Space L Infinity Homework: Open or Closed?

[dirk_mec1]

Homework Statement

http://img254.imageshack.us/img254/121/77689018gc3.png

Homework Equations

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

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?

 

[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?

 

[dirk_mec1]

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?

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?

 

[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.

 

[dirk_mec1]

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

 

[Dick]

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

I think that's correct.

 

[dirk_mec1]

http://img156.imageshack.us/img156/3591/42927439wp4.png

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?

 

[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.

 

[dirk_mec1]

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

 

[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.

 

[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?

 

[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.

 

[dirk_mec1]

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.

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.

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.

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

 

[Dick]

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

 

[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?

 

[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?