Use the <Sierpiński space> $X=\{0,1\}$ with <topology> $\{\varnothing,\{1\},X\}$. The <singleton> $A=\{1\}$ is <compact>: from any <open cover>, choose one member containing $1$. But $X\setminus A=\{0\}$ is not <open>, so $A$ is not <closed>. This also exhibits why the <Hausdorff> hypothesis in the preceding proof matters.

Solution

Past exam of the topology course of the University of Oxford / 2023 / 1 / a / iv / Solution

 Solution (source code)