# Thread: Closure of open ball

1. ## Closure of open ball

How do you show that the closure of an open ball is a closed ball?

2. 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

3. Originally Posted by vinnie100
How do you show that the closure of an open ball is a closed ball?
Originally Posted by southprkfan1
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.

4. oh i was convinced, but i can see plato's point... hmmmm

5. 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.

6. yep that is exactly what I was asking to prove...but its false now I see! sorry for the confusion! thanks for all your help everyone