# Problem on compact and Hausdorff topological spaces.

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?

(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.

• @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). Jul 3 at 8:19
• Oopsy!!! . I did not see that you assumed compactness at the very beginning . Sorry my bad . (+1) Jul 3 at 8:21

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 $$T_{1}=T_{2}$$.