In a metric space the intersection of nested closed balls is empty.

Question

Recently an answer was given to show that in a metric space, the intersection of nested closed balls is empty. I have doubts about the answer.

For the metric space $\mathbb{N}$, of natural numbers, the answer proposed construction of the closed balls as follows:

The answer is : let $X$ be the space $\mathbb{N}$ with metric
$$d(m,n) = 1 + 1/\min(m,n),\quad m\neq n$$

The closed ball with centre $n$ and radius $n + 1/n$ is $\{ n, n+1, n+2, .....\}$ .

My queries: 1. $\{ n, n+1, n+2, .....\}$ looks like a closed unbounded set not closed balls with finite radius. Writing $A_n = \{ n, n+1, n+2, .....\}$, $n= 1,2,.$, then $A_n$ is a system of closed nested sets.

2. If the radius is $1 + 1/n$, then $$A_n = \{n -1, n, n +1 \} , \quad A_{n+1} = \{ n, n + 1, n+ 2 \}$$ $$A_{n+2} = \{ n+1, n+2, n+3 \}.$$

The above $A_n$'s are not nested closed balls. But their intersection is empty. Can somebody clear my doubts ??

Answer 1

The set $A_n$ looks unbounded to you because you are looking at it from the point of view of the natural metric on $\mathbb{N}$, but in fact, in the metric that you wrote down in your question, the entire space $\mathbb{N}$ is bounded, because for any two distinct natural numbers, $d(m,n)<2$. Thus, $A_n$ as it is written in the first place, is indeed closed and bounded, decreasing with respect to inclusion, whose intersection is empty.

Answer 2

$k \in B(n, 1+\frac 1 n)$ iff $d(k,n) <1+\frac 1 n$ iff $1+\frac 1 {\min \{k,n\}} <1+\frac 1 n$ iff $k \geq n$. Hence $B(n, 1+\frac 1 n)=\{n,n+1,...\}$. You are thinking of balls in usual metric but one is talking about balls w.r.t. the metric $d$ here.

Answer 3

Recently an answer was given to show that in a metric space, the intersection of nested closed balls is empty.

The example you gave shows that there exists some metric spaces where the intersection of some nested closed balls is empty, from which all we can conclude is: in a metric space, the intersection of nested closed balls might be empty.

looks like a closed unbounded set not closed balls with finite radius

You are thinking of unbounded in the sense of the usual distance $d(m,n)=|m-n|$. However the distance you gave is different, and yup using different metric the closed balls of finite radius might seem weird, but it's nothing wrong. One classical distance is the discrete distance defined by $d(x,x)=0$ and $d(x,y)=1$ for every $x\neq y$. Then you can see that, on $\mathbb N$, the closed ball centered at any point of $\mathbb N$ and of radius $1$ is all of $\mathbb N$.

The above $A_n$'s are not nested closed balls. But their intersection is empty. Can somebody clear my doubts ??

You should verify first that the $A_n$'s you gave are indeed closed balls. I didn't verify, but there are other simpler examples: take $\mathbb N$ with the usual distance $|m-n|$. Then ${0}$ and ${1}$ are closed sets, yet their intersection is empty, and they're not nested. This shows that the intersection of non nested closed balls can indeed be empty. This is a perfectly possible and fine thing.

Another more visual example is to take in $\mathbb R^2$ with the Euclidean distance two balls of radius 1 centered at $(-2,-2)$ and $(2,2)$ respectively.