Cauchy Sequence That Does Not Converge In A Space

In summary: To see this, let e > 0 and let N be such that n, m > N implies |1/n - 1/m| < e. Then take n = N+1 and m = N+2. Then n, m > N and yet |1/(N+1) - 1/(N+2)| = 1/(N+1)(N+2) > 1/N^2 > e. This sequence is in X which is not complete because it does not contain its limit.
  • #1
azdang
84
0
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.
 
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  • #2
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
[tex] x {:=} \lim_{j \to \infty} x_j [/tex]
has infinitely many (i.e. there are infinitely many i such that (x)i is not 0).
 
  • #3
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
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
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
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 [itex]x_j[/itex] converges.
 
  • #7
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
That's a sequence I had in mind... showing that it converges and works for your problem is up to you :smile:

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
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
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
l^inf is the space of bounded sequences, so (1, 1, 1, ...) is in l^inf.
 
  • #12
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
Ah yes, you're right.
 
  • #14
CompuChip said:
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).


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
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
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, ...
[tex]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}[/tex]

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:
  • #17
Gotcha. Thank you SO much for all of your help. I really appreciate it.
 
  • #18
CompuChip said:
That is, for a sequence x with elements x1, x2, x3, ..., xi, ...
[tex]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}[/tex]

Careful. Having infinitely many zeros doesn't imply having finitely many nonzero terms. Consider the sequence (0, 1, 0, 1, 0, 1, ...)
 
  • #19
Consider the sequence 1, 1/2, 1/3, ... in the space X = R \ {0}. This is a Cauchy sequence in X which does not converge in X.
 

Related to Cauchy Sequence That Does Not Converge In A Space

1. What is a Cauchy sequence that does not converge in a space?

A Cauchy sequence is a series of numbers where the terms get closer and closer together as the sequence progresses. In a space, this means that the sequence should have a limit or converge to a single point. However, there are cases where a Cauchy sequence does not converge to a point in a given space.

2. How can a Cauchy sequence still not converge in a space?

In order for a sequence to converge in a space, the space must be complete. This means that all Cauchy sequences in the space must have a limit or converge to a single point. If the space is not complete, there may be Cauchy sequences that do not have a limit and therefore do not converge.

3. Can a Cauchy sequence that does not converge in a space still be bounded?

Yes, it is possible for a Cauchy sequence that does not converge in a space to still be bounded. Boundedness means that the terms in the sequence do not exceed a certain value. Even though the sequence may not have a limit, it can still be bounded within a certain range.

4. What is the significance of a Cauchy sequence that does not converge in a space?

A Cauchy sequence that does not converge in a space is significant because it demonstrates that the space is not complete. Incomplete spaces have important implications in mathematics and can affect the validity of certain theorems and proofs.

5. Can a Cauchy sequence that does not converge in a space be made to converge by changing the space?

No, a Cauchy sequence that does not converge in a space cannot be made to converge by changing the space. Convergence of a sequence depends on the properties of the space itself and cannot be altered by changing the space.

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