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Let $(X,d)$ be a metric space, let $B(x,r)$ be the open ball of radius $r$ about $x$ and $N(x,r)$ be the set of elements $y\in X$ such that $d(x,y)\leq r$. It is well-known that it is not always true that $N(x,r)$ is the closure of $B(x,r)$.

I need, for some research, to restrict my attention to metric spaces for which that property is true, i.e. $N(x,r)$ is the closure of $B(x,r)$. Do they have a particular name in literature?

Thanks in advance,

Valerio

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  • $\begingroup$ Fix your typo "d(x,r) \leq r" since r is a real number, not a point in X. $\endgroup$
    – KConrad
    Sep 6, 2011 at 15:35
  • $\begingroup$ I've fixed the typo and improved the title. Feel free to revert either change if you don't think they are appropriate. $\endgroup$
    – j.c.
    Sep 6, 2011 at 15:39
  • $\begingroup$ Note that every metric space that is not a single point has a uniformly equivalent distance for which that property does not hold, that is, a truncated distance $(x,y)\mapsto\min(d(x,y),r)$. Just to point out that it is really a property of the distance function. $\endgroup$ Sep 6, 2011 at 15:48
  • $\begingroup$ @Pietro: Out of curiosity: is it known which metric spaces have an equivalent distance function for which the property does hold? $\endgroup$ Sep 6, 2011 at 15:59
  • $\begingroup$ Thanks for the correctio of the typo and also for improving the title. $\endgroup$ Sep 6, 2011 at 16:59

1 Answer 1

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I'm not sure what they're called, but according to this site an equivalent characterization of spaces $X$ where $\overline{B(x,r)} = N(x,r)$ is: for all $p\in X$, the only local minimum of the function $x \rightarrow d(x,p)$ is at $x=p$. The proof is also there.

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  • $\begingroup$ nice discussion! $\endgroup$ Sep 6, 2011 at 16:58
  • $\begingroup$ This equivalent condition is interesting. At the beginning it looks some kind of convexity, but it is not: also concave subspace of $\mathbb R^2$ can have that property, if the concaveness is not too strong. This property says that I can move from a point towards another point without getting back. How can I call this space? Let's try.. "spaces with no pits". How does that sound? $\endgroup$ Sep 7, 2011 at 9:13
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    $\begingroup$ Sounds wordy. Pitless? ~~~~ $\endgroup$ Sep 7, 2011 at 10:28
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    $\begingroup$ So would a space that does not have this property be called 'pitiful'? $\endgroup$
    – LSpice
    Sep 7, 2011 at 16:14

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