1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Oct 1, 2019 at 0:11
  • $\begingroup$ So, I think using proof by contradiction is not useful. $\endgroup$
    – Ahmed
    Commented Oct 1, 2019 at 0:32
  • $\begingroup$ 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 $n.$ $\endgroup$ Commented Oct 1, 2019 at 5:47

3 Answers 3

4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – Ahmed
    Commented Oct 1, 2019 at 0:36
0
$\begingroup$

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.

$\endgroup$
3
  • 2
    $\begingroup$ 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 $d(x,y)=\min(|x-y|, 1)$. $\endgroup$ Commented Oct 1, 2019 at 4:33
  • $\begingroup$ @HennoBrandsma Thanks. I see that you're right but I've never heard of the concept "totally bounded." What does it mean? $\endgroup$ Commented Oct 1, 2019 at 4:36
  • $\begingroup$ 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. $\endgroup$ Commented Oct 1, 2019 at 4:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .