Completeness of R^2 with sup norm

[Fredrik]

Let x_n=(x_n(1), x_n(2)) be an arbitrary Cauchy sequence in (R^2, || ||_\infty). Need to show x_n is convergent in (R^2 || ||_\infy)

Claim: x_n(1) and x_n(2) are Cauchy sequences in (R, | |).

Since they are Cauchy,

As I said (many, many times), the last line in the quote above is the single biggest mistake you can possibly make in a proof. Please read my previous post again.

 

[bugatti79]

As I said (many, many times), the last line in the quote above is the single biggest mistake you can possibly make in a proof. Please read my previous post again.

Is it correct to 'claim' that they are Cauchy and then prove this? If so, I don't know how to prove they are Cauchy in R apart from what I have shown.

 

[Fredrik]

It's certainly OK to say that you're going to prove something, and then do it. I like statements like that in proofs, because they make the proofs easier to read.

I always interpreted your "Claim:" as "I'm going to prove that x_n(1) and x_n(2) are Cauchy".

Surely you must have some idea about how to do it. You seem understand what the statements "x_n(1) is Cauchy" and "x_n(2) is Cauchy" mean. Do you really see no way to use the assumption about x_n to prove that those statements are true? If you don't see it immediately, I suggest that you just write down the meaning of the statements

(a) x_n is Cauchy with respect to $\|\ \|_\infty$.
(b) x_n(1) is Cauchy with respect to | |.
(c) x_n(2) is Cauchy with respect to | |.​

on a piece of paper, and than stare at what you wrote until you see how similar (a) is to the other two. When you understand what you're looking at, it will be easy to prove that (a) implies (b), and that (a) implies (c).

 

[bugatti79]

It's certainly OK to say that you're going to prove something, and then do it. I like statements like that in proofs, because they make the proofs easier to read.

I always interpreted your "Claim:" as "I'm going to prove that x_n(1) and x_n(2) are Cauchy".

Surely you must have some idea about how to do it. You seem understand what the statements "x_n(1) is Cauchy" and "x_n(2) is Cauchy" mean. Do you really see no way to use the assumption about x_n to prove that those statements are true? If you don't see it immediately, I suggest that you just write down the meaning of the statements

(a) x_n is Cauchy with respect to $\|\ \|_\infty$.
(b) x_n(1) is Cauchy with respect to | |.
(c) x_n(2) is Cauchy with respect to | |.​

on a piece of paper, and than stare at what you wrote until you see how similar (a) is to the other two. When you understand what you're looking at, it will be easy to prove that (a) implies (b), and that (a) implies (c).

No its not obvious to me so I will write down the meaning of each later. In the meantime however, what about thread 13 starting from 'suppose...' to prove that x_n(1) and x_n(2) are Cauchy?

 

[Fredrik]

No its not obvious to me so I will write down the meaning of each later.

I'll write down the meaning now. I'll ignore (c) for now (because it's dealt with in essentially the same way as (b)).

Let ε>0 be arbitrary.

(a) There's a positive integer N such that for all n,m≥N, max{|x_n(1)-x_n(1)|,|x_n(2)-x_n(2)|}<ε.
(b) There's a positive integer N such that for all n,m≥N, |x_n(1)-x_m(1)|<ε.

You just need to prove that if (a) is true, then so is (b).

what about thread 13 starting from 'suppose...' to prove that x_n(1) and x_n(2) are Cauchy?

That's a good way to prove that if x_n(1) and x_n(2) is convergent, they are also Cauchy. This is of no use to us here, since we won't obtain the result that they're convergent until after we have already proved that they're Cauchy.

 

[bugatti79]

I'll write down the meaning now. I'll ignore (c) for now (because it's dealt with in essentially the same way as (b)).

Let ε>0 be arbitrary.

(a) There's a positive integer N such that for all n,m≥N, max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε.
(b) There's a positive integer N such that for all n,m≥N, |x_n(1)-x_m(1)|<ε.

You just need to prove that if (a) is true, then so is (b).

1) I find it counter intuitive that we use 'assumption about x_n' to prove that x_n(1) and x_n(2) are Cauchy in R when we actually need to prove that x_n is convergent in (R^2,|| ||_infty)
I thought one would be using the facts we obtained about x_n(1) and x_n(2) in R to prove x_n is convergent in R^2.

2) I don't know how to prove (a) is true despite you illustrating its meaning unfortunately...?

