Complete Metric Subspaces: Are These Metric Subspaces Complete?

[BSCowboy]

Homework Statement

Determine whether the following metric subspaces are complete:
a) the set E of sequences containing only entries 0 & 1 in [tex](m,||\cdot||_{\infty})[/tex]
b) the unit sphere in any Banach Space

Homework Equations

a) for [tex]x=\{\lambda_1,\lambda_2,\ldots,\lambda_n,\ldots \}[/tex]
[tex]||x||_{\infty}=sup\{|\lambda_n|:n=1,2,\ldots\}[/tex]

b)[tex]\{x\in X:||x-x_0||=1\}[/tex]

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, [tex]||x||_{\infty}=sup\{|\lambda_n|:n=1,2,\ldots\}=1[/tex]
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?

 

[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.

 

[BSCowboy]

I was thinking of:
[tex]x_1=(1,0,\ldots) [/tex]
[tex]x_2=(1,1,0,\ldots) [/tex]
[tex]\vdots [/tex]
[tex]x_n=(1,1,\ldots,1,0,\ldots)[/tex]

 

[Dick]

I was thinking of:
[tex]x_1=(1,0,\ldots) [/tex]
[tex]x_2=(1,1,0,\ldots) [/tex]
[tex]\vdots [/tex]
[tex]x_n=(1,1,\ldots,1,0,\ldots)[/tex]

Is that Cauchy?

 

[BSCowboy]

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

 

[Dick]

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

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.

 

[BSCowboy]

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

 

[BSCowboy]

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

 

[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?

 

[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?

 

[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?

 

[BSCowboy]

You can say that [tex]||a_n-a_m||_{\infty}<\epsilon[/tex] is true

 

[Dick]

You can say that [tex]||a_n-a_m||_{\infty}<\epsilon[/tex] is true

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

 

[BSCowboy]

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

 

[Dick]

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

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.

 

[BSCowboy]

In this case:

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

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

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

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

 

[BSCowboy]

In this case:

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

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

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

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

 

[Dick]

In this case:

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

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

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

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

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}??

 

[BSCowboy]

It would be the sequence containing only "c"?

 

[Dick]

It would be the sequence containing only "c"?

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?

 

[BSCowboy]

That is my understanding

 

[Dick]

That is my understanding

Ok, so doesn't the sequence have a limit?? And isn't that limit x_(N+1)?? What does that tell you about completeness?

 

[BSCowboy]

Are you saying that every sequence in E is it's own limit and that since any sequence I construct in E will be it's own limit then E is complete?

 

[Dick]

Are you saying that every sequence in E is it's own limit and that since any sequence I construct in E will be it's own limit then E is complete?

That sounds pretty garbled again. Remember a single sequence x_n is a POINT in E. The sort of sequences you need to talk about for convergence are sequences of sequences {x_n}. Got that? {x_n} is a whole bunch of sequences.

 

[BSCowboy]

You're right, I am having a hard time understanding sequences in E. Can you give me an example.

See in [tex]E_o[/tex] (the set of all sequences with a finite number of non-zero terms) an example would be:
[tex]x_1=(1,0,\ldots)[/tex]
[tex]x_2=(1,\frac{1}{2},\ldots)[/tex]
[tex]x_n=(1,\frac{1}{2},\ldots, \frac{1}{n},\ldots)[/tex]

This sequence converges to
[tex]x=(1,\frac{1}{2},\ldots)[/tex]
and I understand this sequence is not a part of [tex]E_o[/tex]
Therefore, [tex]E_o[/tex] is not complete.

I am not understanding the argument in this space E

 

[Dick]

E is a very different space from E_0. E is an example of a discrete metric space, if that helps. If x_n and x_m are ANY two different points in E then ||x_n-x_m||=1. Please, tell me you understand that. Please? That makes "Cauchy" sequences in the two spaces very different. E IS COMPLETE. Try to approach the problem with that attitude instead of assuming it's not.

An example of a convergent sequence in E has to look like, basically,

x_1=(1,0,1,0,0,0...)
x_2=(1,0,1,0,0,0...)
x_3=(1,0,1,0,0,0...)
x_4=(1,0,1,0,0,0...)
x_5=(1,0,1,0,0,0...)

I've omitted the terms where x_m is not equal to x_n. They are ALL the SAME.

 

[BSCowboy]

Yes, I think I've got it. As usual..it's seems pretty simple, now.

Also, I appreciate you stubbornly sticking with me and my stubborn self.

 

[BSCowboy]

I've been thinking about this some more;
If [tex]B(x,1/2)\cap(E\setminus \{x\})[/tex]
[tex]\Rightarrow ||y-x||_{\infty}<1/2\quad \text{and} \quad y\not=x[/tex]

This would be a contradiction since
[tex]||x-y||_{\infty}=1 \text{ if } x\not= y[/tex]

 

[Dick]

I've been thinking about this some more;
If [tex]B(x,1/2)\cap(E\setminus \{x\})[/tex]
[tex]\Rightarrow ||y-x||_{\infty}<1/2\quad \text{and} \quad y\not=x[/tex]

This would be a contradiction since
[tex]||x-y||_{\infty}=1 \text{ if } x\not= y[/tex]

Well, yes. That's the point, right? A sequence can only converge to x by eventually taking the constant value of x. There's no way for it to get close to x without being x.