PARTITIONS AND EQUIVALENCE RELATIONS ROHAN RAMCHAND, MICHAEL MIYAGI Definition 1. A partition of a set X is a set P = fCi ⊆ X ji 2 Ig such that [ Ci = X (covering property) i2I 8i 6= s Ci \ Cs = ? (mutual disjointness) In essence, a set is completely divided into mutually disjoint partitions { no two partitions share any elements. Moreover, there is no element of X that is not contained in one of its partitions. The concept of a partition is illustrated by the following examples. Example. Let X = f1; 2; 3g. The set P1 = ff1g; f2; 3gg is a valid partitioning of X, since f1g and f2; 3g share no elements; moreover, every element of X is contained in P1. Let P2 = ff1g; f2gg. P2 is not a valid partitioning of X; even though f1g and f2g share no elements and P2 is therefore mutually disjoint, P2 does not contain the element 3 and is therefore not covering. Let P3 = ff1g; f2g; f3g; f1; 2; 3gg. P3 covers X, unlike P2, but is not mutually disjoint; therefore, P3 is not a valid partitioning of X. There are two special partitions of any set X. The first is the minimal partition: Pmin = fXg: The second is the maximal partition: Pmax = ffxg j x 2 Xg: These are valid partitions for any set. Example. Let X = f1; 2; 3g. Then Pmin = ff1; 2; 3gg and Pmax = ff1g; f2g; f3gg: 1 PARTITIONS AND EQUIVALENCE RELATIONS 2 Partitions of Z Let n 2 N for some n ≥ 2. Let Pn = fC0;C1; :::; Cn−1g, where Cr = fa 2 Z j nj(a − r) g: | {z } n divides a − r Example. Let n = 3. Then C0 = f3k j k 2 Zg = f0; 3; 6; :::g C1 = f3k + 1 j k 2 Zg = f1; 4; 7; :::g C2 = f3k + 2 j k 2 Zg = f2; 5; 8; :::g Note that C0 [ C1 [ C2 = Z and C0 \ C1 = C1 \ C2 = C0 \ C2 = ?. We now state the following theorems of Pn. Theorem 1. Let Cr be defined as above. Then n−1 [ Cr = Z r=0 8r1 6= r2 Cr1 \ Cr2 = ? Then Pn is a valid partitioning of Z. The proof of this theorem is left as an exercise to the reader. Equivalence Relations Definition 2. Let X be a set. Then a relation on X is a subset R ⊆ X × X: Let a; b 2 X and let (a; b) 2 R. Then a is related to b via R; this is denoted aRb: Relations are somewhat general, and don't say very much about sets; therefore, we intro- duce the concept of the equivalence relation, which is a slightly more specifically-defined relation. Definition 3. Let X be a set and let R ⊆ X × X. Then R is an equivalence relation on X if it satisfies the following properties. Reflexivity: For every x 2 X, (x; x) 2 R: PARTITIONS AND EQUIVALENCE RELATIONS 3 Symmetry: Let a; b 2 X. Then (a; b) 2 R , (b; a) 2 R: Transitivity: Let a; b; c 2 X. Then (a; b) 2 R ^ (b; c) 2 R , (a; c) 2 R: If R is an equivalence class, a ∼ b is a more common way of denoting (a; b) 2 R and will be used from here on. The most famous example of an equivalence relation on practically any set is equality; the proof is trivial and relies more on definition than any actual algebra. (We will revisit equality later.) Equivalence relations are demonstrated in the following example. Example. Let X = f1; 2; 3g. Consider the set R1 = ?. Since there are no elements in R1, reflexivity fails and therefore R1 is not an equivalence relation. Consider the set R2 = f(1; 1); (1; 2)g. Although 1 ∼ 1, 2: ∼ 2 and theref<F5>ore reflexivity fails. (Symmetry and transitivity also fail, but it is not necessary to show this.) Then R2 is not an equivalence relation. Consider the set R3 = f(1; 1); (2; 2); (3; 3); (1; 2); (2; 1)g. Every element is related to itself, and therefore R3 satisfies reflexivity. Since both 1 ∼ 2 and 2 ∼ 1, R3 satisfies symmetry. (Note that it is not a requirement that 3 be related to anything other than itself; however, if it is, it must be symmetrically related.) There are no transitive relations in R3, but again, this is not necessary; however, if for example, (1; 3) 2 R3, then (2; 3) (and (3; 1)) should also be in R3. Therefore, all three conditions are satisfied and R3 is an equivalence relation. As with most other structures previously explored, there are two canonical equivalence relations for any set X. Definition 4. Let X be a set. Then the maximal equivalence relation is the set R = X × X: Definition 5. Let X be a set. Then the minimal equivalence relation is the set R = f(x; x) j x 2 Xg: This relation is also referred to as equality and is denoted in set form by ∆. We now return to the divisibility partition above. Recall that njk is shortform for \n divides k". Then consider the equivalence relation ∼N = Z × Z PARTITIONS AND EQUIVALENCE RELATIONS 4 ≥2 for some N 2 N . Let a; b 2 Z; then a ∼N b , nj(b − a): We will now prove that ∼N is an equivalence relation. Proof. Reflexivity: Let a 2 Z: a ∼N a , nj0 Since n always divides 0, ∼N satisfies reflexivity. Symmetry: Let a; b 2 Z: a ∼N b , nj(b − a) , nj(a − b) , b ∼N a Then ∼N satisfies symmetry. Transitivity: Let a; b; c 2 Z and let a ∼N b and b ∼N c: a ∼N b , nj(b − a) ! b − a = k1n b ∼N c , nj(c − b) ! c − b = k2n c − a = (k1 + k2)n ! a ∼N c Then ∼N satisfies transivity. Therefore, ∼N is an equivalence relation. .