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P ROOFS OF T HEOREMS 9.7, 9.8, AND 9.11 1 Proof of Theorem 9.7

We first establish some lemmas.

LEMMA14.1. Let P be substitutable. Letµ and µ0be pre-matchings. If µ ≤Bµ0then, for all w ∈ W and f ∈ F , U (f , µ) ⊆ U (f , µ0) and V (w,µ) ⊇ V (w,µ0).

PROOF. We prove thatV(w,µ) ⊇ V (w,µ0). The proof that U(f ,µ) ⊆ U(f ,µ0) is analo-gous.

First, ifV(w,µ0) = {;}, then there is nothing to prove, as {;} = V (w,µ0) ⊆ V (w,µ).

Suppose thatV(w,µ0) 6= {;}, and let f ∈ V (w,µ0). Then, w ∈ Ch(µ0(f ) ∪ w,P(f )).

Butµ ≤Bµ0, so the definition of≤Bimplies that, for allf ∈ F , either µ0(f ) = µ(f ) so w ∈ Ch (µ(f ) ∪ w, P(f )), or µ0(f ) = Ch(µ0(f ) ∪ µ(f ),P(f )). Then w ∈ Ch(µ0(f ) ∪ w ,P(f )) implies that

w∈ Ch (µ0(f ) ∪ w,P(f ))

= Ch(Ch(µ0(f ) ∪ µ(f ),P(f )) ∪ w,P(f ))

= Ch(µ0(f ) ∪ µ(f ) ∪ {w },P(f )).

The second equality above is from Proposition 2.3 in Blair(1988). (Blair proves that ifP is substitutable, then Ch(A ∪ B,P(f )) = Ch(Ch(A,P(f )) ∪ B,P(f )) for all A and B.) Substitutability ofP implies that w ∈ Ch (µ(f ) ∪ w, P(f )). Then f ∈ V (w, µ) and thus

V(w,µ) ⊇ V (w,µ0). ƒ

LEMMA14.2. Let P be strongly substitutable. Let µ and µ0be pre–matchings. If µ ≤ µ0 then, for all w ∈ W and f ∈ F , U (f , µ) ⊆ U (f , µ0) and V (w,µ) ⊇ V (w,µ0).

PROOF. We prove thatV(w,µ) ⊇ V (w,µ0). The proof that U(f ,µ) ⊆ U(f ,µ0) is analo-gous.

First, ifV(w,µ0) = {;}, then there is nothing to prove. Suppose that V (w,µ0) 6= {;}, and let f ∈ V (w, µ0). Then, w ∈ Ch(µ0(f ) ∪ w,P(f )). Strong substitutability implies then w∈ Ch (µ(f ) ∪ w, P(f )), as µ0(f ) R(f ) µ(f ) because µ ≤ µ0. ƒ LetV0= {ν ∈ V : ν(s ) R(s ) ; for all s ∈ F ∪ W }. We need to work on the set V0instead ofV because ν0andν1are the smallest and largest, respectively, elements ofV0. Note thatT(V ) ⊆ V0, so there is no loss in working withV0.

LEMMA14.3. If P is substitutable, then T|V0is increasing whenV0is endowed withB. If P is strongly substitutable, then T|V0is increasing whenV0is endowed with≤.

PROOF. We show thatT|V0is increasing whenV0 is endowed with order≤B. That is, wheneverµ ≤Bµ0we have(T µ) ≤B(T µ0). The proof for ≤ follows along the same lines, usingLemma 14.2instead of14.1.

Letµ ≤B µ0and fixf ∈ F and w ∈ W . Lemma 14.1says thatU(f ,µ) ⊆ U(f ,µ0). We first show that

Ch(U(f ,µ0),P(f )) = Ch([Ch(U(f ,µ0),P(f )) ∪ Ch(U(f ,µ),P(f ))],P(f )). (39) To see this, letS⊆ Ch (U (f , µ0),P(f )) ∪ Ch(U(f ,µ),P(f )). Then S ⊆ U(f ,µ) ∪ U(f ,µ0) = U(f ,µ0), so Ch(U(f ,µ0),P(f )) R(f ) S. But Ch(U(f ,µ0),P(f )) ⊆ Ch(U(f ,µ0),P(f )) ∪ Ch(U(f ,µ),P(f )), so we have established (39).

Now,(T µ0)(f ) = Ch(U(f ,µ0),P(f )) and (T µ)(f ) = Ch(U(f ,µ),P(f )), so (39) implies that

(T µ0)(f ) = Ch([(T µ0)(f ) ∪ (T µ)(f )],P(f ))).

