Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Give an example of open, nested sets such that the intersection is closed nonempty.

I will ask questions if I am in doubt of the example provided. Thank you!

share|improve this question
    
take $(-1/n, 1/n)$ over the natural numbers $n > 0$ –  user29743 Oct 29 '12 at 6:47

3 Answers 3

up vote 4 down vote accepted

Let $A_n = \left(-1-\dfrac1n,1+\dfrac1n \right)$ and $$A = \displaystyle \bigcap_{n=1}^{\infty} A_n = [-1,1]$$

share|improve this answer

More generally, if $F \subseteq \mathbb{R}$ is nonempty closed, for each $n$ define $$U_n = \left\{ x \in \mathbb{R} : ( \exists y \in F ) ( | x - y | < 1/n ) \right\}.$$ Then each $U_n$ is an open subset of $\mathbb{R}$, $U_{n+1} \subseteq U_n$, and $$\bigcap_{n=1}^\infty U_n = F.$$

share|improve this answer
    
Nice. And it works in any metric space. –  Did Oct 29 '12 at 8:48

If you experiment even a little you should find one of the many possible examples. However, if you really don’t see one right away, you can attack the problem systematically. The simplest open sets in $\Bbb R$ are open intervals, so it makes sense to try to construct an example using them.

Suppose that you have nested open intervals $(a_n,b_n)$ for $n\in\Bbb N$, so that $(a_{n+1},b_{n+1})\subseteq(a_n,b_n)$ for each $n\in\Bbb N$. Then you have

$$a_0\le a_1\le a_2\le\ldots\le\ldots b_2\le b_1\le b_0\;.$$

The sequence $\langle a_n:n\in\Bbb N\rangle$ is bounded and monotonically non-decreasing, so it converges to some limit $a$. Similarly, $\langle b_n:n\in\Bbb N\rangle$ converges to some limit $b$. Moreover, it’s clear that $a\le b$, and you shouldn’t have much trouble seeing that $\bigcap_{n\in\Bbb N}(a_n,b_n)$ must be one of the intervals $(a,b)$, $[a,b)$, $(a,b]$, and $[a,b]$. Since you want a closed, non-empty intersection, you want to arrange matters so that $\bigcap_{n\in\Bbb N}(a_n,b_n)=[a,b]$.

Suppose that the sequence $\langle a_n:n\in\Bbb N\rangle$ is eventually constant: from some point on all of the $a_n$’s are the same. Then $\langle a_n:n\in\Bbb N\rangle$ converges to that constant value, and we have $a_n=a$ for all sufficiently large $n$. You should have no trouble seeing that $a\notin\bigcap_{n\in\Bbb N}(a_n,b_n)$ under these circumstances. Since we want $a\in\bigcap_{n\in\Bbb N}(a_n,b_n)$, we need to make sure that $\langle a_n:n\in\Bbb N\rangle$ is not eventually constant. The easiest way to ensure this is to make it strictly increasing: $a_0<a_1<a_2<\ldots$.

Essentially the same reasoning shows that if we’re to get $b\in\bigcap_{n\in\Bbb N}(a_n,b_n)$, we must ensure that the sequence $\langle b_n:n\in\Bbb N\rangle$ is not eventually constant, which we can do by making it strictly decreasing: $b_0>b_1>b_2>\ldots$. Thus, if we choose the intervals so that

$$a_0<a_1<a_2<\ldots<\ldots b_2<b_1<b_0\;,\tag{1}$$

we guarantee that $\bigcap_{n\in\Bbb N}(a_n,b_n)=[a,b]$.

You should have no difficulty at all coming up with intervals $(a_n,b_n)$ for $n\in\Bbb N$ such that $(1)$ is satisfied and $a$ and $b$ are easy to find.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.