5. Partitions and Relations Ch.22 of PJE.

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1 5. Partitions and Relations Ch. of PJE. We now generalize the ideas of congruence classes of Z to classes of any set X. The properties of congruence classes that we start with here are that they are disjoint and that their union is the whole of Z. Definition Let X be a non-empty set. A partition of X is a subset Π P (X), i.e. a collection of subsets, such that (i) the sets in Π are non-empty, A Π A, (ii) the sets in Π are disjoint, so A, A Π : A A A A =, (iii) the sets cover X, i.e. X = A Π A, or x X, A Π : x A. We call the sets in Π the parts of the partition. Examples (a) Possible partitions of Z are (i) Π = {odd integers, even integers} = {[0], [] } = Z. In general, for m N, Z m is a partition of Z. (ii) Π= {{n Z : n < 0}, {0}, {n Z : n > 0}}. (b) But {{n Z : n < 0}, {n Z : n > 0}} is not a partition of Z since 0 is in no part. Similarly {{n Z : n 0}, {n Z : n 0}} is not a partition of Z since the parts are not disjoint. (c) If A = {a, b, c, d, e, f} then Π = {{a, c}, {b, d}, {e}, {f}} is a partition of A. Definition A relation on a set X is a subset R X X, i.e. a collection of ordered pairs. If (a, b) R we say that a is related to b and write arb or a b. If (a, b) / R we say that a is not related to b and write a b.

2 Note we have two ways of writing a relation, either as R, a set of ordered pairs, or using. We will use both ways. Example (i) Three different relations on Z could be (a) x < y, so R = {(, ), (, 9), (, 0),...} (b) x = y so R = {(, ), (00, 00), (, ),...} (c) x y mod 7 so R = {(, 8), (8, ), ( 5, 6), (, 0),...}. (ii) If A = {a, b, c, d, e, f} then R = {(a, a), (a, c), (c, a), (c, c), (b, b), (b, d), is a relation on A. (iii) R = {(x, x ) : x R} is a relation on R. (d, b), (d, d), (e, e), (f, f)} This relation is also the graph of the function f : R R, x x. This can be represented pictorially as x So R is represented by all the points on the parabola, a subset of the Cartesian plane. In fact, the graph of any function f : X X, namely is a relation. G (f) = {(x, f (x)) : x X} X X But the converse is not true, not all relations are graphs of functions. Example If X = {,, } then (a) R = {(, ), (, )} is not the graph of a function since is not related to anything, i.e. it has no image,

3 (b) R = {(, ), (, ), (, ), (, )} is not a graph of a function since has two images. Equivalence Relations Definition Suppose that is a relation on a set X. Then (i) is reflexive if a X, a a, (ii) is symmetric if a, b X, a b b a, (iii) is transitive if a, b, c X, a b and b c a c. If satisfies all three parts then we say that is an equivalence relation. Note that in (ii) or (iii) the elements a, b X or a, b, c X need not be different. Example Let X = {,, }. (i) R = {(, ), (, ), (, )}. Is not reflexive since (, ) / R, Is symmetric Is not transitive since (, ), (, ) R but (, ) / R (so a =, b = and c = in the definition of transitive, highlighting the point above that a, b, c need not be different). (ii) R = {(, ), (, ), (, )}. Is reflexive, Is symmetric, Is transitive. Hence R is an equivalence relation. Equivalence classes on a general set X are generalisations of congruence classes on Z. Example We have already seen that for m then x y defined by x y mod m is an equivalence relation on Z. We can give some examples of equivalence relations not based on congruences in Z. ) Example Let X = Z and be given by x y if, and only if, (x y) (x + y) is even.

4 Solution Reflexive: Let x Z for which (x x) (x + x) = 0 is even, so x x. Symmetric: Let x, y Z and assume x y. Then (x y) (x + y) = n, even. Thus ( n) = (y x) (y + x) is even and hence y x. Transitive: Let x, y, z Z and assume x y, y z. Then (x y) (x + y) = n while (y z) (y + z) = m for some integers n and m. Alternatively, n = x y and m = y z. Add together and get (m + n) = x z = (x z) (x + z). Hence (x z) (x + z) is even and thus x z. ) Example Let X = R and be given by (x, y ) (x, y ) if, and only if, x x = y y. Solution Left to student. ) Example Let X = R \ {(0, 0)} and be given by (x, y ) (x, y ) if, and only if, x y = x y. Solution Left to student. From partitions to relations. Partitions lead to relations in the following way. Given a partition Π of X define a relation Π by a Π b, a, b X if, and only if, A Π : a, b A, i.e. a and b lie in the same part of Π. Then R Π = {(a, b) : a, b X, a Π b}. Example Let A = {a, b, c, d, e, f} and Π = {{a, c}, {b, d}, {e}, {f}} be a partition of A. Then R Π = {(a, a), (a, c), (c, a), (c, c), (b, b), (b, d) (b, d), (d, b), (d, d), (e, e), (f, f)}. Theorem Let Π be a partition of X and Π the associated relation. Then Π is an equivalence relation. Proof p.65 From relations to partitions. We can go back from an equivalence relation to a partition, again simply by generalizing the ideas seen for congruence classes. 4

