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Question: For a metric space $(X,d)$ is it always true that the closure of $B_r(a)$ is equal to $\{{ x \in X:d(x,a)\leq r}\}$?

My attempt at a counter example:

Take the discrete metric, $d_0(x,y)=\begin{cases} 1 & \text{if} \hspace{0.2cm}x\neq y\\ 0 & \text{if} \hspace{0.2cm} x=y \end{cases}$ for all $x,y \in X$. Now consider the open disk of radius $1$ around $a \in X$, $B_1(a)=\{{ x \in X:d_0(x,a)< 1}\}=\{a\}$. I believe that the closure of this singleton set in $X$ is just the set itself, $\{a\}$, but $\{{ x \in X:d_0(x,a)\leq 1}\}=X$. It is not necessarily true that the entire set $X$ is equal to the singleton set $\{a\}$, completing the counterexample.

I'm usually wrong for some reason or the other, so please correct me!

piece_and_love
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    It is just fine. – José Carlos Santos Mar 25 '19 at 19:53
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    In the discrete topology all sets are both open and closed so for any set $A$ we know $\overline A = A$. So you just need to find a case in the discrete topology where $B_r(a) = {x\in X|d(a,x)< r} \ne {x\in X|d(a,x) \le r}$. $B_r(a)= {x}$ if $r \le 1$ and $B_r(a) = X$ if $r>1$. And $ {x\in X|d(a,x) \le r} = {x}$ if $r < 1$ and $ {x\in X|d(a,x) \le r} = X$ if $r \ge 1$... so you can see you picked a good counter example. – fleablood Mar 25 '19 at 20:52

1 Answers1

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As I wrote in the comments, your example is just fine. I will provide a distinct one. Take $X=[0,1]\cup[2,3]$, endowed with its usual metric. Then $2\notin\overline{B_1(1)}$, but $2\in\{x\in X\,|\,d(x,1)\leqslant1\}.$

J. W. Tanner
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José Carlos Santos
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