Completeness of R^2 with sup norm

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

Oster

Start with a Cauchy sequence in R^2?

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

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

bugatti79

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

Oster

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

bugatti79

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

Fredrik

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?

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!...

Fredrik

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

bugatti79

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

Fredrik

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?

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?

Fredrik

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.

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.

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..?

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.