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.

Answer 1

Answer on Question #76239 – 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 $x \in X$. Prove that if $x \in X$, then there is one and only one equivalence class which contains $x$.

Solution

Since $R$ is an equivalence relation, $R$ is reflexive. Then $xRx$ for each $x \in X$. Then for each $x \in X$ there is the equivalence classes $[x]$ such that $x \in [x]$.

Suppose that $x \in [a]$ and $x \in [b]$ with not $aRb$. Since $x \in [a]$ and $x \in [b]$, $xRa$ and $xRb$. By symmetry of $R$ it follows that $aRx$. By transitivity of $R$ we have that $aRb$ ($aRx$ and $xRb$). We got a contradiction with the assumption.

Hence, for all $x \in X$ there is one and only one equivalence class which contains $x$.

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