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bugatti79
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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
bugatti79 said: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 said:Start with a Cauchy sequence in R^2?
Oster said:Where did your x come from?
Oster said:Where did your x come from?
bugatti79 said: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
Fredrik said: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?
Fredrik said: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.
Fredrik said: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?
Fredrik said: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 said: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.
LCKurtz said: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 said:Assume an arbitrary sequence ##x_n \to x ## in ##\mathbb{R^2}##.
LCKurtz said: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.
Fredrik said: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.
Fredrik said: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 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.
There are three mistakes here.bugatti79 said: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)
Fredrik said: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.
The stuff in this quote is fine.bugatti79 said: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, || ||).
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?bugatti79 said:Since we know x_n \to x \in R^2, || ||_\infty
Fredrik said: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.
That last thing should be (R, | |), since the norm on ℝ is just the absolute value.bugatti79 said: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, || ||).
No. Since they are Cauchy, for all ε>0, there exists a positive integer N such that...bugatti79 said:Since x_n(1) and x_n(2) are Cauchy then there exist an epsilon>0 s.t.
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.bugatti79 said:||?||_\infty = sup |?| < ε for all ? >= ?
(The above line, I am not sure what to put in, confused with the lettering etc.
You can't say that ##x_n\to x(1)## until you have defined what x(1) is.bugatti79 said: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 said: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.
Fredrik said: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.
Fredrik said:Yes, both of those are correct. The first one is the one we want. Note that it follows from the second one. (We can define N=max{N_1,N_2}).
Instead of "there exists a positive integer N_(1,2)", it would have been better to say "there exist positive integers N_1 and N_2".
Not really important, but perhaps still worth mentioning: If you need more than two symbols for integers, and you have already used n and m, I think you should use i,j or k. This is more of a tradition than anything else.
Fredrik said:Oh, wait, when I said those two were correct, I didn't see that you had written things like |x_m-x_n(1)|. This clearly doesn't make sense. x_m is in ℝ2, and x_n(1) is in ℝ, so the difference x_m-x_n(1) is nonsense. I thought you had written |x_m(1)-x_n(1)|.
There are several other (enormous) problems here. I may have caused some of your confusion, because I sort of forgot what we were doing for a while, and only looked at whether your statements made sense individually. You were supposed to do what I called "step 1" in post #25, i.e. you were supposed to prove that prove that x_n(1) and x_n(2) are Cauchy with respect to the standard metric on R. Since I had temporarily forgotten that, I didn't even realize that you opened with "Since x_n(1) and x_n(2) are Cauchy...".
I don't even know what to say. How many times have you been told that you can't assume what you're trying to prove? Is this concept hard to understand? Consider this "proof" of the claim that the moon is made of cheese:Since the moon is made of cheese, [insert anything you want here]. Therefore, the moon is made of cheese.Don't you see how absurd this is? If you assume the thing you're trying to prove, you can prove anything. You can prove that every statement is true, even the ones that are known to be false. You need to continue to think about this until you are absolutely sure that you understand it perfectly.
Fredrik said: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.