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

  • 2
    $\begingroup$ so what is the question? $\endgroup$
    – erz
    Commented Sep 28, 2018 at 1:17
  • $\begingroup$ @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). $\endgroup$
    – Brian
    Commented Sep 28, 2018 at 1:34
  • 1
    $\begingroup$ Isn't this just Cantor's lemma once you look at the closed ball centred on $x_{i+1}$ with radius $1+\varepsilon/2$? $\endgroup$
    – Yemon Choi
    Commented Sep 28, 2018 at 1:44
  • $\begingroup$ @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. $\endgroup$
    – Brian
    Commented Sep 28, 2018 at 2:00
  • 1
    $\begingroup$ @M.Dus Please refer to the link on the last line in the original post. $\endgroup$
    – Brian
    Commented Sep 29, 2018 at 19:04

1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.