# Cauchy sequence proof , help

1. Dec 6, 2008

### lion0001

1. The problem statement, all variables and given/known data

Every convergent sequence is a cauchy sequence

2. Relevant equations

Proof: Suppose {a_n} converges to A, choose e > 0 , then e/2 > 0

there is a positive integer N such that n>=N implies abs[a_n - A ] < e/2

the difference between a_n , a_m was less than twice the original choice of e

Now if m , n >= N , then abs[a_n - A ] < e/2 and abs[a_m - A] < e/2 , hence

abs[a_n - a_m ] = abs[ a_n - A + A - a_m ] <= abs[a_n - A ] + abs[ A - a_m ]
= abs[a_n - A ] + abs[ a_m - A ] < e/2 + e/2 = e
thus {a_n} is cauchy

3. The attempt at a solution

the part where i have trouble understanding this proof is , where does the e/2 comes from?
in other words how does e/2 appears in the proof, how can i represent it in the real number graph?
if the answer is this " the difference between a_n , a_m was less than twice the original choice of e " i would like to see the arithmetic that produces e/2
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 6, 2008

### mutton

The sentence "the difference between a_n , a_m was less than twice the original choice of e " is false; get rid of it.

In the first sentence, e is fixed.

The convergence of $$\{a_n\}$$ to A means that for any $$\varepsilon > 0$$, there exists an N such that $$n \ge N$$ implies $$|a_n - A| < \varepsilon$$.

So we can pick any $$\varepsilon$$. In the proof, $$\varepsilon = e/2$$ is picked. $$\varepsilon = e/4$$ could have been picked and it would still work. $$\varepsilon = 123$$ could have been picked and it wouldn't work.

3. Dec 6, 2008

### lion0001

the sentence " the difference between a_n , a_m was less than twice the original choice of e " is false; get rid of it. " is not false , look , a_n and a_m are in the set ( A - e , A + e ) , and this is an interval of length 2e , and in the proof the choice of e/2 was not a coincidence it was because ( A +e ) - ( A -e ) = 2e , i dont know if this is right, but it comes straight from my book

4. Dec 6, 2008

### mutton

In this proof, a_n and a_m are in (A - e/2, A + e/2) because |a_n - A| < e/2 and |a_m - A| < e/2.

The goal was to show that |a_n - a_m| < e, not 2e. See the second last line of the proof.

5. Dec 6, 2008

6. Dec 6, 2008

### lion0001

how can e/4 work ?

could you please give me your version of this cauchy sequence ( explaining the e/2 's or 2e's . My book is confusing me more.

This is what it is in my book, i dont know how it went from saying that the difference is 2e and then using e/2 . The part that confuses me , is why the book outlines (1) to then use it in the proof, it looks that (1) doesnt even matter because since e > 0 then e/2 will do just fine.

(1) Suppose { a_n } converges to A. Choose e > 0, THere is a positive integer N such that, if
n, m >= N , then A - e < a_n < A + e and A - e < a_m < A + e
Thus for all n, m >= N we find a_n ∈ ( A - e , A + e ) and
a_m ∈ ( A - e , A +e ) . the set ( A - e, A +e ) is an interval of length 2e ,
hence the difference between a_n, and a_m is less then 2e

we will now state a theorem , the proof of which we have just outlined

THEOREM :Every convergent sequence is a cauchy sequence

Proof: Suppose {a_n} converges to A, choose e > 0 , then e/2 > 0

there is a positive integer N such that n>=N implies abs[a_n - A ] < e/2

by (1) the difference between a_n , a_m was less than twice the original choice of e

Now if m , n >= N , then abs[a_n - A ] < e/2 and abs[a_m - A] < e/2 , hence

abs[a_n - a_m ] = abs[ a_n - A + A - a_m ] <= abs[a_n - A ] + abs[ A - a_m ]
= abs[a_n - A ] + abs[ a_m - A ] < e/2 + e/2 = e
thus {a_n} is cauchy

Last edited: Dec 6, 2008
7. Dec 6, 2008

### mutton

Just replace e/2 in the proof with e/4. Then near the end, e/4 + e/4 = e/2 < e.

8. Dec 7, 2008

### lion0001

Wait, in the original cauchy proof, e/2 + e/ 2 is e , but at the end it becomes
e < e ?? isnt this false

| a_n - a_m | < e , we have to satisfy this
with e /2 we reach e < e ??

9. Dec 7, 2008

### mutton

Nowhere is e < e implied. |a_n - a_m| is less than or equal to the sum of two expressions. Each expression is less than e/2, so their sum is less than e.

|a_n - a_m| $$\le$$ |a_n - A| + |a_m - A| < e/2 + e/2 = e