# Separable Metric Spaces And Lindelöf's Covering Theorem

Prove that Lindelöf's Covering Theorem is valid in any separable metric space. Please check my proof: Let $$X$$ be a separable metric space then by the definition there is a countable dense subset say $$E$$ of $$X$$. Now let $$F$$ be a collection of open sets that cover $$X$$. Now chose any open set $$S$$ in $$F$$ which contains some $$x \in X$$. Now as $$E$$ is dense in $$X$$ and as $$S$$ is open so there is a $$x_E \in E$$ such that $$x_E \in S$$. Now for any other open set $$S'$$ in $$F$$ which contains $$x' \in X$$ and $$x'$$ is not the member of $$S$$. Now as $$x' \in S'$$ and it doesn't lie in $$S$$, so there is an open n-ball $$B(x', r)$$ of $$x'$$ such that $$B(x', r) \in S'$$ and this open n-ball doesn't contain $$x_E$$. Now as $$E$$ is dense in $$X$$, so there is a $$x_E' \in E$$ which lies in this open n-ball and hence lies in $$S'$$. Also we have $$d(x_E', x_E)>0$$ and hence for every $$S$$ in $$F$$ which contains some member of $$X$$ there is a unique member of $$E$$ which is a countable set, and hence there is a countable subcollection of $$F$$ which covers $$X$$ Is The Proof Correct??

• I don't think it is clear what you want to show. How do you define the subcovering? How do you show it satisfies the required property? Commented Oct 11, 2020 at 13:18
• Your proof appears to claim that $F$ is countable. Could you give more detail about what the elements of your subcovering are? Commented Oct 11, 2020 at 13:19
• I have shown that there for every $S$ in $F$ there is one and only one member of $E$ which is countable and hence number of all such sets is countable and hence there is a countable sub collection of $F$ which covers $X$ Commented Oct 11, 2020 at 13:32
• You have shown no such thing. You chose $x$ and $S$ arbitrarily and then some other $s'$ and $x' \in S'\setminus S$. We also have an $x_E \in E\cap S$. You take a ball inside $S'$ that does not contain $x_E$. Then you choose $x'_E$. You have not shown $x_E$ is unique nor $x'_E$, far from it, even. Commented Oct 11, 2020 at 13:42
• But i have shown that for any two $S \neq S'$ there exists two different $x_E \neq x_E'$ and hence the number of such sets is coubtable Commented Oct 11, 2020 at 13:46

No. Your ideas do not show in any way how to select the countable subcover for $$F$$. Also you don't really use the metric $$d$$, which already indicates it cannot be correct. There are separable hereditarily normal spaces that are not Lindelöf. The metric is essential...
You could show first that $$X$$ has a countable base (if $$D$$ is dense, $$\{B(d,q)\mid d \in D, q \in \Bbb Q\}$$ will do as a base, say write it as $$\{B_n: n \in \Bbb N\}$$) and then show that if $$\mathcal{U}$$ is any open cover of $$X$$, for each $$n$$ such that some $$U_n \in \mathcal{U}$$ obeys $$B_n \subseteq U_n$$, we pick one such $$U_n$$, otherwise we set $$U_n = \emptyset$$ (or leave $$U_n$$ undefined). The non-empty (or defined) $$U_n$$ then form a countable subcover of $$\mathcal{U}$$. (if $$x \in X$$, $$x$$ is in some $$U \in \mathcal{U}$$ and so for some $$m$$ we have $$x \in B_m \subseteq U$$ as the $$(B_n)$$ form a base for $$X$$, but then for $$m$$ we have indeed chosen some $$U_m \in \mathcal{U}$$ and so $$x \in B_m \subseteq U_m$$ and $$x$$ is indeed covered by the $$(U_n)_n$$.) This uses the countable version of the Axiom of Choice, but that is known to be unavoidable.
As a general fact consider this answer. It shows among others that if $$X$$ has a countable dense subset, every open cover of $$X$$ has a countable subcover ($$7 \to 3$$ for the case $$\kappa=\aleph_0$$).
• I have shown that there for every $S$ in $F$ there is one and only one member of $E$ which is countable and hence number of all such sets is countable and hence there is a countable sub collection of $F$ which covers $X$, then what is the fault?? Commented Oct 11, 2020 at 13:33
• @user771946 Try your argument for a space $X$ that is separable and not Lindelöf like the Sorgenfrey plane and see it fail. Commented Oct 11, 2020 at 20:50