An Example of a Nested Decreasing Sequence of Bounded Closed Sets with Empty Intersection

Could someone provide me with an example of a metric space having a nested decreasing sequence of bounded closed sets with empty intersection? I first thought of Cantor set but the intersection is not empty!

Solution 1:

Let $\mathbb N$ be endowed with the discrete metric. In this metric space, every subset is bounded (although not necessarily totally bounded) and closed. Moreover, the subsets \begin{align*} A_1\equiv&\,\{1,2,3,4,\ldots\},\\ A_2\equiv&\,\{\phantom{1,\,}2,3,4,\ldots\},\\ A_3\equiv&\,\{\phantom{1,2,\,}3,4,\ldots\},\\ \vdots&\, \end{align*} are nested, and their intersection is empty.

However, if you stay within the realm of $\mathbb R$ endowed with the usual Euclidean metric, than you can't have a situation like the one above:

Claim: Suppose that $$A_1\supseteq A_2\supseteq A_3\supseteq\ldots$$ is a countable family of non-empty, closed, bounded subsets of $\mathbb R$. Then, $\bigcap_{n=1}^{\infty} A_n\neq\varnothing$.

Proof: By the Heine–Borel theorem, $A_n$ is compact for each $n\in\mathbb N$. For the sake of contradiction, suppose that $\bigcap_{n=1}^{\infty} A_n=\varnothing$. This is equivalent to $\bigcup_{n=1}^{\infty} A_n^{\mathsf c}=\mathbb R$. In particular, $$A_1\subseteq\bigcup_{n=1}^{\infty} A_n^{\mathsf c}.$$ Since $A_1$ is compact and the sets $(A_n^{\mathsf c})_{n=1}^{\infty}$ form an open cover of it, there must exist a finite subcover. That is, there exists some $m\in\mathbb N$ such that $$A_1\subseteq\bigcup_{n=1}^m A_n^{\mathsf c}=A_m^{\mathsf c},$$ where the second equality follows from the fact that $$A_1^{\mathsf c}\subseteq A_2^{\mathsf c}\subseteq A_3^{\mathsf c}\subseteq\ldots.$$ Now, $A_1\subseteq A_m^{\mathsf c}$ means that if a point is in $A_1$, then it must not be in $A_m$, so that $A_1\cap A_m=\varnothing$. But $A_m\subseteq A_1$ (given that the sets are nested), so that $A_1\cap A_m= A_m=\varnothing$, which contradicts the assumption that $A_m$ is not empty. This contradiction reveals that the intersection $\bigcap_{n=1}^{\infty} A_n$ must not be empty. $\quad\blacksquare$

Solution 2:

Another simple example is to look at the "punctured line": $(-\infty, 0) \cup (0, \infty)$, which is just the real numbers with $0$ removed. The sets $A_n = \{ x \ \in \Bbb R \,|\, |x| \le 1/n \text{ and } x \ne 0 \}$ are closed and bounded in the punctured line, but their intersection is empty.

Solution 3:

Do you know a theorem about nested bounded closed sets having non-empty intersection? If you do then you would need to find a metric space that does not satisfy the conditions of that theorem. $\mathbb{R}$ satisfies that theorem and hence you're not going to find a counter-example there. But there is a smaller metric space sitting inside it, namely $\mathbb{Q}$, that will give you a counter-example.

Solution 4:

I initially missed the fact that bounded sets were desired.

Let the metric space be $(0,1)$, the set of all real numbers between $0$ and $1$, not including the endpoints, with the usual metric $d(x,y)=|x-y|$.

Then the example can be that the $n$th set is $(0,\ 1/n]$. This is closed within this space; it contain all of its limit points in the space.

The first thing I thought of was $\displaystyle \bigcap_{n=0}^\infty [n,+\infty)$.

If the object called $+\infty$ were included in the space, with the appropriate topology so that $n\to\infty$, then these sets would not be closed, but would become closed if one added $+\infty$ to them as a new member, and then the intersection would not be empty because $+\infty$ would be a member of it.