MATH 308 — Lec 14 — Equivalence Relations

Lecture 14 — Equivalence Relations and Equivalence Classes
MATH 308  Bridge to Advanced Mathematics  Chartrand §9.3–9.4

Goals for today. By the end of this lecture you should be able to:

• check whether a relation is reflexive, symmetric, and transitive, and so decide whether it is an equivalence relation;

• compute the equivalence class $[a]$[a] of an element and list the distinct classes;

• state and prove the Fundamental Theorem: the equivalence classes of an equivalence relation on $S$S form a partition of $S$S, and conversely every partition arises this way;

• recognize equivalence as “sameness up to something” across the rest of mathematics.

Equivalence relations (§9.2)

Recall from Lecture 13 that a relation $R$R on a set $S$S is just a subset $R \subseteq S \times S$R⊆S×S; we write $a\, R\, b$aRb to mean $(a,b) \in R$(a,b)∈R. Three properties of a relation will matter today.

Definition 1. A relation $R$R on a set $S$S is

1. reflexive if $a\, R\, a$aRa for every $a \in S$a∈S;

2. symmetric if $a\, R\, b \Rightarrow b\, R\, a$aRb⇒bRa for all $a,b \in S$a,b∈S;

3. transitive if $\bigl(a\, R\, b \text{ and } b\, R\, c\bigr) \Rightarrow a\, R\, c$(aRb and bRc)⇒aRc for all $a,b,c \in S$a,b,c∈S.

An equivalence relation is a relation that is reflexive, symmetric, and transitive. When $R$R is an equivalence relation we usually write $a \sim b$a∼b in place of $a\, R\, b$aRb, read “$a$a is equivalent to $b$b.”

The three axioms are exactly the properties of the most familiar relation of all — equality. “$a=a$a=a” (reflexive), “$a=b$a=b forces $b=a$b=a” (symmetric), “$a=b$a=b and $b=c$b=c force $a=c$a=c” (transitive). An equivalence relation is what you get when you keep these three rules but relax “identical” to “the same in some chosen respect.”

Example 1 (The seed from Lecture 2: same parity). Define $a \sim b$a∼b on $\mathbb{Z}$Z to mean $a$a and $b$b have the same parity (both even or both odd). This is an equivalence relation:

• Reflexive: $a$a has the same parity as itself.

• Symmetric: if $a$a and $b$b have the same parity, so do $b$b and $a$a.

• Transitive: if $a,b$a,b agree in parity and $b,c$b,c agree in parity, then $a,c$a,c agree.

There turn out to be exactly two “kinds” of integer under $\sim$∼: the evens and the odds — precisely the partition $\{E,O\}${E,O} of $\mathbb{Z}$Z we met in Lecture 2. Hold onto this; §3 explains why it is no coincidence.

Example 2 (Congruence modulo $n$n). Fix $n \in \mathbb{N}$n∈N. For $a,b \in \mathbb{Z}$a,b∈Z define

$a \equiv b \pmod{n} \quad\Longleftrightarrow\quad n \mid (a-b),$a≡b(modn)⟺n∣(a−b),

read “$a$a is congruent to $b$b modulo $n$n” (recall $n \mid m$n∣m from Lecture 7 means $m = nk$m=nk for some $k \in \mathbb{Z}$k∈Z). This is an equivalence relation on $\mathbb{Z}$Z:

• Reflexive: $n \mid (a-a) = 0$n∣(a−a)=0 since $0 = n\cdot 0$0=n⋅0.

• Symmetric: if $a-b = nk$a−b=nk then $b-a = n(-k)$b−a=n(−k), so $n \mid (b-a)$n∣(b−a).

• Transitive: if $a-b = nk$a−b=nk and $b-c = n\ell$b−c=nℓ, then $a-c = (a-b)+(b-c) = n(k+\ell)$a−c=(a−b)+(b−c)=n(k+ℓ), so $n \mid (a-c)$n∣(a−c).

Same parity is exactly the case $n=2$n=2. We develop congruence modulo $n$n in full next time (Lecture 15).

Example 3 (A non-example). On $\mathbb{Z}$Z let $a\, R\, b$aRb mean $a \le b$a≤b. This is reflexive ($a \le a$a≤a) and transitive, but not symmetric: $2 \le 3$2≤3 yet $3 \not\le 2$3≤2. So “$\le$≤” is not an equivalence relation. (It is an order relation — a different, equally important species.) The lesson: all three axioms must be checked; losing even one changes the whole character of the relation.

Remark 1. A common trap: some relations look reflexive but fail because a lone element has no partner. On $S=\{1,2,3\}$S={1,2,3} the relation $R=\{(1,1),(1,2),(2,1),(2,2)\}$R={(1,1),(1,2),(2,1),(2,2)} is symmetric and transitive, but not reflexive, since $(3,3)\notin R$(3,3)∈/R. Reflexivity is a promise about every element of $S$S.

