ECON 831
Solution Midterm
1.
(a) Consider the following sequence of subsets
S
k
in the metric space
(
R
,d
1
)
:
S
k
= (1

1
k
+1
,
1 +
1
k
+1
)
. The following conditions are satis ed:
(i) For any
k
, consider the real number
1
. It is such that, for any
k <
∞
,
1
∈
S
k
because
1

1
k
+1
<
1
<
1 +
1
k
+1
. Hence,
S
k
6
=
∅
.
(ii) For any
k
:
S
k
+1
= (1

1
k
+2
,
1 +
1
k
+2
)
with
1
k
+2
<
1
k
+1
. Hence, we have, for any
k
,
1

1
k
+1
<
1

1
k
+2
<
1 +
1
k
+2
<
1 +
1
k
+1
. Hence,
S
k
+1
⊆
S
k
.
(iii) We have
diam
(
S
k
) = 1 +
1
k
+1

(
1

1
k
+1
)
=
1
2(
k
+1)
→
0
.
(iv)
T
∞
k
=1
S
k
=
T
∞
k
=1
(1

1
k
+1
,
1 +
1
k
+1
) = (1

lim
k
(
1
k
+1
)
,
1 + lim
k
1
k
+1
) = (1
,
1) =
∅
.
(b) The above sequence is not closed in the metric space
((0
,
1)
,d
1
)
. Consider now the
sequence of subsets
S
k
of
(0
,
1)
:
S
k
= (0
,
1
k
+1
]
.
It is easy to see that conditions (i) to (iv) are still satis ed.
(v) For any
k
, the complement of
S
k
in (0,1) is
S
c
k
= (
1
k
,
1)
. We can easily show that
this set is open in
(0
,
1)
because for any point in
S
c
k
, I can nd an
²
neighborhood in
(0,1) s.t. every point in that neighborhood remains in
S
c
k
. Hence, the complement of
S
c
k
in (0,1) is closed in (0,1) ie
S
k
closed in
(0
,
1)
.
2.
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 Fall '09
 Antoine
 Economics, Order theory, Monotonic function, Convex function, associated value function, c Sk

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