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I am having some problems with the proof of the following Theorem:

"Let $E$ be a set in a metric space $\mathscr{X}$. Then $E$ has the Lindelöf property provided there exists a countable set $D$ which is dense in $E$ ".

A set $E\in\mathscr{X}$ has the Lindelöf property if every open cover has a countable subcollection that covers $E$. The proof I am using can be found on page 80 of the book "Topological Ideas" by K.G. Binmore. I have a real problem only with one part of the proof, but I show the whole proof as given in the text for completeness (I have made some notational changes). For a proof for $\mathscr{X}=\mathbb{R}^{n}$ see How to prove that if $D$ is countable, then $f(D)$ is either finite or countable?.

Let $\mathscr{U}$ be any collection of open sets that covers $E$. We need to show that a countable subcollection of $\mathscr{U}$ covers $E$. Let $\mathscr{B}$ be the class of open balls $B_{q}(d)$ with centers $d\in D$, and rational radii $q$ such that $B_{q}(d)\subset U$ for at least one $U\in\mathscr{U}$ (I will use $B\in\mathscr{B}$ for a general element). Then $\mathscr{B}$ is countable (I have omitted some stuff which shows this). The following statement is then proven (this is the bit I am having a problem with):


The above is proven in the following way: Let $u\in U$ for any $U\in\mathscr{U}$. Since $U$ is open, it is contained in an open ball $B_{\epsilon}(u)$ such that $B_{\epsilon}(u)\subset U$. Since $E$ is dense in $D$ we have $d(e,D)=0$ for each $e\in E$, and so we can always find a point $d\in D$ such that $d(u,d)<\frac{1}{3}\epsilon$ (my problem is here: I think this only holds if $u\in E$ correct?). We can also choose a rational number $q$ such that $\frac{1}{3}\epsilon<q<\frac{2}{3}\epsilon$ (a Theorem is referenced for this result). Then because $d(u,d)<\frac{1}{3}\epsilon<q$, we have $u\in B_{q}(d)\in\mathscr{B}$. Now let $y\in B_{q}(d)$. We have $d(u,y)\leq d(u,d)+d(d,y)<\frac{1}{3}\epsilon+q<\epsilon$. Thus $y\in B_{\epsilon}(u)$, and so $u\in B_{q}(d)\subset B_{\epsilon}(u)\subset U$. Thus every $U\in\mathscr{U}$ must be a union of all the $B_{q}(d)\subset B_{\epsilon}(u)$ that contain each $u\in U$. These open balls are a subset of $\mathscr{B}$, which proves (1)

So my question is, for the point $u\in U$ in the above paragraph, how can we always find a point $d\in D$ such that $d(u,d)<\frac{1}{3}\epsilon$ if $u\not\in E$? From the hypotheses of the Theorem I do not see why every $u\in U$ has to be in $E$. I now finish the proof as it is given in the book.

Proof continued: Let $f:\mathscr{B}\rightarrow\mathscr{U}$ be chosen so that $B\subset f(B)$ for each $B\in\mathscr{B}$, then $f(\mathscr{B})$ is a countable subcollection of $\mathscr{U}$ (see How to prove that if $D$ is countable, then $f(D)$ is either finite or countable? for a proof of this one). Thus

$E\subset\bigcup_{B\in\mathscr{B}}B\subset\bigcup_{B\in\mathscr{B}}f(B)\subset\bigcup_{U\in f(\mathscr{B})}U\subset\mathscr{U}$.

This shows that $f(\mathscr{B})$ is a countable subcollection of $\mathscr{U}$.

I have one question regarding the above paragraph: Can we always choose the function $f$ so that $B\subset f(B)$ for each $B\in\mathscr{B}$? I suppose $f(b)=b$ for all $b\in B$ does the trick? Any help on these questions would be greatly appreciated.

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Regarding your problem (that you bolded): If $u \in U$, then $u\in E$ since $E$ is the ambient space. – Quinn Culver Jun 1 '12 at 16:32
@QuinnCulver Thanks very much for your prompt response. What do you mean by the ambient space? My assumption was, since nothing was stated to the contrary in the proof, only that $\mathscr{U}\subset\mathscr{X}$. As far as I can tell this does not preclude $u\not=E$ for a $u\in U$ for any particular $U\in\mathscr{U}$. – dandar Jun 1 '12 at 16:59
As Brian said below, you're just interested in $E$ with the subspace topology. – Quinn Culver Jun 2 '12 at 23:42
up vote 2 down vote accepted

You have in fact found a hole in the proof as given, though it’s easily patched. One fix is the one suggested by Quinn Culver in the comments: throw away $\mathscr{X}\setminus E$ and take $E$ to be the metric space in which you’re working (the ambient space), replacing every open set in $\mathscr{X}$ with its intersection with $E$.

