loops (since we know it is reflexive, so there is a self loop on every vertex).

Closure

If we have a relation \(R\) that doesn't satisfy a property \(P\) (such as reflexivity or symmetry), we can add edges until it does. This is called the \(P\) closure of \(R\). Formally:

Definition: the if \(P\) is a property of relations, \(P\) closure of \(R\) is the smallest relation containing \(R\) that satisfies property \(P\).

For example, to take the reflexive closure of the above relation, we need to add self loops to every vertex (this makes it reflexive) and nothing else (this makes it the smallest reflexive relation). Here is a picture of the reflexive closure:

reflexive closure of R (click for LaTeX source)

reflexive closure of R (click for LaTeX source)

Similarly for the transitive and symmetric closures:

symmetric closure of R (click for LaTeX source)

symmetric closure of R (click for LaTeX source)

transitive closure of R (click for LaTeX source)

transitive closure of R (click for LaTeX source)

We can take the reflexive symmetric transitive closure of \(R\) to turn \(R\) into an equivalence relation:

reflexive symmetric transitive closure of R (click for LaTeX source)

reflexive symmetric transitive closure of R (click for LaTeX source)

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:

equivalence classes (click for LaTeX source)

equivalence classes (click for LaTeX source)

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

Foreshadowing

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

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)

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. We'll discuss this further next lecture.

Here is the result:

\(f\) 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.