Completeness of R^2 with sup norm

[bugatti79]

Homework Statement

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

Homework Equations

The Attempt at a Solution

How may I tackle this?

Thanks

 

[Oster]

Start with a Cauchy sequence in R^2?

 

[bugatti79]

Homework Statement

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

Homework Equations

The Attempt at a Solution

How may I tackle this?

Thanks

Start with a Cauchy sequence in R^2?

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]

Where did your x come from?

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]

Where did your x come from?

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

ie $x_n \to x$ if its cauchy.

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

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]

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?

Yes, that's my approach.

Now, that you mention it, I guess I don't 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]

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.

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]

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?

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]

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.

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.

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]

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.

Assume an arbitrary sequence $x_n \to x$ in $\mathbb{R^2}$.

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.

 

[Fredrik]

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.

 

[bugatti79]

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, 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.

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...?

 

[Fredrik]

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.

I have added some colored comments to your proof below.

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[/color] Need to show x_n is [strike]Cauchy[/strike] convergent[/color] 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 | |[/color]. 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)|[/color] = [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.[/color], 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[/color]

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).[/color]
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).[/color] 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.[/color]

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)|[/color] < 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.[/color]

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[/color]

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.[/color]
​

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.

 

[bugatti79]

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.

I have added some colored comments to your proof below.

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[/color] Need to show x_n is [strike]Cauchy[/strike] convergent[/color] 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 | |[/color]. 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)|[/color] = [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.[/color], 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[/color]

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).[/color]
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).[/color] 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.[/color]

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)|[/color] < 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.[/color]

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[/color]

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.[/color]
​

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.

ok, another attempt..

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, || ||). To see this note

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

So x_n is Cauchy sequence in (R^2, || ||_\infty) which we know is complete (told this)
By definition of completeness, there exists an x_n(1) s.t x_n \to x.

Therefore it is sufficient 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_i|x_n(i)-x_m(i)|< ε/2 for all n,m >= n_0

Letting m \to \infty, we have |x_n(i)-x| < ε/2 for n >= n_0

ie ||x_n-x||_\infty < ε/2 < ε for n>=n_0

implies x_n \to x in (R^2, || ||_\infty)..?

 

[Fredrik]

Much better, but still not right.

ok, another attempt..

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, || ||). To see this note

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

So x_n is Cauchy sequence in (R^2, || ||_\infty)

There are three mistakes here.

1. The <= should actually be =, since that's just the definition of the sup norm. (I guess I should have mentioned that last time).
2. You didn't actually prove the claim you set out to prove, since you didn't write down the inequalities that prove that x_n(1) and x_n(2) are Cauchy.
3. After that incomplete attempt to prove the claim, you say "so", followed by the assumption you started with. This makes it look like you think you have just proved your starting assumption. When you start a sentence with "so", the next thing you say should be a consequence of the statements made before the "so". It shouldn't be a repeat of the starting assumption.

I didn't read the rest, since I had already found three mistakes.

 

[bugatti79]

Much better, but still not right.

There are three mistakes here.

1. The <= should actually be =, since that's just the definition of the sup norm. (I guess I should have mentioned that last time).
2. You didn't actually prove the claim you set out to prove, since you didn't write down the inequalities that prove that x_n(1) and x_n(2) are Cauchy.
3. After that incomplete attempt to prove the claim, you say "so", followed by the assumption you started with. This makes it look like you think you have just proved your starting assumption. When you start a sentence with "so", the next thing you say should be a consequence of the statements made before the "so". It shouldn't be a repeat of the starting assumption.

I didn't read the rest, since I had already found three mistakes.

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 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

So x_n is Cauchy sequence in (R^2, || ||_\infty) which we know is complete (told this)

By definition of completeness, there exists an x_n(1) s.t x_n \to x.

||x_n-x_m||_\infty = sup_i(|x_n(i)-x_m(i)|) = sup(|x_n(1)-x_m(1)|,|x_n(2)-x_m(2)|)

Therefore it is sufficient 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_i|x_n(i)-x_m(i)|< ε/2 for all n,m >= n_0

Letting m \to \infty, we have |x_n(i)-x| < ε/2 for n >= n_0

ie ||x_n-x||_\infty < ε/2 < ε for n>=n_0

implies x_n \to x in (R^2, || ||_\infty)..?

