Spatial bargaining in rectilinear facility location problem

Abstract

We consider a spatial bargaining model where players collectively choose a facility location on a two-dimensional rectilinear distance space through bargaining using the unanimity rule. We show that as players become infinitely patient, their stationary subgame perfect equilibrium utilities converge to the utilities that satisfy the lexicographic maximin utility criterion introduced by Sen (Collective choice and social welfare, 1970).

Appendix

In this appendix, we prove Theorems 1 and 2. Hereafter, since we fix the value of $\delta$, we abbreviate $G\left( {\delta }\right)$, $r^M\left( {\delta }\right)$, $r^L\left( {\delta }\right)$, and $A_*\left( {\delta }\right)$ as G, $r^M$, $r^L$, and $A_*$, respectively.

1.1 Appendix 1: Preliminary lemmas for Theorems 1 and 2

If $m_1^{\mathrm{L}}< y_1^{\gamma } < m_1^{\mathrm{R}}$,⁸ define

$$\begin{aligned} n^{\mathrm{V}}:={\left\{ \begin{array}{ll} \left( {y_1^{\gamma },y_1^{\gamma }+\frac{b_3+b_4}{2}}\right) &{}\text { if}\ d^+<d^-\\ \left( {y_1^{\gamma },-y_1^{\gamma }+\frac{b_1+b_2}{2}}\right) &{}\text { if}\ d^+>d^-. \end{array}\right. } \end{aligned}$$

Then, $n^{\mathrm{V}}$ denotes the intersection of the vertical line through player $\gamma$’s location (that is, $\left\{ {x \in X}\mid {x_1=y_1^{\gamma }}\right\}$) and the set of minimax distance locations except for the two endpoints (that is, $M {\setminus } \left\{ {m^{\mathrm{L}},m^{\mathrm{R}}}\right\}$).⁹ Furthermore, if $\min \left\{ {m_2^{\mathrm{L}},m_2^{\mathrm{R}}}\right\}<y_2^{\gamma }<\max \left\{ {m_2^{\mathrm{L}},m_2^{\mathrm{R}}}\right\}$,¹⁰ define

$$\begin{aligned} n^{\mathrm{H}}:= {\left\{ \begin{array}{ll} \left( {y_2^{\gamma }-\frac{b_3+b_4}{2},y_2^{\gamma }}\right) &{}\text { if}\ d^+<d^-\\ \left( {-y_2^{\gamma }+\frac{b_1+b_2}{2},y_2^{\gamma }}\right) &{}\text { if}\ d^+>d^-. \end{array}\right. } \end{aligned}$$

Then, $n^{\mathrm{H}}$ denotes the intersection of the horizontal line through player $\gamma$’s location (that is, $\left\{ {x \in X}\mid {x_2=y_2^{\gamma }}\right\}$) and the set of minimax distance locations except for the two endpoints (that is, $M {\setminus } \left\{ {m^{\mathrm{L}},m^{\mathrm{R}}}\right\}$).¹¹ Remember that $\ell ^{\mathrm{L}}$ and $\ell ^{\mathrm{R}}$ denote the locations that satisfy $\ell ^{\mathrm{L}}, \ell ^{\mathrm{R}} \in \arg \min _{x \in M} d^{\gamma }\left( {x}\right)$ and $\ell _1^{\mathrm{L}} \le x_1 \le \ell _1^{\mathrm{R}}$ for each $x \in \arg \min _{x \in M} d^{\gamma }\left( {x}\right)$. Furthermore, remember that $\underline{d}=\min _{x \in M} d^{\gamma }\left( {x}\right)$. Then, we obtain the following lemma from Figures 12 and 13 based on Elzinga and Hearn’s (1972) construction of minimax distance locations (or tedious calculations).

Fig. 12

Lexicographic minimax distance locations in the case where $d^+ < d^-$. The small square represents player $\gamma$’s location $y^{\gamma }$. The solid line represents the set M of minimax distance locations. The thick part of the solid line represents the set L of lexicographic minimax distance locations

Fig. 13

Lexicographic minimax distance locations in the case where $d^+ > d^-$. The small square represents player $\gamma$’s location $y^{\gamma }$. The solid line represents the set M of minimax distance locations. The thick part of the solid line represents the set L of lexicographic minimax distance locations

Lemma 1

(1) $\ell ^{\mathrm{L}}$, $\ell ^{\mathrm{R}}$, and $\underline{d}$ are given in Table 1if $d^+<d^-$. (2) $\ell ^{\mathrm{L}}$, $\ell ^{\mathrm{R}}$, and $\underline{d}$ are given in Table 2if $d^+>d^-$. (3) $\ell ^{\mathrm{L}}=m^{\mathrm{L}}$, $\ell ^{\mathrm{R}}=m^{\mathrm{R}}$, and $\underline{d}=d^{\gamma }\left( {m^{\mathrm{L}}}\right) =d^{\gamma }\left( {m^{\mathrm{R}}}\right)$ if $d^+=d^-$.

Table 1 $\ell ^{\mathrm{L}}$, $\ell ^{\mathrm{R}}$, and $\underline{d}$ in the case where $d^+<d^-$

Table 2 $\ell ^{\mathrm{L}}$, $\ell ^{\mathrm{R}}$, and $\underline{d}$ in the case where $d^+>d^-$

For each $Z \in 2^X$ and $r \in {\mathbb {R}}_+$, define $O\left( {Z;r}\right) :=\cup _{x' \in Z}\left\{ {x \in X}\mid {d\left( {x,x'}\right) < r}\right\}$. Then, $O\left( {Z;r}\right)$ denotes the r-open neighborhood of Z.

Lemma 2

(1) Suppose that $x \in C\left( {M;r^M}\right)$. Then, for each $i \in I$, $d^{i}\left( {x}\right) \le \overline{d}+r^M$. (2) Suppose that $x \in C\left( {L;r^L}\right)$. Then, $d^{\gamma }\left( {x}\right) \le \underline{d}+r^L$. (3) Suppose that $x \in O\left( {M;r^M}\right)$. Then, for each $i \in I$, $d^{i}\left( {x}\right) <\overline{d}+r^M$. (4) Suppose that $x \in O\left( {L;r^L}\right)$. Then, $d^{\gamma }\left( {x}\right) <\underline{d}+r^L$. (5) Suppose that $x \in X {\setminus } C\left( {M;r^M}\right)$. Then, for some $i \in I$, $d^{i}\left( {x}\right) >\overline{d}+r^M$. (6) Suppose that $x \in C\left( {M;r^M}\right) {\setminus } C\left( {L;r^L}\right)$. Then, $d^{\gamma }\left( {x}\right) >\underline{d}+r^L$.

Proof

Statements (1) and (3) immediately follow from Proposition 1, and Statements (2) and (4) immediately follow from Corollary 1.

As shown in Fig. 14, note that for each $z^{\mathrm{L}}, z^{\mathrm{R}} \in X$ such that $z_1^{\mathrm{L}} \le z_1^{\mathrm{R}}$ and $|z_1^{\mathrm{L}}-z_1^{\mathrm{R}}|=|z_2^{\mathrm{L}}-z_2^{\mathrm{R}}|$ and $r \in {\mathbb {R}}_+$,

$$\begin{aligned} C\left( {\left[ {z^{\mathrm{L}},z^{\mathrm{R}}}\right] ;r}\right)&= \left\{ {x \in X}\mid {x_2 \ge -x_1+z_1^{\mathrm{L}}+z_2^{\mathrm{L}}-r}\right\} \\&\quad \cap \left\{ {x \in X}\mid {x_2 \le -x_1+z_1^{\mathrm{R}}+z_2^{\mathrm{R}}+r}\right\} \\&\quad \cap \left\{ {x \in X}\mid {x_2 \le x_1-z_1^{\mathrm{L}}+z_2^{\mathrm{L}}+r}\right\} \\&\quad \cap \left\{ {x \in X}\mid {x_2 \ge x_1-z_1^{\mathrm{R}}+z_2^{\mathrm{R}}-r}\right\} . \end{aligned}$$

Fig. 14

Form of $C\left( {\left[ {z^{\mathrm{L}},z^{\mathrm{R}}}\right] ;r}\right)$ such that $z_1^{\mathrm{L}} \le z_1^{\mathrm{R}}$ and $|z_1^{\mathrm{L}}-z_1^{\mathrm{R}}|=|z_2^{\mathrm{L}}-z_2^{\mathrm{R}}|$. The lower left-, upper right-, upper left-, and lower right-hand side broken lines represent $\left\{ {x \in X}\mid {x_2=-x_1+z_1^{\mathrm{L}}+z_2^{\mathrm{L}}-r}\right\}$, $\left\{ {x \in X}\mid {x_2=-x_1+z_1^{\mathrm{R}}+z_2^{\mathrm{R}}+r}\right\}$, $\left\{ {x \in X}\mid {x_2=x_1-z_1^{\mathrm{L}}+z_2^{\mathrm{L}}+r}\right\}$, and $\left\{ {x \in X}\mid {x_2=x_1-z_1^{\mathrm{R}}+z_2^{\mathrm{R}}-r}\right\}$, respectively. The rectangle formed by the four broken lines represents $C\left( {\left[ {z^{\mathrm{L}},z^{\mathrm{R}}}\right] ;r}\right)$

