# Cauchy sequence

1. Dec 6, 2008

### lion0001

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

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
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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,
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

2. Dec 6, 2008

### matticus

Now if m , n >= N , then abs[a_n - A ] < e' and abs[a_m - A] < e'
Since {a_n} is convergent, you know that the above statement is true for all e', therefore it is certainly true when e' = e/2. It is just a convenient choice for e' since at the end when you add the two e/2's together you get e, which is what you needed.

3. Dec 6, 2008

### lion0001

thank you very much, i thought this was derived in a more complicated way

ANY OTHER ANSWERS ARE WELCOME

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

### poutsos.A

The law of logic that allow MATTICUS to do what he did is called Universal Elimination.

Since the sequence converges ,then for all Ε>o there exists an N......e.t.c,e.t.c.

But ε/2,ε/3, ε/4 ,ε/5 ..........................ε/n are all E"s >0, and according to the above law ,we can have for ε/2>0, an N such ......e.t.c,e.t.c.

If all people in a village are poor then Tom ,who stays in the village is poor

5. Dec 6, 2008

### poutsos.A

Note the ε>0 DENOTES a positive quantity ,hence any quantity which is positive could apply.

For example ε^4 +4 ,could apply since ε^4 + 4 is bigger than 0.

ε^2 + 2ε + 3 ,is applicable as well

6. Dec 6, 2008

### lion0001

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

7. Dec 6, 2008

### poutsos.A

What is the name of your book?

You must realize that many analysis books ,apart from the terrible mistakes they have,

are written in "a bent minded" manner

8. Dec 7, 2008

### lion0001

the name of the book is , Introduction to Analysis , by Edward D. Gaughan 5th edition
ISBN 0534351778

i have a question , can we just forget about part (1) where the author gets 2e , and just go straight to the proof, because the proof seems that doesnt need the part where the author gets 2e

9. Dec 7, 2008

### JinM

Well, I'm taking Analysis now, and that's exactly how you show convergence implies Cauchy--that is, without appealing to (1).

10. Dec 7, 2008

### matticus

The definition for limit of a sequence {Sn} says for all e there exists a N such that n>N implies |Sn-n| < e. It doesn't say |Sn-n| < 2e. To use 2e you would have to prove that if there is an M such that m > M implies |Sm - m| < 2e then there must also be an N such that n>N implies |Sn-n| < e.

He mentions it because as he says, it is an outline of his proof. Note that in the outline, when he picks the number e, it guarantees |an - am| < 2e. He's trying to explain how he came up with the seemingly random substitution. If using e resulted in 2e, then using e/2 would result in e, which implies the sequence is Cauchy.

So yes he could have left it out, but he probably thought it was beneficial for the reader to see how he miraculously got the e/2.

Last edited: Dec 7, 2008
11. Dec 9, 2008

### poutsos.A

In analysis ,there is a great difference between suppose or given and choose

Choose stands for there exists . So you can say choose an N or k belonging to natural Nos

or a δ>0 in the limits of a function ,and you can say given or suppose ε>0.

When you say given ε>0 ,you start a hypothesis,which in conjunction with the definition of a limit and the law of logic called M.Ponens will enable to choose an N or k.

The following is a more detailed proof of the above theorem:

Suppose ε>0,then ε/2>0.

Now since $$a_{n}$$ converges to A,for all θ>0 there exists an N belonging to natu ral Nos such that

.......................l$$a_{k}$$-Al <θ....for all k, k>=N.

Now using then Universal Elimination law put θ=ε/2,and we have:

If ε/2>0 then there exists an N belonging to natural Nos such that:

.............for all k, k>=N then l$$a_{k}$$-Al<e/2..............

But since ε/2>0 by the law of M.Ponens indeed we have that there exists an N belonging to natural Nos such that:

.....................for all k, k>=N then l$$a_{k}$$-Al<e/2......................

Now suppose ,n>=N AND m>=N.

Since now for all,k k>=N then l$$a_{n}$$-Al<e/2........

if again by using the law Universal Elimination we put k=n firstly and then k=m we have:

if n>=N THEN l$$a_{n}$$-Al<e/2 and

if m>=N THEN l$$a_{m}$$-Al<e/2

And using now the supposition ,n>=N AND m>=N,the above and M.Ponens we conclude:

l$$a_{n}$$-Al<e/2 and l$$a_{m}$$-Al<e/2

............and.............

l$$a_{n}$$-$$a_{m}$$l <ε by using the above and the triangular inequality

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