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
 A: Zorn's lemma should give you a maximal disjoint subcollection, and your space being separable would guarantee that such a subcollection must be countable.
A: 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.
