# Exercise 9, Section 30 of Munkres’ Topology

(a) Let $$A$$ be a closed subspace of $$X$$. Show that if $$X$$ is lindelof, then $$A$$ is lindelof.

(b) Let $$A$$ be a open subspace of $$X$$. Show that if $$X$$ is separable, then $$A$$ is separable.

My attempt: (a)

Approach(1): Let $$U=\{ U_\alpha \in \mathcal{T}_A|\alpha \in J\}$$ be an open cover of $$A$$. Since $$U_\alpha \in \mathcal{T}_A$$, we have $$U_\alpha =V_\alpha \cap A$$; $$V_\alpha \in \mathcal{T}_X$$. $$A=\bigcup_{\alpha \in J} (A\cap V_\alpha)=A \cap (\bigcup_{\alpha \in J} V_\alpha)$$. Which implies $$A\subseteq \bigcup_{\alpha \in J} V_\alpha$$. Since $$A$$ is closed in $$X$$, $$X-A\in \mathcal{T}_X$$. So $$V=\{V_\alpha |\alpha \in J\} \cup \{X-A\}$$ is an open cover of $$X$$. Since $$X$$ is lindelof, $$\exists B$$ countable subcover of $$V$$. If $$X-A \notin B$$, then $$B=\{ V_i| i\in I\}$$, where $$I\subseteq J$$ is countable. So $$X=\bigcup_{i\in I}V_i$$. By elementary set theory, $$A= X\cap A= (\bigcup_{i\in I} V_i)\cap A= \bigcup_{i\in I} (V_i \cap A)=\bigcup_{i\in I} U_i$$. Thus $$\{ U_i| i\in I\}$$ is countable subcover of $$U$$. If $$X-A\in B$$, then $$B=\{ V_i |i\in I\} \cup \{X-A\}$$, where $$I\subseteq J$$ is countable. So $$X=(\bigcup_{i\in I}V_i)\cup (X-A)$$. By elementary set theory, $$A=X\cap A=[(\bigcup_{i\in I}V_i)\cup (X-A)]\cap A=[A\cap (\bigcup_{i\in I}V_i)] \cup \emptyset=\bigcup_{i\in I}(A\cap V_{i})=\bigcup_{i\in I}U_i$$. Thus $$\{ U_i| i\in I\}$$ is countable subcover of $$U$$. Is this proof correct?

Approach(2): https://math.stackexchange.com/a/244709/861687. Claim: $$A$$ is lindelof $$\iff$$ Every open cover of $$A$$ in $$X$$ has countable subcover. Proof: Proof is very similar to lemma 26.1 of Munkres’ topology.

Let $$U=\{ U_{\alpha}\in \mathcal{T}_X|\alpha \in J\}$$ be an open cover of $$A$$ in $$X$$. Since $$A$$ is closed, $$X-A \in \mathcal{T}_X$$. So $$(\bigcup_{\alpha \in J}U_\alpha)\cup (X-A)=X$$. $$V=\{ U_\alpha| \alpha \in J\} \cup \{X-A\}$$ is an open cover of $$X$$. Since $$X$$ is lindelof, $$\exists B$$ countable subcover of $$V$$. $$X=A \cup (X-A)=\bigcup B$$. so $$B\setminus \{X-A\}=\{ U_{\alpha_n}|n\in \Bbb{N}\}$$ is countable subcover of $$U$$. Is this proof correct?

Approach(3): https://math.stackexchange.com/a/244680/861687. Claim: $$X$$ is lindelof $$\iff$$ If $$\{ A_\alpha \subseteq X| A_\alpha$$ is closed in $$X$$, $$\alpha \in J\}$$ have countable intersection property, then $$\bigcap_{\alpha \in J} A_\alpha \neq \emptyset$$. Proof: Proof is very similar to theorem 26.9 of Munkres’ topology or proposition 2.4 page no. 20.

Let $$\{ U_\alpha \subseteq A| U_\alpha$$ is closed in $$A$$, $$\alpha \in J\}$$ with countable intersection property. $$A$$ is closed in $$X$$. By exercise 2 section 17, $$U_\alpha$$ is closed in $$X$$, $$\forall \alpha \in J$$. Since $$X$$ is lindelof, $$\bigcap_{\alpha \in J} U_\alpha \neq \emptyset$$. Our desired result.

(b) since $$X$$ is separable, $$\exists D\subseteq X$$ such that $$D$$ is countable and $$\overline{D}=X$$. Since $$D$$ is countable, $$D\cap A \subseteq D$$ is countable, by subset of countable set is countable. It’s easy to check $$(\overline{D\cap A})_A=A$$. Hence $$D\cap A$$ is countable dense subset of $$A$$. Is this proof correct? for more detail proof look.

• The only difference I see between approaches (1) and (2) is that (1) interprets "$A$ is compact" to mean the subspace $A$ is compact under the relative topology, while (2) interprets it as "$A$ is a compact set in the topology of $X$". The method of proof is the same, but because (1) is working with the subspace topology, it must take extra steps. Since it should have been proven very quickly that the two intepretations are equivalent, I personally prefer the second form that doesn't get bogged down in subspace details. Apr 30, 2022 at 16:15
• @PaulSinclair yup. I agree with you. Apr 30, 2022 at 16:18

Assume that $$X$$ is Lindelöf, and let $$\mathcal{B}$$ be a open cover of $$A$$. Then for each $$B \in \mathcal{B}$$, we have $$B=U \cap A$$ for some set $$U$$ which is open in $$X$$. If we denote the collection of these $$U$$ by $$\mathcal{U}$$, then $$\mathcal{U} \cup\{X-A\}$$ is an open cover of $$X$$. Since $$X$$ is Lindelöf, we can choose a countable subcover $$U_{1}, U_{2}, \cdots$$. Throwing out $$X-A$$ if necessary, and letting $$B_{k}=U_{k} \cap A$$, we conclude that $$B_{1}, B_{2}, \cdots$$ is a countable subcover of $$A$$. Thus $$A$$ is Lindelöf.