Every metric space contains a discrete, coarsely dense subset I'm wondering on how to prove the following: Let $(X,d) $ be a metric space. We say that a subset $ A \subset X $ is coarsely dense iff $ \exists_{C > 0} \forall_{x \in X} \exists_{a \in A} d(x,a) < C$. Every metric space contains a discrete coarsely dense subset. I found this statement in this paper: < and it says there that it's a straightforward application of Zorn's lemma. I guess that by _discrete_ we mean a space in which singletons are open (this should be equivalent to the condition: all balls are finite. Am I making sense?) Then we look at the family $ \\{D \subset X~ | ~ D \text{ is discrete}\\} $. It doesn't make much sense, since I don't know how every chain could be bounded in this family. What should I use? Which family should I consider? I'd appreciate some help

A different family will work. Pick a number $\epsilon > 0$. Use the family of all sets $D$ such that for all $x \
e y \in D$ we have $d(x,y) > \epsilon$. Every chain will be bounded, because the union of each chain still satisfies the property that for all $x,y$ in the union we have $d(x,y) > \epsilon$. A maximal set $D$ will have the property that each $x \in X$ has distance $\le \epsilon$ from some point in $D$, or else $D \cup \\{x\\}$ violates maximality of $D$.