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Consider $([0, 1], T_1)$, where $T_1$ is the subspace topology induced by the Euclidean topology on $\mathbb{R}$, and let $T_2$ be any topology on $[0, 1]$. Show that the following statements are true:

(i) If $T_1$ is a proper subset of $T_2$, then $([0, 1], T_2)$ is not compact.

(ii) If $T_2$ is a proper subset of $T_1$, then $([0, 1], T_2)$ is not Hausdorff.

My attempt: Suppose that $([0, 1], T_2)$ is compact. This means that every open cover of $T_2$ has a finite subcover. This implies that $([0, 1], T_1)$ as $T_1$ is a proper subset of $T_2$. How to think ahead?

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2 Answers 2

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(i) Suppose $([0,1],T_2)$ is compact. Let $I: ([0,1],T_2) \to ([0,1],T_1)$ be the identity map. This is continuous because $T_1$ is a subset of $T_2$. Now let $U \in T_2$ and $C=[0,1]\setminus U$. Then $C$ is closed, hence compact in $([0,1],T_2)$. [ $T_1$ contained in $T_2$ implies that $T_2$ is Hausdorff. `In a compact Hausdorff space closed subsets are compact]. Since continuous image of a compact set is compact we see that $C=I(C)$ is compact in $([0,1],T_1)$. This implies that $C$ is closed in $([0,1],T_2)$ and its complement $U$ is open in $([0,1],T_2)$. We have proved that $T_1=T_2$. This contradiction finishes the proof.

(ii) is very similar, so I will leave the details to you.

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  • $\begingroup$ @Mr.GandalfSauron $T_1$ contained in $T_2$ implies that $T_2$ is Hausdorff. `In a compact Hausdorff space closed subsets are compact. Your example doe snot satisfy the hypotheiss of (i). $\endgroup$ Jul 3 at 8:19
  • $\begingroup$ Oopsy!!! . I did not see that you assumed compactness at the very beginning . Sorry my bad . (+1) $\endgroup$ Jul 3 at 8:21
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The essence of the argument lies in the more general fact that a continuous bijection from a compact hausdorff space to a hausdorff space is a homeomorphsim.

Proof: If $X$ be a compact hausdorff space and $Y$ be hausdorff and if $f:X\to Y$ be a continuous bijection . Then we have for a closed subset $C\subset X$ , is compact and hence $f(C)$ is compact as it is the continuous image of a compact set and hence is closed in $Y$ as $Y$ is Hausdorff. Thus $f$ is a closed map and a continuous bijection which means that $f$ is a homeomorphism.

So in the first case , if $([0,1],T_{2})$ is assumed to be compact then the identity map becomes a homeomorphism from $([0,1],T_{2})$ to $([0,1],T_{1})$ and hence $T_{1}=T_{2}$.

Alternatively we can use the fact that if a set $C$ is compact in a finer topology then it is compact in the coarser one. That is we take an open set $U$ which is in $T_{2}$ but not in $T_{1}$ and we consider the closed set $[0,1]\setminus U$ which is closed in $[0,1]$. If we assume $([0,1],T_{2})$ is compact then a closed subset of a compact set is compact and hence $[0,1]\setminus U$ is compact. Hence it is also compact in $T_{1}$ which implies that $[0,1]\setminus U$ is closed in $T_{1}$ and hence $U$ is open in $T_{1}$ which is a contradiction.

In the second case if $([0,1],T_{2})$ is assumed to be hausdorff then again the identity map $\text{id}:([0,1],T_{1})\to([0,1],T_{2})$ becomes a homeomorphism and hence again $T_{1}=T_{2}$.

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