Why don't all open balls have a finite diameter? This is probably really basic, but I can't seem to find where my thinking is going wrong.  My book says that for a subspace $S$, if there exists a supremum of $d(a,b)$ such that $d(a,b)\in S$, then $S$ has a finite diameter.  This makes sense, but for open balls, $d(a,b)<r$.  According to the least upper bound axiom, if a non-empty set is bounded above, it has a supremum.  Why doesn't it follow that for open (and closed) balls, $d(a,b)$ is bounded above and thus has a supremum and is always finite?
 A: If $\langle X,d\rangle$ is a metric space, $r>0$, and $B(x,r)=\{y\in X:d(x,y)<r\}$, is the open ball of radius $r$ centred at $x$, then for all $y,z\in B(x,r)$ we have
$$d(y,z)\le d(y,x)+d(x,z)<r+r=2r$$
by the triangle inequality. Thus,
$$\operatorname{diam}B(x,r)=\sup\{d(y,z):y,z\in B(x,r)\}\le 2r\;:$$
every open ball in every metric space has a diameter that is at most twice its radius. In particular, it has a finite diameter. An arbitrary open set, however, can be unbounded: the set $P$ of positive reals, for instance, is an open subset of $\Bbb R$ that does not have a finite diameter, because for any positive integer $n$, the points $1$ and $n+1$ are points of $P$ that are $n$ units apart.
Note, by the way, that an open ball can have a diameter smaller than twice its radius. In fact, it can have a diameter smaller than its radius. Let $X$ be any set, and for $x,y\in X$ let 
$$d(x,y)=\begin{cases}0,&\text{if }x=y\\1,&\text{if }x\ne y\;;\end{cases}$$
it’s easy to check that $d$ is a metric on $X$. However, it’s also easy to check that if $0<r\le 1$, then $B(x,r)=\{x\}$ has diameter $0$ for any $x\in X$, and that if $r>1$, then $B(x,r)=X$ has diameter $1$ for any $x\in X$.
A: It isn't entirely clear from your question, but it sounds as if this is what is going on:  
Assume the subspace is the positive real line with the standard metric (within the entire real line with the same metric). Then each ball $B_{\epsilon}(x) := [x - \epsilon, x + \epsilon] \cap \mathbb{R^+}, x, \epsilon > 0$, has by definition finite radius. But when you look at the sequence of balls $B_n(n), n \in \mathbb{Z^+}$, then the supremum over the radius of such balls goes to $\infty$. 
As said, I'm not quite clear from your description if this answers your question, but it came to mind.
