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I was solving a question and I came upon a statement which I can't really prove. I know that this is indeed true when our metric space is $\mathbb{R}^2$ with the euclidean metric.

Let $(X,d),$ be a metric space. Take $\overline{B}_r(x)$ be the closed ball around a point $x\in X$ of radius $r.$ Take another point $x'\in \overline{B}_r(x).$ Suppose $\textrm{dist}(x',\partial \overline{B}_r(x))=k,$ where $\textrm{dist}(x,A)=\inf\limits_{y\in A} d(x,y), A\subseteq X.$ Take $r'\leq k,$ and consider the closed ball $\overline{B}_{r'}(x').$ Then we have $\overline{B}_{r'}(x')\subseteq \overline{B}_r(x).$

I am wondering if there exists a metric/metric space in which this property does not satisfy...

I tried using the triangle inequality but I am not getting anything useful, and I am starting to wonder if this property only follows when the ball $\overline{B}_r(x)$ is path-connected, as then we can apply the property, though I am not sure if this holds in all spaces:

For a point $x$ in a closed ball, the shortest path from $x$ to the boundary of the disc is the path from $x,$ along the radius of the disc (that passes through $x)$, to the boundary circle.

Any help regarding this would be much appreciated!

PCeltide
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1 Answers1

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There is, as other have noticed, the $0$-$1$ distance, where $\partial\overline B_{r}(x)=\emptyset$ for all $r$, thus incurring at least in definitory issues. For another example, we can observe that, by Fermat-Wiles, in the metric space $(\Bbb Q^2,d_3)$ with $d_3(x,y)=\sqrt[3]{\lvert x_1-y_1\rvert^3+\lvert x_2-y_2\rvert^3}$ it holds that $\partial\overline B_1(0)=\{(1,0),(0,1),(-1,0),(0,-1)\}$, and therefore, by simple geometric considerations, there are several $x\in B_1(x)$ and $r<d\left(x,\partial \overline B_1(0)\right)$ such that $B_{r}(x)\nsubseteq \overline B_1(0)$.

This can be adapted to a a path-connected counterexample by considering the metric space $X=\{x\in\Bbb R^2\,:\, \lVert x\rVert_3\ne 1\lor x\in\Bbb Q^2\}$ with the distance induced by the ambient space $(\Bbb R^2,\lVert \bullet\rVert_3)$. Then, the same considerations as before on $\overline B_1(0)$ hold, and that specific ball is path-connected. Unfortunately, I cannot think of a complete counterexample, or of a counterexample where all balls are path-connected, and I don't know if they are possible.

  • Thank you this is brilliant! It's very insightful! – PCeltide Jul 27 '20 at 06:31
  • @PratikApshinge Thank you, but looking back it could have been simplified to avoid Fermat-Wiles. For instance, I could have just removed the boundary leaving only one point. –  Jul 27 '20 at 06:33
  • What do you exactly mean by removing the boundary? – PCeltide Jul 27 '20 at 06:37
  • Oh, this is rich! – AlvinL Jul 27 '20 at 06:45
  • @PratikApshinge It is quite clear that the third counterexample is $\Bbb R^2$, removed the boundary of a ball, plus adding back four points that were on that boundary. There is no need to state this in a way that passes through Fermat-Wiles. –  Jul 27 '20 at 06:51