1.5 Cantor’s Intersection or Nested Interval theorem
If \([a_{n + 1}, b_{n + 1}]\subseteq [a_n, b_n]\) for every \(n \in \mathbb {N}\), then the intersection of the intervals \([a_n, b_n]\), \(n\in \mathbb {N}\), is non-empty. When the lengths also shrink to zero we can say more.
Theorem 1.5.1 (Cantor’s Intersection or Nested Interval Theorem). Let \(\{A_n \}^{\infty }_{n = 1}\) be a sequence of closed intervals such that \(A_{n + 1} \subseteq A_n\) for every \(n\) and \(\lim _{n\rightarrow \infty }(\text {length of } A_n) = 0\). Then \(\bigcap ^{\infty }_{n=1} A_n\) consists of exactly one point.
Proof. Let \(A_n = [a_n, b_n]\). Since \(A_{n + 1} \subseteq A_n\), we have \(a_n \leq a_{n + 1} \leq b_{n + 1} \leq b_n\) for every \(n\). So \(\{a_n\}\) is increasing, \(\{b_n\}\) is decreasing, and \(a_m \leq b_n\) for all \(m\) and \(n\). Hence \(L = \sup \{ a_n : n \in \mathbb {N}\}\) exists and satisfies \(a_n\leq L \leq b_n\) for every \(n\), that is, \(L \in \bigcap ^{\infty }_{n = 1} A_n\).
If \(L^*\) is another point of \(\bigcap ^{\infty }_{n = 1} A_n\), then \(L\) and \(L^*\) both lie in \([a_n,b_n]\), so \(|L - L^*| \leq b_n - a_n\) for every \(n\). Since \(b_n - a_n \longrightarrow 0\), this forces \(L^* = L\). Hence the intersection consists of exactly one point. □
Example 1.5.2. Show by examples that each hypothesis of the Nested Interval Theorem is needed: the intervals must be closed, and they must be bounded.
Solution. The intervals \(A_n = \left (0,\frac {1}{n}\right ]\) are nested and their lengths tend to \(0\), but they are not closed, and \(\bigcap _n A_n = \emptyset \): any \(x>0\) fails to lie in \(A_n\) once \(\frac {1}{n}<x\), and \(0\) lies in none of them.
The intervals \(A_n = [n,\infty )\) are closed and nested but unbounded, and again \(\bigcap _n A_n = \emptyset \), since any real \(x\) fails to lie in \(A_n\) once \(n>x\).
By contrast, \(A_n = \left [0,\frac {1}{n}\right ]\) satisfies all the hypotheses, and \(\bigcap _n A_n = \{0\}\).
Example 1.5.3. Use nested intervals to locate \(\sqrt {2}\), and show that the method determines it uniquely.
Solution. Start with \(A_1 = [1,2]\), since \(1^2<2<2^2\). At each step, bisect the current interval \([a_n,b_n]\) at its midpoint \(m\) and keep the half \([a_n,m]\) or \([m,b_n]\) whose end points satisfy \(a^2\leq 2\leq b^2\). The first few intervals are \begin {gather*} [1,2],\quad [1,1.5],\quad [1.25,1.5],\quad [1.375,1.5],\\ [1.375,1.4375],\quad [1.40625,1.4375],\quad [1.40625,1.421875],\ \dots \end {gather*} For example, \(1.5^2 = 2.25>2\) gives \([1,1.5]\), and then \(1.25^2 = 1.5625<2\) gives \([1.25,1.5]\). Each interval is closed, contains the next, and has length \(2^{-(n-1)}\longrightarrow 0\). By the Nested Interval Theorem they have exactly one common point \(L\). Since \(a_n^2\leq 2\leq b_n^2\) and \(a_n, b_n\longrightarrow L\), we get \(L^2 = 2\), so \(L = \sqrt {2}\). After \(n\) steps the midpoint approximates \(\sqrt {2}\) with error at most \(2^{-n}\); for example ten steps give three correct decimal places.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.