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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?

InsideOut
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user66503
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3 Answers3

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Consider the identity map $e$ from $\langle A,\tau_2\rangle$ to $\langle A,\tau_1\rangle$. Since $\tau_2\supseteq\tau_1$, $e$ is continuous. Since $\tau_2\supsetneqq\tau_1$, there is a set $K\subseteq A$ such that $K$ is closed in $\tau_2$ but not in $\tau_1$. If $\langle A,\tau_2$ were compact, $K$ would be compact in $\tau_2$, and since $e$ is continuous, $K=e[K]$ would be compact in $\tau_1$. But $K$ is not $\tau_1$-closed, and $\tau_1$ is Hausdorff, so ... ?

Brian M. Scott
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  • Right, so this forms a contradiction because $K$ is open, and any compact subset of a Hausdorff space is closed. – user66503 Mar 13 '13 at 00:19
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    @user66503: Almost, but not quite. You can’t say that $K$ is open in $\tau_1$; you can say only that it’s not closed. But that’s good enough, because, as you say, every compact subset of a Hausdorff space is closed. – Brian M. Scott Mar 13 '13 at 00:21
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Hint: Consider $X\subseteq A$ not closed in $(A,\tau_1)$ but closed in $(A,\tau_2)$, then $X$ is not compact in $(A,\tau_1)$ since $(A,\tau_1)$ is Hausdorff, and thus not compact in $(A,\tau_2)$, then since $X^{c}$ is open in $(A,\tau_2)$...

Camilo Arosemena Serrato
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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 $\tau_1$.

Stefan Hamcke
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