# Maximal subcollection

I get a trouble with the following: Let $E$ be any open subset of a separable metric space $(X,d)$. We consider a collection of ball $\{B(x,\delta_x):\;x\in E\}$. In a book it states that we can chose a maximal disjoint subcollection $B_1,B_2,\ldots$, ie.. for any $x\in E$ then $B(x,\delta_x)\cap B_k\neq\emptyset$ for some index $k$.

I am not sure that if it is true or not and how to prove this?

Thanks

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I wonder what gives the example $E=X=\mathbb{R}$ with the usual distance $d(x,y)=|y-x|$ and $\delta_x=|x|+1$ for every $x$. – Did May 25 '11 at 12:20
@Didier: If I understand the statement correctly, any one-element collection $\{B(x, \delta_x)\}$ is a maximal disjoint subcollection, since any two-element subcollection is not disjoint. I think the notation "$B_1, B_2, \dots$" should be read as promising that the collection is at most countable; clearly it can't guarantee that the collection is infinite. – Nate Eldredge May 25 '11 at 12:32

Thank you Didier, Nate and jorki. I follow the ideas from jorki: Let chose a sequence $\{r_k\}$ so that it dense in $X$. For $x_1$, we chose any ball so that it contains $x_1$. Assume that $x_1,\ldots,x_n$ are chosen with disjoint balls. If there is a ball contains $x_{n+1}$ which disjoint from chosen balls, we chose its. If not, we will remove $x_{n+1}$. From that, it is easy to see that sequence of chosen balls is satisfied the requirement. But I am not sure how make it maximal. How should we understand the word "maximal subcollection" correctly in this sense? How we use Zorn lemma in here? – Jie Fan May 26 '11 at 0:57
@HDHung: "Maximal" generally has the sense of "cannot be made any larger". The subcollection we seek should be maximal in the sense that, if even one more $B(x, \delta_x)$ is added to it, it ceases to be disjoint. As for Zorn's lemma, it should be pretty standard: consider the set of all disjoint subcollections of $\{B(x,\delta_x) : x \in E\}$, partially ordered by inclusion. If you haven't worked with Zorn's lemma before, it may seems mysterious: a good example to compare is the proof that every vector space has a basis. – Nate Eldredge May 26 '11 at 3:38
Since the space is separable, there is a sequence of points such that every nonempty open subset of the space contains at least one element of the sequence. You can construct a maximal sequence of disjoint balls by going through the sequence of points and, whenever you encounter a point contained in a ball that is disjoint from the balls in the sequence so far, take that ball as the next ball in the sequence. (There may be more than one such ball for a given point, so this may require choice.) Since every nonempty open subset of the space contains at least one of the points in the sequence, every ball $B$ contains at least one such point $x$. Either $x$ was covered by a ball $B'$ in the sequence, or it was skipped because none of the balls containing $x$ was disjoint from the balls already in the sequence. In the first case, $B$ is not disjoint from $B'$, and in the second case $B$ was one of the non-disjoint balls causing $x$ to be skipped; in either case, $B$ is not disjoint from all balls in the sequence.