Coalition-proofness in a class of games with strategic substitutes

Abstract

We examine the coalition-proofness and Pareto properties of Nash equilibria in pure strategy $\sigma$-interactive games with strategic substitutes and increasing/decreasing externalities. For this class of games: (i) we prove the equivalence among the set of Nash equilibria, the set of coalition-proof Nash equilibria under strong Pareto dominance and the set of Nash equilibria that are not strongly Pareto dominated by other Nash equilibria; (ii) we prove that the fixpoints of some “ extremal” selections from the joint best reply correspondence are both coalition-proof Nash equilibria under weak Pareto dominance and not weakly Pareto dominated by other Nash equilibria. We also provide an order-theoretic characterization of the set of Nash equilibria and show various applications of our results.

Appendix

Fact 1

Let $\Gamma =(M,(S_{i})_{i\in M},(u_{i})_{i\in M})$ be a real $\sigma$ -interactive game with strategic substitutes. We can define a game

$$\begin{aligned} \Gamma ^{\leftrightarrows }=(M,(S_{i}^{\leftrightarrows })_{i\in M},(u_{i}^{\leftrightarrows })_{i\in M}) \end{aligned}$$

such that $S_{i}^{\leftrightarrows }=-S_{i}$ and $u_{i}^{\leftrightarrows }:S_{M}^{\leftrightarrows }\rightarrow \mathbb {R}$ , $u_{i}^{\leftrightarrows }:s\mapsto u_{i}\left( -s\right)$ , for all $i\in M$.²¹ Besides we can define the family

$$\begin{aligned} \sigma ^{\leftrightarrows }=\{\sigma _{i}^{\leftrightarrows }\}_{i\in M} \end{aligned}$$

such that $\sigma _{i}^{\leftrightarrows }:S_{M}^{\leftrightarrows }\rightarrow \mathbb {R}$, $\sigma _{i}^{\leftrightarrows }:s\mapsto -\sigma _{i}\left( -s\right)$ for all $i\in M$ . Indeed, also $\Gamma ^{\leftrightarrows }$ is a real $\sigma ^{\leftrightarrows }$ -interactive game with strategic substitutes.

Proof

Since $\Gamma$ is a real $\sigma$-interactive game with strategic substitutes, there exists $\upsilon _{i}:S_{i}\times \sigma _{i}\left[ S_{M} \right] \rightarrow \mathbb {R}$ such that $u_{i}\left( s\right) =\upsilon _{i}\left( s_{i},\sigma _{i}\left( s\right) \right)$ at all $s\in S_{M}$, for all $i\in M$. Letting $\upsilon _{i}^{\leftrightarrows }:S_{i}^{\leftrightarrows }\times \sigma _{i}^{\leftrightarrows }\left[ S_{M}^{\leftrightarrows }\right] \rightarrow \mathbb {R}$, $\upsilon _{i}^{\leftrightarrows }:\left( x,y\right) \mapsto \upsilon _{i}\left( -x,-y\right)$ for all $i\in M$, it can be easily verified that $\Gamma ^{\leftrightarrows }$ is a real $\sigma ^{\leftrightarrows }$-interactive game with strategic substitutes. $\square$

Fact 2

Let $\Gamma =(M,(S_{i})_{i\in M},(u_{i})_{i\in M})$ be a real $\sigma$ -interactive game with strategic substitutes and increasing (resp. decreasing) externalities, and define $\Gamma ^{\leftrightarrows }$ and $\sigma ^{\leftrightarrows }$ as in Fact 1. Then $\Gamma ^{\leftrightarrows }$ is a real $\sigma ^{\leftrightarrows }$ -interactive game with strategic substitutes and decreasing (resp. increasing) externalities.

Proof

Define $\upsilon _{i}^{\leftrightarrows }$ for all $i\in M$ as in the proof of Fact 1. Then Fact 2 is an immediate consequence of Fact 1 and of the decreasingness (resp. increasingness) of each $\upsilon _{i}^{\leftrightarrows }$ in the second argument. $\square$

Proof of Observation I

Consider the game $\Gamma$ with $M=\left\{ 1,2,3\right\}$, $S_{1}=S_{2}= \left[ 0,1\right]$, $S_{3}=\left\{ 0\right\}$, $u_{i}:s\mapsto 0$, $u_{2}:s\mapsto 0$ and $u_{3}:s\mapsto 1_{A}\left( s\right)$ (i.e., $u_{3}$ is the indicator function $1_{A}$ of $A\subset S_{M}$) where

$$\begin{aligned} B=\left\{ s\in S_{M}:\max \left\{ s_{1},s_{2}\right\} \ne 1\right\} \cup \left\{ \left( 1,0,0\right) \right\} \text { and }A=S_{M}\backslash B . \end{aligned}$$

