# Understanding closed and open balls

Let $$E$$ be a metric space where $$p_ 0 \in E$$ is the centre of an open ball with radius $$r>0$$. Then the open ball is the set $$\{ p\in E : d(p_ 0,p) and similarly the closed ball is the set $$\{ p\in E : d(p_ 0,p)\leq r \}$$. I'm having trouble actually understanding what the open/closed balls are geometrically. First of all, is the centre point $$p_ 0$$ arbitrarily chosen or is it the actual fixed centre point of some space? Does a ball necessarily have to be a circle or its equivalent?

Say that we have some large circle centred in the first quadrant with overlapping in all the other quadrants. Then define the boundary $$x,y\geq 0$$ so that the ball is cut off a bit and only present in the first quadrant. Is the set of all points in this cut off circle a closed ball? I'm thinking of saying yes since the distance between every element in the set and the centre is less than or equal to the radius. But if a closed ball is a closed set. However, I can pick any point in the closed ball and form an open ball, which means the closed ball is an open set? Where is my error?

• Can you pick a point on the boundary of the closed ball and form an open set?` – Edward Evans Dec 24 '16 at 23:59
• This example (the ball with a portion cut off) would not be a closed ball (it is no longer a ball at all). EDIT: unless the portion you are cutting off is actually removed from you metric space, and not just the ball. – Morgan Rodgers Dec 25 '16 at 0:01
• Can't you pick a point on the boundary and surround it with an arbitrary shape so that no point in the said shape is more than r units away? – Ayumu Kasugano Dec 25 '16 at 0:27

The plane should provide your geometric intuition. The inequality $$(x-a)^2 + (y -b)^2 < r^2$$ defines the open ball about $p=(a,b)$ with radius $r$. There are lots of these - one for each choice of $p$ and $r$. Every open ball has lots of smaller open balls inside it.
You get closed balls with $\le$.