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I know that for a metric space $(X,d)$ and $\emptyset\ne A\subset X$ if we consider $A$ as a submetric space induced by $d$ then for any ball $B_A(x,r)$ in $A,$ $B_A(x,r)=B(x,r)\cap A.$ Now is it true that for any ball $B(x,r)$ of $X$ which has a non-empty intersection with $A,$ $B(x,r)\cap A$ is a ball in $X?$

Edoardo Lanari
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Sriti Mallick
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  • No. If the centre of the ball doesn't lie in $A$, then in general the trace of the ball in $A$ is not an $A$-ball. Consider the intersection of two disks in $\mathbb{R}^2$ to convinceyourself of that. Sometimes, consider $A$ a plane in $3$-dimesnional space, for example, all the traces of balls in $A$ are also $A$-balls even if the centre does not lie in $A$. – Daniel Fischer Aug 07 '13 at 13:02
  • I'm not sure what "ball in $A$" means, but it is not true in general that an intersection of an open set in $X$ with $A$ is open in $X$. However, it is defined open in $A$. – Tunococ Aug 07 '13 at 13:10

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No. Remember that $A $ can be any arbitrary subset of $X$. If $X = \mathbb{R}^n$ and $A$ consists of three points of modulus less than $1$, then $B(x,1) \cap A = A$, which is surely not a ball in $X$.

Eric Auld
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No, it isn't true. You can consider the closed ball $B(0,1)=[0,1]$ in $\mathbb{R}$ and take $A=[-\frac{3}{2},-\frac{1}{2}] \cup [\frac{1}{2},\frac{3}{2}]$. Then $A \cap [0,1] \neq \emptyset$ but $A \cap [0,1]$ is not of the form $B(x,r)$ for $x \in \mathbb{R}$ and $r > 0$.

pitchounet
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