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Let $(M,d)$ be a metric space and $B(x, \epsilon) = \{ y \in M \mid d(x,y) < \epsilon \}$ and $\bar{B}(x, \epsilon) = \{ y \in M \mid d(x,y) \leqslant \epsilon \}$. In general, it is not true that $\bar{B}(x, \epsilon) = \overline{B(x, \epsilon)}$ (closure of $B(x, \epsilon)$).

The metric space $M$ is said to have:

  1. nice closed balls if and only if $\bar{B}(x, \epsilon) = M$ or $\bar{B}(x, \epsilon)$ is compact for every $x\in M$ and $\epsilon \in \mathbb{R}^+$.
  2. Compact closed balls if and only if the ball $\bar{B}(x, \epsilon)$ is compact for every $x\in M$ and $\epsilon \in \mathbb{R}^+$.

    Let $N$ be a metrizable topological manifold (no additional structure). Questions:

    It is true in $N$ that $\bar{B}(x, \epsilon) = \overline{B(x, \epsilon)}$ ? Does $N$ has nice closed balls and compact closed balls ?

    Thanks in advance !! Cheers...

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    Certainly compact closed balls is too much to ask for. Consider $(0,1)$ with its usual metric and the ball of radius $\frac12$ around $\frac13$.2012-01-16
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    It is true, for open manifold, we cannot have in general compact closed balls. But what about compact manifolds ? Do they have nice and compact closed balls ?2012-01-17
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    I have found out that every locally compact metric space has nice closed balls, so that every metrizable manifold has nice closed balls. It still remains to know if compact manifolds have compact closed balls2012-01-18
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    No, it is not the case that every locally compact metric space has "nice closed balls". Consider my example above. $\overline B(\frac13,\frac12) = (0,\frac56]$ which is certainly not compact.2012-01-18
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    Also, closed balls are in fact closed (but they are not necessarily the closure of the open ball with the same centre and radius), and closed subspaces of compact spaces are compact. Here's a sketch of the argument: Take an open cover of $V$ ($V$ being closed in $X$ which is compact). Use the definition of the subspace topology to produce an open cover of $X$ which has a finite subcover and then pass back to $V$.2012-01-18
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    It seems i was missing something in the comment about locally compact metric spaces and nice closed balls. The result i was taking about can be found in the book of Beer. He proved that a metric space is locally compact if and only if it has a compatible metric with nice closed balls. So this says indeed that the result is not true with the same metric, but with a compatible one.2012-01-18
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    The book is: G. Beer, Topologies on Closed and Closed Convex Sets, Kluwer Academic, Dordrecht, 1993.2012-01-18
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    I saw the affirmation and the reference to Beer's book in a paper. I do not know if some extra hypothesis is missing (like compactness of the whole space)2012-01-18
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    You're quoting correctly. Indeed $(0,1)$ does have a compatible metric with nice closed balls, *but* it's not the standard metric. Instead you have to do something like this: there exists a homeomorphism $f: (0,1) \to \mathbb R$ inducing a new metric on $(0,1)$: $d(x,y) := |f(x)-f(y)|$. $((0,1),d)$ will then have nice closed balls and be homemorphic to $((0,1),d_{\text{usual}})$2012-01-18

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