Can the intersection of open or closed balls be empty, if their radii are bounded from below?

I am wondering about the following question:

Given a (countable) sequence of nested open balls:

$$ B_1 \supseteq B_2 \supseteq \cdots $$ Not necessarily having the same same center. All having radius bounded from below, say by $r > 0$. Then can we say that $$\bigcap_{i=1}^{\infty} B_i \neq \varnothing$$ it is certainly true in the reals, as one can simply go to the point where the radii are close to $r$ then take the center of that ball. However, I'm having trouble seeing whether or not it is true in the general case. So thanks in advance for proof or counterexample.

Solution 1:

Let $N$ be the set of positive integers. Define a metric on $N$ as follows: $$d(m,n)=\left\{\begin{array}{ll} 1+ \frac{1}{mn}& \mbox{if }m \neq n\\ 0 &\mbox{if }m=n \end{array}\right.$$ It is straightforward to check that this is a complete metric on $N$ (complete because this metric is discrete).

The closed balls $$B(n, 1+1/n^2)=\{m: m\ge n\}, n=1,2,\cdots$$ are decreasing and have empty intersection. Of course, the corresponding open balls also have empty intersection.

Solution 2:

The below argument is incorrect as pointed out by TCL. I'll leave the answer here anyway, just incase.

Note that $\bigcap_i^n B_i=B_n$ because each $B_i$ is a subset of $B_{i+1}$. Let $d_i=\mbox{diam}(B_i)$. It should be clear that if $\bigcap_i^{\infty} B_i=\emptyset$, then $\lim_{i\rightarrow\infty}d_i=0$ and so there exists some $i\in\mathbb{N}$ such that $d_i< 2r$, but then $B_i$ is an open ball of radius less than $r$ which contradicts the assumption that each $B_i$ has radius bounded below by some non-zero $r$.

I should add that i'm using the definition of the diameter of a subset $A$ of a metric space $X$ with metric $d$ to be $\mbox{diam}(A)=\sup\{d(x,y)|x,y\in A\}$.