{c}

\b[Yes, $[0,1]$ is <closed>.] Its <complement> is $(-\infty,0)\cup(1,\infty)$. Every $x<0$ has the basic <open> <neighbourhood> $[x,0)$ contained in this <complement>, and every $x>1$ has $[x,x+1)$ contained in it. Hence the <complement> is <open> in $\mathbb R_{\mathcal T}$.

Solution

Past exam of the topology course of the University of Oxford / 2021 / 1 / b / i / Solution

 Solution (source code)