Then, by Proposition 1, $C\left( {M;r^M}\right) =C\left( {\left[ {m^{\mathrm{L}},m^{\mathrm{R}}}\right] ;r^M}\right) =\cap _{j=1}^4 M_j$ where

$$\begin{aligned} M_1&:=\left\{ {x \in X}\mid {x_2 \ge -x_1+m_1^{\mathrm{L}}+m_2^{\mathrm{L}}-r^M}\right\} ,\\ M_2&:=\left\{ {x \in X}\mid {x_2 \le -x_1+m_1^{\mathrm{R}}+m_2^{\mathrm{R}}+r^M}\right\} ,\\ M_3&:=\left\{ {x \in X}\mid {x_2 \le x_1-m_1^{\mathrm{L}}+m_2^{\mathrm{L}}+r^M}\right\} ,\\ M_4&:=\left\{ {x \in X}\mid {x_2 \ge x_1-m_1^{\mathrm{R}}+m_2^{\mathrm{R}}-r^M}\right\} . \end{aligned}$$

Furthermore, by Corollary 1, $C\left( {L;r^L}\right) =C\left( {\left[ {\ell ^{\mathrm{L}},\ell ^{\mathrm{R}}}\right] ;r^L}\right) =\cap _{j=1}^4 L_j$ where

$$\begin{aligned} L_1&:=\left\{ {x \in X}\mid {x_2 \ge -x_1+\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}-r^L}\right\} ,\\ L_2&:=\left\{ {x \in X}\mid {x_2 \le -x_1+\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}+r^L}\right\} ,\\ L_3&:=\left\{ {x \in X}\mid {x_2 \le x_1-\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}+r^L}\right\} ,\\ L_4&:=\left\{ {x \in X}\mid {x_2 \ge x_1-\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}-r^L}\right\} . \end{aligned}$$

Remember that $C\left( {M;r^M}\right) =\cap _{j=1}^4 M_j$. Thus, $X {\setminus } C\left( {M;r^M}\right) =\cup _{j=1}^4 M_j^c$.

Suppose that $x \in X {\setminus } C\left( {M;r^M}\right)$. Then, $x \in \cup _{j=1}^4 M_j^c$. Note that since

$$\begin{aligned} M_1^c&=\left\{ {x \in X}\mid {x_2< -x_1+b_1-\overline{d}-r^M}\right\} ,\\ M_2^c&=\left\{ {x \in X}\mid {x_2> -x_1+b_2+\overline{d}+r^M}\right\} ,\\ M_3^c&=\left\{ {x \in X}\mid {x_2 > x_1+b_3+\overline{d}+r^M}\right\} ,\\ M_4^c&=\left\{ {x \in X}\mid {x_2 < x_1+b_4-\overline{d}-r^M}\right\} , \end{aligned}$$

it follows that $d\left( {x,x'}\right) >\overline{d}+r^M$ for each $x' \in \left\{ {x \in X}\mid {x_2=-x_1+b_1}\right\}$ if $x \in M_1^c$, for each $x' \in \left\{ {x \in X}\mid {x_2=-x_1+b_2}\right\}$ if $x \in M_2^c$, for each $x' \in \left\{ {x \in X}\mid {x_2=x_1+b_3}\right\}$ if $x \in M_3^c$, and for each $x' \in \left\{ {x \in X}\mid {x_2=x_1+b_4}\right\}$ if $x \in M_4^c$. Furthermore, note that $\left\{ {y^i}\right\} _{i \in I} \cap \left\{ {x \in X}\mid {x_2=-x_1+b_1}\right\} \not =\emptyset$, $\left\{ {y^i}\right\} _{i \in I} \cap \left\{ {x \in X}\mid {x_2=-x_1+b_2}\right\} \not =\emptyset$, $\left\{ {y^i}\right\} _{i \in I} \cap \left\{ {x \in X}\mid {x_2=x_1+b_3}\right\} \not =\emptyset$, and $\left\{ {y^i}\right\} _{i \in I} \cap \left\{ {x \in X}\mid {x_2=x_1+b_4}\right\} \not =\emptyset$. Thus, for some $i \in I$, $d^{i}\left( {x}\right) >\overline{d}+r^M$. Hence, Statement (5) is true.

Remember that $C\left( {L;r^L}\right) =\cap _{j=1}^4 L_j$. Let

$$\begin{aligned} G_1&:=\left\{ {x \in X}\mid {x_2 \ge -x_1+m_1^{\mathrm{L}}+m_2^{\mathrm{L}}-r^L}\right\} ,\\ G_2&:=\left\{ {x \in X}\mid {x_2 \le -x_1+m_1^{\mathrm{R}}+m_2^{\mathrm{R}}+r^L}\right\} ,\\ G_3&:=\left\{ {x \in X}\mid {x_2 \le x_1-m_1^{\mathrm{L}}+m_2^{\mathrm{L}}+r^L}\right\} ,\\ G_4&:=\left\{ {x \in X}\mid {x_2 \ge x_1-m_1^{\mathrm{R}}+m_2^{\mathrm{R}}-r^L}\right\} . \end{aligned}$$

Then, since $r^M \le r^L$, for each $j \in \left\{ {1,2,3,4}\right\}$, $M_j \subset G_j$, and thus, $M_j \cap G_j^c=\emptyset$. Let

$$\begin{aligned} H_1&:=\left\{ {x \in X}\mid {x_2 \ge -x_1+y_1^{\gamma }+y_2^{\gamma }-\underline{d}-r^L}\right\} ,\\ H_2&:=\left\{ {x \in X}\mid {x_2 \le -x_1+y_1^{\gamma }+y_2^{\gamma }+\underline{d}+r^L}\right\} ,\\ H_3&:=\left\{ {x \in X}\mid {x_2 \le x_1-y_1^{\gamma }+y_2^{\gamma }+\underline{d}+r^L}\right\} ,\\ H_4&:=\left\{ {x \in X}\mid {x_2 \ge x_1-y_1^{\gamma }+y_2^{\gamma }-\underline{d}-r^L}\right\} . \end{aligned}$$

