# 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. – Dan Rust Jun 26 '14 at 0:55
• What about completeness? – mfl Jun 26 '14 at 0:58
• Fine, remove a small open ball. – Dan Rust Jun 26 '14 at 1:01
• @mfl: the circle is a closed subset of $\mathbb{R}^{2}$, hence complete; or am I misunderstanding the question. – Matt Rosenzweig Jun 26 '14 at 1:06
• @Matt when you remove a point it is no longer complete, so removing an open ball is necessary to preserve completeness. – Dan Rust Jun 26 '14 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.