# Homework Help: Completeness of R^2 with sup norm

1. Mar 10, 2012

### bugatti79

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

Given that R is complete, prove that R^2 with the sup norm is complete

2. Relevant equations

3. The attempt at a solution

How may I tackle this?

Thanks

2. Mar 10, 2012

### Oster

3. Mar 10, 2012

### bugatti79

Let $x_n=(x_n(1), x_n(2))$ be an arbitrary Cauchy sequence $\in R^2, || ||_\infty$

by definition we have $||x_n-x||_\infty=sup(|x_n(1)-x(1)|,|x_n(2)-x(2)|)$ for $n \in \mathbb{N}$

ie $||x_n-x||_\infty \to 0$ as $n \to \infty$.

We need to show that

$x_n(1) \to x(1)$.

Since $x_n$ is Cauchy,

$\exists n_0 \in N$ s.t. $||x_n-x||_\infty < \epsilon \forall n \ge n_0$

we have that

$|x_n(1)-x(1)| < \epsilon/2 \forall n_1 \ge n_0$ and

$|x_n(2)-x(2)| < \epsilon/2 \forall n_2 \ge n_0$

$\implies |x_n-x| < \epsilon/2 + \epsilon/2 < \epsilon$...........? Not sure if this right or how to complete it?

Thanks

4. Mar 10, 2012

### Oster

Where did your x come from?
Cauchy means for all r>0, there exists a natural number p such that for all m,n>p, d(x_m,x_n) < r.
In this case, max{|xn(1)-xm(1)|,|xn(2)-xm(2)|}<r for all n,m>p

Last edited: Mar 10, 2012
5. Mar 12, 2012

### bugatti79

I am defining x as an arbitrary sequence in R^2 to start off the proof, am I not?

6. Mar 12, 2012

### Oster

I thought x_n was your arbitrary Cauchy sequence. You have an x_n and an x in your post.

7. Mar 12, 2012

### bugatti79

I should have inserted what is highlighted in red above. Does it makes sense now?

8. Mar 12, 2012

### Fredrik

Staff Emeritus
What is x(1) and x(2)? It looks like you're thinking that $x(1)=\lim_n x_n(1)$ and similarly for (2). But what makes you think that these limits exist?

Last edited: Mar 12, 2012
9. Mar 13, 2012

### bugatti79

Yes, thats my approach.

Now, that you mention it, I guess I dont know whether they exist but its seems to be a pattern I've seen when writing these proofs. Ie, assume x_n converges to x and continue. That's all I've done here!...

10. Mar 13, 2012

### Fredrik

Staff Emeritus
So you think you can prove that xn→x by assuming that xn→x? If this had made sense, we could prove any statement by just assuming that it's true.

11. Mar 13, 2012

### bugatti79

but if the sequence is Cauchy then this assumption is valid, right?

12. Mar 13, 2012

### Fredrik

Staff Emeritus
Which sequence? If you mean $\langle x_n\rangle$, then no, it's not valid, because you're assuming what you're trying to prove. If you mean $\langle (x_n)_1\rangle$ and $\langle (x_n)_2\rangle$, then my question is, have you proved that these sequences are Cauchy?

13. Mar 13, 2012

### bugatti79

Let $x_n$ be an arbitrary Cauchy sequence in R

Proof. given $\epsilon >0$
$\exists n_0 \in \mathbb{N}$ s.t. $\forall n,m > n_0$ then
$||x_n-x_m|| < \epsilon$

Suppose x_n converges to L and let ε>0 be given. Then $\exists n_0 in N$ s.t. $|x_n-L| < \epsilon/2 \forall n \ge n_0$

Let any integer $m \ge n$ be given. since $m \ge n \ge n_0$ then $|x_m-L| < \epsilon/2$

By the triangle inequality $|x_n-x_m|=|(x_n-L)+(L-x_m)| \le |x_m -L|+|x_n-L| < \epsilon/2+\epsilon/2 < \epsilon \implies x_n$ is Cauchy

If I have correctly proven this is Cauchy, what is the next step?

14. Mar 13, 2012

### Fredrik

Staff Emeritus
If you delete everything before the word "suppose" (in particular the part where you assume what you want to prove), then what we have left is a correct proof of the claim that every convergent sequence in ℝ is Cauchy. But this doesn't seem to be relevant to the problem you're trying to solve.