 

[bugatti79]

1) I find it counter intuitive that we use 'assumption about x_n' to prove that x_n(1) and x_n(2) are Cauchy in R when we actually need to prove that x_n is convergent in (R^2,|| ||_infty)
I thought one would be using the facts we obtained about x_n(1) and x_n(2) in R to prove x_n is convergent in R^2.

2) I don't know how to prove (a) is true despite you illustrating its meaning unfortunately...?

I'll write down the meaning now. I'll ignore (c) for now (because it's dealt with in essentially the same way as (b)).

Let ε>0 be arbitrary.

(a) There's a positive integer N such that for all n,m≥N, max{|x_n(1)-x_n(1)|,|x_n(2)-x_n(2)|}<ε.
(b) There's a positive integer N such that for all n,m≥N, |x_n(1)-x_m(1)|<ε.

You just need to prove that if (a) is true, then so is (b).

if we use use the assumption that x_n is Cauchy wrt sup norm then we can say

There's a positive integer N such that for all n,m≥N, max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε then we have that

|x_n(1)-x_m(1)|<=max{|x_n(1)-x_n(1)|,|x_n(2)-x_n(2)|}<ε and
|x_n(2)-x_m(2)|<=max{|x_n(1)-x_n(1)|,|x_n(2)-x_n(2)|}<ε...?

 

[Fredrik]

1) I find it counter intuitive that we use 'assumption about x_n' to prove that x_n(1) and x_n(2) are Cauchy in R when we actually need to prove that x_n is convergent in (R^2,|| ||_infty)
I thought one would be using the facts we obtained about x_n(1) and x_n(2) in R to prove x_n is convergent in R^2.

We will do that too. That's step 3. But to prove that x_n is convergent, we need to first guess what member of ℝ² it converges to. That member will have two components, that we are choosing to call x(1) and x(2). So I find it very intuitive that we will use the two "component sequences" x_n(1) and x_n(2) to define x(1) and x(2).

Is it also intuitive that, we should use the properties of x_n and the sup norm to do that? Maybe not, but the problem tells you to do that. We can't possibly expect to prove an implication $A\Rightarrow B$ without using the definition of A.

Recall that this is the plan for the proof:

Assumption: x_n is Cauchy with respect to the sup norm.

Step 1: Prove that x_n(1) and x_n(2) are Cauchy with respect to the standard metric on R.

Step 2: Conclude that x_n(1) and x_n(2) are convergent with respect to the standard metric on R, and define x(1) and x(2) as the limits of those sequences.

Step 3: Guess that x_n converges to (x(1),x(2)) with respect to the sup norm, and prove that your guess is correct.

I added the colored part now, to improve clarity.

This comment breaks down step 1 further:

Let ε>0 be arbitrary.

(a) There's a positive integer N such that for all n,m≥N, max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε.
(b) There's a positive integer N such that for all n,m≥N, |x_n(1)-x_m(1)|<ε.

You just need to prove that if (a) is true, then so is (b).

This time the colored stuff is a correction of a typo. x_n(1)-x_n(1) would of course be =0.

2) I don't know how to prove (a) is true despite you illustrating its meaning unfortunately...?

Why do you think you need to prove (a)?

if we use use the assumption that x_n is Cauchy wrt sup norm then we can say

There's a positive integer N such that for all n,m≥N, max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε

Exactly. (a) is true by assumption.

|x_n(1)-x_m(1)|<=max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε and
|x_n(2)-x_m(2)|<=max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε...?

Yes, these are the inequalities you need to use. (Note the colored corrections. When you copied and pasted from my post, you included the typo). You just need to be more precise about what we're actually doing here. This is how I would say it:

Let ε>0 be arbitrary. Let N be a positive integer such that for all n,m ≥ N, we have max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε. (The assumption tells us that such an N exists). This choice of N ensures that for all n,m ≥ N, we have |x_n(1)-x_m(1)| ≤ max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε, and |x_n(2)-x_m(2)| ≤ max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε.

That wasn't so hard, was it? That's the entire step 1. Please make sure that you understand what I did here before you move on to step 2.

 

[bugatti79]

I am putting the info together.

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

Assumption: x_n is Cauchy with respect to the sup norm.

Step 1: Prove that x_n(1) and x_n(2) are Cauchy with respect to the standard metric on R

Let ε>0 be arbitrary. Let N be a positive integer such that for all n,m ≥ N, we have max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε. (The assumption tells us that such an N exists).