5 Definition Suppose that is an equivalence relation on a non-empty set X. For each a X define the equivalence class of a to be the set of elements of X related to a. Denote this class by [a] so [a] = {x X : x a}. Denote the set of equivalence classes by X/ = {[a] : a X}. Example When X = Z and was mod m we wrote Z m in place of Z/ ( mod m). What we managed to do for Z m was to define addition and multiplication on Z m, to give it some arithmetic structure. That would be the aim with other examples of X and, which is not achieved in this course. Why do we demand that is an equivalence relation? See the proof of the next result. Theorem Suppose that is an equivalence relation on a non-empty set X. Then for a, b X, (i) If a b then [a] = [b], (ii) If a b then [a] [b] =. Proof p.67 but here I give a slightly different proof. (i) Assume that a b. Let k [a]. Then by definition of the class, k a. Thus we have both k a and a b. So by transitivity k b. Hence k [b] by definition of the class. True for all k [a] means that [a] [b]. Let now l [b]. Then by definition of the class, l b. Use symmetry to write a b as b a. Thus we have both l b and b a. So by transitivity l a. Hence l [a] by definition of the class. True for all l [b] means [b] [a]. Combining [a] [b] and [b] [a] gives [a] = [b]. (ii) Assume a b. Assume, for a contradiction [a] [b]. So there exists c [a] [b]. Here c [a] means c a while c [b] means c b. Use symmetry to write c a as a c. Thus we have both a c and c b. So by transitivity a b. But this contradicts a b. Hence our last assumption is false, thus [a] [b] = as required. Note Since a b is either true or false, i.e. we have either a b or a b, then we have either [a] = [b] or [a] [b] =. Thus two classes are either identical or disjoint. 5

6 Thus [a] = [b] [a] [b] a b, being the contrapositive of (ii). Hence a b [a] = [b]. Theorem Suppose that is an equivalence relation on a non-empty set X. Then Π = X/ is a partition on X. Further, Π is identical to. Proof p.67 Let a X. By reflexivity of we have a a and thus a [a]. This first means that the equivalence classes are non-empty. Secondly every element of X lies in some class, there the collection X/ of classes covers X. Finally, as in the note above, the classes in X/ are disjoint. Hence X/ is a partition of X. Next, for a, b X, a b [a] = [b] by the note above, a, b lie in the same equivalence class in Π = X/, a Π b. Hence and Π are the same. We can give some examples of equivalence classes not based on congruences in Z. ) Example Let X = Z and be given by x y if, and only if, (x y) (x + y) is even. What do the equivalence classes look like? Solution Left to student. ) Example Let X = R and be given by (x, y ) (x, y ) if, and only if, x x = y y. What do the equivalence classes look like? Solution To get started I would suggest choosing a point from R and calculating its equivalence class. For instance, (, 5). Then the class labelled by (, 5) is [(, 5)] = {(x, y) : (x, y) (, 5)} = {(x, y) : x = y 5} = {(x, y) : y = x + }. 6

7 This is the graph of the straight line y = x +, gradient going through the point (, 5). In general [(a, b)] = {(x, y) : y = x + b a} is the graph of the straight line, gradient, going through the point (a, b). Thus R / is a collection of parallel lines that cover the plane. Note how these straight lines have to be parallel. If there were not parallel, they would intersect but we know from above that equivalence classes are either identical or disjoint. Example for student Let X = R \ {(0, 0)} and be given by (x, y ) (x, y ) if, and only if, x y = x y. What do the equivalence classes look like? Solution left to student. 7

