Math 117, F 09, HW#4,5
(sample solutions)
P.57, #35.
For abbreviation’s sake, set
S
=
{
a
n

n
∈
N
}
. We deﬁne a subsequence by
recursion. First, since
x
is an accumulation point of
S
, the neighborhood (
x

1
, x
+ 1)
contains inﬁnitely many points of
S
. Hence, there is a positive integer
n
1
such that

a
n
1

x

<
1.
We claim that there are positive integers
n
1
< n
2
< n
3
<
···
such that

a
n
k

x

<
1
k
for all
k
. We have already chosen
n
1
.
Now suppose that
n
1
<
···
< n
k
have been chosen. Since
x
is an accumulation point of
S
, the neighborhood (
x

1
k
+1
, x
+
1
k
+1
) contains inﬁnitely many points of
S
. Hence, there
are inﬁnitely many positive integers
n
such that

a
n

x

<
1
k
+1
. On the other hand, there
are only ﬁnitely many positive integers less than or equal to
n
k
. Therefore, among the
inﬁnitely many positive integers
n
such that

a
n

x

<
1
k
+1
, there must be some (inﬁnitely
many, actually) which are greater than
n
k
. Choose one, and label it
n
k
+1
. Thus
n
k
+1
> n
k
and

a
n
k
+1

x

<
1
k
+1
. This establishes the recursion.
At this point, we have chosen a subsequence
{
a
n
k
}
∞
k
=1
of the original sequence
{
a
n
}
∞
n
=1
.
It remains to prove that
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 Fall '08
 Akhmedov,A
 Math, positive integers

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