(a) Clearly 0
∈
S
0
.
(b) If
a
(
x
)
,b
(
x
)
∈
S
0
, then this means that
a
(
x
) =
b
(
x
) = 0 for all
x
∈
S
. Therefore so will
(
a
+
b
)(
x
).
(c) If
a
(
x
)
inS
0
so will
ca
(
x
) by the same argument.
2.
If
W
⊂
V
and
x /
∈
W
, then show there exists
f
∈
W
0
such that
f
(
x
)
6
= 0
.
Pick a basis for
W
,
β
W
=
{
v
1
,...,v
k
}
and extend it onto
V
,
β
V
=
{
v
1
,...,v
k
,v
k
+1
,...,v
n
}
. Then
form a dual basis to this basis: (
β
V
)
*
=
{
f
1
,...,f
k
,f
k
+1
,...,f
n
}
. Clearly
W
0
∩{
f
1
,...,f
k
}
since
f
i
(
v
i
) = 1
6
= 0. On the other hand we know that
f
k
+1
,...,f
n
∈
W
0
, since
v
k
+1
,...,v
n
/
∈
W
. But
since
v /
∈
W
, then this means that
v
=
w
+
∑
a
i
v
i
, where
i > k
and at least one of the
a
i
’s is non
zero, let it be
a
l
. But then this means that
f
l
(
v
) =
a
l
6
= 0, therefore we have found one.