This choice of N ensures that for all n,m ≥ N, we have

|x_n(1)-x_m(1)| ≤ max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε, and

|x_n(2)-x_m(2)| ≤ max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε

Step 2: Conclude that x_n(1) and x_n(2) are convergent with respect to the standard metric on R, and define x(1) and x(2) as the limits of those sequences.

I am a little confused here. In step 1, we were to prove x_n(1) and x_n(2) are Cauchy but here in step 2 we have to conclude that these 2 are convergent. I am unable to see the link between 'Cauchy' and 'convergent'. Looking at my notes I just come across the following definition
'A metric space or nls in which all Cauchy sequences converge is called a complete metric or nls'
But I can't make any further conclusion.

When we do prove these 2 are convergent I know we can write

x(1)=lim x_(1) for n to \infty
x(2)=lim x_(2) for n to \infty

 

[Fredrik]

I am a little confused here. In step 1, we were to prove x_n(1) and x_n(2) are Cauchy but here in step 2 we have to conclude that these 2 are convergent. I am unable to see the link between 'Cauchy' and 'convergent'. Looking at my notes I just come across the following definition
'A metric space or nls in which all Cauchy sequences converge is called a complete metric or nls'
But I can't make any further conclusion.

Here you just have to know that ℝ with the standard metric is complete. That's an entirely different theorem, so you wouldn't be expected to prove it as a part of this problem. So you can immediately conclude that x_n(1) and x_n(2) are convergent. Then you define x(1)=lim_n x_n(1) and x(2)=lim_n x_n(2), and move on to step 3.

 

[bugatti79]

Here you just have to know that ℝ with the standard metric is complete. That's an entirely different theorem, so you wouldn't be expected to prove it as a part of this problem. So you can immediately conclude that x_n(1) and x_n(2) are convergent. Then you define x(1)=lim_n x_n(1) and x(2)=lim_n x_n(2), and move on to step 3.

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

Assumption: x_n is Cauchy with respect to the sup norm.

Step 1: Prove that x_n(1) and x_n(2) are Cauchy with respect to the standard metric on R

Let ε>0 be arbitrary. Let N be a positive integer such that for all n,m ≥ N, we have max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε. (The assumption tells us that such an N exists).

This choice of N ensures that for all n,m ≥ N, we have

|x_n(1)-x_m(1)| ≤ max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε, and

|x_n(2)-x_m(2)| ≤ max{|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|}<ε

Step 2: Conclude that x_n(1) and x_n(2) are convergent with respect to the standard metric on R, and define x(1) and x(2) as the limits of those sequences.

Since we are given that R is complete with the standard metric we can say that x_n(1) and x_n(2) are convergent. Hence
x(1)=lim_n x_n(1) and x(2)=lim_n x_n(2)

Step 3: Guess that x_n converges to x, ie x_n=(x_n(1),x_n(2)) converges to x=(x(1),x(2)) and prove that your guess is correct.

let ε>0 be arbitrary. From step 2 we have that x_n(1) converges to x(1), then there exists n_1 in N s.t |x_n(1)-x(1)| < ε/2 for all n >=n_1

Similarly

we have that x_n(2) converges to x(2), then there exists n_2 in N s.t |x_n(2)-x(2)| < ε/2 for all n >=n_2

Therefore for all n>n_0=max{n_1,n_2} we have that
||x_n-x||_\infty = ||(x_n(1)-x(1)), (x_n(2)-x(2))||_\infty< ε/2+ε/2<ε

Therefore
x_n converges to x. Since x_n was Cauchy this implies R^2 with the sup norm is complete...?

 

[Fredrik]

Assumption: x_n is Cauchy with respect to the sup norm.

This statement is OK, but I would rather say "Let x_n be an arbitrary sequence in ℝ² that's Cauchy with respect to the sup norm". (I just think it sounds a bit weird to suggest that we're making an assumption, when we're just trying to make it clear that we're going to prove something that holds for all Cauchy sequences in the given metric space).

Step 2: Conclude that x_n(1) and x_n(2) are convergent with respect to the standard metric on R, and define x(1) and x(2) as the limits of those sequences.

Since we are given that R is complete with the standard metric we can say that x_n(1) and x_n(2) are convergent. Hence
x(1)=lim_n x_n(1) and x(2)=lim_n x_n(2)

Here you should say something to indicate how and where you're using the completeness of ℝ. For example, you can change the "since" statement to something like this: "Since ℝ with the standard metric is a complete metric space, the result of step 1 implies that x_n(1) and x_n(2) are convergent."

