Re: Complete metric space

let {x_{n}} be a cauchy sequence in X. by definition for any ε > 0 there is some N such that d(x_{m},x_{n}) < ε for all m,n > N.

so pick ε_{k} = 1/2^{k}, and set U_{k} = cl(B(x_{Nk},ε_{k})).

define X_{k} = U_{1}∩U_{2}∩...∩U_{k}.

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

furthermore since N_{k} < N_{k+1} its clear that for any k:

x_{Nk} is in U_{m}, for all m ≤ k, so each X_{k} is non-empty.

so, we have from (b) that ∩_{k}X_{k} ≠ Ø.

but this means the subsequence {x_{Nk}} of {x_{n}} 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 ∩_{k}X_{k},

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

(i was tempted to use just the 1/2^{k} 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).

Re: Complete metric space

Okay, thank you!

I was thinking the following: let be a Cauchy sequence on then let so is closed and satisfies yet I don't see how to prove that