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

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

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 $$\bigcap_{i=1}^\infty B(x_i, r_i) \neq \emptyset$$.

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 $$r > r'$$.

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

• so what is the question? – erz Sep 28 '18 at 1:17
• @erz Sorry about the confusion. I have edited it to make it clear. Basically, I don't know if the statement given in the beginning is true or not. (I intuitively believe it is true). – Yiran Sep 28 '18 at 1:34
• Isn't this just Cantor's lemma once you look at the closed ball centred on $x_{i+1}$ with radius $1+\varepsilon/2$? – Yemon Choi Sep 28 '18 at 1:44
• @YemonChoi Cantor's lemma requires either the compactness of closed bounded set or diameter of closed sets tend to 0. However, you may not have these based on given information in a general metric space. – Yiran Sep 28 '18 at 2:00
• @M.Dus Please refer to the link on the last line in the original post. – Yiran Sep 29 '18 at 19:04

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.