Balls intersection problem 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!
 A: 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 $A \in \bigcap_n B(a_n,r_n)$.
