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Economics 511
Professor: John Nachbar
Fall 2008
Midterm
Answers
5.
Theorem.
Let
(
X,d
x
)
and
(
Y,d
y
)
be metric spaces and suppose
f
:
X
→
Y
and
g
:
X
→
y
are both continuous. Let
E
⊆
X
be dense in
X
.
(a) If
{
y
t
}
is a sequence in
Y
and
y
t
converges both to
y
*
and to
ˆ
y
then
y
*
= ˆ
y
.
(b) If
f
(
x
) =
g
(
x
)
for
x
∈
E
then
f
=
g
.
(c)
f
(
E
)
is dense in
f
(
X
)
.
Proof.
(a) Suppose
y
t
converges to both
y
*
and ˆ
y
. Take any
ε >
0. Then there is
a
T
*
such that for all
t > T
*
,
y
t
∈
N
ε/
2
(
y
*
) and a
ˆ
T
such that for all
t >
ˆ
T y
t
∈
N
ε/
2
(ˆ
y
). By the triangle inequality, for any
t >
max
{
T
*
,
ˆ
T
}
,
d
y
(
y
*
,
ˆ
y
)
≤
d
y
(
y
*
,y
t
)+
d
y
(
y
t
,
ˆ
y
)
< ε
. Since
ε
was arbitrary,
d
y
(
y
*
,
ˆ
y
) = 0,
hence
y
*
= ˆ
y
.
Here’s an alternative proof. By contraposition. Suppose
y
*
6
= ˆ
y
, hence
d
(
y
*
,
ˆ
y
)
>
0. Suppose also that that
y
t
→
y
*
. Deﬁne 2
ε
=
d
(
y
*
,
ˆ
y
). Since
y
t
→
y
*
, there is a
T
such that for all
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This note was uploaded on 05/20/2010 for the course ECON 511 taught by Professor Johnnachbar during the Fall '08 term at Washington University in St. Louis.
 Fall '08
 JohnNachbar
 Economics

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