How do you proof cantor intersection theorem?

Answer 1

Theorem 4.1.6 Cantor intersection theorem Let (X; d) be a complete metric

space. Let (Fn) be a decreasing sequence of nonempty closed subsets of X s.t. diam(Fn) !

0 in R. Then \nFn contains exactly one point.

Proof: Let F = \nFn. If F contains two points x and y then we have a contradiction

when diam(Fn) < d(x; y). Hence jFj 1.

8n choose xn 2 Fn. diam(Fn) ! 0 ) (xn) is Cauchy.

Hence 9x 2 X s.t. (xn) ! x. We show that x 2 Fn 8n. If fxng is nite then

xn = x for innitely many n, so that x 2 Fn for innitely many n. Since Fn+1 Fn

this implies x 2 Fn 8n. So suppose fxng is innite. 8m, (xm; xm+1; : : : ; xm+k; : : : ) is

a sequence in Fm converging to x. Since fxngnm is innite, this implies x is a limit

point of Fm. But Fm is closed, so x 2 Fm.