Is there condition on a Metric spaces which is equivalent to: The closure of all open balls are the closed balls?

Asked Apr 06 '19 at 12:22

Active Apr 06 '19 at 12:22

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For many of the spaces we work in the closure of open balls are the closed balls, however, I had an example in a class recently that there are some spaces in which this is not true. This is easily proved using either $\mathbb{Z}$ with the euclidean metric or a non-trivial set with the discrete metric.

I was wondering if there is a condition on metric spaces which is equivalent to this intuitive statement being true. I define "the closed balls" as follows:

The closed ball around $z$ of radius $r$, $\hat{B}_r(z) = \{x\in X:d(x,z)\le r\}$

I thought convexity might be a sufficient condition but that's just postulation, don't know how I'd prove it.

asked Apr 06 '19 at 12:22

Simon Goodwin

True in any normed linear space. – Kavi Rama Murthy Apr 06 '19 at 12:35
$\mathbb{Z}$ seems kind of pathological. Singletons are open balls of radius 1 and closed balls of a given radius $r$ are open balls of radius $r+1$. – JMJ Apr 06 '19 at 12:39
If $X$ has this property, and $Y \subseteq X$ is an open (not necessarily connected) subspace of $X$, then $Y$ has this property too. – Theo Bendit Apr 06 '19 at 12:47
If $X$ is crowded, it obeys this property. I think the reverse need not hold. – Henno Brandsma Apr 06 '19 at 12:51

Is there condition on a Metric spaces which is equivalent to: The closure of all open balls are the closed balls?

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