# Question about nested sequence in complete metric space

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.

• Thinks about your argument in $\mathbb{R}$, for example if $A_n=[-1n,1/n]$, and see where things go wrong: We have $A_1\supseteq A_2\supseteq\cdots$ and $d(A_n)=2/n$ converges to $0$. Your 'big equation' (right after "Note that"), becomes $$0=\lim_n d(A_n)=d(\left\{0\right\}),$$which is alright. But why do you conclude that just because a set has zero diameter then it is empty? Just as we saw, we have $d(\left\{0\right\})=0$, but $\left\{0\right\}$ is nonempty. That is where your mistake is. Commented Oct 1, 2019 at 0:11
• So, I think using proof by contradiction is not useful. Commented Oct 1, 2019 at 0:32
• If $R$ is the set of reals and $\rho (x,y)=|x-y|$ and $A_n=[n.\infty)$ then we don't have $\lim_{n\to \infty}d(A_n)=0.$ We have instead $d(A_n)=\infty$ for each $n.$ Commented Oct 1, 2019 at 5:47

## 3 Answers

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 $$n$$: $$x \in \bigcap_n A_n \neq \emptyset$$

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.

• I like that thank you.. after proving that this sequence converge? how do we know that $\lim_{n \rightarrow \infty}x_n$ in the intersection? Commented Oct 1, 2019 at 0:36

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.

• A bounded closed subset of a complete metric space need not be compact, you need totally bounded for that. Easy counterexample: $\Bbb R$ in the truncated metric $d(x,y)=\min(|x-y|, 1)$. Commented Oct 1, 2019 at 4:33
• @HennoBrandsma Thanks. I see that you're right but I've never heard of the concept "totally bounded." What does it mean? Commented Oct 1, 2019 at 4:36
• It means that for every $r>0$ we can find finitely many open balls of radius $r$ that cover the set. It's a uniform version (same ball radius) of compactness in a sense. See Wikipedia and many textbooks. Commented Oct 1, 2019 at 4:47