Showing that Closure of a Ball is Not the Closed Ball

Question

Let $\overline{B_d (x, r)}$ be the closure of a ball in some metric space ($X, d$), and let $\hat{B}_d (x, r)$ be the closed ball in some metric space ($X, d$). Prove, or disprove, that $\overline{B_d (x, r)}$ = $\hat{B}_d (x, r)$ in any metric space.

I have a sense the answer is false when we use the discrete metric (professor told us if we ever want to "break" open/closed relationship, we should start with the discrete metric), but can't really understand how it would be false.

I was thinking maybe considering some radius $r > 1$. Then, $\overline{B_d (x, r)}$ would be just $x$ itself (i.e. {$x$}), since we can't have a radius greater than $1$ with the discrete metric, right? But then wouldn't $\hat{B}_d (x, r)$ be the same thing since we can't have distances greater than $1$..? I don't think I'm really understanding the discrete metric well. How would this statement be false wrt. the discrete metric? Any explanation/clarity would be very helpful. Thank you.

Answer 1

In discrete topology, for $r=1,$ the closure $$\overline{B_d (x, r)}$$ is the intersection of all closed sets which include $ B_d (x, r)$ and that is the singleton{$x$}.

On the other hand the closed ball, $$ \hat{B}_d (x, r)$$ is the whole space.

Thus the two sets are not necessarily the same.

Answer 2

Instead of considering some radius $r>1$, why don't you consider the more blatantly problematic radius $r=1$? As a hint that you are in the right track, you should arrive in a point/whole space situation.

Answer 3

Let $$ d(x,y)=\begin{cases} 0, & \mbox{ if }x=y\\ 1, & \mbox{ if }x\neq y \end{cases}. $$ We can verify that the axioms of metric are satisfied. That is, $d(x,x)=0$, $d(x,y)=d(y,x)$, and $d(x,z)\leq d(x,y)+d(y,z)$. Let $X=\mathbb{R}$ as a set, but equipped with metric $d$. Pick any $x_{0}\in X$, then $B(x_{0},1)=\{x_{0}\}$, $\hat{B}(x_{0},1)=X$. Obviously $B(x_{0},1)$ is closed, so $\overline{B(x_{0},1)}=B(x_{0},1)\neq\hat{B}(x_{0},1)$.