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18.100B,
Fall
2002,
Homework
5
Due
by
Noon,
Tuesday
October
8.
Rudin:
(1)
Chapter
3,
Problem
1
Solution:
The
sequence
is
supposed
to
be
in
R
n
.
We
use
the
triangle
inequality
in
the
form

b

=

b
−
a
+
a
 ≤ 
a
−
b

+

a

which
implies
that

b
 − 
a
 ≤ 
a
−
b

.
Reversing
the
roles
of
a
and
b
we
also
see
that

a
 − 
b
 ≤

a
−
b

and
so

a
 − 
b

≤

a
−
b

.
If
{
s
n
}
converges
to
s
then
given
>
0 there
exists
N
such
that
n > N
implies

s
n
−
s

<
.
By
the
triangle
inequality

s
n
−
s

≤

s
n
−
s

so
{
s
n
}
converges
to

s

.
(2)
Chapter
3,
Problem
20
Solution:
Let
{
p
n
}
be
a
Cauchy
sequence
in
a
metric
space
X.
By
as
sumption,
some
subsequence
{
p
n
(
k
)
}
converges
to
p
∈
X.
Thus,
given
>
0
there
exits
K
such
that
k >
K
implies
that
d
(
p, p
n
(
k
)
<
/
2
for
all
k >
K.
By
the
Cauchy
condition,
given
>
0
there
exists
M
such
that
n, m
>
M
implies
d
(
x
n
, x
m
)
<
/
2
.
Now,
consider
N
=
n
(
l
) for
some
l
≥
K
such
that
n
(
l
)
> M,
which
exists
since
n
(
k
)
→ ∞
with
k.
For
this
choice,
n > N
=
⇒
d
(
p
n
, p
)
≤
d
(
p
n
, p
n
(
l
)
) +
d
(
p
n
(
l
)
, p
)
<
shows
that
{
p
n
}
converges
to
p.
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 Fall '09
 Frade
 en, Metric space, pn, complete metric space, metric space X., k Gk

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