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Math 310 
hw
6 solutions
11.6, 11.7, 11.10;
Wednesday, 21 Oct 2009
11.6 Show that every subsequence of a subsequence of a given sequence is itself a subsequence of the given sequence
Hint:
Deﬁne subsequence as in (3) of Deﬁnition 11.1.
Let (
s
n
)
n
be a sequence which we regard as a function
s
:
N
→
R
. A subsequence (
s
n
k
)
k
may be regarded as
in (3) of Deﬁnition 11.1, as a composition
s
◦
σ
:
N
→
R
where
σ
:
N
→
N
, given for
k
∈
N
, by
σ
(
k
) =
n
k
, is
increasing, i.e.
σ
(
k
)
< σ
(
k
+ 1). A subsequence of this subsequence is then given by (
s
◦
σ
)
◦
τ
where
τ
:
N
→
N
is another increasing function.
We need to show that
σ
◦
τ
:
N
→
N
is also an increasing function. Let
m
∈
N
. We ﬁrst prove by induction
that for all
n
∈
N
,
τ
(
m
)
< τ
(
m
+
n
). Since
σ
is increasing,
σ
(
m
)
< σ
(
m
+ 1), so the base case holds. Next
n
∈
N
and suppose that
σ
(
m
)
< σ
(
m
+
n
). Since
σ
is increasing,
σ
(
m
+
n
)
< σ
(
m
+
n
+ 1). Thus by transitivity
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 Spring '10
 janeday

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