Question

Let X be a non-empty set, and let R be an equivalence relation on X. Let C be the set of all equivalence classes of R. So C={A⊆X such that A=[x] for some x ∈ X}.

Now, define f : X → C by the rule f(x) = [x] for all x ∈ X.

Prove that if x ∈ X, then there is one and only one equivalence class which contains x.

Suppose X = {1, 2, 3, 4, 5} and that R is an equivalence relation for which 1 R 3, 2 R 4 but 1 R̸ 2,1 R̸ 5,and 2 R̸ 5.

Write down the equivalence classes of R and draw a diagram to represent the function f.

Answer 1

Answer on Question #76317 - Math - Discrete Mathematics

Question

Let $X$ be a non-empty set, and let $R$ be an equivalence relation on $X$ . Let $C$ be the set of all equivalence classes of $R$ . So $C = \{A \subseteq X : \text{such that } A = [x] \text{ for some } x \in X\}$ .

Now, define $f: X \to C$ by the rule $f(x) = [x]$ for all $x \in X$ .

Suppose $X = \{1,2,3,4,5\}$ and that $R$ is an equivalence relation for which $1R3, 2R4$ but $1R2, 1R5$ , and $2R5$ .

Write down the equivalence classes of $R$ and draw a diagram to represent the function $f$ .

Solution

Since $1R3$ , $[1] = \{1,3\}$ . Since $2R4$ , $[2] = \{2,4\}$ . Note that $[1] \neq [2]$ because $1R2$ . Also $5 \notin [1]$ and $5 \notin [2]$ because $1R5$ , and $2R5$ . Since $R$ is reflexive, $5R5$ . Thus $5 \in [5]$ . Therefore there are three equivalence classes: $[1]$ , $[2]$ and $[5]$ .

Checking: $[1]\cup [2]\cup [5] = \{1,3\} \cup \{2,4\} \cup \{5\} = \{1,2,3,4,5\} = X$

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