Step 3: Guess that x_n converges to x, ie x_n=(x_n(1),x_n(2)) converges to x=(x(1),x(2)) and prove that your guess is correct.

let ε>0 be arbitrary. From step 2 we have that x_n(1) converges to x(1), then there exists n_1 in N s.t |x_n(1)-x(1)| < ε/2 for all n >=n_1

Similarly

we have that x_n(2) converges to x(2), then there exists n_2 in N s.t |x_n(2)-x(2)| < ε/2 for all n >=n_2

Therefore for all n>n_0=max{n_1,n_2} we have that
||x_n-x||_\infty = ||(x_n(1)-x(1)), (x_n(2)-x(2))||_\infty< ε/2+ε/2<ε

Therefore
x_n converges to x. Since x_n was Cauchy this implies R^2 with the sup norm is complete...?

Most of this looks good, but there are some problems with the calculation ||x_n-x||_\infty = ||(x_n(1)-x(1)), (x_n(2)-x(2))||_\infty< ε/2+ε/2<ε. The first equality is good (it shows that you're using the definition of the - symbol correctly), but then you need to show that you're using the definition of the sup norm, and the properties of the n_0 you defined, to turn $\|\text{something}\|_\infty$ into something that involves ε. If you do this, the result will not be ε/2+ε/2.

 

[bugatti79]

Most of this looks good, but there are some problems with the calculation ||x_n-x||_\infty = ||(x_n(1)-x(1)), (x_n(2)-x(2))||_\infty< ε/2+ε/2<ε. The first equality is good (it shows that you're using the definition of the - symbol correctly), but then you need to show that you're using the definition of the sup norm, and the properties of the n_0 you defined, to turn $\|\text{something}\|_\infty$ into something that involves ε. If you do this, the result will not be ε/2+ε/2.

||x_n-x||_\infty = ||(x_n(1)-x(1)), (x_n(2)-x(2))||_\infty =max{|x_n(1)-x(1)|,|x_n(2)-x(2)|}<ε

If this is correct, I am not sure why we couldn't use epsilon over 2 ..?

 

[Fredrik]

That's correct. If you choose n_0 such that |x_n(1)-x(1)|<ε and |x_n(2)-x(2)|<ε for all n ≥ n_0, then the equalities and inequality that you just posted hold for all n ≥ n_0. If you instead choose n_0 such that |x_n(1)-x(1)|<ε/2 and |x_n(2)-x(2)|<ε/2 for all n ≥ n_0, then you get,

||x_n-x||_\infty = ||(x_n(1)-x(1)), (x_n(2)-x(2))||_\infty =max{|x_n(1)-x(1)|,|x_n(2)-x(2)|}<ε/2<ε.

for all n ≥ n_0. So it doesn't matter which of these two definitions of n_0 you use. What I was concerned about was that you wrote ε/2+ε/2. Such expressions show up when we use the triangle inequality, but we didn't do that here. So I was worried that you didn't even use the definition of the norm, and did something else (perhaps something involving a triangle inequality) to get ε/2+ε/2.

 

[bugatti79]

Such expressions show up when we use the triangle inequality, but we didn't do that here. So I was worried that you didn't even use the definition of the norm, and did something else (perhaps something involving a triangle inequality) to get ε/2+ε/2.

Yes, I was mixing this up with what I did in post 13 which involved the triangle inequality.

Just one final question,

From your post #43 you state

'Is it also intuitive that, we should use the properties of x_n and the sup norm to do that? Maybe not, but the problem tells you to do that.'

Where in the question in post 1 does it specifically tell you to do this?

Thanks for you perseverance, much appreciated.

 

[Fredrik]

From your post #43 you state

'Is it also intuitive that, we should use the properties of x_n and the sup norm to do that? Maybe not, but the problem tells you to do that.'

Those two sentences ended up kind of weird. It's not at all clear what I meant. I'll try to be more clear. The problem asked us to prove the following theorem:

The normed space $(\mathbb R^2,\|\ \|_\infty)$ is complete.​

(It also told us that you're allowed to use the theorem that says that $(\mathbb R,|\ |)$ is complete). Since a normed space is complete if and only if every Cauchy sequence in its underlying set is convergent, the theorem is equivalent to the following:

For all sequences $\langle x_n\rangle_{n=1}^\infty$ in $\mathbb R^2$, if $\langle x_n\rangle_{n=1}^\infty$ is Cauchy with respect to the sup norm, then $\langle x_n\rangle_{n=1}^\infty$ is convergent with respect to the sup norm.​

It should be obvious that it's impossible to prove this without using the definitions of "sup norm" and "Cauchy". OK, I suppose we could also do it by referring to some other theorem, which could be referring to yet another theorem, and so on, but the definitions of "sup norm" and "Cauchy" must both be used somewhere in the chain of proofs that ensure that all these theorems hold. If they're not used, we can't possibly claim that we have proved something about sequences that are Cauchy with respect to the sup norm.

