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Not a homework question, came across this exercise in Churchill's complex analysis. I haven't done any analysis before so I'm not sure how to answer it.

We have a closed interval $$a_0 \leq b_0$$ which we subdivide into $a_1 \leq b_1$, which can be either the left hand or right hand half of $a_0 \leq b_0$. This is then subdivided into $a_2 \leq b_2$, which again can be either the left or right hand half of $a_1 \leq b_1$. This is continued ad infinitum.

Prove that there is a point $x_0$ common to all intervals.

How do I do this? I've got the inequalities $$a_0 \leq a_n \leq a_{n+1} \leq b_0 $$ and $$a_0 \leq b_{n+1} \leq b_n \leq b_0 $$ so $a_n$ and $b_n$ are bounded with $$\lim_{n\rightarrow \infty} a_n \leq b_0 $$ and $$\lim_{n\rightarrow \infty} b_n \geq a_0 $$ but I'm not sure how to show that the limits of $a_n$ and $b_n$ exist and that they both equal the same number. Any help?

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up vote 1 down vote accepted

Completeness of $\Bbb R$ ensures that every bounded monotone sequence converges, so $\lim\limits_{n\to\infty}a_n$ and $\lim\limits_{n\to\infty}b_n$ certainly exist; call them $a$ and $b$, respectively. To show that $a=b$, note that by construction $$b_{n+1}-a_{n+1}=\frac12(b_n-a_n)$$ for each $n$, so $\lim\limits_{n\to\infty}(b_n-a_n)=0$. And $a_n\le a\le b\le b_n$ for each $n$, so ... ?

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excellent, thanks! – user46080 Dec 20 '12 at 5:52
@user46080: You’re welcome. – Brian M. Scott Dec 20 '12 at 6:03

Here is another proof that relies on compactness and nesting.

Let $I_n = [a_n,b_n]$. Note that $I_{n+1} \subset I_n \subset I_0$. If $J$ is a finite set of indices, then $\cap_{j \in J} I_j = I_{\max(J)}$, which is non-empty. Since $I_0=[a,b]$ is compact, and the $I_n$ are closed, the finite intersection property gives $\cap_{n=0}^\infty I_n \neq \emptyset$, which is the desired result.

Note: Since in this case, $\text{diam} I_n \to 0$, we can additionally show that $\cap_{n=0}^\infty I_n = \{x_0\}$ for some $x_0 \in I_0$.

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Alternatively, suppose, for contradiction, that the intersection is empty.

Then let $S_i$ be the complement of the $i$th interval. $S_i$ is open, and $\{S_i\}$ is thus an open cover of the compact interval $[a_0,b_0]$. There thus must exist a finite subcover $\{S_{i_1},S_{i_2},\ldots,S_{i_n}\}$, a contradiction since none of those sets contains the point $\frac{1}{2}(a_{i_n}+b_{i_n})$.

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This is not true in general. For example, a nested sequence of closed subsets of $\mathbb{Q}$ that have an empty intersection is: $$ A_n =\mathbb{Q}\cap [\sqrt{2} -\frac{1}{n},\sqrt{2} + \frac{1}{n}] $$ but this is true in $\mathbb{R}$ by the nested intervals lemma.

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Of course, we would need the space to be complete. – toypajme Dec 21 '12 at 16:21

Here's another proof: The sequence $(a_i)_{i=0}^\infty$ is bounded above by every $b_i$, so it has a least upper bound, call it $a$. The sequence $(b_i)_{i=0}^\infty$ is bounded below by $a$, so it has a greatest lower bound $b \ge a$. The point $\frac{a+b}{2}$ satisfies $$a_i \le a \le \frac{a+b}{2} \le b \le b_i, \quad \forall \; i \in \{0,1,2,..\},$$

and therefore $\frac{a+b}{2} \in \cap_{i=0}^{\infty} [a_i, b_i] \ne \varnothing$.

Note that all this proof requires is that $[a_{i+1}, b_{i+1}] \subseteq [a_i, b_i] \subset \mathbb{R}, \; \forall \; i \in \{0,1,2,..\},$ which is a bit more general than your statement.

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