# cantor intersection theorem - exercise

Can anybody give an example of a complete metric space and a sequence of nested closed balls, which has empty intersection. (Exercise from Kolmogorov, Fomin: Elements of the Theory of Functions and Functional Analysis)

What is the answer to a similar question, where just the complete metric space is changed by the complete normed space?

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Your follow-up question (about Banach spaces - complete normed spaces) is identical to this question: Nested sequences of balls in a Banach space. –  Martin Sleziak Sep 25 '12 at 9:33

Let's ignore the ball condition at first, and just look for a complete metric space and a nested sequence of nonempty closed sets with empty intersection. This is clearly impossible for a finite metric space, so let's try the simplest possible infinite metric space: a countable set, which might as well be $\mathbb{N}$, with the discrete metric $d(m,n)=1 \Leftrightarrow m \neq n$. Every set in this space is closed, so it's easy to find the sequence we want: for example, we can take $A_n$ to be $\{ n, n+1, n+2, \ldots\}$.

This is all very well, but of course these sets aren't actually closed balls yet. This doesn't mean we should give up on our example, though. Rather, we try to tweak the metric to make each $A_n$ into a closed ball. We also need our metric to stay complete, and the simplest way to ensure this is to keep the condition that all nonzero distances are $\ge 1$.

A ball is defined by two things: a centre and a radius. The simplest choice for centre is to have $A_n$ centred at $n$. The radii will have to be decreasing (since we want $n+1 \in A_n$ but not $n \in A_{n+1}$) as well as being at least $1$, so let's try to make the radius of $A_n$ be $1+\frac1n$.

So for $m<n$, we want $d(m,n)\le 1+\frac1m$ (so that $n$ is in $A_m$) but $d(m,n)>1+\frac1n$ (so that $m$ is not in $A_n$). The simplest way to do this is to just let $d(m,n)=1+\frac1m$.

All the above is just a way of motivating the following example: let $X$ be the space $\mathbb N$ with metric $d(m,n)=1+\frac{1}{\min (m,n)}$ for $m \neq n$. It's not hard to check that this is indeed a complete metric space, and the closed ball with centre $n$ and radius $1+\frac1n$ is $\{n, n+1, n+2, \ldots\}$.

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+1. Very nice explanation how it is possible to find this example. Basically the same example is given in Shilov's Elementary Real and Complex Analysis on p.85. –  Martin Sleziak Sep 25 '12 at 9:05