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.

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 #76240 – 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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