The proof for(T µ0)(w ) is analogous. ƒ

NowT|V0:V0→ V0is increasing andV0is a lattice (Remark 9.1). We haveT(V ) ⊆ V0 soE (P) ⊆ V0, andE (P) equals the set of fixed points of T |V0. So Tarski’s fixed point theo-rem implies that〈E (P), ≤B〉 and 〈E (P), ≤〉 are nonempty lattices. Item (ii) inTheorem 9.7 follows from standard results (Topkis 1998, Chapter 4).

This finishes the proof ofTheorem 9.7.

14.2 Proof ofTheorem 9.8 We first prove item (i).

Let ν, ν0 ∈ E (P) be such that ν0(w ) R(w ) ν(w ) for all w ∈ W . Suppose, by way of contradiction, that there is some f ∈ F such that ν0(f ) P(f ) ν(f ). Let C = Ch(ν(f ) ∪ ν0(f ),P(f )), so C R(f ) ν0(f ) P(f ) ν(f ). But ν ∈ E (P) implies that ν(f ) = Ch(ν(f ),P(f )) (Lemma 11.4), so C * ν(f ). Hence there is w ∈ C \ ν(f ); note that w∈ ν0(f ). Now

w∈ C = Ch (ν(f ) ∪ ν0(f ) ∪ w ,P(f ))

and the substitutability ofP(f ) imply that w ∈ Ch(ν(f ) ∪ w ,P(f )). So f ∈ V (w ,ν).

Now,w ∈ ν0(f ) \ ν(f ) implies f ∈ ν0(w ) \ ν(w ). Then ν0(w ) R(w ) ν(w ) implies that ν0(w ) P(w ) ν(w ), as P(w ) is strict. But ν0(w ) = Ch(ν0(w ),P(w )) = Ch(ν0(w ) ∪ f ,P(w )) by Lemma 11.4. So strong substitutability implies that f ∈ Ch (ν(w ) ∪ f , P(w )). Since f /∈ ν(w ), we obtain ν(w ) ∪ f P(w ) ν(w ). This contradicts ν ∈ E (P), since we showed f ∈ V (w , ν) and ν ∈ E (P) imply ν(w ) = Ch (V (w , ν), P(w )).

To prove item (ii) in the theorem, note that whenP(F ) is strongly substitutable the model is symmetric, and the argument above holds with firms in place of workers, and

workers in place of firms. ƒ

14.3 Proof ofTheorem 9.11

We first prove that〈E (P), ≤〉 is a sublattice of 〈V , ≤〉. That 〈E (P), ≤〉 is distributive follows then immediately. We need to verify that the lattice operations∨ and ∧ in V are the lattice operations in〈E (P), ≤〉.

Letν1,ν2∈ E (P). Let ν = ν1∨ν2inV . We prove that ν is the join of ν1,ν2in〈E (P), ≤〉.

The proof forν1∧ ν2is analogous.

By hypothesisν is a matching; so

w∈ ν(f ) → f ∈ ν(w ).

We prove that ν ∈ E (P). Suppose, by way of contradiction, that there is f such that (T ν)(f ) 6= ν(f ). Without loss of generality, say that ν(f ) = ν1(f ) R(f ) ν2(f ). Since ν1E (P), ν1is individually rational (Lemma 11.4), sof ∈ Ch (ν1(w ),P(w )) = Ch(ν1(w ) ∪ f , P(w )) for all w ∈ ν1(f ). For all w , on the other hand, ν1(w ) R(w ) ν(w ). So strong sub-stitutability gives f ∈ Ch (ν(w ) ∪ f , P(w )) for all w ∈ ν1(f ). Thus ν1(f ) ⊆ U(f ,ν). Since (T ν)(f ) = Ch(U(f ,ν),P(f )) and ν1is individually rational,(T ν)(f ) \ ν(f ) 6= ;.

Letw ∈ (T ν)(f ) \ ν(f ). By substitutability, w ∈ Ch (ν1(f ) ∪ w ,P(f )). Strong substi-tutability andν1(f ) R(f ) ν2(f ) then imply w ∈ Ch(ν2(f ) ∪ w ,P(f )). So

f ∈ V (w , νi) (40)

fori= 1,2.

On the other handw∈ (T ν)(f ) implies w ∈ U (f , ν), so

f ∈ Ch (ν(w ) ∪ f , P(w )). (41)

Leti be such thatν(w ) = νi(w ). Then (40) andνi ∈ E (P) imply νi(w ) ∪ f ∈ V (w ,νi).

But we assumedw /∈ ν(f ), so f /∈ νi(w ), as ν is a matching. Then νi(w ) ∪ f 6= νi(w ), which contradictsνi ∈ E (P), given (41) andν(w ) ∪ f ∈ V (w ,νi).

For the rest of the theorem, we need a lemma.

