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Let $B(a,r)$ be the ball of center $a$ and radius $r$ in $\mathbb{R}^n$. I have to show two obvious things, but I am having trouble formalising it:

i) If $B(a,r) \subset B(b,r') \Rightarrow ||a-b|| < r'-r$.

ii) Using i), prove that if $B(a_{n+1},r_{n+1}) \subset B(a_n,r_n) \; \forall n\Rightarrow \bigcap_{n} B(a_n,r_n) \neq \emptyset.$

Thank you in advance!

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1 Answer 1

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For part (i), consider the line segment between $a$ and $b$. This line segment can be extended to a radius $L$ of $B(b,r')$. Let $x$ be the point where $L$ intersects the boundary of $B(b,r')$. Then we can denote $L$ as $\overline{b x}$. We note that $\overline{b x}$ is the union of the two segments $\overline{b a}$ and $\overline{a x}$. What we know:

  • the length of $\overline{b x}$ is $r'$
  • the length of $\overline{b a}$ is $\|a - b\|$
  • the length of $\overline{a x}$ is greater than $r$, since $x$ is on the boundary of $B(b,r') \supset B(a,r)$

So we get that $r' > \|a-b\| + r$.

For part (ii), assume that $B(a_{n+1}, r_{n+1}) \subset B(a_n, r_n)$ for all $n$. We consider the sequence $\{r_i\}_{n \in \mathbb{N}}$. By assumption, this sequence is decreasing. If $R>0$ is a lower bound for the sequence, then there exists $b \in \mathbb{R}^n$ such that $B(b,R) \subset \cap_n B(a_n,r_n)$ and so the intersection is non-empty.

Assume that no such $R$ exists. Then $\{r_i\}$ is a decreasing sequence bounded below by 0. So $\lim_{i \rightarrow \infty} r_i = 0$. We will use this and part (i) to show that $\lim_{i \rightarrow \infty} a_i$ exists and is in our intersection.

Let $\epsilon > 0$. Since $\lim_{i \rightarrow \infty} r_i = 0$, there exists $N$ such that $r_n < \epsilon$ for all $n > N$. If $n,m > N$, then $\|a_n - a_m\| < |r_n - r_m| < \epsilon$, which shows that $\{a_i\}$ is Cauchy. Since $\mathbb{R}^n$ is complete, this proves that the sequence converges. Let $A = \lim_{i \rightarrow \infty} a_i$. By construction, $A \in B(a_n,r_n)$ for all $n$, so $A \in \bigcap_n B(a_n,r_n)$.

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Thank you for your help! Got the first part! For (ii), what you're saying would make me think that the balls are concentric, when they might not be...right? –  Nico May 14 '13 at 20:05
    
I read the indices backwards. I'll add an explanation for the second part. –  Corey Harris May 14 '13 at 20:28
    
Hey, thanks again for the explanation...I was just looking at it today a little more carefully and I was wondering: isn't the triangle inequality you used the opposite? Meaning that $| ||b-x||-||a-x|| |< ||b-x-(a-x)||$? –  Nico May 15 '13 at 17:02
    
You're right. I couldn't fix the argument, so I replaced it with a more geometric proof. –  Corey Harris May 16 '13 at 1:59
    
Thanks!!!!!!!!! –  Nico May 16 '13 at 18:04

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