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Cantor's Intersection Theorem 📂Analysis

Cantor's Intersection Theorem

Definition1

A sequence {Sn}n=1\left\{ S_{n} \right\}_{n=1}^{\infty} of a set is said to be nested if for every natural number nn, Sn+1SnS_{n+1} \subset S_{n} holds.

Explanation

The translation of nested might not be smooth, but since there is no better alternative, it is recommended to just memorize it as “Nested.”

Theorem

For the nested interval [an,bn][a_{n}, b_{n}], the following holds:

(a) n=1[an,bn]\displaystyle \bigcap_{n=1}^{\infty} [a_{n}, b_{n}] \ne \emptyset

(b) Specifically, if limn(bnan)=0\displaystyle \lim_{n \to \infty} (b_{n} - a_{n}) = 0 then n=1[an,bn]\displaystyle \bigcap_{n=1}^{\infty} [a_{n}, b_{n}] is a singleton set.

A singleton set is a set that contains only one element.

Proof

(a)

Given that for all natural numbers nn

[an+1,bn+1][an,bn]a1anbnb1 [a_{n+1} , b_{n+1} ] \subset [a_{n} , b_{n} ] \\ a_{1} \le a_{n} \le b_{n} \le b_{1}

by the axiom of completeness, there are two numbers

a:=sup{an}b:=inf{bn} a:=\sup \left\{ a_{n} \right\} \\ b:=\inf \left\{ b_{n} \right\}

Since for all natural numbers, anabbna_{n} \le a \le b \le b_{n} holds, it follows that [a,b][an,bn][a,b] \subset [a_{n} , b_{n} ], hence

n=1[an,bn] \bigcap_{n=1}^{\infty} [a_{n}, b_{n}] \ne \emptyset

(b)

Assuming limn(bnan)=0\displaystyle \lim_{n \to \infty} (b_{n} - a_{n}) = 0, since a=ba=b

n=1[an,bn]={a}={b} \bigcap_{n=1}^{\infty} [a_{n}, b_{n}] = \left\{ a \right\} = \left\{ b \right\}

See Also


  1. William R. Wade, An Introduction to Analysis (4th Edition, 2010), p55 ↩︎