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$B_k$ sequence of closed balls in $\mathbb{R}^n$ such that $B_{k+1}\subset B_k$, show $\bigcap_k B_k$ is either a point or a closed ball.

Please help me check the proof, thanks!

Define $x_k$ to be the center of each closed ball, and $r_k$ to be the radius of each closed ball.

First if $r_k \rightarrow 0$, we show that $\bigcap_k B_k$ is a point. By nested set theorem, $\bigcap_k B_k$ is nonempty, suppose $x,y\in \bigcap_k B_k$ and $x\neq y$, we have $|x-y| = \epsilon$. We can take a ball say $B_k^*$ with radius less than $\frac{\epsilon}{3}$ since $r_k \rightarrow 0$, clearly we can not have $x,y \in B_k^*$, contradiction. The intersection is a singleton.

Second if $r_k \rightarrow r$ with $r>0$, we claim that $x_k$ is Cauchy, thus converges to a point $x \in \mathbb{R}^n$, and $\bigcap_k B_k = \overline{B(x,r)}$. Let $\epsilon$ be given, take $N$ such that for each $n\geq N$ we have $r_n -r \leq \epsilon$, and observe that for each $p \geq q \geq N$ we have $$|x_p - x_q| \leq r_q - r \leq r_N - r\leq \epsilon,$$ therefore $x_k$ is Cauchy and converges to some $x$.

Finally to show $\bigcap_k B_k = \overline{B(x,r)}$,

  • let $b\in\bigcap_k B_k$, for each $\epsilon > 0$, take $B_k^*$ such that $|x-x_k^*| \leq \frac{\epsilon}{3}$ and $|r-r_k^*| \leq \frac{\epsilon}{3}$, we then have $b\in B_k^* \subset B(x, r+\epsilon)$, therefore $b\in \overline{B(x,r)}$.
  • let $b\in \overline{B(x,r)}$, clearly $b\in B_k$ for each $k$, thus $b\in\bigcap_k B_k$.
Xiao
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1 Answers1

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Since you only operate with closed balls here, why use the overline on $B(x,r)$? I'd omit it.

Your proof is fine, provided that you understand what's behind "clearly" in the last line. To prove inclusion $B(x,r)\subset B_k$ you need to show that $ |x_k-x| \le r_k-r$ for every $k$. Which follows by taking $p\to \infty$ in your inequality $|x_p - x_k| \leq r_k - r $.