- #1
lion0001
- 21
- 0
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 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
(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