Prove that l^1 is complete

[winter85]

1. The problem statement, all variables and given/known data

Prove that \ell^1, the space of all (real) sequences v = \{v_k\} such that \sum|x_k| < \infty , is complete.

2. Relevant equations

\ell^1 is a normed space with the norm ||x|| = \sum |x_k|

3. The attempt at a solution

Let v_n be a Cauchy sequence of sequences in \ell^1. Then for all \epsilon > 0 there exists N > 0 such that for all n,m > N we have \sum |v_{n,k} - v_{m,k}| < \epsilon (here v_{n,k} means the kth term of the nth sequence)

in particular this means that |v_{n,k} - v_{m,k}| < \epsilon so we can define a sequence u = \{u_k\} as u_k = \lim v_{n,k} as n goes to infinity.

Now i think the sequence u would be the limit of v_n as n goes to inifnity, but i'm not sure how to prove it. Firstly, I dont know how to prove that u converges absolutely. the problem is by the definition of u, given \epsilon I can find a sequence v_n whose terms are each within [tex]\epsilon[\tex] from the corresponding term in u, but when summing, this is like summin [tex]\epsilon[\tex] infinitly many times.. so how can I do it? any hint would be appreciated :)
Thanks.

[winter85]

if we do this:
\lim \sum |v_{n,k} - v_{m,k}| = \sum \lim |v_{n,k} - v_{m,k}| where the limit is taken as m -> infinity, then we have \sum |v_{n,k} - u_{k}| < \epsilon for sufficiently large n. But how can I justify interchanging the sum and the limit?

[Billy Bob]

Your u is good.

There is a trick to interchanging the sum and the limit, which roughly speaking is this:

(1) use epsilon,
(2) change the bound on the infinite sum to a bound on a finite sum from k=1 to j,
(3) note that the bound is independent of j,
(4) take your limit as m approaches infinity (OK to do, since sum is finite),
(5) bound is still independent of j, so now let j approach infinity.

Here are the details, applied to the problem of showing u converges absolutely.

(1) v_n is Cauchy so it's bounded; thus ||v_n||\le M (using M instead of epsilon for this example)

(2) for any j, \sum_{k=1}^j |v_{n,k}|\le M

(3) previous bound is independent of j

(4) take limit as n (using n instead of m) approaches infinity, \sum_{k=1}^j |u_k|\le M

(5) bound is still independent of j, so let j approach infinity, ||u||\le M.

Now you can try it for the convergence.

[winter85]

oh, I see! It's now very clear to me. Thank you so much. :)