Economics 511
Professor John H. Nachbar
Fall 2002
Midterm
Answers
5. (a) Consider the sequence
{
1
/
2
,
1
/
4
,
1
/
2
,
3
/
4
,
1
/
8
,
1
/
4
,
3
/
8
,
1
/
2
,...
}
. The set
of subsequential limits is [0
,
1].
A more sophisticated approach is to note the following theorem, which is
a relative of Theorem 6 in the Sequences and Series notes.
Theorem.
Given a sequence
{
x
t
}
, let
F
be the image of
{
x
t
}
and let
F
0
be the set of limit points of
F
. If
x
∈
F
0
then
x
is a subsequential limit
of
{
x
t
}
.
Proof.
Since
x
is a limit point of
F
there is an
x
t
1
∈
N
1
(
x
). Similarly,
since
x
is a limit point of
F
there are inﬁnitely many points of
F
in
N
1
/
2
(
x
). Therefore, choose
x
t
2
∈
N
1
/
2
(
x
) with
t
2
> t
1
. And so on. By
construction, this is a subsequence converging to
x
.
±
It follows that if
F
is any countable subset of [0
,
1] with the property
that
F
0
= [0
,
1] then
any
sequence formed from all of the members of
F
has [0
,
1] as the set of its subsequential limits. Perhaps surprisingly,
order does not matter
. Nor is it necessary for any of the elements of
F
to
appear more than once. Thus, for example, one could take
any
sequence
for which
F
=
Q
∩
[0
,
1].
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 Fall '08
 JohnNachbar
 Economics, Topology, Universal quantification, Closed set, subsequence, Reductio ad absurdum, Nε

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