Solution
= Solution
\b[The union need not be an <equivalence relation>], because <transitivity> can mix the two relations. Take $X=\{1,2,3\}$. Let $R$ have <equivalence classes> $\{1,2\}$ and $\{3\}$, and $S$ have classes $\{1\}$ and $\{2,3\}$. Then $1W2$ and $2W3$, but neither component relates $1$ to $3$, so $1W3$ fails. The <union of equivalence relations> is reflexive and symmetric; it is precisely <transitivity> that may fail.