Exist $\delta>0$ such that the ball of ratio $\delta $ is contained in the intersection? My problem is: Given a Normed space $X$ and an sequence of closed balls such that
$$\overline{B}(x_1,r_1)\supset \overline{B}(x_1,r_1)\supset\ldots$$
with $r_1\geq r_2\geq ...>r>0$ to show that $\cap_{i=1}^{\infty}\overline{B}(x_i,r_i)\neq \emptyset$ and the second question is Exist $\delta>0$ such that the ball of ratio $\delta $ is contained in all the intersection? . I proved yet that $\cap_{i=1}^{\infty}\overline{B}(x_i,r_i)\neq \emptyset$ but for the second question i do know how to building the ratio if the answer is yes.
In my prove i showed that the sequence $r_i$ is Cauchy and therefore converge to one $r^{*}\geq r$ and with this forall $\epsilon>0$ exist $N_{\epsilon}\in \mathbb N$ such that $||x_n-x_{N_{\epsilon}}||<\epsilon$ for $n\geq N_{\epsilon}$ wich implies that
$$x_{N_{\epsilon}}\in \cap_{i=N_{\epsilon}}^{\infty}\overline{B}(x_i,r_i)\subset \cap_{i=1}^{\infty}\overline{B}(x_i,r_i) $$
If i suppose that the answer is "yes" is because perhaps is posible to build this number $\delta>0$ such that $ B_{\delta}(x_{N_{\epsilon}})\subset \overline{B}(x_i,r_i)$ for all $i\in\mathbb N$ and because  because  in $\mathbb R$ we have that the intersection is just one point if the diameter of the intervals converge to $0$ but for my problem the ratios do not converge to $0$. Is for this reason that i "believe" that the answer is yes.
then if i suppose that my tool for resolve this problem is with :
$\tilde{x}\in\overline{B}(x_i,r_i)\cap B_{\delta}(x_{N_{\epsilon}})$
$||x_n -\tilde{x}||\leq r_n$ for all $n$. The other side as $r_n\rightarrow r^{*}$ then for all $\epsilon>0$ exist $N_{\epsilon}$ such that $r^{*}-\epsilon<r_n<r^{*}+\epsilon$ then
how can continued , i guest the answer is yesPlease Somebody can help me? Thank you
Best
 A: First we will show that
$$\bar{B}(x,r)\subset \bar{B}(y,s)\quad \iff \|x-y\|\le s-r\quad (*)$$ There claim is clear for $x=y.$  Let $x\neq y.$ Then $$z:=x+{r\over \|x-y\|}(x-y)\in \bar{B}(x,r)\subset \bar{B}(y,s)$$ Hence
$$s\ge \|z-y\|=\left \|\left ({r\over \|y-x\|}+1\right ) (y-x)\right\|=r+\|y-x\|$$
Therefore $$\|x-y\|\le s-r\qquad (**)$$
Assume $(**).$ Let $z\in \bar{B}(x,r).$ Then
$$ \|y-z\|\le \|y-x\|+\|x-z\|\le s-r +r=s$$
Thus $z\in  \bar{B}(y,s)$ and the proof of $(*)$ is complete.
Basing on $(*),$ we obtain
$$\|x_{i}-x_{i+1}\|\le r_i-r_{i+1}$$ The triangle inequality implies
$$\|x_i-x_j\|\le r_i-r_j,\qquad j>i\qquad (***)$$ Therefore the sequence $x_i$ satisfies the Cauchy condition. Let $r=\lim r_i.$
Case $\bf 1$ $X$ is complete
Let $x=\lim x_i.$
The inequality $(***)$ gives
$$\|x_i-x\|\le r_i-r$$
Thus $(*)$ implies
$$\bar{B}(x,r)\subset \bar{B}(x_i,r_i)$$ Hence
$$\bar{B}(x,r)\subset\bigcap_{i=1}^\infty \bar{B}(x_i,r_i)$$
Fix $\delta>0.$ There exists $k$ such that
$ r_k-r<\delta.$
Then $$\|x_k-x\|\le r_k-r<\delta$$
By $(*)$ we get
$$\bar{B}(x_k,r_k)\subset \bar{B}(x,r_k+\delta)\subset \bar{B}(x,r+2\delta)$$
Thus
$$ \bigcap_{i=1}^\infty \bar{B}(x_i,r_i)\subset \bar{B}(x,r+2\delta)$$ As $\delta>0$ is arbitrary we get
$$ \bigcap_{i=1}^\infty \bar{B}(x_i,r_i)\subset \bar{B}(x,r)$$
Summarizing
$$\bigcap_{i=1}^\infty \bar{B}(x_i,r_i)=\bar{B}(x,r)$$
Case $\bf 2$ $X$ is not complete
The proof and the conclusion of Case $1$ remain valid if the sequence $x_n$ is convergent. It remains to consider the case when $x_n$ is not convergent. There are two ways of dealing with that case: directly or by means of the completion of $X.$ We will describe the direct way. There is $k$ such that $r_k-r\le  {r\over 2}.$ Then for any $i\ge k$ we have (see $(***)$)$$\|x_k-x_i\|\le r_k-r_i\le r_k-r\le {r\over 2}$$ hence
$$  \bar{B}(x_k,{\textstyle{r\over 2}})\subset \bar{B}(x_i,{r})\subset \bar{B}(x_i,{r_i}),\quad i\ge k$$
Hence
$$\bar{B}(x_k,{\textstyle{r\over 2}})\subset \bigcap_{i=k}^\infty \bar{B}(x_i,r_i)=\bigcap_{i=1}^\infty \bar{B}(x_i,r_i)$$
Remark The quantity ${r\over 2}$ can be replaced by any $r'$, $0<r'<r,$ i.e. the intersection contains a closed ball with radius $r'$ for any $0<r'<r.$ When $x_n$ is not convergent it can be shown that the intersection does not contain a ball with radius $r.$