You need to start with an arbitrary sequence in $\mathbb R^2$ that's Cauchy with respect to the ∞-norm, and then use the fact that it's Cauchy with respect to the ∞-norm to learn something else about it, something you can use to prove that it's convergent with respect to the ∞-norm.

15. Mar 13, 2012

### LCKurtz

Bugatti79, you need to develop a strategy to work a problem like this. You are trying to show $R^2$ is complete, meaning every Cauchy sequence in $R^2$ converges to a point in $R^2$. What you have to work with is that $R$ is complete. So you start with a Cauchy sequence $\{x_n = (x_n(1),x_n(2))\}$ like you did. You are trying to find an $x=(x(1),x(2))\in R^2$ such that $x_n\to x$. If you could show $x_n(1)$ and $x_n(2)$ converged to something, maybe that something would do for $x$. How might you show they converge? If they do, how could show that their limits could be used for $x$? You have been working on relevant material recently.

16. Mar 14, 2012

### bugatti79

Assume an arbitrary sequence $x_n \to x$ in $\mathbb{R^2}$. We need to show $x_n(1) \to x(1)$ and $x_n(2) \to x(2)$ in $\mathbb{R}$
ie, we need to find $n_0 \in N$ s.t $|x_n(1)-x(1)| < \epsilon \forall n > n_0$

Since we know $x_n \to x \in R^2, || ||_\infty$ we know $\exists n_0 \in N$ s.t

$||x_n-x||_\infty < \epsilon \forall n \ge n_0$ ie

$||(x_n(1)-x(1), x_n(2)-x(2))||_\infty < \epsilon$

We have that $|x_n(1)-x(1)| \le max (|x_n(1)-x(1)|, |x_n(2)-x(2)|) < \epsilon$
This shows $x_n(1) \to x(1)$ as $n \to \infty$

Similarly

We have that $|x_n(2)-x(2)| \le max (|x_n(1)-x(1)|, |x_n(2)-x(2)|) < \epsilon$
This shows $x_n(2) \to x(2)$ as $n \to \infty$

Now, conversely assume $x_n(1) \to x(1)$ and $x_n(2) \to x(2)$
Need to show that $x_n \to x in R^2, || ||_\infty$. Need to find
$n_0 \in N$ s.t. $||x_n-x||_\infty \forall n \ge n_0$

but we know that $x_n(1) \to x(1), \exists n \in N$ s.t.

$|x_n(1)-x(1)| < \epsilon/2 \forall n \ge n_1$ and
$|x_n(2)-x(2)| < \epsilon/2 \forall n \ge n_2$

Take $n_0=max(n_1,n_2)$.

Then $\forall n > n_0$ we have $||x_n-x||_\infty=max(|x_n(1)-x(1)|,|x_n(2)-x(2)|) < \epsilon/2 + \epsilon/2 < \epsilon$

Therefore
$x_n \to x \implies$ R^2 with the sup norm is complete...?
It looks like it is similar to the other question I answered in my other post..?

17. Mar 14, 2012

### LCKurtz

No, no, no. You are trying to show every Cauchy sequence in $R^2$ converges to a point in $R^2$. You don't start with an arbitrary (non Cauchy) sequence and assume it already converges to some $x$. You are trying to prove there is such an $x$. You need to think more about the strategy I outlined and re-quoted above.

18. Mar 14, 2012

### Fredrik

Staff Emeritus
Bugatti79, you need to rethink your entire approach to proofs. Theorems are always implications, i.e. statements of the form $A\Rightarrow B$, but you always ignore that. You always try to avoid making the assumption A, you always try to avoid using the definitions of the terms in A, and most of the time, you even assume B! These are the three biggest mistakes that can possibly be made in a proof. You also often make irrelevant assumptions that have nothing to do with the theorem.

You want to prove that if a sequence is Cauchy with respect to the ∞-norm, it's convergent with respect to the ∞-norm. So A is the statement "$\langle x_n\rangle$ is Cauchy with respect to the ∞-norm", and B is the statement "$\langle x_n\rangle$ is convergent with respect to the ∞-norm". And you start by assuming B, as usual. This is the single biggest mistake that can be made in a proof.

One thing you need to understand is that once a proof of $A\Rightarrow B$ has arrived at the statement B, there's nothing more to say. That's the end of the proof. So if you start by assuming B, nothing more needs to be said. In fact, it wouldn't make any sense to say anything more after that.

