(a) Let $\mathcal{T}$ and $\mathcal{T}’$ be two topologies on the set $X$; suppose that $\mathcal{T}’ \supseteq \mathcal{T}$. What does compactness of $X$ under one of these topologies imply about compactness under the other?
(b) Show that if $X$ is compact Hausdorff under both $\mathcal{T}$ and $\mathcal{T}’$, then either $\mathcal{T}$ and $\mathcal{T}’$ are equal or they are not comparable.
My attempt:
(a) Rephrase this problem. Claim: If $X$ is equipped with $\mathcal{T}$ and $\mathcal{T}’$ topology, $Y\subseteq X$ is compact in $(X,\mathcal{T}’)$ and $\mathcal{T} \subseteq \mathcal{T}’$, then $Y$ is compact in $(X,\mathcal{T})$. Proof: let $U=\{ U_\alpha \in \mathcal{T}|\alpha \in J\}$ be an open cover of $Y$ in $(X,\mathcal{T})$. Since $\mathcal{T} \subseteq \mathcal{T}’$ and $Y$ is compact in $(X,\mathcal{T}’)$, $\exists \{ U_{\alpha_1},..,U_{\alpha_n}\}$ finite subcover of $U$. Hence $Y$ is compact in $(X,\mathcal{T})$.
(b) Approach(1): If $\mathcal{T}= \mathcal{T}’$, then we are done. If $\mathcal{T} \nsubseteq \mathcal{T}’$ and $\mathcal{T} \nsupseteq \mathcal{T}’$ i.e. $\mathcal{T}$ and $\mathcal{T}’$ are not comparable, then we are done. Now assume $\mathcal{T} \subseteq \mathcal{T}’$. Let $U\in \mathcal{T}’$. Then $X-U$ is closed in $(X, \mathcal{T}’ )$. By theorem 26.2, $X-U$ is compact in $(X, \mathcal{T}’ )$. By exercise 1(a) section 26, $X-U$ is compact in $(X, \mathcal{T})$. By theorem 26.3, $X-U$ is closed in $(X, \mathcal{T})$. Thus $U\in \mathcal{T}$. So $\mathcal{T}’ \subseteq \mathcal{T}$. Hence $\mathcal{T} = \mathcal{T}’$. Therefore $\mathcal{T} \subseteq \mathcal{T}’$ $\Rightarrow$ $\mathcal{T} = \mathcal{T}’$. Similarly if $\mathcal{T}’ \subseteq \mathcal{T}$, then $\mathcal{T} = \mathcal{T}’$. Is this proof correct? https://math.stackexchange.com/a/2975675/861687 above proof is “complete”. Note: In proof of $\mathcal{T} \subseteq \mathcal{T}’$ $\Rightarrow$ $\mathcal{T} = \mathcal{T}’$, we didn’t make use of fact $(X, \mathcal{T}’)$ is Hausdorff and $(X, \mathcal{T})$ is compact. Said differently, in proof of $\mathcal{T} \subseteq \mathcal{T}’$ $\Rightarrow$ $\mathcal{T} = \mathcal{T}’$, we only use the fact $(X, \mathcal{T}’)$ is compact and $(X, \mathcal{T})$ is Hausdorff.
Approach(2): https://math.stackexchange.com/a/2442182/861687. $f:(X,\mathcal{T}’) \to (X,\mathcal{T})$ defined by $f(x)=x$, i.e. $f$ is identity map. Let $V\in \mathcal{T}$. It’s easy to check $f^{-1}(V)=V$. By assumption, $ f^{-1}(V)=V \in \mathcal{T}\subseteq \mathcal{T}’$. Thus $f^{-1}(V)\in \mathcal{T}’$. Hence $f$ is continuous. Proof of homeomorphism is in theorem 26.6. So $f^{-1}=g: (X,\mathcal{T})\to (X, \mathcal{T}’)$ is continuous. Let $U\in \mathcal{T}’$. It’s easy to check $g^{-1}(U)=U$. So $g^{-1}(U)=U \in \mathcal{T}$, since $g$ is continuous. Thus $\mathcal{T}’ \subseteq \mathcal{T}$. $ \mathcal{T}’ = \mathcal{T}$. Our desired result.