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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1I 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? – FormulaWriter Oct 11 '20 at 13:18
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Your proof appears to claim that $F$ is countable. Could you give more detail about what the elements of your subcovering are? – preferred_anon Oct 11 '20 at 13:19
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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 $X$ – user771946 Oct 11 '20 at 13:32
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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. – Henno Brandsma Oct 11 '20 at 13:42
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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 – user771946 Oct 11 '20 at 13:46
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BTW you want $7 \to 3$ in this answer. That has a much more general fact. – Henno Brandsma Oct 11 '20 at 13:54
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Well, if you define for $S \neq S'$ a pair $(x_E, x'_E)$ of different values of $E$, this does not show that $F$ has a countable subcover. How then? You must define the subcover somehow. – Henno Brandsma Oct 11 '20 at 13:57
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No i am not claiming that $F$ has a countable sub cover instead i am showing that the number of sets $S$ in $F$ which contain some $x \in X$ is countable and hence there is a countable subcollection of $F$ which covers $X$ – user771946 Oct 11 '20 at 14:00
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You don't show that. Consider $F$ as the set of all open intervals in $\Bbb R$. There are uncountably many members of $F$ that don't contain $0$ e.g. – Henno Brandsma Oct 11 '20 at 14:01
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But $0$ is contained in one of those open intervals – user771946 Oct 11 '20 at 14:12
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I am taking only those open sets in $F$ which contain at least a member of $X$, those which do not contain any member of $X$ can be removed from the covering. – user771946 Oct 11 '20 at 14:14
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Let us continue this discussion in chat. – Henno Brandsma Oct 11 '20 at 14:22
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See I assumed that $F$ is any collection of open sets which cover the mteric space $X$. Now I chose any open set $S$ in $F$. Now if $S \cap X =\phi$, then we can remove $S$ from $F$ as the remaining collection will still be a covering of $X$. And if there is a $x \in X$ such that $x \in S$ then as $E$ is dense in $X$ and $S$ is open so we can chose a $x_E$ sufficiently close to $x$ such that $x_E \in S$. Now chose any other $S'$ in $F$ then again if $S' \cap X=\phi$ or $S' \cap X = S \cap X$ then remove $S'$ and if not then there is a $x' \in S'$ such that $x'$ doesn't lie in $S$ – user771946 Oct 11 '20 at 14:34
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And again as $E$ is dense in $X$ and $S'$ is open so we can chose a $x_E' \in E$ such that $x_E' \in S'$ and $x_E \neq x_E'$. Similarly we can do this for any open set in $F$ and hence we can associate each $S$ to a unique $x_E \in E$ and the result follows as $E$ is countable – user771946 Oct 11 '20 at 14:37
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You nowhere show uniqueness. – Henno Brandsma Oct 11 '20 at 20:51
1 Answers
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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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?? – user771946 Oct 11 '20 at 13:33
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@user771946 Try your argument for a space $X$ that is separable and not Lindelöf like the Sorgenfrey plane and see it fail. – Henno Brandsma Oct 11 '20 at 20:50