Let, $\tau_1, \tau_2,\tau_3$ be three topologies on a set $X$ such that $\tau_1 \subset\tau_2\subset\tau_3$ and $(X,\tau_2)$ be compact $T_2$ space. Then,

(A) $\tau_1=\tau_2$ , if $(X,\tau_1)$ is $T_2$.

(B) $\tau_1=\tau_2$ , if $(X,\tau_1)$ is compact.

(C) $\tau_1=\tau_3$ , if $(X,\tau_3)$ is $T_2$.

(D) $\tau_2=\tau_3$ , if $(X,\tau_3)$ is compact.

I only know that closed subspace of a compact space is compact and $T_2$-ness is a heriditary property. But I have no idea how it depends on topologies , whether the topologies are finer or coarser.



Let's have a look at the identity map $\def\i{\mathrm{id}}\i \colon X \to X$. As $\tau_1 \subseteq \tau_2 \subseteq \tau_3$, the map $\i \colon (X, \tau_{i+1}) \to (X, \tau_i)$ is continuous for $i = 1,2$.

As the continuous image of a compact space is compact, $(X, \tau_2)$ being compact, implies that $(X, \tau_1)$ is compact. If $(X, \tau_1)$ is moreover $T_2$, we have, that $\i \colon (X,\tau_2) \to (X,\tau_1)$ is a homeomorphism (being closed as it maps the $\tau_2$-closed[=compact] subsets onto closed [in the $T_2$-space] subsets). Hence $\tau_1 = \tau_2$.

On the other side, $(X,\tau_2)$ being $T_2$ implies that $(X, \tau_3)$ is $T_2$ (as seperating $\tau_2$-nhoods are $\tau_3$-nhoods also). If moreover $(X,\tau_3)$ is compact, we again have that $\i\colon (X,\tau_3) \to (X, \tau_2) $ is a homeomorphism. Hence $\tau_2 = \tau_3$.

So, (A) and (D) are true, but the others are wrong: Consider $X = [0,1]$, and the indiscrete topology $\tau_1 = \{\emptyset, [0,1]\}$, the Euclidean topology $\tau_2$ and the discrete topology $\tau_3 = \mathfrak P([0,1])$. Then $(X,\tau_1)$ is compact, but not $T_2$, $(X,\tau_2)$ is compact $T_2$, and $(X,\tau_3)$ is not compact, but $T_2$.

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  • $\begingroup$ I know a continuous bijection $f:(X,\tau_1)\to (Y,\tau_2)$ is a homeomorphism iff$X$ is compact and $Y$ is $T_2$. Here you use that $X$ is both compact and $T_2$...Is the $T_2$-ness necessary here ? $\endgroup$ – Empty May 8 '15 at 11:17

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