# Proving the nested interval theorem

Theorem:

Let $$\{I_n\}_{n \in \mathbb N}$$ be a collection of closed intervals with the following properties:

1. $$I_n$$ is closed $$\forall \,n$$, say $$I_n = [a_n,b_n]$$;
2. $$I_{n+1} \subseteq I_n\forall \,n$$. Then $$\displaystyle\bigcap_{n=1}^{\infty} I_n \ne \emptyset$$.

Pf: Let $$I_n$$ be intervals that satisfy 1 and 2. Say $$I_n = [a_n, b_n] \forall n\ge 1$$.

Let the sets $$A$$ and $$B$$ be defined by $$A = \{a_n\}$$ and $$B = \{b_n\}$$. Therefore $$\forall n,k \ge 1$$, $$a_k \le b_n$$.

Case 1: $$k \le n$$

Then $$[a_n,b_n] \subset [a_k, b_k]$$ therefore $$b_n \in [a_k,b_k]$$ and $$a_k \le b_n \le b_k$$ therefore $$a_k \le b_k$$.

Case 2: $$k>n$$

Therefore $$I_k \subset I_n$$. By nestedness, $$[a_k,b_k] \subset [a_n,b_n]$$, therefore $$a_k \le b_n$$. Claim: $$\sup A \le \inf B$$. Proof of claim: Let $$A$$ and $$B$$ be sets such that for all $$a \in A$$ and for all $$b \in B$$, $$a \le b$$ Therefore $$\sup A \le b$$ and $$a \le \inf B$$ therefore $$\sup A \le \inf B$$.

Now we must prove that either $$\bigcap_{n=1}^{\infty} I_n = [\sup A, \inf B]$$ or $$\bigcap_{n=1}^{\infty} I_n = \emptyset$$. First we will show $$[\sup A, \inf B] \subset \bigcap I_n$$. Let $$x \in [\sup A, \inf B]$$. Therefore $$\sup A \le x \le \inf B$$ and $$\forall n$$, $$a_n \le \sup A \le x \le \inf B \le b_n$$ or $$a_n \le x \le b_n$$ and thus $$x \in I_n$$.

Now we will show that $$\bigcap I_n \subset [\sup A, \inf B]$$. Let $$y \in \bigcap I_n$$. Show $$\sup A \le y \le \inf B$$. We know that $$\forall n \ge 1$$, $$a_n \le y \le b_n$$. Since $$a_n \le y$$, we see that $$\sup A \le y$$.

Similarly since $$y \le b_n$$, we see that $$y \le \inf B$$.

As you can see, this proof is very long. Does anyone have any advice to shorten this?

• I think the $\text{ \prod }$ command was the wrong one to use, since it means the Cartesian product of the sets. Instead, I think you mean the intersection, which can be achieved by typing $\text{ \bigcap }$ to get $\bigcap$. Oct 11, 2015 at 23:15
• Also, please check the edits to learn how to improve your proof-writing, besides some LaTeX things (e.g. how to write { in mathmode, or how to write sup and inf). Oct 11, 2015 at 23:28

## 2 Answers

Here is a non-constructive proof.

Construct a sequence by choosing an element $x_n \in I_n$ for every $n$; you can do this however you like. Since this sequence is bounded then the Bolzano-Weierstrass theorem says there exists a convergent subsequence $x_{n_k}$ converging to some real number $c$. Suppose $c \not\in \bigcap_{n=1}^{\infty} I_n$. Then there exists some $N$ for which $c \not\in I_n$ for all $n > N$, and so there is some number $\varepsilon >0$ such that $|x_{n_k} - c| \geq \varepsilon$ for all $n_k > N$ contradicting convergence. Thus $c \in \bigcap_{n=1}^{\infty} I_n$ and so it is nonempty.

As $I_{n+1} \subset I_n$, it follows that any $a_m$ is a lower bound for $\left\{b_n\right\}$ and vice-versa for any $b_m$ being an upper bound for $\left\{a_n\right\}$. $\left\{a_n\right\}$, $\left\{b_n\right\}$ are non-empty so $a= \sup\left\{a_n\right\}$, $b= \inf\left\{b_n\right\}$ exist.

Further, $a\leq b$, as if $b < a$ we can find $a_n, b_m$ arbitrarily close to $a,b$ respectfully, so it would be possible to find $b_m < a_n$ which would be a contradiction. So $[a,b]$ is well defined and $a_n \le a, b \le b_n$ $\forall n$ so $[a,b] \subset I_n$ $\forall n$. So $[a,b] \subset \bigcap_n I_n$.