
#19
Mar1512, 03:28 PM

P: 652

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) <= supx_n(1)x_m(1) = x_n(1)x_m(1)_\infty, similarly x_n(2)x_m(2) <= supx_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_nx_m_\infty =supx_nx_m< ε for all n,m >= n_0 and x in R let a=lim x for n \to infinity by triangle inequality supx_nx_m <= x_na+ax_m<=x_na+x_ma<=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
Mar1512, 04:45 PM

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PF Gold
P: 8,999

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 isEdit: 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. 



#21
Mar1512, 05:52 PM

P: 652

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_nx_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_nx_m_\infty =sup_ix_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_nx_\infty < ε/2 < ε for n>=n_0 implies x_n \to x in (R^2,  _\infty)..? 



#22
Mar1512, 07:08 PM

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P: 8,999

Much better, but still not right.
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. 



#23
Mar1612, 02:40 PM

P: 652

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_nx_\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_nx_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_nx_m_\infty =sup_ix_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_nx_\infty < ε/2 < ε for n>=n_0 implies x_n \to x in (R^2,  _\infty)..? 



#24
Mar1612, 05:12 PM

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PF Gold
P: 8,999

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. 



#25
Mar1612, 05:50 PM

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PF Gold
P: 8,999

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. 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. 



#26
Mar1912, 02:39 PM

P: 652

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




#27
Mar2812, 01:37 PM

P: 652

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...? 



#28
Mar2812, 03:09 PM

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P: 8,999





#29
Mar2912, 12:00 PM

P: 652

Since they are Cauchy, for all ε>0, there exists a positive integer N such that x_mx_n< epsilon for m,n >= N..? 



#30
Mar2912, 02:58 PM

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PF Gold
P: 8,999

The expression x_nx_m doesn't really make sense, does it? For each n, x_n is in ℝ^{2}, 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. 



#31
Mar2912, 03:04 PM

P: 652

x_ax_n(1)< epsilon for a,n >= N and similarly x_bx_n(2)< epsilon for b,n >= N or Since they are Cauchy, for all ε>0, there exists a positive integer N_(1,2) such that x_mx_n(1)< epsilon for m,n >= N_1 and similarly x_mx_n(2)< epsilon for m,n >= N_2 ..? 



#32
Mar2912, 03:39 PM

Emeritus
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PF Gold
P: 8,999

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. 



#33
Mar2912, 04:08 PM

P: 652

x_mx_n(1)< epsilon for m,n >= N_1 and similarly x_mx_n(2)< epsilon for m,n >= N_2. We can define N=(N_1, N_2) so x_mx_n(1) and x_mx_n(2) both tend to 0 when m,n tend to infinity. Hence x_n(1) and x_n(2) are Cauchy, but since we know R is complete, these Cauchy's must converge, therefore let x_(1) and x_(2) be the limits of x_n(1) and x_n(2) respectively...? 



#34
Mar2912, 04:56 PM

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PF Gold
P: 8,999

Oh, wait, when I said those two were correct, I didn't see that you had written things like x_mx_n(1). This clearly doesn't make sense. x_m is in ℝ^{2}, and x_n(1) is in ℝ, so the difference x_mx_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. 



#35
Mar2912, 05:23 PM

P: 652

Claim: x_n(1) and x_n(2) are Cauchy sequences in (R,  ). Since they are Cauchy, for all ε>0, there exists positive integers N_1 and N_2 such that x_m(1)x_n(1)< epsilon for m,n >= N_1 and similarly x_m(2)x_n(2)< epsilon for m,n >= N_2. We can define N=(N_1, N_2) so x_m(1)x_n(1) and x_m(2)x_n(2) both tend to 0 when m,n tend to infinity... 


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