It is immediate that the game is a real $\sigma$-interactive game with strategic substitutes and increasing externalities for the interaction system $\sigma$ such that $\sigma _{i}:s\mapsto u_{i}\left( s\right)$ for all $i\in M$ (just put $\upsilon _{i}:\left( s_{i},t\right) \mapsto t$ for all $i\in M$). On the other hand, it is also quite simple to notice that there cannot exist a continuous function $\varsigma _{3}:S_{M\backslash \left\{ 3\right\} }\rightarrow \mathbb {R}$ such that $u_{3}\left( s\right) = \widehat{u}_{3}\left( s_{3},\varsigma _{3}\left( s_{-3}\right) \right)$ for some $\widehat{u}_{3}$: by way of contradiction suppose on the contrary that such $\varsigma _{3}$ exists; notice that $\varsigma _{3}\left( 1,0\right) \ne \varsigma _{3}\left( 0,1\right)$ (as $\widehat{u}_{3}\left( 0,\varsigma _{3}\left( 1,0\right) \right) \ne \widehat{u}_{3}\left( 0,\varsigma _{3}\left( 0,1\right) \right)$); infer that, by the continuity of $\varsigma _{3}$, there must exist a point $z^{*}\in \left( \left( 0,1\right) \times \left\{ 1\right\} \right) \cup \left\{ \left( 1,1\right) \right\} \cup \left( \left\{ 1\right\} \times \left( 0,1\right) \right)$ such that

$$\begin{aligned} \min \left\{ \varsigma _{3}\left( 1,0\right) ,\varsigma _{3}\left( 0,1\right) \right\} <\varsigma _{3}\left( z^{*}\right) <\max \left\{ \varsigma _{3}\left( 1,0\right) ,\varsigma _{3}\left( 0,1\right) \right\} \end{aligned}$$

and a point $z^{**}\in \{z\in \left( 0,1\right) \times \left( 0,1\right) :z_{1}+z_{2}=1\}$ such that

$$\begin{aligned} \varsigma _{3}\left( z^{*}\right) =\varsigma _{3}\left( z^{**}\right) ; \end{aligned}$$

finally, conclude that we obtained the following impossible equalities

$$\begin{aligned} 1=1_{A}\left( z_{1}^{*},z_{2}^{*},0\right) =\widehat{u}_{3}\left( 0,\varsigma _{3}\left( z^{*}\right) \right) =\widehat{u}_{3}\left( 0,\varsigma _{3}\left( z^{**}\right) \right) =1_{A}\left( z_{1}^{**},z_{2}^{**},0\right) =0\text {.} \end{aligned}$$

This completes the proof (for the case of increasing externalities, clearly Fact 2 guarantees that we can construct an analogous example with decreasing externalities). $\square$

Proof of Observation II

Proof of the if part. Suppose $\Gamma$ has a compatible interaction system $\sigma$ where interaction functions are real-valued and continuous. Let $g:S_{M}\rightarrow \mathbb {R}$, $g:s\mapsto 0$. For all $i\in M$, take an arbitrary $\overline{s}_{i}\in S_{i}$ and let:

• $\varsigma _{i}:S_{M\backslash \left\{ i\right\} }\rightarrow X_{-i}:=\sigma _{i}\left[ S_{M}\right]$, $\varsigma _{i}:s_{-i}\mapsto \sigma _{i}\left( \overline{s}_{i},s_{-i}\right)$;

• $V_{i}:S_{M\backslash \left\{ i\right\} }\rightarrow \mathbb {R}, V_{i}:s_{-i}\mapsto 0$;

• $F_{i}:S_{i}\times \mathbb {R}\rightarrow \mathbb {R}$, $F_{i}:\left( s_{i},x\right) \mapsto 0$.

Finally, for all $i\in M$, let $\widehat{u}_{i}:S_{i}\times X_{-i}\rightarrow \mathbb {R}$ be the function defined by $\widehat{u} _{i}\left( s_{i},\varsigma _{i}\left( s_{-i}\right) \right) =\upsilon _{i}\left( s_{i},\sigma _{i}\left( s\right) \right)$ at all $s\in S_{M}$ and conclude that $\Gamma$ is a generalized quasi-aggregative game with aggregator $g$. Clearly, since each interaction function $\sigma _{i}$ is continuous, also each J-interaction function $\varsigma _{i}$ is continuous.

