{c}

The rational cut is \b[both open and closed in $\mathbb Q$]. It is relatively open as $\mathbb Q\cap(-\infty,\sqrt2)$. Since $\sqrt2$ is irrational, its relative complement is $\mathbb Q\cap(\sqrt2,\infty)$, also relatively open. Thus this cut is a <clopen set> in the <subspace topology> of the rationals.

Solution

Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 4 / i / b / Solution

 Solution (source code)