8 Appendix ) Example Let X = R and be given by (x, y ) (x, y ) if, and only if, x x = y y. Show that is an equivalence relation. Solution Reflexive: For (a, b) R we have a a = 0 = b b and so (a, b) (a, b). Symmetric: Let (a, b), (c, d) R and assume (a, b) (c, d). This means that a c = b d. But this implies c a = d b and so (c, d) (a, b). Transitive: Let (a, b), (c, d), (e, f) R and assume (a, b) (c, d) and (c, d) = (e, f). This means that a c = b d, c e = d f. Add these together to get a e = b f in which case (a, b) (e, f) as required. ) Example Let X = R \ {(0, 0)} and be given by (x, y ) (x, y ) if, and only if, x y = x y. Show that is an equivalence relation. Solution Reflexive: For (a, b) R \ {(0, 0)} we trivially have ab = ab which implies (a, b) (a, b). Symmetric: Let (a, b), (c, d) R \ {(0, 0)} and assume (a, b) (c, d). This means that ad = cb. But this rearranges to cb = ad which implies (c, d) (a, b). Transitive: Let (a, b), (c, d), (e, f) R \ {(0, 0)} and assume (a, b) (c, d) and (c, d) = (e, f). This means that ad = cb and cf = ed. Simply multiply both together to get adcf = cbed. Cancel cd 0 to get af = ed which implies (a, d) (e, f) as required. ) Example Let X = Z and be given by x y if, and only if, (x y) (x + y) is even. What do the equivalence classes look like? Solution With x, y Z, you can check that (x y) (x + y) is even if either x and y are both even or they are both odd. If one of x, y is even and the other odd then (x y) (x + y) is odd. So if a, b Z and b [a] then b has the same parity as a, i.e. b is even if a even, and odd if a odd. This means that (b a) or b a mod. Hence the equivalence classes are [0] and []. 8

9 4) Example Let X = R \ {(0, 0)} and be given by (x, y ) (x, y ) if, and only if, x y = x y. What do the equivalence classes look like? Solution I recommend first looking at some classes. For instance [(, )] = { (x, y) R \ {(0, 0)} : y = x }. Here x = y is the straight line going through the origin and (, ). Thus [(, )] is the graph of this straight line with the origin deleted. What of [(, 4)]? This equals {(x, y) R \ {0} : y = 4 x}. But 4x = y is the straight line going through the origin and (, 4). Thus [(, 4)] is the graph of this straight line with the origin deleted. We can guess that [(a, b)] is the graph of the straight line y = b x, through a (a, b) and the origin with the origin removed. In fact (c, d) [(a, b)] (c, d) (a, b) and (c, d) (0, 0) cb = da and (c, d) (0, 0) d = b c and (c, d) (0, 0) a (c, d) lies on the line y = b x with the origin removed. a 9

10 5) The following section will not be covered in the lectures but it might give you a better understanding of relations. Directed Graphs Given R A A we can denote it by a directed graph or digraph which consists of a set of vertices (or nodes) corresponding to elements of A, and edges (or arcs) that connect vertices v and w if, and only if, (v, w) R with an arrow pointing from v to w. Example If A = {,,, 4} and R = {(, ), (, ), (, ), (4, ), (, )} this relation can be drawn as 4 Example Starting with the digraph a b e c d we see that the relation on {a, b, c, d, e} is {(a, a), (a, b), (a, e), (b, d), (c, b), (d, d), (e, b)}. For R to be reflexive means that in the digraph there is a loop on every vertex. For R to be symmetric means that, in the digraph, on every path between different vertices there will be two arrows. 0

11 For R to be transitive you have to look at every example in the digraph of a path linking three vertices using two line. Then you have to check that there is one line linking the end points (i.e. you have to check that if you can go the long way round then you can go the direct way. Note that in the definition of transitive the vertices x, y and z need not be different. Example Let A = {,, }. (a) Let R = {(, ), (, ), (, )}. This relation is not reflexive, (there is no loop on, say) is symmetric, is not transitive ((, ), (, ) R but (, ) / R ). (b) Let R = {(, ), (, ), (, ), (, ), (, ), (, ), (, )}. This relation is reflexive, is symmetric, is not transitive. ((, ), (, ) R but (, ) / R ) (c) Let R = {(, ), (, ), (, ), (, ), (, ), (, ), (, )}.

12 This relation is reflexive, is not symmetric ((, ) R but (, ) / R ), is transitive. You have to be careful when checking transitivity. (d) Let R 4 = {(, ), (, ), (, )}. This relation is not reflexive, is symmetric, and is not transitive since R but R but NR. (e) Let R 5 = {(, ), (, ), (, )}. This relation is reflexive, is symmetric, is transitive.

2 Equivalence Relations

2 Equivalence Relations In mathematics, we often investigate relationships between certain objects (numbers, functions, sets, figures, etc.). If an element a of a set A is related to an element b of a