Proof of the only if part. Suppose $\Gamma$ is a generalized quasi-aggregative game and let $\sigma _{i}:S_{M}\rightarrow I_{i}:=\mathbb {R }$, $\sigma _{i}:s\mapsto \varsigma _{i}\left( s_{-i}\right)$ for all $i\in M$. Let $\upsilon _{i}:S_{i}\times \sigma _{i}\left[ S_{M}\right] \rightarrow \mathbb {R}$ be the function defined by $\upsilon _{i}\left( s_{i},\sigma _{i}\left( s\right) \right) =\widehat{u}_{i}\left( s_{i},\varsigma _{i}\left( s_{-i}\right) \right)$ at all $s\in S_{M}$, for all $i\in M$. Let $\sigma :=\left\{ \sigma _{i}\right\} _{i\in M}$ and conclude that $\sigma$ is an interaction system which is compatible with $\Gamma$ and that each interaction function $\sigma _{i}$ is real-valued and continuous. The continuity of each $\sigma _{i}$ can be easily verified by the reader considering that $\sigma _{i}$ is, by construction, constant in $s_{i}$ (and not just merely continuous in $s_{i}$) and continuous in $s_{-i}$. $\square$

Proof of Observation III

In fact, the same proof of Observation II (without involving continuity arguments). $\square$

Proof of Observation IV

It is left to the reader to notice that, constructing again each $\sigma _{i}$ as in the proof of the only if part of Observation II, the proof is immediate. $\square$

Proof of Observation V

It is left to the reader to notice that, constructing again each $\varsigma _{i}$ as in the proof of the if part of Observation II, the proof is immediate. $\square$

Proof of Observation VI

For example, construct the following game on network. Put $M=\left\{ 1,2,3\right\}$, $N_{1}=\left\{ 2\right\}$, $N_{2}=\left\{ 3\right\}$, $N_{3}=\left\{ 1\right\}$, $\sigma _{1}:s\mapsto s_{2}$, $\sigma _{2}:s\mapsto s_{3}$, $\sigma _{3}:s\mapsto s_{1}$ and, for all $i\in M$, $S_{i}=\left[ 0,1\right]$, $r_{i}:\left( s_{i},x\right) \mapsto 2\sqrt{ s_{i}+x}$ and $c_{i}:x\mapsto x$. By way of contradiction, suppose there exists an upper semicontinuous function $P:S_{M}\rightarrow \mathbb {R}$ such that

$$\begin{aligned} b_{i}\left( s\right) \supseteq \underset{z\in S_{i}}{\arg \max }P\left( z,s_{-i}\right) \text { at all }s\in S_{M}\text {, for all }i\in M\text {.} \end{aligned}$$

Then, since best-replies are single-valued and since $P$ is an upper semicontinuous function on a compact set, we must have that

$$\begin{aligned} b_{i}\left( s\right) =\underset{z\in S_{i}}{\arg \max }P\left( z,s_{-i}\right) \text { at all }s\in S_{M}\text {, for all }i\in M\text {.} \end{aligned}$$

Note that, for all $i\in M$, $b_{i}\left( s\right) =\left\{ 0\right\}$ if $\sigma _{i}\left( s\right) =1$ and $b_{i}\left( s\right) =\left\{ 1\right\}$ if $\sigma _{i}\left( s\right) =0$. Therefore:

• $P\left( 1,0,0\right) <P\left( 1,1,0\right)$ as $b_{2}\left( 1,x,0\right) =\left\{ 1\right\}$;

• $P\left( 1,1,0\right) <P\left( 0,1,0\right)$ as $b_{1}\left( x,1,0\right) =\left\{ 0\right\}$;

• $P\left( 0,1,0\right) <P\left( 0,1,1\right)$ as $b_{3}\left( 0,1,x\right) =\left\{ 1\right\}$;

• $P\left( 0,1,1\right) <P\left( 0,0,1\right)$ as $b_{2}\left( 0,x,1\right) =\left\{ 0\right\}$;

• $P\left( 0,0,1\right) <P\left( 1,0,1\right)$ as $b_{1}\left( x,0,1\right) =\left\{ 1\right\}$;

• $P\left( 1,0,1\right) <P\left( 1,0,0\right)$ as $b_{3}\left( 1,0,x\right) =\left\{ 0\right\}$.

