# A question about metrizable manifolds

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 ?

Certainly compact closed balls is too much to ask for. Consider $(0,1)$ with its usual metric and the ball of radius $\frac12$ around $\frac13$. –  kahen Jan 16 '12 at 22:33
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. –  kahen Jan 18 '12 at 10:26
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 $V$. –  kahen Jan 18 '12 at 10:31
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