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?

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

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


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