Intersection of nested open ball in complete metric spaces is nonempty?

Question

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

Answer 1

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.