TITLE: Prove that if two open balls in a vector space are equal then their radii and centres are equal
QUESTION [5 upvotes]: Define open balls in a metric space $(X,d)$ with center $a$ and radius $r$: $B_r(a) = \{x\in X: d(x,a) < r\}$
Let $(V, \|*\|)$ be a normed vector space, (with the corresponding metric $d(x,y) := \|x - y\|)$. Let $a,b \in V$ and let $r,s > 0$. Suppose $B_r(a) = B_s(b)$. Prove that $a = b$ and $r = s$.
Hint: In order to give a proof for this problem, the vector
space structure will have to be used in an essential way. In particular, a picture may
help you to come up with the right strategy.)
Not at all sure how to prove this...tried contradiction with a grad student/tutor and got nowhere. Any help is appreciated. 
I would post my attempt at a solution but really I have none and am totally stumped.

REPLY [3 votes]: Suppose $a \neq b$ and suppose without loss of generality that $s \leq r$. Let $\epsilon > 0$ be smaller than $d(a,b)$, and take the point $z = a + (r - \epsilon)\frac{(a-b)}{\|a-b\|}$. Then $d(z,a) = r - \epsilon$, so  $z \in B_r(a)$. However $d(z,b) = \|a-b\| + (r - \epsilon) > r \geq s$, and so $z$ is not in $B_s(b)$.
So the two points $a$ and $b$ must be the same. Now it should be easy to see that $r = s$.