Then, if $d^+ < d^-$, it follows from Lemma 1 that (i) if $y_1^{\gamma } \le m_1^{\mathrm{L}}$ or $y_2^{\gamma } \le m_2^{\mathrm{L}}$, since $\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}=m_1^{\mathrm{L}}+m_2^{\mathrm{L}}$, $L_1=G_1$, and if $m_1^{\mathrm{L}}<y_1^{\gamma }$ and $m_2^{\mathrm{L}}<y_2^{\gamma }$, since $\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}=y_1^{\gamma }+y_2^{\gamma }-\underline{d}$, $L_1=H_1$; (ii) if $m_1^{\mathrm{R}} \le y_1^{\gamma }$ or $m_2^{\mathrm{R}} \le y_2^{\gamma }$, since $\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}=m_1^{\mathrm{R}}+m_2^{\mathrm{R}}$, $L_2=G_2$, and if $y_1^{\gamma }<m_1^{\mathrm{R}}$ and $y_2^{\gamma }<m_2^{\mathrm{R}}$, since $\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}=y_1^{\gamma }+y_2^{\gamma }+\underline{d}$, $L_2=H_2$; (iii) since $-\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}=\frac{b_3+b_4}{2}=-m_1^{\mathrm{L}}+m_2^{\mathrm{L}}$, $L_3=G_3$; and (iv) since $-\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}=\frac{b_3+b_4}{2}=-m_1^{\mathrm{R}}+m_2^{\mathrm{R}}$, $L_4=G_4$. If $d^+ > d^-$, it follows from Lemma 1 that (i) since $\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}=\frac{b_1+b_2}{2}=m_1^{\mathrm{L}}+m_2^{\mathrm{L}}$, $L_1=G_1$; (ii) since $\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}=\frac{b_1+b_2}{2}=m_1^{\mathrm{R}}+m_2^{\mathrm{R}}$, $L_2=G_2$; (iii) if $y_1^{\gamma } \le m_1^{\mathrm{L}}$ or $m_2^{\mathrm{L}} \le y_2^{\gamma }$, since $-\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}=-m_1^{\mathrm{L}}+m_2^{\mathrm{L}}$, $L_3=G_3$, and if $m_1^{\mathrm{L}} < y_1^{\gamma }$ and $y_2^{\gamma } <m_2^{\mathrm{L}}$, since $-\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}=-y_1^{\gamma }+y_2^{\gamma }+\underline{d}$, $L_3=H_3$; and (iv) if $m_1^{\mathrm{R}} \le y_1^{\gamma }$ or $y_2^{\gamma } \le m_2^{\mathrm{R}}$, since $-\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}=-m_1^{\mathrm{R}}+m_2^{\mathrm{R}}$, $L_4=G_4$, and if $y_1^{\gamma } <m_1^{\mathrm{R}}$ and $m_2^{\mathrm{R}} < y_2^{\gamma }$, since $-\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}=-y_1^{\gamma }+y_2^{\gamma }-\underline{d}$, $L_4=H_4$. If $d^+=d^-$, it follows from Lemma 1 that (i) since $\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}=m_1^{\mathrm{L}}+m_2^{\mathrm{L}}$, $L_1=G_1$; (ii) since $\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}=m_1^{\mathrm{R}}+m_2^{\mathrm{R}}$, $L_2=G_2$; (iii) since $-\ell _1^{\mathrm{L}}+\ell _2^{\mathrm{L}}=-m_1^{\mathrm{L}}+m_2^{\mathrm{L}}$, $L_3=G_3$; and (iv) since $-\ell _1^{\mathrm{R}}+\ell _2^{\mathrm{R}}=-m_1^{\mathrm{R}}+m_2^{\mathrm{R}}$, $L_4=G_4$. Thus, for each $j \in \left\{ {1,2,3,4}\right\}$, since $M_j \cap G_j^c=\emptyset$, $\left( {\cap _{j'=1}^4 M_{j'}}\right) \cap L_j^c \subset H_j^c$. Hence, since $C\left( {M;r^M}\right) =\cap _{j'=1}^4 M_{j'}$ and $C\left( {L;r^L}\right) =\cap _{j=1}^4 L_j$, $C\left( {M;r^M}\right) {\setminus } C\left( {L;r^L}\right) =\left( {\cap _{j'=1}^4 M_{j'}}\right) {\setminus } \left( {\cap _{j=1}^4 L_j}\right) =\left( {\cap _{j'=1}^4 M_{j'}}\right) \cap \left( {\cap _{j=1}^4 L_j}\right) ^c =\left( {\cap _{j'=1}^4 M_{j'}}\right) \cap \left( {\cup _{j=1}^4 L_j^c}\right) =\cup _{j=1}^4\left( {\left( {\cap _{j'=1}^4 M_{j'}}\right) \cap L_j^c}\right) \subset \cup _{j=1}^4 H_j^c$.

Suppose that $x \in C\left( {M;r^M}\right) {\setminus } C\left( {L;r^L}\right)$. Then, $x \in \cup _{j=1}^4 H_j^c$. Note that

$$\begin{aligned} H_1^c&=\left\{ {x \in X}\mid {x_2-y_2^{\gamma }< -\left( {x_1-y_1^{\gamma }}\right) -\underline{d}-r^L}\right\} ,\\ H_2^c&=\left\{ {x \in X}\mid {x_2-y_2^{\gamma }> -\left( {x_1-y_1^{\gamma }}\right) +\underline{d}+r^L}\right\} ,\\ H_3^c&=\left\{ {x \in X}\mid {x_2-y_2^{\gamma } > x_1-y_1^{\gamma }+\underline{d}+r^L}\right\} ,\\ H_4^c&=\left\{ {x \in X}\mid {x_2-y_2^{\gamma } < x_1-y_1^{\gamma }-\underline{d}-r^L}\right\} . \end{aligned}$$

Thus, $d^{\gamma }\left( {x}\right) >\underline{d}+r^L$. Hence, Statement (6) is true. $\square$

Lemma 3

(1) For each $i \in \left\{ {\alpha ,\beta }\right\}$, $\min _{x \in C\left( {M;r^M}\right) }d^i\left( {x}\right) =\min _{x \in C\left( {L;r^M}\right) }d^i\left( {x}\right) =\min _{x \in A_*}d^{i}\left( {x}\right) =\max \left\{ {\overline{d}-r^M,0}\right\}$. (2) $\min _{x \in C\left( {M;r^M}\right) }d^{\gamma }\left( {x}\right) =\min _{x \in C\left( {L;r^M}\right) }d^{\gamma }\left( {x}\right) =\min _{x \in A_*}d^{\gamma }\left( {x}\right) =\max \left\{ {\underline{d}-r^M,0}\right\}$.

Proof

(1) First, we show that $\min _{x \in C\left( {M;r^M}\right) }d^i\left( {x}\right) =\max \left\{ {\overline{d}-r^M,0}\right\}$. Consider the case where $\overline{d} \le r^M$. Then, since $y^i \in C\left( {M;r^M}\right)$ by Proposition 1, $\min _{x \in C\left( {M;r^M}\right) }d^i\left( {x}\right) =d^i\left( {y^i}\right) =0=\max \left\{ {\overline{d}-r^M,0}\right\}$. Consider the case where $\overline{d}>r^M$. Suppose that for some $x \in C\left( {M;r^M}\right)$, $d^i\left( {x}\right) <\overline{d}-r^M$. Then, for some $x' \in M$, $d\left( {x,x'}\right) \le r^M$. Note that by Proposition 1, $d^i\left( {x'}\right) \ge \overline{d}$. Furthermore, note that by the triangle inequality, $d^i\left( {x'}\right) \le d^i\left( {x}\right) +d\left( {x,x'}\right)$. Thus, $\overline{d} \le d^i\left( {x'}\right) \le d^i\left( {x}\right) +d\left( {x,x'}\right) <\overline{d}$, which is a contradiction. Hence, for each $x \in C\left( {M;r^M}\right)$, $d^i\left( {x}\right) \ge \overline{d}-r^M$. Let $x \in M$. Then, since $d^i\left( {x}\right) =\overline{d}$ by Proposition 1, $d^i\left( {x}\right) >r^M$. Thus, for some $x' \in X$, $x' \in R\left( {x,y^i}\right) \cap \left\{ {x' \in X}\mid {d\left( {x,x'}\right) =r^M}\right\}$. Note that since $R\left( {x,y^i}\right) \cap \left\{ {x' \in X}\mid {d\left( {x,x'}\right) =r^M}\right\} \subset C\left( {M;r^M}\right)$, $x' \in C\left( {M;r^M}\right)$. Furthermore, note that since $d^i\left( {x}\right) =d^i\left( {x'}\right) +d\left( {x,x'}\right)$ by $x' \in R\left( {x,y^i}\right)$, $d^i\left( {x}\right) =\overline{d}$ by Proposition 1, and $d\left( {x,x'}\right) =r^M$ by $x' \in \left\{ {x' \in X}\mid {d\left( {x,x'}\right) =r^M}\right\}$, $d^i\left( {x'}\right) =d^i\left( {x}\right) -d\left( {x,x'}\right) =\overline{d}-r^M$. Thus, for some $x' \in C\left( {M;r^M}\right)$, $d^i\left( {x'}\right) =\overline{d}-r^M$. Hence, $\min _{x \in C\left( {M;r^M}\right) }d^i\left( {x}\right) =\overline{d}-r^M=\max \left\{ {\overline{d}-r^M,0}\right\}$. Next, we can show that $\min _{x \in C\left( {L;r^M}\right) }d^i\left( {x}\right) =\max \left\{ {\overline{d}-r^M,0}\right\}$ by replacing “Proposition 1” and “M” in the above proof with “Corollary 1” and “L.” Finally, we show that $\min _{x \in A_*}d^i\left( {x}\right) =\max \left\{ {\overline{d}-r^M,0}\right\}$. Note that $L \subset M$, $r^M \le r^L$, and $A_*=C\left( {M;r^M}\right) \cap C\left( {L;r^L}\right)$. Then, $C\left( {L;r^M}\right) \subset A_* \subset C\left( {M;r^M}\right)$. Thus, $\min _{x \in C\left( {M;r^M}\right) }d^i\left( {x}\right) \le \min _{x \in A_*}d^i\left( {x}\right) \le \min _{x \in C\left( {L;r^M}\right) }d^i\left( {x}\right)$. Hence, since $\min _{x \in C\left( {M;r^M}\right) }d^i\left( {x}\right) =\max \left\{ {\overline{d}-r^M,0}\right\}$ and $\min _{x \in C\left( {L;r^M}\right) }d^i\left( {x}\right) =\max \left\{ {\overline{d}-r^M,0}\right\}$, $\min _{x \in A_*}d^i\left( {x}\right) =\max \left\{ {\overline{d}-r^M,0}\right\}$.

