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Let $T_1$, $T_2$ and $T_3$ be topologies on a set X such that is $T_1⊂ T_2 ⊂ T_3$ and $(X,T_2)$ is a compact Hausdorff space. Then which of the followings are true?
1. $T_1$= $T_2$ if $(X, T_1)$ is a Hausdorff space.
2. $T_1$=$T_2$ if $(X,T_1)$ is a compact space.
3. $T_2$ =$T_3$ if $(X, T_3)$ is a Hausdorff space.
4. $T_2$ = $T_3$ if $(X, T_3)$ is a compact space.

How can I solve this problem.can anybody help me.

2 Answers 2

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HINT: The key fact is that every compact Hausdorff topology is both maximal compact and minimal Hausdorff. In other words, if $\tau$ is a compact Hausdorff topology on some set $X$, and $\tau'$ is a topology on $X$, then

  1. if $\tau'\supsetneqq\tau$, then $\tau'$ is not a compact topology on $X$, and
  2. if $\tau'\subsetneqq\tau$, then $\tau'$ is not a Hausdorff topology on $X$.

If you prove these facts, you should be able to answer the question fairly easily.

For (1) you should consider the identity map from $\langle X,\tau'\rangle$ to $\langle X,\tau\rangle$, and for (2) the identity map from $\langle X,\tau\rangle$ to $\langle X,\tau'\rangle$; note that both of these maps are continuous, and recall (a) that continuous maps preserve compactness, and (b) that compact subsets of Hausdorff spaces are closed.

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Here is a hint to help you get started:

No matter what $T_2$ is, you can always let $T_1$ be the "smallest" topology on $X$, and let $T_3$ be the "largest" topology on $X$. One of these is always Hausdorff, and the other is always compact. Since $T_2$ need not be either the largest or smallest topology in general, which statements can you now rule out?