Prove that in $\mathbb{R}^n$, if $B_r(x) = B_s(y)$, then $x = y$ and $r = s$.

Question

This statement is not true if we replace $\mathbb{R}^n$ with a general metric space. We can easily construct a counterexample.

However, why is it true that in $\mathbb{R}^n$, if $B_r(x) = B_s(y)$, then $x = y, r =s$?

Intuitively, one can take a point as close as we'd like to either $x, y$. And by triangle inequality, we would get that $d(x,y)$ would approach zero, which implies that $x =y$. But I can't seem to reason about it in a more proper way.

Any help would be appreciated! Thanks in advance!

Answer 1

First of all, the diameter of $B_r(x)$ is $2r$. Therefore, if $B_x(x)=B_s(y)$, then $2r=2s$ and so $r=s$.

So, $B_r(x)=B_r(y)$. But if $x\neq y$, $x+r'\frac{y-x}{\|y-x\|}\in B_r(x)$ for each $r'\in(-r,r)$. But if $r'$ is close enough to $-r$, $x+r'\frac{y-x}{\|y-x\|}\notin B_r(y)$. THerefore, $x=y$.

Answer 2

Since $\mathrm{B}_r (x)=\mathrm{B}_s (y)$ we obtain $y\in\mathrm{B}_r (x)$ and $x\in\mathrm{B}_s (y)$, therefore $\vert x-y\vert\leq\min\{r,s\} $. Suppose to the contrary that $\vert x-y\vert >0$, then there exist two open balls $\mathrm{B}_k (x),\mathrm{B}_k (y)$ such that $$ \mathrm{B}_k (x)\cap\mathrm{B}_k (y)=\emptyset $$ take $z$ on the line passing through $x,y$ such that $\vert z-x\vert =r$, then $\vert z-y\vert \geq r+2k$, but $z\in\mathrm{B}_s (y)$, therefore $s\geq r+2k>r$, similarly $r\geq s+2k >s$, a contradiction.

Therefore $x=y$, and we henceforth deduce $r=s$.

Answer 3

We prove the claim by induction on the dimension $n$.

• If $ n = 1 $, then $(x-r, x+r) = (y-s, y+s)$ means that $x-r = y-s$ and $x+r = y+r$, from which we immediately deduce $x = y$ and $r = s$.

• Assume that the claim is true in $n$-dimension. For each $i = 1, \cdots, n+1$, let $\pi_i : \mathbb{R}^{n+1} \to \mathbb{R}^n$ be the projection which forgets the $i$-th coordinate. Then $B_r(x) = B_s(x)$ implies that, for each $x_i = \pi_i(x)$,

  $$ B_r(x_i) = \pi_i(B_r(x)) = \pi_i(B_s(y)) = B_s(y_i) $$

  and hence $r = s$ and $x_i = y_i$ by the induction hypothesis. Since this is true for all $i$, we have $x = y$ and the claim is true in $(n+1)$-dimension. as well.