My question is that whether the following statement is true or not.




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*In a complete metric space $(X, d)$, if a sequence of open balls $\{B(x\_i, r\_i)\}\_{i=1}^\infty$ satisfies
$$
\exists \epsilon > 0 ~~s.t.~ B(x\_{i+1}, (1+\epsilon)r\_{i+1}) \subset B(x\_i, r\_i), \forall i \ge 1 \tag{1}
$$
then $\bigcap\_{i=1}^\infty B(x\_i, r\_i) \neq \emptyset$.*




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Here are several related facts:



> 
> 1. If we further require that $r\_i \rightarrow 0$ as $i\rightarrow \infty$, then $\bigcap\_{i=1}^\infty B(x\_i, r\_i) \neq \emptyset$ holds.
> 2. There is a example shows that $\bigcap\_{i=1}^\infty B(x\_i, r\_i) = \emptyset$ if we replace $(1)$ by $\overline{B(x\_{i+1}, r\_{i+1})} \subset B(x\_i, r\_i)$ for all $i\ge 1$ where $\overline{B}$ is the closure of $B$
> 3. In a complete metric space, $B(x, r) \subset B(x', r')$ may hold for $r > r'$.
> 
> 
> 




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As Choi pointed it out, the above statement is similar to Cantor's intersection lemma. Recall that Cantor's intersection lemma in a complete metric space is given as follows:



> 
> In a complete metric space $(X, d)$, if a sequence of closed set $\{C\_i\}\_{i=1}^\infty$ satisfies
> $$
> C\_{i+1} \subset C\_i, \forall i\ge 1 ~~\&~~ diameter(C\_i) \rightarrow 0~ as~ i\rightarrow \infty
> $$
> then $\bigcap\_{i=1}^\infty C\_i \neq \emptyset$
> 
> 
> 


However, we don't require $diameter(B(x\_i, r\_i)) \rightarrow 0$ as $i\rightarrow \infty$. And there is indeed a example shows Cantor's inetersection lemma fails if we drop the diameter restriction. Please refer to [Nested closed balls with empty intersection](https://math.stackexchange.com/questions/201642/example-of-nested-closed-balls-with-empty-intersection?noredirect=1&lq=1)



Intersection of nested open ball in complete metric spaces is nonempty?
I think this statement is true.


Suppose we had a counterexample $\{B(x\_i,r\_i)\}\_{i=1}^\infty$ satisfying condition (1) for some $\epsilon > 0$ but whose intersection was empty. Observe that any subsequence will still be a counterexample.


For each $i$, the point $x\_i$ does not belong to some $B(x\_j,r\_j)$, as otherwise the intersection of all the balls would be nonempty. So by passing to a subsequence we can ensure that $x\_i \not\in B(x\_{i+1},r\_{i+1})$ for all $i$. Note that this forces $r\_{i+1} < r\_i$, as $x\_{i+1}$ does belong to $B(x\_i,r\_i)$.


By Cantor's intersection lemma, the $r\_i$ cannot decrease to zero. So they must decrease to some $r > 0$. Without loss of generality we can now assume that $(1 + \epsilon)r > r\_1$. But for any $i$ we have $x\_i \in B(x\_1,r\_1)$, i.e., $d(x\_i,x\_1) < r\_1 < (1+\epsilon)r$, which means that $x\_1$ belongs to every $B(x\_i,(1+\epsilon)r\_i)$. But then by condition (1) $x\_1$ belongs to every $B(x\_i,r\_i)$, a contradiction.


