Prove intersection of open balls is another open ball I was wondering how I would prove that an intersection of two open balls is also another open ball. The definition I have of an open ball is:
If x $\in X$ and $\epsilon > 0$, $B_{\epsilon}(x) :=$ { $y \in X: d(x,y) < \epsilon$ }
I am trying to prove that the intersection of two open balls is a member of B, which is defined as:
$B = $ { $B_{\epsilon}(x): x \in X, \epsilon > 0$ }.
 A: It is not a ball. I think what you want to prove is that the intersection of two open balls contains an open ball centered at each point inside it. It is a union of open balls.
Proof:
Let $B_1:=B_{\epsilon _1}(x_1)$,$B_2:=B_{\epsilon _2}(x_2)$ be two balls.
if they don't intersect, there is nothing to prove.so assume $x\in B_1\cap B_2$. Define $a:=\min\{\epsilon _1-d(x,x_1),\epsilon _2 -d(x,x_2)\}$, and then if $y\in B_a(x)$, then $d(x,y)<a$, so $d(y,x_1)\leq d(y,x)+d(x,x_1)<a+d(x,x_1)\leq\epsilon _1$, so $y\in B_1$, and the same way for $B_2$. So $B_a(x)\subset B_1\cap B_2$.
This is right for each $x$ in the intersection, so each $x$ has a ball in the intersection, so the intersection is the union of all these balls, and this shows that it is a open set.
A: It isn't true. Take two discs in $\mathbb{R}^2$ of radius 1 with the usual topology (euclidean metric), one centered at the origin, and the other centered at $(\frac{1}{2},0)$ and consider their intersection. This is not a ball in that topology.
However, it is an open set [ by the definition of a topology, the intersection of 2 open sets is an open set ]. 
