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Unformatted text preview: maller than 4 whenever ≥ Nn,m .
(n)
˜
◦ By hypothesis, the sequence {X }n∈IN is Cauchy. So there is a natural number N such that
ε
˜
the third term is smaller than 4 whenever n, m ≥ N .
1
◦ Finally, if ≥ m , the last term is smaller than m .
˜
Choose any natural number N ≥ max N , 4 . I claim that xn − xm V < ε whenever n, m ≥ N .
ε
To see this, let n, m ≥ N . Now choose to be any natural number bigger than max Nn,m , n , m .
ε
Then the four terms in (2) are each smaller than 4 .
(n)
Proof that X = lim X : Let ε > 0. By deﬁnition
n→∞ X − X(n) H = lim m→∞ xm − x(n)
m V ( m) x = lim m→∞ m − x(n)
m V By the triangle inequality
( m) x m − x(n)
m ( m) V ≤x m (n) −x n (n) V +x n − x(n)
m V (3) (n) ◦ Since the sequence xn = x n n∈IN is Cauchy, there is a natural number N such that the
ε
ﬁrst term is smaller than 2 whenever n, m ≥ N .
1
◦ By the construction of n , the second term is smaller than n whenever m ≥ n .
Choose any natural number N ≥ max N , 2 . I claim that X − X(n) H < ε whenever n ≥ N .
ε
To see this, let n ≥ N . Now choose m to be any natural number bigger than max N , n . Then
ε
the two terms in (3) are each smaller than 2 . September 7, 2011...
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 Spring '13
 JoelFeldman

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