n
∈
N
. Negating this,
{
x
n
}
n
is not bounded from above if and only if
∀
M
∈
R
,
∃
n
M
∈
N
,
such that
x
n
M
> M.
(1)
In the above the subscript
M
for
n
M
, just indicates the dependence on
M
of the rank.
Now let us prove the equivalence asked in the statement.
(
⇐
) This implication is easy. Assume that a subsequence
{
x
n
k
}
k
→ ∞
as
k
→ ∞
. By deﬁnition,
∀
M
∈
R
,
∃
K
∈
N
, such that if
k
≥
K
, then
x
n
k
> M
. Thus relation (1) is trivially satisﬁed, taking
n
M
to be, for instance
n
K
.
(
⇒
) This is a bit harder. We assume that relation (1) is true and we’ll construct inductively a subsequence
{
x
n
k
}
k
which will approach
∞
as
k
→ ∞
.
To pick the ﬁrst term of our subsequence, apply ﬁrst (1) with
M
=
M
1
= 1. It follows that there exists
n
1
∈
N
such that
x
n
1
>
1. Now the idea is to apply (1) with a bigger and bigger
M
. But just choosing
M
= 2
,M
= 3, etc. may not work, because we have no guarantee that the ranks
n
2
,
n
3
, etc. which we
get from (1) will be in increasing order.
Here is how we pick the second term of the subsequence. Let
M
2
= max
{
2
,x
1
,x
2
,...,x
n
1
}
and apply (1)
with
M
=
M
2
. It follows that there exists
n
2
∈
N
such that
x
n
2
> M
2
. From the choice of
M
2
we deduce
two things. First,
x
n
2
> M
2
≥
2. Secondly,
x
n
2
> M
2
≥
x
l
, for any
l
∈ {
1
,
2
,...,n
1
}
; in particular, we get
that
n
2
> n
1
because otherwise
x
n
2
would be equal to one of the
x
l
,
l
∈ {
1
,
2
,...,n
1
}
.
Suppose now that we picked
n
1
< n
2
< ... < n
k
, such that
x
n
l
> l
, for any
l
∈ {
1
,
2
,...,k
}
and we’ll
construct the
k
+ 1th term of the subsequence. Let
M
k
+1
= max
{
2
,x
1
,x
2
,...,x
n
k
}
and apply (1) with
M
=
M
k
+1
. It follows that there exists
n
k
+1
∈
N
such that
x
n
k
+1
> M
k
+1
. From the choice of
M
k
+1
it
follows as above that
n
k
+1
> n
k
and that
x
n
k
+1
> k
+ 1.
Thus by induction we construct the subsequence
{
x
n
k
}
k
of
{
x
n
}
n
, with the property that
x
n
k
> k
, for
any
k
∈
N
. The (extended) comparison Theorem implies immediately that
x
n
k
→ ∞
as
k
→ ∞
.
2
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 Fall '12
 Draghic
 Mathematical analysis, Order theory, Xn, 1 2k, Dominated convergence theorem, a2 b2

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