(2) First, we show that $\min _{x \in C\left( {M;r^M}\right) }d^{\gamma }\left( {x}\right) =\max \left\{ {\underline{d}-r^M,0}\right\}$. Consider the case where $\underline{d} \le r^M$. Then, since $y^{\gamma } \in C\left( {M;r^M}\right)$ by Corollary 1, $\min _{x \in C\left( {M;r^M}\right) }d^{\gamma }\left( {x}\right) =d^{\gamma }\left( {y^{\gamma }}\right) =0 =\max \left\{ {\underline{d}-r^M,0}\right\}$. Consider the case where $\underline{d}>r^M$. Suppose that for some $x \in C\left( {M;r^M}\right)$, $d^{\gamma }\left( {x}\right) <\underline{d}-r^M$. Then, for some $x' \in M$, $d\left( {x,x'}\right) \le r^M$. Note that by Proposition 1, $d^{\gamma }\left( {x'}\right) \ge \underline{d}$. Furthermore, note that by the triangle inequality, $d^{\gamma }\left( {x'}\right) \le d^{\gamma }\left( {x}\right) +d\left( {x,x'}\right)$. Thus, $\underline{d} \le d^{\gamma }\left( {x'}\right) \le d^{\gamma }\left( {x}\right) +d\left( {x,x'}\right) <\underline{d}$, which is a contradiction. Hence, for each $x \in C\left( {M;r^M}\right)$, $d^{\gamma }\left( {x}\right) \ge \underline{d}-r^M$. Let $x \in L$. Then, since $d^{\gamma }\left( {x}\right) =\underline{d}$ by Corollary 1, $d^{\gamma }\left( {x}\right) >r^M$. Thus, for some $x' \in X$, $x' \in R\left( {x,y^{\gamma }}\right) \cap \left\{ {x' \in X}\mid {d\left( {x,x'}\right) =r^M}\right\}$. Note that since $R\left( {x,y^{\gamma }}\right) \cap \left\{ {x' \in X}\mid {d\left( {x,x'}\right) =r^M}\right\} \subset C\left( {M;r^M}\right)$, $x' \in C\left( {M;r^M}\right)$. Furthermore, note that since $d^{\gamma }\left( {x}\right) =d^{\gamma }\left( {x'}\right) +d\left( {x,x'}\right)$ by $x' \in R\left( {x,y^{\gamma }}\right)$, $d^{\gamma }\left( {x}\right) =\underline{d}$ by Corollary 1, and $d\left( {x,x'}\right) =r^M$ by $x' \in \left\{ {x' \in X}\mid {d\left( {x,x'}\right) =r^M}\right\}$, $d^{\gamma }\left( {x'}\right) =d^{\gamma }\left( {x}\right) -d\left( {x,x'}\right) =\underline{d}-r^M$. Thus, for some $x' \in C\left( {M;r^M}\right)$, $d^{\gamma }\left( {x'}\right) =\underline{d}-r^M$. Hence, $\min _{x \in C\left( {M;r^M}\right) }d^{\gamma }\left( {x}\right) =\underline{d}-r^M=\max \left\{ {\underline{d}-r^M,0}\right\}$. Next, we can show that $\min _{x \in C\left( {L;r^M}\right) }d^{\gamma }\left( {x}\right) =\max \left\{ {\underline{d}-r^M,0}\right\}$ by replacing “Proposition 1” and “M” in the above proof with “Corollary 1” and “L.” Finally, we show that $\min _{x \in A_*}d^{\gamma }\left( {x}\right) =\max \left\{ {\underline{d}-r^M,0}\right\}$. Note that $L \subset M$, $r^M \le r^L$, and $A_*=C\left( {M;r^M}\right) \cap C\left( {L;r^L}\right)$. Then, $C\left( {L;r^M}\right) \subset A_* \subset C\left( {M;r^M}\right)$. Thus, $\min _{x \in C\left( {M;r^M}\right) }d^{\gamma }\left( {x}\right) \le \min _{x \in A_*}d^{\gamma }\left( {x}\right) \le \min _{x \in C\left( {L;r^M}\right) }d^{\gamma }\left( {x}\right)$. Hence, since $\min _{x \in C\left( {M;r^M}\right) }d^{\gamma }\left( {x}\right) =\max \left\{ {\underline{d}-r^M,0}\right\}$ and $\min _{x \in C\left( {L;r^M}\right) }d^{\gamma }\left( {x}\right) =\max \left\{ {\underline{d}-r^M,0}\right\}$, $\min _{x \in A_*}d^{\gamma } \left( {x}\right) =\max \left\{ {\underline{d}-r^M,0}\right\}$. $\square$

1.2 Appendix 2: Proof of Theorem 1

Let $i \in I$. Then, for each $i' \in \left\{ {\alpha ,\beta }\right\}$,

$$\begin{aligned} u^{i'}\left( {x_*^i}\right)&=\bar{u} -d^{i'}\left( {x_*^i}\right) \quad \text { by definition}\\&\ge \bar{u} -\left( {\overline{d}+r^M}\right) \quad \text { by Lemma}\,2\\&= \delta \left\{ {\bar{u} -\left( {\overline{d}-r^M}\right) }\right\} \quad \text { by definition of}\ r^M\\&\ge \delta \left( {\bar{u} -\max \left\{ {\overline{d}-r^M,0}\right\} }\right) \quad \text { by}\ \overline{d}-r^M \le \max \left\{ {\overline{d}-r^M,0}\right\} \\&=\delta \left( {\bar{u} -d^{i'}\left( {x_*^{i'}}\right) }\right) \quad \text { by Lemma}\,3\\&= \delta u^{i'}\left( {x_*^{i'}}\right) \quad \text { by definition}. \end{aligned}$$

Furthermore

$$\begin{aligned} u^{\gamma }\left( {x_*^i}\right)&=\bar{u} -d^{\gamma }\left( {x_*^i}\right) \quad \text { by definition}\\&\ge \bar{u} -\left( {\underline{d}+r^L}\right) \quad \text { by Lemma}\,2\\&=\bar{u}-r^M-\left( {1-\delta }\right) \overline{d}-\delta \underline{d} -\delta \max \left\{ {r^M-\underline{d},0}\right\} \quad \text { by definition of}\ r^L\\&= \delta \left( {\bar{u} -\max \left\{ {\underline{d}-r^M,0}\right\} }\right) \quad \text { by definition of}\ r^M\\&=\delta \left( {\bar{u} -d^{\gamma }\left( {x_*^{\gamma }}\right) }\right) \quad \text { by Lemma}\,3\\&= \delta u^{\gamma }\left( {x_*^{\gamma }}\right) \quad \text { by definition}. \end{aligned}$$

Thus, for each $i' \in I$, $x_*^i \in A_*^{i'}$. Hence, in $\sigma$, player i’s proposal $x_*^i$ is accepted by all players. Therefore, in $\sigma$, player i’s utility in the subgame beginning with her proposal is $u^i\left( {x_*^i}\right)$.

Consider each proposing node of each player i. Let $x \in X$ such that $u^i\left( {x}\right) > u^i\left( {x_*^i}\right)$. Then, since $d^i\left( {x}\right) <d^i\left( {x_*^i}\right)$ and $x_*^i \in \arg \min _{x \in A_*}d^i\left( {x}\right)$, $x \not \in A_*$. Thus, since $A_*=C\left( {M;r^M}\right) \cap C\left( {L;r^L}\right)$, either $x \in X {\setminus } C\left( {M;r^M}\right)$ or $x \in C\left( {M;r^M}\right) {\setminus } C\left( {L;r^L}\right)$. Hence, by Lemma 2, either (1) for some $i' \in I$, $d^{i'}\left( {x}\right) >\overline{d}+r^M$ or (2) $d^{\gamma }\left( {x}\right) >\underline{d}+r^L$. Suppose that (1) for some $i' \in I$, $d^{i'}\left( {x}\right) >\overline{d}+r^M$. Note that since $d^i\left( {x_*^i}\right) >0$ by $d^i\left( {x_*^i}\right) >d^i\left( {x}\right) \ge 0$ and $d^{i}\left( {x_*^{i}}\right) \le \max \left\{ {\overline{d}-r^M,0}\right\}$ by Lemma 3 and $\underline{d} \le \overline{d}$, $\overline{d}>r^M$. Then, since $\overline{d}>r^M$ and $d^{i'}\left( {x_*^{i'}}\right) \le \max \left\{ {\overline{d}-r^M,0}\right\}$ by Lemma 3 and $\underline{d} \le \overline{d}$, $d^{i'}\left( {x_*^{i'}}\right) \le \overline{d}-r^M$. Thus,

$$\begin{aligned} u^{i'}\left( {x}\right)&= \bar{u} -d^{i'}\left( {x}\right) \quad \text { by definition}\\&< \bar{u} -\left( {\overline{d}+r^M}\right) \quad \text { by supposition}\\&=\delta \left\{ {\bar{u} -\left( {\overline{d}-r^M}\right) }\right\} \quad \text { by definition of}\ r^M\\&\le \delta \left( {\bar{u}-d^{i'}\left( {x_*^{i'}}\right) }\right) \quad \text { by}\ d^{i'}\left( {x_*^{i'}}\right) \le \overline{d}-r^M\\&=\delta u^{i'}\left( {x_*^{i'}}\right) \quad \text { by definition}. \end{aligned}$$

