27 - b R v Now a R x x R b and b R v so a R v by...

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The equivalence classes of a set A can be represented by a Venn diagram: for example, A 1 A 2 A 3 A 4 A 5 A In this case, there are Fve equivalence classes, illustrated by the Fve disjoint subsets. In fact, the equivalence classes always separate the elements into disjoint subsets that cover the whole of the set, as the following proposition states formally. P ROPOSITION 3.15 The set of equivalence classes { [ a ] : a A } forms a partition of A : that is, each [ a ] is non-empty; the classes cover A : that is, A = ± a A [ a ] ; the classes are disjoint (or equal): a, b A. [ a ] [ b ] ± = ∅ ⇒ [ a ] = [ b ] . Proof Given any a A , then a R a by re±exivity and so a [ a ] . Also a ± a A [ a ] , and hence the classes cover A . Suppose [ a ] [ b ] ± = , and let x [ a ] [ b ] . This means that a R x and b R x . It follows that x R b by symmetry. Given any v [ b ] , observe that
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Unformatted text preview: b R v . Now a R x , x R b and b R v , so a R v by transitivity. Therefore v ∈ [ a ] and so [ b ] ⊆ [ a ] . But [ a ] ⊆ [ b ] using a similar argument, so [ a ] = [ b ] . ± 3.7 Transitive Closure Consider the following situation. There are various ±ights between various cities. ²or any two cities, we wish to know whether it is possible to ±y from one to the other allowing for changes of plane. We can model this by deFning a set City of cities and a binary relation R such that a R b if and only if there is a direct ±ight from a to b . This relation may be represented as a directed graph with the cities as nodes, as in the following example: 27...
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