 

[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, || ||).

The stuff in this quote is fine.

Since we know x_n \to x \in R^2, || ||_\infty [/color]

You can't make a statement that involves a variable "x" without first specifying its value, unless the statement is of the type "for all x..." or "there exists an x such that...". So there are three things I could guess that you meant to say here, but all of them suggest that we know that x_n is convergent with respect to the sup norm, and you don't know that. In fact, that's what we're trying to prove. Are you assuming what you're trying to prove again?

I had to stop there, since the mistake is so severe. I only had a quick look at the rest, and it seems like you have now abandoned the correct plan for this proof. Why? You were doing the right things last time. You were just doing them wrong.

I think it would be better if you don't color parts of your proof red. It just makes it harder to read.

 

[Fredrik]

OK, we're clearly not making much progress here. Since your problems with this are so extreme, I will describe the correct plan for this 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)), and prove that your guess is correct.
​

A few tips:

1. Don't ever state the assumption in a way that makes it look like you think it's something you have derived.

2. Don't ever assume the thing you're trying to prove.

3. Don't ever make a statement about a variable that you haven't defined, unless it's a "for all" or "there exists" statement.

4. Don't ever apply a function to something that's not in its domain.

5. Don't ever say that you're going to prove something and then try to prove something else.

6. Don't ever lose track of what your variables represent.​

 

[bugatti79]

ok, I will get back to this asap. I need to cover other material in the mean time. Thanks

 

[bugatti79]

OK, we're clearly not making much progress here. Since your problems with this are so extreme, I will describe the correct plan for this 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 1 and 2 only

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 x_n(1) and x_n(2) are Cauchy then there exist an epsilon>0 s.t.
||?||_\infty = sup |?| < ε for all ? >= ?

(The above line, I am not sure what to put in, confused with the lettering etc.
Assuming I can get the above correct I think I can continue and safely say...)

Since it is given that R is complete, ie the real line is complete, then we can say x_n(1) and x_n(2) converge to x(1) and x(2) respectively...?​

 

[Fredrik]

Step 1 and 2 only

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, || ||).

That last thing should be (R, | |), since the norm on ℝ is just the absolute value.

Since x_n(1) and x_n(2) are Cauchy then there exist an epsilon>0 s.t.

No. Since they are Cauchy, for all ε>0, there exists a positive integer N such that...

||?||_\infty = sup |?| < ε for all ? >= ?

(The above line, I am not sure what to put in, confused with the lettering etc.

The sup norm (which is a norm on ℝ²) has nothing to do with this part, since we're talking about Cauchy sequences in ℝ. What you need to do here is to just use the definition of "Cauchy sequence" to figure out how to end the "for all" statement I started above. You just need to understand what the statements "x_n(1) is a Cauchy sequence" and "x_n(2) is a Cauchy sequence" mean.

Since it is given that R is complete, ie the real line is complete, then we can say x_n(1) and x_n(2) converge to x(1) and x(2) respectively...?

You can't say that $x_n\to x(1)$ until you have defined what x(1) is.

 

[bugatti79]

That last thing should be (R, | |), since the norm on ℝ is just the absolute value.


No. Since they are Cauchy, for all ε>0, there exists a positive integer N such that...


The sup norm (which is a norm on ℝ²) has nothing to do with this part, since we're talking about Cauchy sequences in ℝ. What you need to do here is to just use the definition of "Cauchy sequence" to figure out how to end the "for all" statement I started above. You just need to understand what the statements "x_n(1) is a Cauchy sequence" and "x_n(2) is a Cauchy sequence" mean.

Since they are Cauchy, for all ε>0, there exists a positive integer N such that

|x_m-x_n|< epsilon for m,n >= N..?

 

[Fredrik]

The expression |x_n-x_m| doesn't really make sense, does it? For each n, x_n is in ℝ², but the absolute value is a norm on ℝ. I sugggest you take another look at my list of tips in post #25, in particular tip number 4.

Also, the statement you're trying to use is about x_n(1) and x_n(2). You can't immediately jump to a conclusion about x_n.