Cauchy Sequences Triangle Inequality.

In summary, the problem at hand is to show that cn=|an-bn| is cauchy, given that both an and bn are cauchy. This can be proven using a triangle inequality argument by showing that for any given epsilon, there exists a natural number N for which m,n>=N implies |an-bn|<epsilon. This can be achieved by first setting |a_n-a_m|<epsilon/2 and |b_n-b_m|<epsilon/2, and then using the triangle inequality to show that |(a_n-b_n)-(a_m-b_m)|<epsilon. It is important to note that two separate values of N may be needed for each sequence, so the maximum of these two
  • #1
Enjoicube
49
1

Homework Statement


assuming an and bn are cauchy, use a triangle inequality argument to show that cn=
| an-bn| is cauchy

Homework Equations



an is cauchy iff for all e>0, there is some natural N, m,n>=N-->|an-am|<e

The Attempt at a Solution


I am currently trying to work backwards on this one, since we know
for some m,n>=N where N is a natural
2e>|-an+am|+|bn-bm|
Thus, through the triangle inequality this is >=|am-an+bn-bm|=|(bn-an)-(bm-am)|>=|bn-an|-|bm-am|.

Am I going completely wrong somewhere, because I am getting very stuck here. To anyone who answers, please only hints, not full answers (I am sure this is customary). Again, many thanks.
 
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  • #2
Two quick comments:

I would have said that [itex]|a_n-a_m|< \epsilon / 2[/itex] and [itex]|b_n-b_m|< \epsilon /2[/itex] just to get [itex]|(a_n-b_n)-(a_m-b_m)|<\epsilon[/itex].

Here's a nitpicky thing that your professor might correct (mine would have). Once [itex]\epsilon >0[/itex] is given, there's no reason to think that the same [itex]N[/itex] will work for both sequences. So I would have said that there exists [itex]N_1[/itex] for one sequence and [itex]N_2[/itex] for the other, such that blah blah blah... Then I would have let [itex]N=max\{N_1,N_2\}[/itex].
 
  • #3
Yeah, my instructor does require this also. I can't believe the anguish this is causing me, It must be a very simple problem, and no one asked about it in class. Right now, I am supposed to be reading a history book, but this problem is back there bugging me.
 
  • #4
Not to imply that anyone read this incorrectly, but the problem is cn=abs(an-bn) not just an-bn. Just entered my head that this might not have been emphasized.
 

Related to Cauchy Sequences Triangle Inequality.

1. What is a Cauchy Sequence?

A Cauchy sequence is a sequence of real numbers that converges to a limit. This means that as the sequence progresses, the terms get closer and closer to the limit value, with the distance between consecutive terms getting smaller and smaller.

2. What is the Triangle Inequality?

The Triangle Inequality is a mathematical inequality that states that the sum of any two sides of a triangle must be greater than the third side. In other words, the shortest distance between two points is a straight line.

3. How is the Triangle Inequality related to Cauchy Sequences?

The Triangle Inequality is an essential property of real numbers that is used to prove the convergence of Cauchy sequences. It ensures that the sequence is approaching a finite limit and not diverging to infinity.

4. What are some applications of Cauchy Sequences and the Triangle Inequality?

Cauchy Sequences and the Triangle Inequality are used in various fields of mathematics, such as analysis, topology, and functional analysis. They are also applied in real-world situations, such as in physics, engineering, and economics, to model and solve problems involving limits and distances.

5. Are there any exceptions to the Triangle Inequality for Cauchy Sequences?

Yes, there are exceptions to the Triangle Inequality for Cauchy Sequences in certain mathematical spaces, such as non-Archimedean fields. In these spaces, the Triangle Inequality may not hold, and different definitions of convergence and distance may be used.

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