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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??

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    $\begingroup$ 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? $\endgroup$ Commented Oct 11, 2020 at 13:18
  • $\begingroup$ Your proof appears to claim that $F$ is countable. Could you give more detail about what the elements of your subcovering are? $\endgroup$ Commented Oct 11, 2020 at 13:19
  • $\begingroup$ 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 $X$ $\endgroup$
    – user771946
    Commented Oct 11, 2020 at 13:32
  • $\begingroup$ 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. $\endgroup$ Commented Oct 11, 2020 at 13:42
  • $\begingroup$ 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 $\endgroup$
    – user771946
    Commented Oct 11, 2020 at 13:46

1 Answer 1

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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 $\kappa=\aleph_0$).

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  • $\begingroup$ 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?? $\endgroup$
    – user771946
    Commented Oct 11, 2020 at 13:33
  • $\begingroup$ @user771946 Try your argument for a space $X$ that is separable and not Lindelöf like the Sorgenfrey plane and see it fail. $\endgroup$ Commented Oct 11, 2020 at 20:50

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