Cauchy Sequence Proof: How Does e/2 Appear in the Proof?

[lion0001]

Homework Statement

Every convergent sequence is a cauchy sequence

Homework 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

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

 

[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 &gt; 0, there exists an N such that n \ge N implies |a_n - A| &lt; \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.

 

[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.

In the first sentence, e is fixed.

The convergence of \{a_n\} to A means that for any \varepsilon &gt; 0, there exists an N such that n \ge N implies |a_n - A| &lt; \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.

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 don't know if this is right, but it comes straight from my book

 

[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.

 

[mutton]

The proof at https://www.physicsforums.com/showthread.php?t=277695 is different in this regard.

 

[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.

In the first sentence, e is fixed.

The convergence of \{a_n\} to A means that for any \varepsilon &gt; 0, there exists an N such that n \ge N implies |a_n - A| &lt; \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.

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 don't 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) doesn't 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 outlinedTHEOREM :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

 

[mutton]

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

 

[lion0001]

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

Wait, in the original cauchy proof, e/2 + e/ 2 is e , but at the end it becomes
e < e ?? isn't this false

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

 

[mutton]

Wait, in the original cauchy proof, e/2 + e/ 2 is e , but at the end it becomes
e < e ?? isn't this false

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

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| <[/size] e/2 + e/2 = e