# Exercise 228 218 verify that houthakkers axiom is

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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 ).