# Cauchy Sequence That Does Not Converge In A Space

1. Jan 30, 2009

### azdang

Let M be a subspace of l^infinity consisting of all sequences x = (x_j) with at most finitely many nonzero terms. Find a Cauchy sequence in M which does not converge in M, so that M is not complete.

Does anyone have any ideas what to use for a Cauchy sequence that does not converge in M? I've been trying to think of something, but I can't come up with anything. Thanks.

2. Jan 30, 2009

### CompuChip

If I understand the question correctly, you need a sequence where each element xj has only finitely many non-zero terms (i.e. for every j there are only finitely many i such that (xj)i is not 0), but the limit
$$x {:=} \lim_{j \to \infty} x_j$$
has infinitely many (i.e. there are infinitely many i such that (x)i is not 0).

3. Jan 30, 2009

### azdang

Yes, at least that's what I thought, too. The question is out of my book for Real Analysis. I've been unable to come up with anything though. And also, since M is a subspace of l^infinity, I'm guessing any sequence in M has to be bounded, too? I'm not really sure, I'm reallyyy confused by this question.

4. Jan 30, 2009

### CompuChip

So the obvious sequence with only finitely many non-zero elements is the sequence consisting only of 0: (0, 0, 0, 0, 0, 0, ....), so you can take that as the first element.

A simply sequence with infinitely many non-zero elements would be one that only consists of 1's; (1, 1, 1, 1, 1, 1, 1, 1, ....).

Is it in l_infinity? Then you can take that as your limit... any thoughts as how to get from one in the other (in infinitely many steps, but each step only removing finitely many zeroes)?

5. Jan 30, 2009

### azdang

I think I was trying to think of something along these lines, but I was getting no where because I still was unsure how to construct the sequence. What did you mean by this: Is it in l_infinity? Then you can take that as your limit

Thanks for your help :)

6. Jan 30, 2009

### CompuChip

Sorry, it doesn't need to be in l_infinity. It just has to be outside of M, which is obviously is.

With "then you can take that as your limit" I meant that it is a good candidate for the sequence x to which your sequence of sequences $x_j$ converges.

7. Jan 30, 2009

### azdang

So, if the sequence was like (0,0,0,...), then (1,0,0,0,...), (1,1,0,0,0,...)...does that ultimately converge to (1,1,1,1,...) and work for my problem? I'm not sure I'm getting it right.

8. Jan 31, 2009

### CompuChip

That's a sequence I had in mind... showing that it converges and works for your problem is up to you

Seriously, do you know how to prove convergence with the metric on l_infinity? Or is it the concept of sequences of sequences you're having trouble with?

If you have a sequence of sequences x_i, then what is the difference between two of them, e.g. what is |x_4 - x_2| ?

9. Jan 31, 2009

### azdang

I'm honestly not even sure what's causing me trouble. I think it's the whole idea of trying to show something is incomplete, as oppose to complete, because I have absolutely no trouble proving something is COMPLETE. But thank you again for all of your help, I'm going to try to apply this to my problem.

10. Jan 31, 2009

### CompuChip

Actually showing incompleteness is easier, because instead of taking a general Cauchy sequence and showing that it converges, you only have to find a single Cauchy sequence which does not converge.

Anyway, I think you'd rather use
(0, 0, 0, 0, ....)
(0, 1, 0, 0, ....)
(0, 1, 1/2, 0, ....)
(0, 1, 1/2, 1/3, 0, 0, ...)
or something like that, because when you use just 1's it doesn't converge while this does (as you'll find out).

11. Jan 31, 2009

### Preno

l^inf is the space of bounded sequences, so (1, 1, 1, ...) is in l^inf.

12. Jan 31, 2009

### CompuChip

Yes, but it is not in the set of sequences in l^inf with at most finitely many nonzero terms.

And the reason (1, 1, 1, ...) doesn't work is that the sequence
(0, 0, 0, ...)
(1, 0, 0, ...)
(1, 1, 0, ...)
...
(1, 1, 1, 1...)

doesn't converge in l^inf, I think (the l^inf-distance between any two distinct elements is 1).

13. Jan 31, 2009

### Preno

Ah yes, you're right.

14. Feb 1, 2009

### azdang

I'm a little confused, when a sequence like this converges, isn't it going to converge to a sequence with 0 as its first term. And then, isn't this limit in M? Sorry! I'm having a really hard time visualizing it.

15. Feb 1, 2009

M is the set of sequences with finitely many nonzero terms. In l^inf, it will converge to the sequence (0, 1, 1/2, 1/3, 1/4, ...), which has infinitely many nonzero terms (only the first term is zero).

(The zero term at the beginning isn't necessary; you may as well work with (1, 1/2, 1/3, ...).)

16. Feb 2, 2009

### CompuChip

The kind of formulation used in the question takes some time getting used to.

"At most finitely many non-zero terms" means that there are infinitely many which are zero.
That is, for a sequence x with elements x1, x2, x3, ..., xi, ...
$$x \in M \iff \{ i = 0, 1, 2, \cdots | x_i \neq 0 \} \text{ is finite} \iff \{ i = 0, 1, 2, \cdots | x_i = 0 \} \text{ is infinite}$$

Clearly,
(0, 1, 1/2, 1/3, ...)
(1, 1/2, 1/3, 1/4, ...)
or even
(0, 0, 0, 0, ...., 0, 1, 1/2, 1/3, 1/4) [with any finite number of zeroes added at the beginning)
are not in M then, because you only have finitely many which are zero.

Last edited: Feb 2, 2009
17. Feb 2, 2009

### azdang

Gotcha. Thank you SO much for all of your help. I really appreciate it.

18. Feb 2, 2009