Theorem 0.2 Let K be a nonempty, compact and convex subset of Euclidean space, and f: K= K be a correspondence. If f is nonempty- and convex-valued, and upper hemicon tinuous, then the set of its fixed points is nonempty and compact The trick is contained in the following definition: Definition 5 Fix a finite game G=(N, (Ai, ui)ieN). The mixed extension of G is the game r=(N,(△(A),Ul)ieN), where, for any collection(a)ieN∈IleN△(A), U(a1…,aN)=∑Ia(a)(a1,,ay) (a)ieN∈AieN For every i N, denote by pi Player is Nash best reply correspondence The idea is to consider the modified game in which players have the physical possibility te randomize among their actions, in a stochastically independent fashion. I shall return to the interpretation of this assumption momentarily; formally, note that the set K= IleN A(Ai) as a product of simplices, is a nonempty, compact, convex subset of Euclidean space, as required by Theorem 0. 2. Moreover, define a correspondence f: K= K by f(a1,,aN)={a)ieN:W∈N,a2∈(a-1)} It is easy to see that f( is nonempty, convex-valued and uhc if P is. So, we need Proposition 0. 3 For any finite game G, the Nash best reply correspondence of its mixed extension T is nonempty, convex-valued and upper hemicontinuous Proof: Nonemptyness follows because the best-reply correspondence of the original finite game is nonempty(and A(Ai) contains all degenerate mixed strategies. Now note that ai E P(a-i iff ai assigns positive probability only to actions that are best replies against the belief liti a,(why? ) This immediately implies convexity To prove uhe, note that we can use the characterization in Theorem 0. 1. Consider two convergent sequences(ai)n>1 -a-i and(ai)>1 -a; such that, for every n a∈n(an1, This means that, for every a∈△(A) U(a,a2)≥U(a2,a2) Now note that, by definition, the function U; ( ) is jointly continuous in its arguments Thus,asn→∞,Ul(a2,a2)→Ul(a,a- and vi(a,a2)→Ul(a,a-). The result now follows.■Theorem 0.2 Let K be a nonempty, compact and convex subset of Euclidean space, and f : K ⇒ K be a correspondence. If f is nonempty- and convex-valued, and upper hemicon￾tinuous, then the set of its fixed points is nonempty and compact. The trick is contained in the following definition: Definition 5 Fix a finite game G = (N,(Ai , ui)i∈N ). The mixed extension of G is the game Γ = (N,(∆(Ai), Ui)i∈N ), where, for any collection (αi)i∈N ∈ Q i∈N ∆(Ai), Ui(α1, . . . , αN ) = X (ai)i∈N ∈A Y i∈N αi(ai)ui(a1, . . . , aN ) For every i ∈ N, denote by ρ Γ i Player i’s Nash best reply correspondence. The idea is to consider the modified game in which players have the physical possibility to randomize among their actions, in a stochastically independent fashion. I shall return to the interpretation of this assumption momentarily; formally, note that the set K = Q i∈N ∆(Ai), as a product of simplices, is a nonempty, compact, convex subset of Euclidean space, as required by Theorem 0.2. Moreover, define a correspondence f : K ⇒ K by f(α1, . . . , αN ) = {(α 0 i )i∈N : ∀i ∈ N, α0 i ∈ ρ Γ i (α−i)} It is easy to see that f(·) is nonempty, convex-valued and uhc if ρ Γ i is. So, we need: Proposition 0.3 For any finite game G, the Nash best reply correspondence of its mixed extension Γ is nonempty, convex-valued and upper hemicontinuous. Proof: Nonemptyness follows because the best-reply correspondence of the original finite game is nonempty (and ∆(Ai) contains all degenerate mixed strategies.) Now note that αi ∈ ρ Γ i (α−i) iff αi assigns positive probability only to actions that are best replies against the belief Q j6=i αj (why?). This immediately implies convexity. To prove uhc, note that we can use the characterization in Theorem 0.1. Consider two convergent sequences (α n −i )n≥1 → α−i and (α n i )n≥1 → αi such that, for every n ≥ 1, α n i ∈ ρ Γ i (α n −i . This means that, for every α 0 i ∈ ∆(Ai), Ui(α n i , αn −i ) ≥ Ui(α 0 i , αn −i ) Now note that, by definition, the function Ui(·, ·) is jointly continuous in its arguments. Thus, as n → ∞, Ui(α n i , αn −i ) → Ui(αi , α−i and Ui(α 0 i , αn −i ) → Ui(α 0 i , α−i). The result now follows. 4