Exercise 9, Section 30 of Munkres’ Topology

Question

(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.

Answer 1

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.