Question about nested sequence in complete metric space

Question

Let $(R,\rho)$ be a complete metric space, and $A \subseteq R.$ Define $\displaystyle d(A)=\sup_{x,y \in A} \rho (x,y) \in [0,\infty].$ Let $(A_n)_{n \geq 1}$ be a sequence of nonempty closed sets in $R$ such that $A_1\supset A_2 \supset A_3 \supset ....$, and assume that $\displaystyle \lim_{n \rightarrow \infty}d(A_n)=0.$ Prove that $\displaystyle \bigcap_{n \in \mathbb{N}}A_n \neq \emptyset.$

Note that $$0=\lim_{n \rightarrow \infty}d(A_n)=d(\bigcap_{n \in \mathbb{N}}A_n)$$ so if $\displaystyle \bigcap_{n \in \mathbb{N}}A_n=\emptyset$, then that would mean there is no such an $x \in \displaystyle \bigcap_{n \in \mathbb{N}}A_n.$ I am getting stuck here, what I am thinking is since the intersection is empty, then $d(\bigcap_{n \in \mathbb{N}}A_n)$ is not exists so that contradicts $\displaystyle \lim_{n \rightarrow \infty}d(A_n)=0$?? Is this true?

Also, why this nested sequence $A_n=[n,\infty), n \in \mathbb{N}$ has an empty intersection? what condition does not satisfy in the above question?

Thanks for any help.

Answer 1

Pick $x_n \in A_n$ for each $n$ (the sets are non-empty, so that's no problem).

Now, if $\varepsilon >0$ is given, find $n$ such that $d(A_N)< \varepsilon$ (as $\lim_{n \to \infty} d(A_n)=0$ this is possible). Now if $n,.m \ge N$ we have that $x_n \in A_n \subseteq A_N$ and $x_m \in A_m \subseteq A_N$ (by the nestedness of the sequence). Hence:

$$\rho(x_n, x_m) \le \sup \{\rho(x,y): x,y \in A_n\} = d(A_N) < \varepsilon$$

and this shows that $(x_n)_n$ is a Cauchy sequence in $(X,\rho)$ and by completeness it has a limit $x \in X$.

Now, for every (fixed) $n$, all $x_m \in A_n$ with $m \ge n$ by the nestedness again, and as all but finitely many terms of the sequence lie in $A_n$ for this $A_n$. And the tail $(x_m)_{m \ge n}$ of the sequence is just a subsequence of $(x_n)_n$ and so has the same limit $x$, and this subsequence lies in $A_n$ and $A_n$ is closed so $x \in A_n$.

As this holds for each $n$: $$x \in \bigcap_n A_n \neq \emptyset$$

Answer 2

For $A_n=[n,\infty)$, the limiting radius is $\infty$, not 0.

As for the proof, choose a sequence $x_n$ such that $x_n\in A_n$. What can you say about the sequence? If you can show it's Cauchy, you can show that it converges because $\mathbb{R}$ is complete.

Answer 3

Some $A_N$ must be bounded, and since it's a bounded closed subset of a complete metric space, it must therefore be compact. If $\cap A_n=\emptyset$, then

$$\{A_N \setminus A_n~\vert~ n \gt N\}$$

is an open cover of $A_N$ with no finite subcover, which can't happen.

The nested sequence you propose doesn't have sets going to $0$ diameter. We need at least one set in the sequence to be bounded (in which case all following sets are necessarily bounded) for the proof to work.