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Lecture 19: equivalence classes

• Reading: MCS 10.10-10.11

• Last semester's notes

• $A / R$, properties of equivalence classes, defining functions

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.

Foreshadowing

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.