19. Mar 15, 2012

### bugatti79

Let x_n=(x_n(1), x_n(2)) be an arbitrary Cauchy sequence in R^2. Need to show x_n is Cauchy in (R^2 || ||_\infy)
Claim: x_n(1) and x_n(2) are Cauchy sequences in R wrt || ||_\infty. To see this note

|x_n(1)-x_m(1)| <= sup|x_n(1)-x_m(1) = ||x_n(1)-x_m(1)||_\infty, similarly
|x_n(2)-x_m(2)| <= sup|x_n(2)-x_m(2) = ||x_n(2)-x_m(2)||_\infty

So x_n is Cauchy sequence in R which we know is complete (told this)
By definition of completeness, x_n \to x. Need to show x_n \to x in R^2, || ||_\infty

let ε>0 be given since x_n is Cauchy, there exists n_0 in N s.t. ||x_n-x_m||_\infty =sup|x_n-x_m|< ε for all n,m >= n_0 and x in R

let a=lim x for n \to infinity

by triangle inequality sup|x_n-x_m <= |x_n-a+a-x_m|<=|x_n-a|+|x_m-a|<=ε/2+ε/2<ε for all n,m >= n_0

implies x_n converges to x and therefore R^2 is complete wrt to sup norm...?

20. Mar 15, 2012

### Fredrik

Staff Emeritus
Now we're getting somewhere, but you keep making some pretty strange mistakes. You actually managed to get something wrong in every single statement you made. Somehow, this is still a decent attempt, because it looks like you have the right idea. You're just getting all of the details wrong. You have to be much more careful with the details. You're making it look like you're trying to get everything wrong. If you don't start making a much greater effort to get the details right, I think you will soon find it hard to get people to help you.

Let x_n=(x_n(1), x_n(2)) be an arbitrary Cauchy sequence in R^2. You forgot to specify the norm with respect to which the sequence is Cauchy Need to show x_n is [strike]Cauchy[/strike] convergent in (R^2 || ||_\infy)
Claim: x_n(1) and x_n(2) are Cauchy sequences in R wrt [strike]|| ||_\infty[/strike] That's not even a norm on ℝ. You should have said | |. To see this note

|x_n(1)-x_m(1)| <= [strike]sup|x_n(1)-x_m(1)[/strike] sup_i |x_n(i)-x_m(i)| = [strike]||x_n(1)-x_m(1)||_\infty[/strike] This is wrong and doesn't make sense, since x_n(1) and x_m(1) are in ℝ and the ∞ norm is a norm on ℝ2., similarly
[strike]|x_n(2)-x_m(2)| <= sup|x_n(2)-x_m(2) = ||x_n(2)-x_m(2)||_\infty[/strike] Same comments as for the preceding line

So x_n is Cauchy sequence in R which we know is complete (told this) x_n is a sequence in ℝ2, not ℝ. But you probably meant x_n(1) and x_n(2).
By definition of completeness, x_n \to x. No, by definition of completeness, there exists an x(1) such that x_n(1)→x(1). Need to show x_n \to x in R^2, || ||_\infty That's not obvious at this point. What's obvious is that it's sufficient to show this, i.e. that if you succeed at showing this, you will be done.

let ε>0 be given since x_n is Cauchy, there exists n_0 in N s.t. ||x_n-x_m||_\infty [strike]=sup|x_n-x_m|[/strike] sup_i|x_n(i)-x_m(i)| < ε for all n,m >= n_0 and [strike]x in R[/strike] It's unclear if you meant "for all x in ℝ", or if you just meant to say that x is in ℝ. The former doesn't make sense, and the latter is wrong.

let a=lim x for n \to infinity You still haven't proved that x is convergent. But you probably meant something completely different from what you wrote

by triangle inequality sup|x_n-x_m <= |x_n-a+a-x_m|<=|x_n-a|+|x_m-a|<=ε/2+ε/2<ε for all n,m >= n_0

implies x_n converges to x and therefore R^2 is complete wrt to sup norm...? What are you doing here? Even the first thing you wrote down (sup|x_n-x_m) doesn't make sense.

Edit: I should perhaps have made it more clear that I think that you seem to have found the correct strategy for this proof. It seems that you understand the steps involved in the correct proof. It's just that you're making so many unnecessary mistakes when you try to write it down.

Last edited: Mar 15, 2012