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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)<r\} $ 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?

Paul Sinclair
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Ayumu Kasugano
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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.

If your space is just the first quadrant then the open balls are just the intersections with the first quadrant of the open balls in the plane with centers in the first quadrant. Some of them will have pieces cut off by the axes.

You get closed balls with $\le$.

Ethan Bolker
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