# The space l infinity

1. Sep 29, 2008

### dirk_mec1

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

http://img254.imageshack.us/img254/121/77689018gc3.png [Broken]

2. Relevant equations
A space is open iff there is for each x in A a ball (fully) contained within A.

3. The attempt at a solution
If I consider A in $$l^{\infty }$$ then the sup =1. So for each x there is ball with a radius smaller than 1 contained within A.

Is this correct?

Last edited by a moderator: May 3, 2017
2. Sep 29, 2008

### Dick

Suppose you pick an x such that lim x_n as n->infinity=1. Then what do you use as epsilon? BTW, what is K?

Last edited: Sep 29, 2008
3. Sep 29, 2008

### dirk_mec1

K is C (complex number) or R(real number). So in the sequence you chose I would take for the radius of the ball $$1-x_n$$ since this is the distance between the limit and the sequence. Is this correct Dick?

4. Sep 29, 2008

### Dick

You can't pick a different epsilon for every n. Look at the definition of l_infinity. |x-y|<epsilon means for all n, |xn-yn|<epsilon.

5. Sep 29, 2008

### dirk_mec1

But if that's the case then I cannot find a epsilon so then A cannot be open, right?

6. Sep 29, 2008

### Dick

I think that's correct.

7. Sep 29, 2008

### dirk_mec1

http://img156.imageshack.us/img156/3591/42927439wp4.png [Broken]

So I take the sequence: $$x_n = \frac{1}{2}-\frac{1}{n}$$ This converges with limit 1/2 but we know that $$|x_n|<1$$ so then both the sequence and the limit are in A. But then A is closed.

Is this a correct proof: can I just take a convergent sequence?

Last edited by a moderator: May 3, 2017
8. Sep 29, 2008

### Dick

No, you can't take just any convergent sequence. You have to consider ALL converent sequences. And you can't say anything about just one sequence either. Remember a point in A is itself a sequence. A sequence in A is a sequence of sequences.

9. Sep 30, 2008

### dirk_mec1

A cannot be closed because the sequence that you mentioned is in A but has a limit outside A, right Dick?

10. Sep 30, 2008

### Dick

The 'sequence' I mentioned isn't a sequence in A. It's a point in A. Can you construct a sequence in A that converges to a point that's not in A.

11. Sep 30, 2008

### dirk_mec1

I define the sequence $$x_{nj}$$ , j=1,2... (a sequence of sequences). Each of the $$x_{nj}$$ has the following properties:

1) $$|x_{nj}|<1$$
2) $$\lim_{n \rightarrow \infty} |x_{nj}| = 1$$

but the limit is not in A therefore A cannot be closed. Is it okay now, Dick?

12. Sep 30, 2008

### Dick

Almost. If n in the index within sequence number j, the limit n->infinity=1 is not enough. The limiting sequence could still be in A. You want the limit as j->infinity=1. It might be clearer if you actually give an example. Like a_nj=(1-1/j). So sequence number j is just the constant sequence (1-1/j,1-1/j,...). The limiting sequence is (1,1,1,...) which is NOT in A. Try writing down the sequences explicitly like this to make sure you are saying what you mean.

13. Oct 1, 2008

### dirk_mec1

So I made the mistake of the fact that for each j the sequence converges but what we want is that the entire sequence converges to one.

So in this example $$a_{nj} = 1 - \frac{1}{j}$$ for each j the sequence is just a constant for every natural number, right?

14. Oct 1, 2008

### Dick

Yes, each sequence is constant. The limit of the constants is 1.

15. Oct 2, 2008

### dirk_mec1

Ok finally I got for the closure:

$$A:= \{ x: \mathbb{N} \rightarrow \mathbb{K} |\ \forall n \in \mathbb{N}\ |x_n| \leq 1 \}$$

But how can I prove that there are accumalation points around x_n =1?

16. Oct 2, 2008

### Dick

If you have sequence where some of the x_i's are one, isn't it pretty straightforward to construct a sequence of sequences in A that converges to it?