Exercise 2.28 (2.18).
Verify that Houthakker’s Axiom is equivalent to the follow-
ing: (*) if
A
∩ C
(
B
)
negationslash
=
∅
, then
C
(
A
)
∩
B
⊆ C
(
B
).
Proof.
First we show that (*) implies HA. Suppose that (*) is true and that
x, y
∈
A
∩
B
,
x
∈ C
(
A
), and
y
∈ C
(
B
) all hold. Note that
y
∈
A
∩C
(
B
). Thus
A
∩C
(
B
)
negationslash
=
∅
,
so (*) implies that
C
(
A
)
∩
B
⊆ C
(
B
). Since
x
∈ C
(
A
)
∩
B
, we have
x
∈ C
(
B
), and
Houthakker’s axiom holds.
Now we show that HA implies (*). Suppose that HA is true and that
A
∩ C
(
B
)
negationslash
=
∅
.
Then there exists
y
∈
A
∩ C
(
B
)
⊆
A
∩
B
; in particular,
y
∈
A
∩
B
and
y
∈ C
(
B
).
Take any
x
∈ C
(
A
)
∩
B
. Since
x
∈
A
∩
B
and
x
∈ C
(
A
), HA implies that
x
∈ C
(
B
)
as well. Hence
C
(
A
)
∩
B
⊆ C
(
B
), and (*) holds.

Section 2: Choice, Preferences, and Utility
2-19
Exercise 2.29 (2.19).
Prove that if
C
is rationalizable, then
C
satisfies Houthakker’s
Axiom.
Proof.
Suppose that
C
is rationalizable. Let
x, y
∈
A
∩
B
be given, and suppose that
x
∈ C
(
A
) and
y
∈ C
(
B
). Because
C
is rationalizable, there exists a preference relation
followsorequal
for which
C
=
C
followsorequal
. Since
x
∈ C
(
A
) =
C
followsorequal
(
A
) and
y
∈
A
, we know that
x
followsorequal
y
. Since
y
∈ C
(
B
) =
C
followsorequal
(
B
), we have
y
followsorequal
z
for all
z
∈
B
. By transitivity,
x
followsorequal
z
for all
z
∈
B
.
That is,
x
∈ C
followsorequal
(
B
) =
C
(
B
). Hence
C
satisfies Houthakker’s Axiom.
Exercise 2.30 (2.20).
Suppose
X
=
{
a, b, c
}
and assume
C
(
{
a, b
}
) =
{
a
}
,
C
(
{
b, c
}
) =
{
b
}
, and
C
(
{
a, c
}
) =
{
c
}
. Prove that if
C negationslash
=
∅
, then it must violate Houthakker’s ax-
iom.
Proof.
We are going to prove the contrapositive. Suppose that Houthakker’s axiom
holds.
Assume, first, that
a
∈ C
(
{
a, b, c
}
).
Note that
a, c
∈ {
a, b, c
} ∩ {
a, c
}
and
c
∈ C
(
{
a, c
}
). Houthakker’s axiom implies that
a
∈ C
(
{
a, c
}
) as well, which is not
the case.
Thus
a /
∈ C
(
{
a, b, c
}
).
Similar reasoning shows that
b /
∈ C
(
{
a, b, c
}
) and
c /
∈ C
(
{
a, b, c
}
); thus
C
(
{
a, b, c
}
) =
∅
if Houthakker’s axiom holds.
Hence, if
C
is
nonempty, then it violates Houthakker’s axiom.
Exercise 2.33 (2.23).
Prove that Sen’s
α
is equivalent to the following: (*) if
B
⊆
A
then
C
(
A
)
∩
B
⊆ C
(
B
).
Proof.
First we prove that (*) implies
α
. Assume that (*) holds and let
x
∈
B
⊆
A
and
x
∈ C
(
A
). Then
x
∈ C
(
A
)
∩
B
. Applying (*), which tells us that
C
(
A
)
∩
B
⊆ C
(
B
),
yields
x
∈ C
(
B
). Thus
α
is satisfied.
Now we show that
α
implies (*). Suppose that
α
holds, and let
B
⊆
A
. Take any
x
∈ C
(
A
)
∩
B
.
Note that
x
∈
B
⊆
A
and
x
∈ C
(
A
), so, by Sen’s
α
,
x
∈ C
(
B
).
Therefore
C
(
A
)
∩
B
⊆ C
(
B
), which means that (*) holds.
Exercise 2.34 (2.24).
Suppose that
C
satisfies Sen’s
α
. Verify that if
C
(
B
) =
B
,
then
C
(
A
) =
A
for all
A
⊆
B
.
Proof.
The inclusion
C
(
A
)
⊆
A
holds by construction, so it remains to show that
C
(
A
)
⊇
A
.
Let
x
∈
A
be given.
Because
A
⊆
B
, we have
x
∈
B
=
C
(
B
).
Since
x
∈
A
⊆
B
and
x
∈ C
(
B
), Sen’s
α
implies that
x
∈ C
(
A
). Hence
A
⊆ C
(
A
).