Equivalence classes (§9.3)

An equivalence relation sorts $S$S into groups of mutually equivalent elements. Each group is named by any one of its members.

Definition 2. Let $\sim$∼ be an equivalence relation on $S$S and let $a \in S$a∈S. The equivalence class of $a$a is

$[a] \;=\; \left\{\, x \in S \;:\; x \sim a \,\right\},$[a]={x∈S:x∼a},

the set of all elements equivalent to $a$a. Any element $b \in [a]$b∈[a] is called a representative of the class. The set of all equivalence classes is written $S/\!\sim$S/∼ (read “$S$S mod $\sim$∼”).

Example 4. For same parity on $\mathbb{Z}$Z there are just two classes:

$[0] = \left\{\, x \in \mathbb{Z} \;:\; x \text{ even} \,\right\} = E, \qquad [1] = \left\{\, x \in \mathbb{Z} \;:\; x \text{ odd} \,\right\} = O.$[0]={x∈Z:x even}=E,[1]={x∈Z:x odd}=O.

Note $[0]=[2]=[-4]=E$[0]=[2]=[−4]=E: different representatives can name the same class. And $E \cup O = \mathbb{Z}$E∪O=Z with $E \cap O = \varnothing$E∩O=∅ — the classes cover $\mathbb{Z}$Z and are disjoint.

Example 5 (Congruence mod $3$3). For $\equiv \pmod 3$≡(mod3) on $\mathbb{Z}$Z, sort integers by their remainder on division by $3$3:

$[0]=\{\dots,-3,0,3,6,\dots\},\quad [1]=\{\dots,-2,1,4,7,\dots\},\quad [2]=\{\dots,-1,2,5,8,\dots\}.$[0]={…,−3,0,3,6,…},[1]={…,−2,1,4,7,…},[2]={…,−1,2,5,8,…}.

Every integer lands in exactly one of these three classes, and $[3]=[0]$[3]=[0], $[4]=[1]$[4]=[1], and so on. There are exactly three distinct classes; we write $\mathbb{Z}/3\mathbb{Z}= \{[0],[1],[2]\}$Z/3Z={[0],[1],[2]}.

Congruence mod $3$3 colors the integers with three repeating labels. Filled dots are $[0]$[0], open circles $[1]$[1], triangles $[2]$[2]: three interleaved classes that together tile $\mathbb{Z}$Z.

The two examples share a striking feature: the distinct classes never overlap, and together they exhaust $S$S. That is the content of §3.

The Fundamental Theorem: classes are a partition (§9.3)

Recall from Lecture 2 that a partition of a nonempty set $S$S is a collection of nonempty, pairwise-disjoint subsets whose union is all of $S$S — equivalently, every element of $S$S lies in exactly one cell. We now show that equivalence relations and partitions are two names for the same thing. Everything rests on one lemma.

Lemma 1. Let $\sim$∼ be an equivalence relation on $S$S, and let $a,b \in S$a,b∈S. Then:

1. $a \in [a]$a∈[a];

2. $a \sim b \Leftrightarrow [a] = [b]$a∼b⇔[a]=[b];

3. if $a \not\sim b$a∼b, then $[a] \cap [b] = \varnothing$[a]∩[b]=∅.

Proof. (a) By reflexivity $a \sim a$a∼a, so $a \in [a]$a∈[a]. In particular every class is nonempty.

(b) ($\Rightarrow$⇒) Suppose $a \sim b$a∼b. We show $[a] \subseteq [b]$[a]⊆[b] and $[b] \subseteq [a]$[b]⊆[a] (double containment, Lecture 1). Let $x \in [a]$x∈[a], so $x \sim a$x∼a. Together with $a \sim b$a∼b, transitivity gives $x \sim b$x∼b, i.e. $x \in [b]$x∈[b]; hence $[a] \subseteq [b]$[a]⊆[b]. For the reverse, $a \sim b$a∼b gives $b \sim a$b∼a by symmetry, and the same argument with roles swapped yields $[b] \subseteq [a]$[b]⊆[a]. Therefore $[a]=[b]$[a]=[b].

($\Leftarrow$⇐) Suppose $[a]=[b]$[a]=[b]. By part (a), $a \in [a] = [b]$a∈[a]=[b], which means $a \sim b$a∼b.