A fix that stays closer to the author’s apparent intent goes as follows:

Let $u\in E$. Then $u\in U$ for some $U\in\mathscr{U}$. Since $U$ is open, there is some $\epsilon>0$ such that $u\in B_\epsilon(u)\subseteq U$. Since $E$ is dense in $D$, there is a point $d\in D\cap B_{\epsilon/3}(u)$. We can also choose a rational number $q$ such that $\frac{\epsilon}3<q<\frac{2\epsilon}3$. Then because $d(u,d)<\frac{\epsilon}3<q$, we have $u\in B_{q}(d)\in\mathscr{B}$. Now let $y\in B_{q}(d)$. We have $d(u,y)\leq d(u,d)+d(d,y)<\frac{1}{3}\epsilon+q<\epsilon$. Thus $y\in B_{\epsilon}(u)$, and so $u\in B_{q}(d)\subset B_{\epsilon}(u)\subset U$. Thus, for each $U\in\mathscr{U}$, $$E\cap U=\bigcup\{B\in\mathscr{B}:B\subseteq U\}\subseteq U\;.$$ If $\mathscr{B}_0=\{B\in\mathscr{B}:B\subseteq U\text{ for some }U\in\mathscr{U}\}$, this shows that $E\subseteq\bigcup\mathscr{B}_0$. Now for each $B\in\mathscr{B}_0$ choose $U_B\in\mathscr{U}$ such that $B\subseteq U_B$; by the definition of $\mathscr{B}_0$ such a $U_B$ must exist. $\mathscr{B}_0$ is a subset of the countable collection $\mathscr{B}$, so $\mathscr{B}_0$ is countable, and therefore $\mathscr{U}_0=\{U_B:B\in\mathscr{B}_0\}$ is countable. But $E\subseteq\bigcup\mathscr{B}_0\subseteq\bigcup\mathscr{U}_0$, so $\mathscr{U}_0$ is a countable subset of $\mathscr{U}$ that covers $E$.

Notice that in this version I’m not claiming that you can find points of $D$ arbitrarily close to every point of $U$, but only to those that are in $E$. You were right to worry about it, because in fact you can’t guarantee that an arbitrary point of $U$ is in the closure of $D$. As an example, take $\mathscr{X}$ to be $\Bbb R^2$ with the usual topology, and let $E=\{\langle x,0\rangle:0<x<1\}$, the open unit interval on the $x$-axis. The points of $E$ with rational first coordinate are a countable dense subset. However, if $U$ is any open set in $\Bbb R^2$, $U\setminus E\ne\varnothing$, and for each $p\in U\setminus E$ there is an $\epsilon>0$ such that $B_\epsilon(p)\cap D=\varnothing$.

From your final question I suspect that you may have misunderstood the nature of the function $f$, which is one reason I’ve used a different notation above. The domain of $f$ is my $\mathscr{B}_0$, not some set of points of $\mathscr{X}$: the elements of the domain are the sets $B_q(d)$, not the points in those sets. For each $B\in\mathscr{B}$ that’s a subset of some member of $U$, $f$ picks a member of $U$ containing $B$. I tried to make that a little more clear by using $U$ as the name of the function instead of $f$ and writing the argument as a subscript: his $f(B)$ is my $U_B$, which I think is more obviously a single member of $\mathscr{U}$.

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Brian thank-you so much for taking the time to answer so clearly. I see that we do indeed need to restrict our attention to all points $E\cap U$ for a given $U\in\mathscr{U}$. Also your construction of the function $f$ is very good, and clearly answers my last question. Am I right in saying for a $B\in\mathscr{B}$ you find any $U\in\mathscr{U}$ such that $B\subseteq U$ (there may be more than one?), and so there may be many $\mathscr{B}_{0}'s$? This would then explain your notation $\cup\mathscr{B}_{0}$ which means the union over all of these $\mathscr{B}_{0}'s$? – dandar Jun 2 '12 at 9:36
Brian is there a typo in your first line? You say "Let $u\in E$. Then $x\in U$ for some...". Should that say "Let $u\in E$. Then $u\in U$ for some..."? Thanks again. – dandar Jun 2 '12 at 9:38
@dandar: You’re quite right about the typo. Thanks for catching it; I’ve now fixed it. Yes, for a given $B\in\mathscr{B}$ there may be many possible choices for $U_B$, and therefore many different possible countable subcovers $\mathscr{U}_0$. For any family of sets $\mathscr{A}$, $\bigcup\mathscr{A}$ means the same thing as $\bigcup_{A\in\mathscr{A}}A$: $\{x:\exists A\in\mathscr{A}(x\in A)\}$. – Brian M. Scott Jun 2 '12 at 9:43
Thank-you again Brian. In words the notation $\bigcup\mathscr{B}_{0}$ then means a union over all the sets in $\mathscr{B}_{0}$, rather than a union over lots of $\mathscr{B}_{0}'s$. – dandar Jun 2 '12 at 15:03
@dandar: Yes, that’s exactly right. – Brian M. Scott Jun 2 '12 at 20:50

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