Let $C$ be any <equivalence class>. Since $A$ is both open and closed, $C\cap A$ is both open and closed in the <subspace topology> on $C$. Its <connectedness> makes $C\cap A$ either empty or all of $C$. Thus an <equivalence class> meeting $A$ lies wholly in $A$. Hence \b[$A$ is saturated], a <union> of whole equivalence classes.

Solution

Past exam of the options II course of the University of Oxford / 2009 / E / E4 / b / i / Solution

 Solution (source code)