MATH 117 LECTURE NOTES FEBRUARY 19, 2009
DR. JULIE ROWLETT
1.
Subsequences and Cauchy Sequences
Definition 1.1.
A
subsequence
of a sequence
{
s
n
}
n
∈
N
of real numbers is an ordered
subset of
{
s
n
}
n
∈
N
indexed by
n
1
< n
2
< n
3
< . . . < n
k
< n
k
+1
. . .
∈
N
,
such that
n
k
→ ∞
.
1.1.
Examples of sequences and subsequences.
(1) Let’s take our favorite sequence, the harmonic one with
s
n
=
1
n
.
Then, a subsequence is
s
k
=
1
2
k
.
However,
u
1
=
1
5
, u
2
=
1
4
, u
3
=
1
6
, u
3+
k
=
s
k
for
k
≥
1
,
is
not
a subsequence since the order of terms is not preserved.
(2) The main idea in the preceding example is that, although a subsequence
may
omit
(or skip) terms, it may not rearrange them.
(3)
Exercise:
Construct a sequence which is not monotone but which has a
monotone subsequence.
1.2.
Subsequence Theorems.
Theorem 1.2.
If a sequence
{
s
n
}
∞
n
=1
converges to a real number
s,
then every
subsequence also converges to
s.
Proof:
Let
{
s
n
k
}
be a subsequence of the original sequence
{
s
n
}
.
Let
>
0
.
Then, since
s
n
→
s
(this is another way to write “
s
n
converges to
s
”), there exists
N
∈
N
such that if
n
≥
N,

s
n

s

<
.
By the definition of subsequence,
n
k
→ ∞
.
Hence, for all
n
k
≥
N,

s
n
k

s

<
.
This is the definition of convergence for the subsequence, so
s
n
k
→
s
also.
♥
Theorem 1.3.
If every subsequence of a sequence of real numbers
{
s
n
}
n
∈
N
con
verges to a real number
s,
then
s
n
→
s.
Proof:
The proof is by contradiction. Assume that
s
n
s.
Then, there exists
>
0
,
such that for any
N
∈
N
,
there exists
n > N
such that

s
n

s
 ≥
.
So, there
exists
n >
1 such that
(1.4)

s
n

s
 ≥
.
1