Since our plan for the proof doesn't involve any references to other theorems about sequences that are Cauchy with respect to the sup norm, we definitely have to use the definitions of those terms in a step that derives something from the assumption that $\langle x_n\rangle_{n=1}^\infty$ is Cauchy with respect to the sup norm. Our step 1 is such a step. It proves that $\langle x_n\rangle_{n=1}^\infty$ (an arbitrary sequence in $\mathbb R^2$ that's Cauchy with respect to the sup norm) has the property that the sequences $\langle x_n(1)\rangle_{n=1}^\infty$ and $\langle x_n(2)\rangle_{n=1}^\infty$ are Cauchy with respect to the absolute value function on ℝ. So it wouldn't make any kind of sense to try to do step 1 without using the definitions of "sup norm" and "Cauchy".

That's how the theorem we're supposed to prove "tells you" to use the definitions of those terms in the first step. It tells you, simply by being a statement of the form "For all x, P(x) implies Q(x)", where P(x) is a statement about x that involves those terms. (In this case, P(x) is the statement that x is a sequence in ℝ² that's Cauchy with respect to the sup norm).

Since the problem explicitly tells us that we can use the theorem that says that $(\mathbb R,|\ |)$ is complete, I think it's very close to obvious that the person who wrote the problem intended us to solve it the way we did. The part of the problem statement I just mentioned is telling us to do steps 1 and 2, and step 3 is just to use what we found in steps 1-2 to complete the proof of the theorem.

 

[bugatti79]

Those two sentences ended up kind of weird. It's not at all clear what I meant. I'll try to be more clear. The problem asked us to prove the following theorem:

The normed space $(\mathbb R^2,\|\ \|_\infty)$ is complete.​

(It also told us that you're allowed to use the theorem that says that $(\mathbb R,|\ |)$ is complete). Since a normed space is complete if and only if every Cauchy sequence in its underlying set is convergent, the theorem is equivalent to the following:

For all sequences $\langle x_n\rangle_{n=1}^\infty$ in $\mathbb R^2$, if $\langle x_n\rangle_{n=1}^\infty$ is Cauchy with respect to the sup norm, then $\langle x_n\rangle_{n=1}^\infty$ is convergent with respect to the sup norm.​

It should be obvious that it's impossible to prove this without using the definitions of "sup norm" and "Cauchy". OK, I suppose we could also do it by referring to some other theorem, which could be referring to yet another theorem, and so on, but the definitions of "sup norm" and "Cauchy" must both be used somewhere in the chain of proofs that ensure that all these theorems hold. If they're not used, we can't possibly claim that we have proved something about sequences that are Cauchy with respect to the sup norm.

Since our plan for the proof doesn't involve any references to other theorems about sequences that are Cauchy with respect to the sup norm, we definitely have to use the definitions of those terms in a step that derives something from the assumption that $\langle x_n\rangle_{n=1}^\infty$ is Cauchy with respect to the sup norm. Our step 1 is such a step. It proves that $\langle x_n\rangle_{n=1}^\infty$ (an arbitrary sequence in $\mathbb R^2$ that's Cauchy with respect to the sup norm) has the property that the sequences $\langle x_n(1)\rangle_{n=1}^\infty$ and $\langle x_n(2)\rangle_{n=1}^\infty$ are Cauchy with respect to the absolute value function on ℝ. So it wouldn't make any kind of sense to try to do step 1 without using the definitions of "sup norm" and "Cauchy".

That's how the theorem we're supposed to prove "tells you" to use the definitions of those terms in the first step. It tells you, simply by being a statement of the form "For all x, P(x) implies Q(x)", where P(x) is a statement about x that involves those terms. (In this case, P(x) is the statement that x is a sequence in ℝ² that's Cauchy with respect to the sup norm).

Since the problem explicitly tells us that we can use the theorem that says that $(\mathbb R,|\ |)$ is complete, I think it's very close to obvious that the person who wrote the problem intended us to solve it the way we did. The part of the problem statement I just mentioned is telling us to do steps 1 and 2, and step 3 is just to use what we found in steps 1-2 to complete the proof of the theorem.

Fredrik, this is real good information. Thank you for taking the time out to write this.
It is really appreciated.