I saw following claim in example 3 section 30 of Munkres’ topology, How to show $R_l$ is Lindelof space?
Claim: let $(X,\mathcal{T}_X)$ be a topological space and $\mathcal{B}$ is a basis of $\mathcal{T}_X$. $X$ is lindelof $\iff$ Every basis cover of $X$ has countable subcover.
My attempt: ($ \Rightarrow$) trivially holds, because every basis cover is also an open cover. ($\Leftarrow$) Let $U=\{ U_\alpha |\alpha \in J\}$ be an open cover of $X$. Since $U_\alpha \in \mathcal{T}_X$, $U_\alpha = \bigcup _{i\in I_\alpha} B_{\alpha, i}$, where $B_{\alpha, i} \in \mathcal{B}$, $\forall i\in I_\alpha$. By elementary set theory, $X= \bigcup_{\alpha \in J} U_\alpha =\bigcup_{\alpha \in J}(\bigcup_{i\in I_\alpha} B_{\alpha ,i})=\bigcup_{\alpha \in J, i\in I_\alpha} B_{\alpha ,i}$. So $B= \{ B_{\alpha ,i}| \alpha \in J, i\in I_\alpha\} \subseteq \mathcal{B}$ is basis cover of $X$. By hypothesis, $\exists B’=\{B_{\alpha_n ,i_k}| n,k \in \Bbb{N}\}$ countable subcover of $B$. Fix $n\in \Bbb{N}$. We have $\bigcup_{k\in \Bbb{N}} B_{\alpha_n ,i_k} \subseteq U_{\alpha _n} =\bigcup_{i\in I_{\alpha_n}} B_{\alpha_n ,i}$ (i.e. $i_k \in I_{\alpha_n}$, $\forall k\in \Bbb{N}$). Since $n$ was arbitrary, $\bigcup_{k\in \Bbb{N}} B_{\alpha_n ,i_k} \subseteq U_{\alpha _n}$, $\forall n\in \Bbb{N}$. So $\bigcup_{n\in \Bbb{N}} (\bigcup_{k\in \Bbb{N}} B_{\alpha_n ,i_k}) =\bigcup_{n,k \in \Bbb{N}} B_{\alpha_n ,i_k}= X \subseteq \bigcup_{n\in \Bbb{N}} U_{\alpha_n}$. Hence $X= \bigcup_{n\in \Bbb{N}}U_{\alpha_n}$. Thus $\{ U_{\alpha_n}|n\in \Bbb{N}\}$ is countable subcover of $U$. Is this proof correct?
Best Answer
It is correct but I would add a comment:
Once you arrive to: "By elementary set theory, $$X=\bigcup_{\alpha\in J,i\in I\alpha}B_{\alpha,i}$$ So B is a basis cover of X", you say that, by hypothesis, there is a countable subcover of $B$. What we know is that there is a a countable subcover $B^\prime \subseteq B$ of $X$. And now we have finished, because for each $G\in B^\prime$, there exists $U_\alpha\in U$ such that $G\subseteq U_\alpha$, so the union of those countable many $U_\alpha$'s covers $X$.