How do you show that the closure of an open ball is a closed ball?
Let A be the the closure of the open ball B
Let x be in $\displaystyle A^c $
We want to show $\displaystyle A^c $ is open (because a set is closed if its complement is open)
Suppose $\displaystyle A^c $ is not open, then for all r > 0, the open ball $\displaystyle B_r(x) $ intersects A. But then x is in the closure of A which equals A. Contradiction
southprkfan1, does that show that the closure is a closed ball?
Oh, is does show that the complement of a closed set is open. BUT?
Here is a hint: In a metric space the closure of a set is the set of points a distance of zero from the set.
I'm not sure exacdtly what you're asking but if it's to prove that $\displaystyle B_{\delta}[x]=\overline{B_{\delta}(x)}$ ([] means closed and () means open) then this isn't always true. Consider the three point discrete space $\displaystyle \{0,1,2\}$. Then, $\displaystyle B_{1}[1]=\{0,1,2\}$ but $\displaystyle \overline{B_{1}[1]}=\{1\}$ since the discrete space has no limit points.