# Complete Metric Subspaces

1. Feb 26, 2010

### BSCowboy

1. The problem statement, all variables and given/known data
Determine whether the following metric subspaces are complete:
a) the set E of sequences containing only entries 0 & 1 in $$(m,||\cdot||_{\infty})$$
b) the unit sphere in any Banach Space

2. Relevant equations
a) for $$x=\{\lambda_1,\lambda_2,\ldots,\lambda_n,\ldots \}$$
$$||x||_{\infty}=sup\{|\lambda_n|:n=1,2,\ldots\}$$

b)$$\{x\in X:||x-x_0||=1\}$$

3. The attempt at a solution
I think:
A complete space is one in which all Cauchy sequences converges to a sequence (of points) in the space

a) it seems that if I construct whatever sequence I construct will always have zeros, but my limit will be a sequence of only 1's, so it will not be in the space.
That is, $$||x||_{\infty}=sup\{|\lambda_n|:n=1,2,\ldots\}=1$$
If this is correct, how do I show that?

b) It seems this space is complete by the same reasoning above, but again, how do I show that?

2. Feb 26, 2010

### Dick

You are thinking about the space E wrong. Your sequence is a single point in E. To discuss convergence you need a sequence of points in E, i.e. a sequence of sequences.

3. Feb 26, 2010

### BSCowboy

I was thinking of:
$$x_1=(1,0,\ldots)$$
$$x_2=(1,1,0,\ldots)$$
$$\vdots$$
$$x_n=(1,1,\ldots,1,0,\ldots)$$

4. Feb 26, 2010

### Dick

Is that Cauchy?

5. Feb 26, 2010

### BSCowboy

I don't think so, since it doesn't have the property:
$$|x_n-x_{n+1}|\rightarrow0 \text{ as }n\rightarrow\infty$$

Last edited: Feb 26, 2010
6. Feb 26, 2010

### Dick

Right. It's not. Look at the definition of Cauchy in terms of epsilon and think about what happens if you put epsilon=1/2.

7. Feb 26, 2010

### BSCowboy

That would mean:
$$||x_n-x_m||_{\infty}<\epsilon=1/2$$

Last edited: Feb 26, 2010
8. Feb 26, 2010

### BSCowboy

I'm still confused, and we haven't even started talking about B yet.
known knowns:
$$(m,||\cdot||_{\infty})$$ is complete since it contains all bounded sequences.

Last edited: Feb 26, 2010
9. Feb 26, 2010

### Dick

If x_n and x_m are in E and |x_n-x_m|<1/2, doesn't that mean x_n=x_m?

10. Feb 26, 2010

### BSCowboy

Yes it does. So, are you saying that my sequence I thought of is not in E or that it's in E and E it's not Cauchy so therefore E is not complete?

11. Feb 26, 2010

### Dick

Your sequence is in E. But it's not Cauchy and it doesn't converge. So it doesn't tell you anything about whether E is complete or not. Having thought about the epsilon=1/2 thing, if {a_n} is a Cauchy sequence in E, what can you say about the sequence?

12. Feb 26, 2010

### BSCowboy

You can say that $$||a_n-a_m||_{\infty}<\epsilon$$ is true

13. Feb 26, 2010

### Dick

Apply what we decided in posts 9 and 10 to that.

14. Feb 26, 2010

### BSCowboy

I don't understand where you are heading with this. Are you saying that we cannot construct a Cauchy sequence in E?

15. Feb 26, 2010

### Dick

No. I'm saying if you put the definition of a Cauchy sequence together with the fact that "if |x_n-x_m|<1/2 then x_n=x_m", then it's easy to show all Cauchy sequences converge. Look up the definition of Cauchy sequence again and think about that.

16. Feb 26, 2010

### BSCowboy

In this case:

A sequence $$x_n$$ in $$\left(X,||\cdot||_{\infty}\right)$$ is Cauchy iff

$$\forall \epsilon >0\quad \exists N\quad \ni \quad ||x_n-x_m||_{\infty}<\epsilon \quad \forall m,n\geq N$$

You are saying if we let $$\epsilon = 1/2$$ then:
$$||x_n-x_m||_{\infty}<1/2 \quad \Rightarrow \quad x_n=x_m$$

So, what insight is this providing in E (the set of sequences containing only entries 0 & 1)?

17. Feb 26, 2010

### BSCowboy

In this case:

A sequence $$x_n$$ in $$\left(X,||\cdot||_{\infty}\right)$$ is Cauchy iff

$$\forall \epsilon >0\quad \exists N\quad \ni \quad ||x_n-x_m||_{\infty}<\epsilon \quad \forall m,n\geq N$$

You are saying if we let $$\epsilon = 1/2$$ then:
$$||x_n-x_m||_{\infty}<1/2 \quad \Rightarrow \quad x_n=x_m$$

So, what insight is this providing in E (the set of sequences containing only entries 0 & 1)?

18. Feb 26, 2010

### Dick

Ok, so that says that if {x_n} is Cauchy, then there exists an N such that for all n,m>N, x_n and x_m are the SAME SEQUENCE. Doesn't it?? The sequence {x_n} has to eventually be constant. Call that constant sequence 'c'. What's the limit of the sequence {x_n}??

Last edited: Feb 26, 2010
19. Feb 26, 2010

### BSCowboy

It would be the sequence containing only "c"?

20. Feb 26, 2010

### Dick

Hmmm. I'm not sure what that means. I don't think I'm getting through. All of the elements in the sequence {x_n} are equal to x_(N+1) for n>N, right?