Hence, $x \not \in A_*^{i'}$. Suppose that (2) $d^{\gamma }\left( {x}\right) >\underline{d}+r^L$. Then,

$$\begin{aligned} u^{\gamma }\left( {x}\right)&= \bar{u} -d^{\gamma }\left( {x}\right) \quad \text { by definition}\\&< \bar{u} -\left( {\underline{d}+r^L}\right) \quad \text { by supposition}\\&=\bar{u}-r^M-\left( {1-\delta }\right) \overline{d}-\delta \underline{d} -\delta \max \left\{ {r^M-\underline{d},0}\right\} \quad \text { by definition of}\ r^L\\&= \delta \left( {\bar{u} -\max \left\{ {\underline{d}-r^M,0}\right\} }\right) \quad \text { by definition of}\ r^M\\&=\delta \left( {\bar{u} -d^{\gamma }\left( {x_*^{\gamma }}\right) }\right) \quad \text { by Lemma}\,3\\&= \delta u^{\gamma }\left( {x_*^{\gamma }}\right) \quad \text { by definition}. \end{aligned}$$

Thus, $x \not \in A_*^{\gamma }$. Hence, in $\sigma$, x is rejected by some player. Therefore, by a one-stage deviation of proposing x such that $u^i\left( {x}\right) > u^i\left( {x_*^i}\right)$, player i obtains $\delta u^i\left( {x_*^{i''}}\right) \le \max _{x \in A_*}u^i\left( {x}\right) = u^i\left( {x_*^i}\right)$ for some $i'' \in I$. By a one-stage deviation of proposing x such that $u^i\left( {x}\right) \le u^i\left( {x_*^i}\right)$, player i obtains $u^i\left( {x}\right) \le u^i\left( {x_*^i}\right)$ if x is accepted by all players, and $\delta u^i\left( {x_*^{i''}}\right) \le \max _{x \in A_*}u^i\left( {x}\right) = u^i\left( {x_*^i}\right)$ for some $i'' \in I$ if x is rejected by some player. Thus, in $\sigma$, player i’s proposal $x_*^i$ is optimal.

Consider each responding node of each player i following each proposal x. By rejecting x, player i obtains $\delta u^i\left( {x_*^i}\right)$. By accepting x, player i obtains $u^i\left( {x}\right) \ge \delta u^i\left( {x_*^i}\right)$ if $x \in A_*^i$, $u^i\left( {x}\right) \le \delta u^i\left( {x_*^i}\right)$ if $x \in \left( {\cap _{\iota \succ i} A_\dag ^\iota }\right) {\setminus } A_\dag ^i$, and $\delta u^i\left( {x_*^{i'}}\right) \le \delta \max _{x \in A_*}u^i\left( {x}\right) = \delta u^i\left( {x_*^i}\right)$ for some $i' \in I$ if $x \notin \cap _{\iota \succ i} A_\dag ^\iota$. Thus, in $\sigma$, player i’s response is optimal.

Hence, $\sigma$ is an SPE in G. Therefore, since $\sigma$ is stationary, $\sigma$ is an SSPE in G. $\square$

1.3 Appendix 3: Proof of Theorem 2

Let $\sigma$ be an SSPE in G. For each $i \in I$, let $x^i$ be player i’s proposal in $\sigma$. For each $i \in I$, let $\pi ^i$ be player i’s utility in the subgame beginning with her proposal in $\sigma$. For each $i \in I$, let $A^i$ be the set of locations that player i accepts in $\sigma$. For each $i \in I$, let $A_O^i := \left\{ {x \in X}\mid {u^i\left( {x}\right) > \delta \pi ^i}\right\}$ and $A_C^i := \left\{ {x \in X}\mid {u^i\left( {x}\right) \ge \delta \pi ^i}\right\}$. Let $A := \cap _{i \in I} A^i$, $A_O := \cap _{i \in I} A_O^i$, and $A_C := \cap _{i \in I} A_C^i$.

Lemma 4

$A_O \subset A \subset A_C$.

Proof

It suffices to show that for each $i \in I$, $\cap _{\iota \succsim i} A_O^\iota \subset \cap _{\iota \succsim i} A^\iota \subset \cap _{\iota \succsim i} A_C^\iota$. We show this by induction. (1) Let $i \in I$. Suppose that player i is the last responder. Let $x \in X$. Consider a response of player i to proposal x. If she accepts it, she obtains $u^i\left( {x}\right)$. If she rejects it, she obtains $\delta \pi ^i$. Since $\sigma$ is an SPE in G, $x \in A^i$ if $u^i\left( {x}\right) > \delta \pi ^i$, and $x \notin A^i$ if $u^i\left( {x}\right) < \delta \pi ^i$. Then, $A_O^i \subset A^i \subset A_C^i$. Thus, $\cap _{\iota \succsim i} A_O^\iota \subset \cap _{\iota \succsim i} A^\iota \subset \cap _{\iota \succsim i} A_C^\iota$. (2) Let $i,i' \in I$ and suppose that player i is the next responder of player $i'$. Suppose that $\cap _{\iota \succsim i} A_O^\iota \subset \cap _{\iota \succsim i} A^\iota \subset \cap _{\iota \succsim i} A_C^\iota$. Let $x \in \cap _{\iota \succsim i'} A_O^\iota$. Consider a response of player $i'$ to proposal x. If she rejects it, she obtains $\delta \pi ^{i'}$. Since $x \in \cap _{\iota \succsim i'} A_O^\iota \subset \cap _{\iota \succsim i} A_O^\iota \subset \cap _{\iota \succsim i} A^\iota$, if she accepts it, she obtains $u^{i'}\left( {x}\right)$. Note that since $x \in \cap _{\iota \succsim i'} A_O^\iota \subset A_O^{i'}$, $u^{i'}\left( {x}\right) > \delta \pi ^{i'}$. Then, since $\sigma$ is an SPE in G, $x \in A^{i'}$. Note that $x \in \cap _{\iota \succsim i'} A_O^\iota \subset \cap _{\iota \succsim i} A_O^\iota \subset \cap _{\iota \succsim i} A^\iota$. Then, $x \in \cap _{\iota \succsim i'} A^\iota$. Thus, $\cap _{\iota \succsim i'} A_O^\iota \subset \cap _{\iota \succsim i'} A^\iota$. Let $x \in \cap _{\iota \succsim i'} A^\iota$. Consider a response of player $i'$ to proposal x. If she rejects it, she obtains $\delta \pi ^{i'}$. Since $x \in \cap _{\iota \succsim i'} A^\iota \subset \cap _{\iota \succsim i} A^\iota$, if she accepts it, she obtains $u^{i'}\left( {x}\right)$. Note that $x \in \cap _{\iota \succsim i'} A^\iota \subset A^{i'}$. Then, since $\sigma$ is an SPE in G, $u^{i'}\left( {x}\right) \ge \delta \pi ^{i'}$, and thus, $x \in A_C^{i'}$. Note that $x \in \cap _{\iota \succsim i'} A^\iota \subset \cap _{\iota \succsim i} A^\iota \subset \cap _{\iota \succsim i} A_C^\iota$. Then, $x \in \cap _{\iota \succsim i'} A_C^\iota$. Thus, $\cap _{\iota \succsim i'} A^\iota \subset \cap _{\iota \succsim i'} A_C^\iota$. Hence, $\cap _{\iota \succsim i'} A_O^\iota \subset \cap _{\iota \succsim i'} A^\iota \subset \cap _{\iota \succsim i'} A_C^\iota$. $\square$

Lemma 5

For each $i \in I$, $x^i \in A$, $\pi ^i = u^i\left( {x^i}\right)$, and $x^i \in \arg \max _{x \in A} u^i\left( {x}\right)$.

Proof

Suppose that for each $i \in I$, $x^i \notin A$. Let $i' \in I$. Then, for each $i \in I$, since $\pi ^i = 0$, $u^{i}\left( {x^{i'}}\right) > 0 = \delta \pi ^{i}$. Thus, $x^{i'} \in A_O$. Hence, by Lemma 4, $x^{i'} \in A$, which is a contradiction. Suppose that for some $i \in I$, $x^i \notin A$. If $\pi ^i = 0$, since $x^{i'} \in A$ for some $i' \in I$, player i can improve her utility from 0 to $u^i\left( {x^{i'}}\right) > 0$ by proposing $x^{i'}$, which is a contradiction. If $\pi ^i > 0$, since $\pi ^i = \delta ^t u^i\left( {x}\right)$ for some $x \in A$ and $t \in {\mathbb {N}}$, player i can improve her utility from $\delta ^t u^i\left( {x}\right)$ to $u^i\left( {x}\right) >\delta ^t u^i\left( {x}\right)$ by proposing x, which is a contradiction. Thus, for each $i \in I$, $x^i \in A$. Hence, for each $i \in I$, $\pi ^i = u^i\left( {x^i}\right)$. Furthermore, since $\sigma$ is an SPE in G, for each $i \in I$, $x^i \in \arg \max _{x \in A} u^i\left( {x}\right)$. $\square$

Lemma 6

Suppose that $\max _{i \in I} d^i\left( {x^i}\right) =0$. Then, for each $i \in I$, $x^i \in \arg \min _{x \in A_*}d^i\left( {x}\right)$.

