TITLE: Compactness and Strictly Finer Topologies.
QUESTION [7 upvotes]: If $(A,\tau{_1})$ is a compact Hausdorff space  and $\tau{_2}$ is a strictly finer topology on $X$, can $(A, \tau_{2})$ be compact?

REPLY [2 votes]: The finer topology $\tau_2$ not only isn't compact, it is even equal to $\tau_1$. Consider the identity map $(A,\tau_2)\to(A,\tau_1)$. It is continuous if and only if $\tau_1\subseteq\tau_2$. It is an easy exercise to prove that a continuous map from a compact to a Hausdorff space is a closed map. This means the identity is a homeomorphism, i.e. the topologies agree. The same argument can be applied to show that a coarser Hausdorff topology is equal to $\tau_1$.