# Complete metric space

• Dec 16th 2012, 10:06 AM
Ustanak
Complete metric space
Let $\displaystyle X_{n+1}\subset X_n$ for each $\displaystyle n\in\mathbb N.$ Let $\displaystyle (M,d)$ be a metric space. Prove that the following assertions are equivalent:

a)$\displaystyle (M,d)$ is a complete metric space.
b) For each $\displaystyle X_{n+1}\subset X_n$ being closed and non-empty where $\displaystyle \text{diam}(X_n)\to0$ as $\displaystyle n\to\infty,$ we have $\displaystyle \bigcap_{n=1}^\infty X_n\ne\varnothing.$

I already did a) $\displaystyle \implies$ b) but for the other implication, I have a long solution, is there a short way to do it?
• Dec 16th 2012, 11:21 AM
Deveno
Re: Complete metric space
let {xn} be a cauchy sequence in X. by definition for any ε > 0 there is some N such that d(xm,xn) < ε for all m,n > N.

so pick εk = 1/2k, and set Uk = cl(B(xNkk)).

define Xk = U1∩U2∩...∩Uk.

it's clear that diam(Xk) ≤ 1/2k-1 and that each Xk is closed being the intersection of finitely many closed sets.

furthermore since Nk < Nk+1 its clear that for any k:

xNk is in Um, for all m ≤ k, so each Xk is non-empty.

so, we have from (b) that ∩kXk ≠ Ø.

but this means the subsequence {xNk} of {xn} converges to some point x of X (the sole element of the infinite intersection, since if d(x,y) = 0, y = x, and for any y in ∩kXk,

we have d(y,xNk) < 1/2k-1 for ALL k in N). hence {xn} also converges to x, so X is complete.

(i was tempted to use just the 1/2k balls at first, but i realized that the 2nd ball could contain points OUTSIDE the first ball, but their intersection has no such defect. the center of the 3rd ball has to lie within the intersection of the first 2, and the center of the 4th ball has to lie within the intersection of the first 3, etc. the diameter of the intersections can be no larger than the diameter of the ball, which is twice its radius).
• Dec 16th 2012, 12:30 PM
Ustanak
Re: Complete metric space
Okay, thank you!

I was thinking the following: let $\displaystyle x_n$ be a Cauchy sequence on $\displaystyle X,$ then let $\displaystyle X_n=\overline{\{x_m:m\ge n\}},$ so $\displaystyle X_n$ is closed and satisfies $\displaystyle X_{n+1}\subset X_n,$ yet I don't see how to prove that $\displaystyle \text{diam}(X_n)\to0.$