Proof

Let $i \in I$. Suppose that $y^i \not \in A_*$. Let $x \in L$. Note that since $y^i \not \in A_*$, $r^M<d^i\left( {x}\right)$. Furthermore, note that by Corollary 1, $d^i\left( {x}\right) \le \overline{d}$. Then, since $r^M=\frac{1-\delta }{1+\delta }\left( {\bar{u}-\overline{d}}\right)$ by the definition of $r^M$, $\left( {1 - \delta }\right) \bar{u} =\left( {1 +\delta }\right) r^M+\left( {1 - \delta }\right) \overline{d}<\left( {1 +\delta }\right) d^i\left( {x}\right) +\left( {1 - \delta }\right) \overline{d} \le 2 \overline{d}$. Note that since $\max _{i \in I} d^i\left( {x^i}\right) =0$, $x^{\alpha } = y^{\alpha }$ and $x^{\beta } = y^{\beta }$. Furthermore, note that by Proposition 1 and the definition, $d^{\beta }\left( {y^{\alpha }}\right) = 2 \overline{d}$ and $d^{\beta }\left( {y^{\beta }}\right) =0$. Then, since $u^{\beta }\left( {x^{\alpha }}\right) \ge \delta u^{\beta }\left( {x^{\beta }}\right)$ by Lemmas 4 and 5, $\left( {1 - \delta }\right) \bar{u} \ge d^{\beta }\left( {x^{\alpha }}\right) -\delta d^{\beta }\left( {x^{\beta }}\right) =d^{\beta }\left( {y^{\alpha }}\right) -\delta d^{\beta }\left( {y^{\beta }}\right) =2 \overline{d}$, which is a contradiction. Thus, $y^i \in A_*$. Hence, since $d^i\left( {y^i}\right) =0$ by the definition, $y^i \in \arg \min _{x \in A_*}d^i\left( {x}\right)$. Therefore, since $x^i = y^i$ by $\max _{i \in I} d^i\left( {x^i}\right) =0$, $x^i \in \arg \min _{x \in A_*}d^i\left( {x}\right)$.

Lemma 7

Suppose that $\max _{i \in I} d^i\left( {x^i}\right) >0$. Then, for each $i \in I$, $x^i \in \arg \min _{x \in A_*}d^i\left( {x}\right)$.

Proof

Let $\tilde{\alpha } \in I$ such that

$$\begin{aligned} \tilde{\alpha } \in \arg \max _{i \in I} d^i\left( {x^i}\right) . \end{aligned}$$

(1)

Suppose that $x^{\tilde{\alpha }} \in A_O$. Then, since $x^{\tilde{\alpha }} \not =y^{\tilde{\alpha }}$ and $u^i\left( {x}\right)$ is continuous in X for each $i \in I {\setminus } \left\{ {\tilde{\alpha }}\right\}$, there exists $x \in X$ such that $u^{\tilde{\alpha }}\left( {x}\right) >u^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right)$ and $x \in A_O^i$ for each $i \in I {\setminus } \left\{ {\tilde{\alpha }}\right\}$. Note that since $x \in A_O$ and $A_O \subset A$ by Lemma 4, $x \in A$. Then, there exists $x \in A$ such that $u^{\tilde{\alpha }}\left( {x}\right) >u^{\tilde{\alpha }} \left( {x^{\tilde{\alpha }}}\right)$. This contradicts Lemma 5. Thus, $x^{\tilde{\alpha }} \not \in A_O$. Note that by Lemmas 4 and 5, $x^{\tilde{\alpha }} \in A_C$. Furthermore, note that by Lemma 5, $x^{\tilde{\alpha }} \in A_O^{\tilde{\alpha }}$. Then, for some $i',i'' \in I{\setminus } \left\{ {\tilde{\alpha }}\right\}$ such that $i' \not =i''$, either (i) $x^{\tilde{\alpha }} \in A_O^{\tilde{\alpha }}$, $x^{\tilde{\alpha }} \in A_C^{i'} {\setminus } A_O^{i'}$, and $x^{\tilde{\alpha }} \in A_O^{i''}$, or (ii) $x^{\tilde{\alpha }} \in A_O^{\tilde{\alpha }}$, $x^{\tilde{\alpha }} \in A_C^{i'} {\setminus } A_O^{i'}$, and $x^{\tilde{\alpha }} \in A_C^{i''} {\setminus } A_O^{i''}$. Consider the case where (i) $x^{\tilde{\alpha }} \in A_O^{\tilde{\alpha }}$, $x^{\tilde{\alpha }} \in A_C^{i'} {\setminus } A_O^{i'}$, and $x^{\tilde{\alpha }} \in A_O^{i''}$. Suppose that $x^{\tilde{\alpha }} \not \in R\left( {y^{\tilde{\alpha }},y^{i'}}\right)$. Then, since $x^{\tilde{\alpha }} \not \in R\left( {y^{\tilde{\alpha }},y^{i'}}\right)$ and $u^{i''}\left( {x}\right)$ is continuous in X, there exists $x \in X$ such that $u^{\tilde{\alpha }}\left( {x}\right) >u^{\tilde{\alpha }} \left( {x^{\tilde{\alpha }}}\right)$, $u^{i'}\left( {x}\right) >u^{i'} \left( {x^{\tilde{\alpha }}}\right)$, and $x \in A_O^{i''}$. Note that since $x \in A_O$ and $A_O \subset A$ by Lemma 4, $x \in A$. Then, there exists $x \in A$ such that $u^{\tilde{\alpha }}\left( {x}\right) >u^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right)$. This contradicts Lemma 5. Thus, $x^{\tilde{\alpha }} \in R\left( {y^{\tilde{\alpha }},y^{i'}}\right)$. Consider the case where (ii) $x^{\tilde{\alpha }} \in A_O^{\tilde{\alpha }}$, $x^{\tilde{\alpha }} \in A_C^{i'} {\setminus } A_O^{i'}$, and $x^{\tilde{\alpha }} \in A_C^{i''} {\setminus } A_O^{i''}$. Suppose that $x^{\tilde{\alpha }} \not \in R\left( {y^{\tilde{\alpha }},y^{i'}}\right)$, $x^{\tilde{\alpha }} \not \in R\left( {y^{\tilde{\alpha }},y^{i''}}\right)$, and $x^{\tilde{\alpha }} \not \in R\left( {y^{i'},y^{i''}}\right)$. Then, since $x^{\tilde{\alpha }} \not \in R\left( {y^{\tilde{\alpha }},y^{i'}}\right) \cup R\left( {y^{\tilde{\alpha }},y^{i''}}\right) \cup R\left( {y^{i'},y^{i''}}\right)$, there exists $x \in X$ such that $u^{\tilde{\alpha }}\left( {x}\right) > u^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right)$, $u^{i'}\left( {x}\right) >u^{i'}\left( {x^{\tilde{\alpha }}}\right)$, and $u^{i''}\left( {x}\right) >u^{i''}\left( {x^{\tilde{\alpha }}}\right)$. Note that since $x \in A_O$ and $A_O \subset A$ by Lemma 4, $x \in A$. Then, there exists $x \in A$ such that $u^{\tilde{\alpha }}\left( {x}\right) >u^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right)$. This contradicts Lemma 5. Suppose that $x^{\tilde{\alpha }} \not \in R\left( {y^{\tilde{\alpha }},y^{i'}}\right)$, $x^{\tilde{\alpha }} \not \in R\left( {y^{\tilde{\alpha }},y^{i''}}\right)$, and $x^{\tilde{\alpha }} \in R\left( {y^{i'},y^{i''}}\right)$. Then,

$$\begin{aligned} u^{i'}\left( {x^{i''}}\right)&=\bar{u}-d^{i'}\left( {x^{i''}}\right) \quad \text {by definition}\\&\le \bar{u}-d\left( {y^{i'},y^{i''}}\right) +d^{i''}\left( {x^{i''}}\right) \quad \text {by triangle inequality}\\&< \bar{u}-d\left( {y^{i'},y^{i''}}\right) +d^{i''}\left( {x^{\tilde{\alpha }}}\right) \quad \text {by Lemma}\,5\ \text {and} \ x^{\tilde{\alpha }} \in A_C^{i''} {\setminus } A_O^{i''}\\&= \bar{u}-d^{i'}\left( {x^{\tilde{\alpha }}}\right) \quad \text {by}\ x^{\tilde{\alpha }} \in R\left( {y^{i'},y^{i''}}\right) \\&= u^{i'}\left( {x^{\tilde{\alpha }}}\right) \quad \text {by definition}\\&= \delta \pi ^{i'}\quad \text {by}\ x^{\tilde{\alpha }} \in A_C^{i'} {\setminus } A_O^{i'}. \end{aligned}$$

