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Let $X$ be a metric space such that the closure of any open ball coincide with the closed ball. I'm trying to prove (or disprove), that given $x \in X$, $R>0$ and $\varepsilon >0$, there exists an $\delta>0$ such that $B(x,R+\delta)\subset U(B(x,R),\varepsilon)$, where $U(B(x,R),\varepsilon) = \bigcup_{p \in B(x,R)} B(p,\varepsilon)$ is the $\varepsilon$-neighborhood of $B(x,R)$.

Taking out an annulus from $\mathbb{R}^2$, we can see that the claim can fail for $X$ not complete. If the metric is intrinsic to a length structure, then the result follows from the $\varepsilon$-midpoint property of the space, in fact $\overline{B(x,R)}=B[x,R]$ is a consequence of the structure. I was unable to prove that $X$ has such a property when the length structure is unavailable and the result from this answer didn't help me either. The question then is:

If $X$ is a complete metric space such that the closure of any ball coincides with the closed ball, then is it true that we can put the open ball with an extra little radius inside the $\varepsilon$-neighborhood? If so, how can I show this? Do we need to put another hypothesis into the space $X$?

Thanks in advance.

Kaitei
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  • "If the metric is intrinsic to a length structure" - what does it mean? Is there "length structure" separate from metric? – markvs Jan 09 '22 at 02:17
  • @markvs I'm talking about an arc-length functional $L$ which takes a curve and return its "length". In general metric spaces you can always define $L$ through the rectification of the curve using the metric. In this way, the metric $d$ is intrinsic if $d = \inf_{\gamma} L(\gamma)$ in a very similar way that happens in riemannian geometry. This is all really well treated in the book 'A Course in Metric Geometry' by Burago, Yuri and Ivanov. – Kaitei Jan 09 '22 at 03:36

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