ECON 831
Solution Homework #1
1.
(i)
(ii)  Study at 0:
Γ(0) = (0
,
1]
.
Preliminary remarks:
NB1: there are only 2 open sets
O
in
[0
,
1]
s.t.
(0
,
1]
⊆
O
, (1)
[0
,
1]
(which is open and
closed); (2)
(0
,
1]
.
NB2: for any
² >
0
,
N
²,
[0
,
1]
(0) =
{
y
∈
[0
,
1]
/ d
(0
,y
)
< ²
}
= [0
,²
)
(when
²
≤
1
) and
[0,1] (when
² >
1
).
Proof: for any
y
∈
(0
,
1]
,
Γ(
y
) = (0
,y
)
⊆
Γ(0)
. Hence,
Γ
is upperhemicontinuous at 0.
 Study at
y
∈
(0
,
1)
:
Γ(
y
) = (0
,y
)
and for
²
"small enough"
N
²,
[0
,
1]
(
y
) = (
y

²,y
+
²
)
.
We prove that
Γ
is NOT upper hemicontinuous. Consider
(0
,y
)
: it is an open set in
[0
,
1]
and it contains the image of
y
by
Γ
. However, for any
² >
0
, there exist a point
y
*
(to the right of
y
) s.t.
1
≥
y
*
> y
and
y
*
∈
N
²,
[0
,
1]
(
y
)
and
Γ(
y
*
) = (0
,y
*
)
*
(0
,y
)
.
Hence,
Γ
is not upper hemicontinuous at
y
.
 Study at 1:
Γ(1) = (0
,
1)
.
Preliminary remark: an
²
neighborhood around 1 in
[0
,
1]
takes the form
N
²,X
(1) =
(1

²,
1]
for
²
"small enough".
Proof: for any
y
*
s.t.
1

² < y
*
< y
,
Γ(
y
*
) = (0
,y
*
)
⊆
(0
,y
)
. Hence,
Γ
is upper
hemicontinous at 1.
1