# Connected complete metric spaces with more than one point.

Does every connected complete metric space with more than one point have infinitely many closed balls? And is any closed ball in a connected complete metric space connected?

• The answer to the second question is no. Remove a single point from a circle and consider a small ball near the removed point. Jun 26, 2014 at 0:55
– mfl
Jun 26, 2014 at 0:58
• Fine, remove a small open ball. Jun 26, 2014 at 1:01
• @mfl: the circle is a closed subset of $\mathbb{R}^{2}$, hence complete; or am I misunderstanding the question. Jun 26, 2014 at 1:06
• @Matt when you remove a point it is no longer complete, so removing an open ball is necessary to preserve completeness. Jun 26, 2014 at 1:10

1. Yes. Take distinct points $a,b\in X$. For $0<r<d_X(a,b)$ there is a point $c\in X$ such that $d(a,c)=r$; indeed, otherwise $X$ would be the union of disjoint open sets $\{x:d(a,x)<r\}$ and $\{x:d(a,x)> r\}$. Therefore, all closed balls $\{x:d(a,x)\le r\}$, $0<r<d_X(a,b)$ are distinct sets.

2. No. An example was given in comments. For another example, remove the open rectangle $(-10,10)\times (0,1)$ from $\mathbb R^2$; keep the metric the same. The closed ball of radius $2$ centered at $(0,0)$ is not connected.

By the way, completeness was not used in the proof of 1.