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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.

JohanLiebert
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2 Answers2

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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 $r = s$.

Michael Biro
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  • How did you come up with that z? – JohanLiebert Sep 18 '13 at 03:14
  • The intuitive idea is that if I take a point in $B_r(a)$ that's as far as possible from $b$ I can come up with a contradiction - that $s > r$. The farthest point in $B_r(a)$ from $b$ is just under $r$ in distance from $a$ in the direction directly away from $b$, which is given by $a - b$. Dividing by $|a - b|$ is to normalize so I get a distance of $r - \epsilon$. – Michael Biro Sep 18 '13 at 03:26
  • How exactly is it easy to see that r=s? – JohanLiebert Sep 18 '13 at 16:33
  • If $r < s$ then points at distance $\frac{r + s}{2}$ from the center are in $B_s(a)$ but not in $B_r(a)$. – Michael Biro Sep 18 '13 at 21:43
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Without loss of generality, assume $r \leq s$. Now look at $b + (s-\epsilon)\frac{b-a}{\left\lVert b-a\right\rVert}$ for arbitrarily small $\epsilon$.

ronno
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