Solving Completion Problem: Showing Null Sequences Converge

[homology]

I'm finding that I'm not very good at showing that a space is complete. I was wondering if you could help me out.

Consider the space of null sequences. That is, sequences of complex numbers that converge to zero. It has a max norm. So if x is a sequence the norm of x, |x| = max |a_n| where the max ranges over n.

So we need to take a cauchy seqence of elements and show that it converges to a null sequence. I'm just not sure how to do it. Any hints?

 

[matt grime]

this isn't a linear algebra problem, post it in analysis.

 

[homology]

sure it is

Does it really matter?

 

[Hurkyl]

Sounds more like it belongs in homework to me!

Anyways, have you tried anything at all?

 

[matt grime]

Does it really matter?

it isn't an algebra question, it is analysis, it deals with the completeness in an analytical sense of the word, it doesn't get to be more analytic than that. and if you post it in the 'wrong' thread people may choose not to read it though they may be able to supply an answer.

 

[homology]

Yeah I've at least applied the definitions of cauchy sequences. I'm new at this completion deal. I've never proved anything like it before, so I'm not sure how to show that there is something in the space to which this sequence converges.
So the elements of the sequence, which are sequences, are getting closer together in the sense of the max norm. I can write that down but what next? There doesn't seem to be anything preceding the problem in the text that helps.


homology

 

[matt grime]

Let x(m)_n be the n'th entry in the m'th sequence.

we know that for all n |x(p)_n-x(q)_n| is less than the max over n and that the max is cuachy, that tells us that if we fix an n, the n'th terms of the sequences are cauchy hence they converge to y_n since C is complete. Now you must show that y_n tends to zero as n tends to infinity.

 

[homology]

Okay so now we have this sequence {y_n} and we need to show that it converges to zero. how about the following:

For any number e>0 there exists an N such that for all p>N
|y_n - x(p)_n|<e

Now if you took the limit of both sides with respect to n, you'd get

lim |y_n| < e

But e is arbitrary so lim|y_n| = 0.

Feel free to shoot me down, I appreciate the help.

homology