LEMMA14.4. Let P be strongly substitutable. For all f and w , for any ν and ν0 in V , U(f ,ν ∨ ν0) = U(f ,ν) ∪U(f ,ν0), U(f ,ν ∧ ν0) = U(f ,ν) ∩U(f ,ν0), V (w,ν ∨ ν0) = V (w,ν) ∩ V(w,ν0), and V (w,ν ∧ ν0) = V (w,ν) ∪ V (w,ν0).

PROOF. We prove only thatU(f ,ν ∨ ν0) = U(f ,ν) ∪ U(f ,ν0) and that V (w,ν ∨ ν0) = V(w,ν) ∩ V (w,ν0). The proof of the other statements is symmetric.

We first prove thatU(f ,ν ∨ν0) ⊆ U(f ,ν)∪U(f ,ν0). Let w ∈ U(f ,ν ∨ν0), so f ∈ Ch((ν ∨ ν0)(w ) ∪ f ,P(w )). Now, (ν ∨ ν0)(f ) equals either ν(f ) or ν0(f ). If (ν ∨ ν0)(w ) = ν(w ), then f ∈ Ch (ν(w )∪ f , P(w )); so w ∈ U (f , ν). Similarly, if (ν ∨ν0)(w ) = ν0(w ), then w ∈ U(f ,ν0).

This proves thatU(f ,ν ∨ ν0) ⊆ U(f ,ν) ∪U(f ,ν0).

Second, we prove thatU(f ,ν) ∪ U(f ,ν0) ⊆ U(f ,ν ∨ ν0). Let w ∈ U(f ,ν), so f ∈ Ch(ν(w ) ∪ f ,P(w )). Now ν(w ) R(w ) (ν ∨ ν0)(w ), so strong substitutability implies f ∈ Ch((ν ∨ ν0)(w ) ∪ f ,P(w )). Hence w ∈ U(f ,ν ∨ ν0). This proves that U(f ,ν) ∪ U(f ,ν0) ⊆ U(f ,ν ∨ ν0). So, U(f ,ν ∨ ν0) = U(f ,ν) ∪U(f ,ν0).

We now prove thatV(w,ν ∨ ν0) = V (w,ν) ∩ V (w,ν0). First we prove V (w,ν ∨ ν0) ⊆ V(w,ν) ∩ V (w,ν0). Let f ∈ V (w,ν ∨ ν0), so

w∈ Ch ((ν ∨ ν0)(f ) ∪ w,P(f )). (42) Without loss of generality, say(ν ∨ν0)(f ) = ν(f ) R(f ) ν0(f ). Then (ν ∨ν0)(f ) = ν(f ) implies thatf ∈ V (w, ν). Statement (42) and strong substitutability imply thatw∈ Ch (ν0(f ) ∪ w , P(f )), as (ν ∨ ν0)(f ) R(f ) ν0(f ). Thus f ∈ V (w,ν), and we obtain V (w,ν ∨ ν0) ⊆ V (w,ν) ∩ V(w,ν0).

Finally, we prove thatV(w,ν) ∩ V (w,ν0) ⊆ V (w,ν ∨ ν0). Let f ∈ V (w,ν) ∩ V (w,ν0), so w ∈ Ch (ν(f ) ∪ w, P(f )) and w ∈ Ch (ν0(f ) ∪ w,P(f )). Now, (ν ∨ ν0)(w ) equals either ν(w ) orν0(w ), so either way w ∈ Ch((ν ∨ ν0)(f ) ∪ w,P(f )). Hence f ∈ V (w,ν ∨ ν0). ƒ Lemma 14.4implies immediately thatψ is a lattice homomorphism: Let ν0,ν ∈ V . For anyf and w ,

(ψ(ν ∨ ν0))(f ) = U(f ,ν ∨ ν0) = U(f ,ν) ∪U(f ,ν0) = (ψν)(f ) ∪ (ψν0)(f ) (ψ(ν ∨ ν0))(w ) = V (w,ν ∨ ν0) = V (w,ν) ∩ V (w,ν0) = (ψν)(f ) ∩ (ψν0)(f ).

Soψ(ν ∨ ν0) = ψν t ψν0. Thatψ(ν ∧ ν0) = ψν u ψν0is also trivial fromLemma 14.4.

We now show thatψ|E (P) is an isomorphism onto its range. Letν, ν0 ∈ E (P). Let ψν = ψν0. Then for all f , U(f ,ν) = U(f ,ν0) so (T ν)(f ) = (T ν0)(f ). Similarly (T ν)(w ) = (T ν0)(w ) for all w . So T ν = T ν0. Thenν, ν0∈ E (P) implies ν = ν0, asv= T ν and v0= T ν0. Henceψ is one-to-one, as ψν = ψν0impliesν = ν0.

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Submitted 2005-9-1. Final version accepted 2005-12-16. Available online 2005-12-16.

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