(c) We prove the contrapositive: if $[a] \cap [b] \neq \varnothing$[a]∩[b]=∅ then $a \sim b$a∼b. Pick $x \in [a] \cap [b]$x∈[a]∩[b]. Then $x \sim a$x∼a and $x \sim b$x∼b. By symmetry $a \sim x$a∼x, and with $x \sim b$x∼b transitivity gives $a \sim b$a∼b. Contrapositively, $a \not\sim b$a∼b forces $[a] \cap [b] = \varnothing$[a]∩[b]=∅. ◻

Theorem 2 (Equivalence classes partition $S$S). Let $\sim$∼ be an equivalence relation on a nonempty set $S$S. Then the set of equivalence classes $S/\!\sim$S/∼ is a partition of $S$S.

Proof. We verify the two defining properties of a partition.

The classes are nonempty and cover $S$S. By Lemma 1(a), $a \in [a]$a∈[a] for every $a$a, so each class is nonempty and every element of $S$S lies in some class; hence $\bigcup_{a \in S}[a] = S$⋃a∈S​[a]=S.

Distinct classes are disjoint. Take two classes $[a]$[a] and $[b]$[b]. If they are not disjoint, then by Lemma 1(c) (contrapositive) $a \sim b$a∼b, and then Lemma 1(b) gives $[a]=[b]$[a]=[b]. So any two classes are either equal or disjoint. Thus the distinct classes are pairwise disjoint.

Both conditions hold, so $S/\!\sim$S/∼ is a partition of $S$S. ◻

The converse is equally true and just as useful: any way of chopping $S$S into cells produces an equivalence relation.

Theorem 3 (Partitions give equivalence relations). *Let $\mathcal{P}$P be a partition of $S$S. Define a relation on $S$S by

$a \sim b \quad\Longleftrightarrow\quad a \text{ and } b \text{ lie in the same cell of } \mathcal{P}.$a∼b⟺a and b lie in the same cell of P.

Then $\sim$∼ is an equivalence relation, and its equivalence classes are exactly the cells of $\mathcal{P}$P.*

Proof. Because $\mathcal{P}$P is a partition, each $a \in S$a∈S lies in exactly one cell; call it $C_a$Ca​.

• Reflexive: $a$a lies in the same cell as itself, so $a \sim a$a∼a.

• Symmetric: “$a$a and $b$b share a cell” is unchanged by swapping $a,b$a,b.

• Transitive: if $a,b$a,b share a cell and $b,c$b,c share a cell, both cells contain $b$b; since cells are disjoint or equal, they are the same cell, so $a,c$a,c share it.

Finally, for any $a$a, its class is $[a] = \left\{\, x \;:\; x \in C_a \,\right\} = C_a$[a]={x:x∈Ca​}=Ca​, exactly the cell containing $a$a. ◻

The Fundamental Theorem as a dictionary. On the left, $\sim$∼ links $a,b,c$a,b,c together and $d,e$d,e together; on the right, those links become the cells of a partition of $S$S. Theorems 2 and 3 say the two pictures carry exactly the same information.

Corollary 4. Equivalence relations on $S$S and partitions of $S$S are in one-to-one correspondence: passing to classes and passing to “same-cell” are inverse operations.

Proof. Theorem 2 sends each relation to a partition; Theorem 3 sends each partition to a relation and shows its classes are the original cells. Running the two constructions in either order returns the object you started with, so each is the inverse of the other. ◻

Homework

1. For each relation on $\mathbb{Z}$Z, say which of reflexive/symmetric/transitive hold, and whether it is an equivalence relation: (a) $a\, R\, b \Leftrightarrow a-b$aRb⇔a−b is even;  (b) $a\, R\, b \Leftrightarrow ab \ge 0$aRb⇔ab≥0;  (c) $a\, R\, b \Leftrightarrow a=b$aRb⇔a=b or $a=-b$a=−b.

2. For $\equiv \pmod 5$≡(mod5) on $\mathbb{Z}$Z, list the equivalence class $[2]$[2] (give five elements) and state how many distinct classes there are.

3. On $S=\{1,2,3,4,5,6\}$S={1,2,3,4,5,6}, the partition $\bigl\{\{1,4\},\{2,3,5\},\{6\}\bigr\}${{1,4},{2,3,5},{6}} defines an equivalence relation by Theorem 3. Write $[3]$[3] and $[6]$[6], and decide: is $2 \sim 4$2∼4?

4. Define $\sim$∼ on $\mathbb{R}$R by $x \sim y \Leftrightarrow x-y \in \mathbb{Z}$x∼y⇔x−y∈Z. Prove $\sim$∼ is an equivalence relation, and describe $[0]$[0] and $[\tfrac12]$[21​].

5. Prove or disprove: if $R$R is symmetric and transitive on $S$S, then $R$R is reflexive. (Where does the “proof” break?)

Suggested textbook exercises (Chartrand §9.2–9.3): 9.14, 9.16, 9.20, 9.24, 9.28, 9.32.