For each <continuous map> to a Hausdorff target, equality of its values is reflexive, symmetric and transitive. The relation $R$ requires all these equalities simultaneously. Explicitly, $f(x)=f(x)$ gives reflexivity; $f(x)=f(y)$ implies $f(y)=f(x)$; and $f(x)=f(y)=f(z)$ gives $f(x)=f(z)$. These arguments apply to every map in the defining family, proving \b[$R$ is an <equivalence relation>].

Solution

Past exam of the options I course of the University of Oxford / 2010 / E / E2 / b / i / Solution

 Solution (source code)