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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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    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

0 Answers 0