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I've seen the proof that if the closure of each open ball in compact metric space is the closed ball with the same center and radius, then any ball in this space is connected. Does this statement hold also in the other direction, aka: Is it true that:

If any ball in a compact metric space is connected, then the closure of each open ball in compact metric space is the closed ball with the same center and radius.

Also what can we say about non-compact metric spaces?

John
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1 Answers1

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Your claim is not true, basically because, while the distance tells everything about the topology, the topology does not tell much on the numerical values of the distance. For instance, consider $(X,d')$, with $X=[0,2]$ and $d'(x-y)=\min\{1,\lvert x-y\rvert\}$.

$(X,d')$ is homeomorphic to standard $[0,2]$, the open balls are some intervals, the closed balls are some other intervals, but you run into the same well-known issue you have in discrete spaces:

$$B(0,1)=[0,1)\qquad E(0,1)=[0,2]$$

In general, any metric space $(X,d)$ with at least two points $\{x_0,x_1\}$ can be endowed with a new metric $d'$ such that:

  1. $(X,d)$ and $(X,d')$ are weakly equivalent
  2. every open/closed ball in $(X,d')$ is an open/closed ball in $(X,d)$
  3. there is $r>0$ such that $E_{d'}(x,r)=X$ for all $x$ but $B_{d'}(x_0,r)$ is not dense

You just need to consider $d'(x,y)=\min\{\alpha,d(x,y)\}$, where $\alpha<\frac13 d(x_1,x_0)$. In this case, $E_{d'}(x_0,\alpha)=X$, but $B_{d'}(x_0,\alpha)\cap B_{d'}(x_1,\alpha)=\emptyset$ (therefore, $B_{d'}(x_0,\alpha)$ is not dense).

For the second question, i.e. if there is a non-compact metric space where the closure of any open ball is the closed ball with the same center and radius but some balls are not connected, consider $\Bbb Q$ with the standard metric: it is totally disconnected, but the closure of any open ball is the closed ball of the same center and radius. I can't think of an example which is also complete, though.