Thus, $x^{i''} \not \in A_C$. Hence, since $A \subset A_C$ by Lemma 4, $x^{i''} \not \in A$. This contradicts Lemma 5. Thus, either $x^{\tilde{\alpha }} \in R\left( {y^{\tilde{\alpha }},y^{i'}}\right)$ or $x^{\tilde{\alpha }} \in R\left( {y^{\tilde{\alpha }},y^{i''}}\right)$. Hence, for some $i \in I{\setminus } \left\{ {\tilde{\alpha }}\right\}$, $x^{\tilde{\alpha }} \in R\left( {y^{\tilde{\alpha }},y^i}\right)$ and $x^{\tilde{\alpha }} \in A_C^i {\setminus } A_O^i$. Note that $x^{\tilde{\alpha }} \in R\left( {y^{\tilde{\alpha }},y^i}\right)$ if and only if $d\left( {y^{\tilde{\alpha }},y^i}\right) =d^{\tilde{\alpha }} \left( {x^{\tilde{\alpha }}}\right) +d^i\left( {x^{\tilde{\alpha }}}\right)$. Furthermore, note that since $\pi ^i=u^i\left( {x^i}\right)$ by Lemma 5, $x^{\tilde{\alpha }} \in A_C^i {\setminus } A_O^i$ if and only if $u^i\left( {x^{\tilde{\alpha }}}\right) =\delta u^i\left( {x^i}\right)$. Therefore, for some $i \in I{\setminus } \left\{ {\tilde{\alpha }}\right\}$, $d\left( {y^{\tilde{\alpha }},y^i}\right) =d^{\tilde{\alpha }} \left( {x^{\tilde{\alpha }}}\right) +d^i\left( {x^{\tilde{\alpha }}}\right)$ and $u^{i}\left( {x^{\tilde{\alpha }}}\right) =\delta u^i\left( {x^i}\right)$.

Let $\tilde{\beta } \in I{\setminus }\left\{ {\tilde{\alpha }}\right\}$ such that

$$\begin{aligned} d\left( {y^{\tilde{\alpha }},y^{\tilde{\beta }}}\right)&=d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) +d^{\tilde{\beta }}\left( {x^{\tilde{\alpha }}}\right) , \end{aligned}$$

(2)

$$\begin{aligned} u^{\tilde{\beta }}\left( {x^{\tilde{\alpha }}}\right)&=\delta u^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) . \end{aligned}$$

(3)

Then,

$$\begin{aligned} d\left( {y^{\tilde{\alpha }},y^{\tilde{\beta }}}\right)&=d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) +d^{\tilde{\beta }}\left( {x^{\tilde{\alpha }}}\right) \quad \text {by Statement}\ (2)\\&=d^{\tilde{\alpha }}\left( {x^{\tilde{\beta }}}\right) +d^{\tilde{\beta }} \left( {x^{\tilde{\beta }}}\right) +u^{\tilde{\alpha }}\left( {x^{\tilde{\beta }}}\right) \\&\quad -u^{\tilde{\beta }}\left( {x^{\tilde{\alpha }}}\right) +d^{\tilde{\alpha }} \left( {x^{\tilde{\alpha }}}\right) -d^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) \quad \text {by definition}\\&\ge d^{\tilde{\alpha }}\left( {x^{\tilde{\beta }}}\right) +d^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) +\delta u^{\tilde{\alpha }} \left( {x^{\tilde{\alpha }}}\right) -\delta u^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) +\delta \left( {d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) -d^{\tilde{\beta }} \left( {x^{\tilde{\beta }}}\right) }\right) \\&\quad \text {by Lemmas}\,4, 5\ \text {and Statements} \ (1), (3)\\&= d^{\tilde{\alpha }}\left( {x^{\tilde{\beta }}}\right) +d^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) \quad \text {by definition}. \end{aligned}$$

Note that by the triangle inequality, $d\left( {y^{\tilde{\alpha }},y^{\tilde{\beta }}}\right) \le d^{\tilde{\alpha }}\left( {x^{\tilde{\beta }}}\right) +d^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right)$. Thus,

$$\begin{aligned} d\left( {y^{\tilde{\alpha }},y^{\tilde{\beta }}}\right) =d^{\tilde{\alpha }} \left( {x^{\tilde{\beta }}}\right) +d^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) . \end{aligned}$$

(4)

Furthermore

$$\begin{aligned} u^{\tilde{\alpha }}\left( {x^{\tilde{\beta }}}\right)&=u^{\tilde{\beta }}\left( {x^{\tilde{\alpha }}}\right) +d^{\tilde{\beta }} \left( {x^{\tilde{\beta }}}\right) -d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) \quad \text {by Statements} \ (2), (4)\\&=\delta u^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) +d^{\tilde{\beta }} \left( {x^{\tilde{\beta }}}\right) -d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) \quad \text {by Statement}\ (3)\\&=\delta u^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) +\left( {1-\delta }\right) \left( {d^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) -d^{\tilde{\alpha }} \left( {x^{\tilde{\alpha }}}\right) }\right) \quad \text {by definition}\\&\le \delta u^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) \quad \text {by Statement}\ (1). \end{aligned}$$

Note that by Lemmas 4 and 5, $u^{\tilde{\alpha }}\left( {x^{\tilde{\beta }}}\right) \ge \delta u^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right)$. Thus,

$$\begin{aligned} u^{\tilde{\alpha }}\left( {x^{\tilde{\beta }}}\right) =\delta u^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) . \end{aligned}$$

(5)

By Statements (2), (3), (4), and (5),

$$\begin{aligned} d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) =d^{\tilde{\beta }} \left( {x^{\tilde{\beta }}}\right) =\frac{d\left( {y^{\tilde{\alpha }}, y^{\tilde{\beta }}}\right) -\left( {1-\delta }\right) \bar{u}}{1+\delta }. \end{aligned}$$

(6)

Thus, by Statements (1) and (6),

$$\begin{aligned} \tilde{\beta } \in \arg \max _{i \in I} d^i\left( {x^i}\right) . \end{aligned}$$

(7)

Hence, for each $i,i' \in I$,

$$\begin{aligned} d\left( {y^i,y^{i'}}\right)&\le d^i\left( {x^i}\right) +d^{i'}\left( {x^i}\right) \quad \text {by triangle inequality}\\&= d^i\left( {x^i}\right) +\bar{u}-u^{i'}\left( {x^i}\right) \quad \text {by definition}\\&\le d^i\left( {x^i}\right) +\bar{u}-\delta u^{i'}\left( {x^{i'}}\right) \quad \text {by Lemmas}\,4, 5\\&=d^i\left( {x^i}\right) +\left( {1-\delta }\right) \bar{u}+\delta d^{i'}\left( {x^{i'}}\right) \quad \text {by definition}\\&\le d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) +\left( {1-\delta }\right) \bar{u}+\delta d^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) \quad \text {by Statements}\ (1), (7)\\&= d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) +\bar{u} -\delta u^{\tilde{\beta }}\left( {x^{\tilde{\beta }}}\right) \quad \text {by definition}\\&= d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) +\bar{u}-u^{\tilde{\beta }} \left( {x^{\tilde{\alpha }}}\right) \quad \text {by Statement} (3)\\&= d^{\tilde{\alpha }}\left( {x^{\tilde{\alpha }}}\right) +d^{\tilde{\beta }} \left( {x^{\tilde{\alpha }}}\right) \quad \text {by definition}\\&=d\left( {y^{\tilde{\alpha }},y^{\tilde{\beta }}}\right) \quad \text {by Statement}\ (2). \end{aligned}$$

Therefore

$$\begin{aligned} d\left( {y^{\tilde{\alpha }},y^{\tilde{\beta }}}\right) =\max _{i,i' \in I}d\left( {y^i,y^{i'}}\right) . \end{aligned}$$

(8)

