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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!

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take $(-1/n, 1/n)$ over the natural numbers $n > 0$ – user29743 Oct 29 '12 at 6:47
up vote 4 down vote accepted

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

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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 $$\bigcap_{n=1}^\infty U_n = F.$$

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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.

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