Note that for each $i \in I$,

$$\begin{aligned} \bar{u}-\overline{d}-r^M&=\delta \left( {\bar{u}-\frac{2\overline{d} -\left( {1-\delta }\right) \bar{u}}{1+\delta }}\right) \quad \text {by definition of}\ r^M\\&=\delta \left( {\bar{u}-\frac{d\left( {y^{\tilde{\alpha }},y^{\tilde{\beta }}}\right) -\left( {1-\delta }\right) \bar{u}}{1+\delta }}\right) \quad \text {by Proposition}\,1 \ \text {and Statement} (8)\\&\le \delta \left( {\bar{u}-d^{i}\left( {x^{i}}\right) }\right) \quad \text {by Statements} (1), (6), (7)\\&= \delta u^i\left( {x^i}\right) \quad \text {by definition}\\&=\delta \pi ^i\quad \text {by Lemma}\,5, \end{aligned}$$

where equality holds if $i \in \left\{ {\tilde{\alpha },\tilde{\beta }}\right\}$. Suppose that $x \not \in C\left( {M;r^M}\right)$, that is, $x \in X {\setminus } C\left( {M;r^M}\right)$. Then, for some $i \in I$, $d^i\left( {x}\right) >\overline{d}+r^M$ by Lemma 2, and thus, $u^i\left( {x}\right) =\bar{u}-d^i\left( {x}\right) <\bar{u}-\overline{d}-r^M \le \delta \pi ^i$, which implies that $x \not \in A_C$. Thus,

$$\begin{aligned} A_C \subset C\left( {M;r^M}\right) . \end{aligned}$$

(9)

Suppose that $x \in O\left( {M;r^M}\right)$. Then, for each $i \in \left\{ {\tilde{\alpha },\tilde{\beta }}\right\}$, $d^i\left( {x}\right) <\overline{d}+r^M$ by Lemma 2, and thus, $u^i\left( {x}\right) =\bar{u}-d^i\left( {x}\right) >\bar{u}-\overline{d}-r^M=\delta \pi ^i$, which implies that $x \in A_O^i$. Thus, for each $i \in \left\{ {\tilde{\alpha },\tilde{\beta }}\right\}$,

$$\begin{aligned} O\left( {M;r^M}\right) \subset A_O^i. \end{aligned}$$

(10)

Let $\tilde{\gamma } \in I {\setminus } \left\{ {\tilde{\alpha },\tilde{\beta }}\right\}$. Note that by definition and Statement (8), $\left\{ {\tilde{\alpha }, \tilde{\beta }, \tilde{\gamma }}\right\} = I$ and $d\left( {y^{\tilde{\alpha }},y^{\tilde{\beta }}}\right) =\max _{i,i' \in I}d\left( {y^i,y^{i'}}\right)$. Suppose that $d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) <\underline{d} -r^M$. Then, for each $x \in M$, since $\underline{d} \le d^{\tilde{\gamma }}\left( {x}\right)$ by Proposition 1 and $d^{\tilde{\gamma }}\left( {x}\right) \le d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) +d\left( {x^{\tilde{\gamma }},x}\right)$ by the triangle inequality, $r^M < \underline{d}-d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) \le d^{\tilde{\gamma }}\left( {x}\right) -d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) \le d\left( {x^{\tilde{\gamma }},x}\right)$, and thus, $x^{\tilde{\gamma }} \not \in C\left( {M;r^M}\right)$. However, since $x^{\tilde{\gamma }} \in A$ by Lemma 5, $A \subset A_C$ by Lemma 4, and $A_C \subset C\left( {M;r^M}\right)$ by Statement (9), $x^{\tilde{\gamma }} \in C\left( {M;r^M}\right)$, which is a contradiction. Thus, $d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) \ge \underline{d} -r^M$. Hence, since $d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) \ge 0$, $d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) \ge \max \left\{ {\underline{d}-r^M,0}\right\}$. Suppose that $d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) > \max \left\{ {\underline{d}-r^M,0}\right\}$. Note that by Lemma 3, $\min _{x \in C\left( {M;r^M}\right) }d^{\tilde{\gamma }}\left( {x}\right) =\max \left\{ {\underline{d}-r^M,0}\right\}$. Then, $\min _{x \in C\left( {M;r^M}\right) }d^{\tilde{\gamma }}\left( {x}\right) <d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right)$. Thus, since $d^{\tilde{\gamma }}\left( {x}\right)$ is continuous in X, there exists $x \in O\left( {M;r^M}\right)$ such that $d^{\tilde{\gamma }}\left( {x}\right) <d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right)$. Note that for each $i \in \left\{ {\tilde{\alpha },\tilde{\beta }}\right\}$, since $O\left( {M;r^M}\right) \subset A_O^i$ by Statement (10), $x \in A_O^i$. Furthermore, note that since $u^{\tilde{\gamma }}\left( {x}\right)>u^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) >\delta u^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) =\delta \pi ^{\tilde{\gamma }}$ by Lemma 5, $x \in A_O^{\tilde{\gamma }}$. Then, $x \in A_O$. Thus, since $A_O \subset A$ by Lemma 4, $x \in A$. Hence, there exists $x \in A$ such that $d^{\tilde{\gamma }}\left( {x}\right) <d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right)$. This contradicts Lemma 5. Therefore

$$\begin{aligned} d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) = \max \left\{ {\underline{d}-r^M,0}\right\} . \end{aligned}$$

(11)

Note that

$$\begin{aligned} \bar{u}-\underline{d}-r^L&=\bar{u}-\underline{d}-r^M -\left( {1-\delta }\right) \left( {\overline{d}-\underline{d}}\right) -\delta \max \left\{ {r^M-\underline{d},0}\right\} \quad \text {by definition of}\ r^L\\&=\delta \left( {\bar{u}-\max \left\{ {\underline{d}-r^M,0}\right\} }\right) \quad \text {by definition of}\ r^M\\&=\delta \left( {\bar{u}-d^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) }\right) \quad \text {by Statement}\ (11)\\&=\delta u^{\tilde{\gamma }}\left( {x^{\tilde{\gamma }}}\right) \quad \text {by definition}\\&=\delta \pi ^{\tilde{\gamma }}\quad \text {by Lemma}\,5. \end{aligned}$$

Suppose that $x \not \in X {\setminus } \left( {C\left( {M;r^M}\right) {\setminus } C\left( {L;r^L}\right) }\right)$, that is, $x \in C\left( {M;r^M}\right) {\setminus } C\left( {L;r^L}\right)$. Then, $d^{\tilde{\gamma }}\left( {x}\right) >\underline{d}+r^L$ by Lemma 2, and thus, $u^{\tilde{\gamma }}\left( {x}\right) =\bar{u}-d^{\tilde{\gamma }}\left( {x}\right) <\bar{u}-\underline{d}-r^L=\delta \pi ^{\tilde{\gamma }}$, which implies that $x \not \in A_C$. Thus,

$$\begin{aligned} A_C \subset X {\setminus } \left( {C\left( {M;r^M}\right) {\setminus } C\left( {L;r^L}\right) }\right) . \end{aligned}$$

(12)

Suppose that $x \in O\left( {L;r^L}\right)$. Then, $d^{\tilde{\gamma }}\left( {x}\right) <\underline{d}+r^L$ by Lemma 2, and thus, $u^{\tilde{\gamma }} \left( {x}\right) =\bar{u}-d^{\tilde{\gamma }}\left( {x}\right) >\bar{u} -\underline{d}-r^L=\delta \pi ^{\tilde{\gamma }}$, which implies that $x \in A_O^{\tilde{\gamma }}$. Thus,

$$\begin{aligned} O\left( {L;r^L}\right) \subset A_O^{\tilde{\gamma }}. \end{aligned}$$

(13)

Let $i \in I$. Suppose that $x^i \not \in \arg \min _{x \in A_*} d^i\left( {x}\right)$. Note that since $A_*=C\left( {M;r^M}\right) \cap C\left( {L;r^L}\right)$ by the definition of $A_*$, $x^i \not \in \arg \min _{x \in C\left( {M;r^M}\right) \cap C\left( {L;r^L}\right) } d^i\left( {x}\right)$. Furthermore, note that since $x^i \in A$ by Lemma 5, $A \subset A_C$ by Lemma 4, and $A_C \subset C\left( {M;r^M}\right) \cap C\left( {L;r^L}\right)$ by Statements (9) and (12), $x^i \in C\left( {M;r^M}\right) \cap C\left( {L;r^L}\right)$. Then, $\min _{x \in C\left( {M;r^M}\right) \cap C\left( {L;r^L}\right) } d^i\left( {x}\right) <d^i\left( {x^i}\right)$. Thus, since $d^i\left( {x}\right)$ is continuous in X, there exists $x \in O\left( {M;r^M}\right) \cap O\left( {L;r^L}\right)$ such that $d^i\left( {x}\right) <d^i\left( {x^i}\right)$. Note that since $O\left( {M;r^M}\right) \cap O\left( {L;r^L}\right) \subset A_O$ by Statements (10) and (13), and $A_O \subset A$ by Lemma 4, $x \in A$. Then, there exists $x \in A$ such that $d^i\left( {x}\right) <d^i\left( {x^i}\right)$. This contradicts Lemma 5. Thus, $x^i \in \arg \min _{x \in A_*} d^i\left( {x}\right)$. $\square$

The statement of Theorem 2 follows from Lemmas 5, 6, and 7. $\square$