But this is impossible because we obtain $P\left( 1,0,0\right) <P\left( 1,0,0\right)$. $\square$

Proof of Claim 1

A consequence of Proposition 1 below and of Fact 2. $\square$

Proposition 1

The following statements are true:

• (i) there exists a real $\Sigma$-interactive game with strategic substitutes and decreasing externalities where $E_{wCPN}^{\Gamma }=wF_{N}^{\Gamma }\subset E_{N}^{\Gamma }$;

• (ii) there exists a real $\Sigma$-interactive game with strategic substitutes and decreasing externalities where $\underline{E}_{N}^{\Gamma }\subset E_{wCPN}^{\Gamma }=wF_{N}^{\Gamma }$;

• (iii) there exists a real $\Sigma$-interactive game with strategic substitutes and decreasing externalities where $wF_{N}^{\Gamma }\backslash E_{wCPN}^{\Gamma }\ne \emptyset$;

• (iv) there exists a real $\Sigma$-interactive game with strategic substitutes and decreasing externalities where $E_{wCPN}^{\Gamma }\backslash wF_{N}^{\Gamma }\ne \emptyset$.

Proof

• (i) See Example 1 in Sect. 4.

• (ii) See Example 2 in Sect. 4.

• (iii) See Example 3 in Sect. 4.

• (iv) Consider the game $\Gamma \$with $M=\left\{ 1,2,3\right\}$, $S_{1}=S_{2}=\left[ 0,1\right]$, $S_{3}=\left[ 0,1\right] \cup \left[ 7,8 \right]$,

$$\begin{aligned} u_{i}\left( s\right) =\left\{ \begin{array}{l@{\quad }l} s_{i}\left( 1-{\Sigma }_{i}\left( s\right) \right) -1 &{} \text {if }{\Sigma } _{i}\left( s\right) \ge \frac{3}{2} \\ s_{i}\left( 1-{\Sigma }_{i}\left( s\right) \right) &{} \text {if }{\Sigma } _{i}\left( s\right) <\frac{3}{2} \end{array} \right. \end{aligned}$$

for $i\in \left\{ 1,2\right\}$ and

$$\begin{aligned} u_{3}\left( s\right) =\left\{ \begin{array}{l@{\quad }l} -1 &{} \text {if }s_{3}=0\text { and }{\Sigma }_{3}\left( s\right) >1 \\ -\frac{1}{2} &{} \text {if }s_{3}=0\text { and }{\Sigma }_{3}\left( s\right) =1\\ 0 &{} \text {if }s_{3}=0\text { and }{\Sigma }_{3}\left( s\right) <1 \\ s_{3}\left( 1-{\Sigma }_{3}\left( s\right) \right) -1 &{} \text {if }s_{3}\in \left( 0,1\right) \cup \left\{ 8\right\} \text { and }{\Sigma }_{3}\left( s\right) >1 \\ s_{3}\left( 1-{\Sigma }_{3}\left( s\right) \right) &{} \text {if }s_{3}\in \left( 0,1\right) \cup \left\{ 8\right\} \text { and }{\Sigma }_{3}\left( s\right) \le 1 \\ -9 &{} \text {if }s_{3}\in \left\{ 1\right\} \cup \left[ 7,8\right) . \end{array} \right. \end{aligned}$$

Define $Q_{1}:=\left\{ \left( \frac{1}{2},\frac{1}{2},\frac{1}{2}\right) \right\}$, $Q_{2}:=\left\{ \left( 0,0,8\right) \right\}$, $Q_{3}:=\left\{ \left( 1,0,x\right) :0<x<1\right\}$, $Q_{4}:=\left\{ \left( 0,1,x\right) :0<x<1\right\}$, $Q_{5}:=\left\{ \left( x,1,0\right) :0<x<1\right\}$ and $Q_{6}:=\left\{ \left( 1,x,0\right) :0<x\le 1\right\}$. Noting that

$$\begin{aligned} b_{i}\left( s\right) =\left\{ \begin{array}{l@{\quad }l} \left\{ 0\right\} &{} \text {if }{\Sigma }_{i}\left( s\right) >1 \\ \left[ 0,1\right] &{} \text {if }{\Sigma }_{i}\left( s\right) =1 \\ \left\{ 1\right\} &{} \text {if }{\Sigma }_{i}\left( s\right) <1 \end{array} \right. \end{aligned}$$

for $i\in \left\{ 1,2\right\}$ and

$$\begin{aligned} b_{3}\left( s\right) =\left\{ \begin{array}{l@{\quad }l} \left\{ 0\right\} &{} \text {if }{\Sigma }_{3}\left( s\right) >1 \\ \left( 0,1\right) \cup \left\{ 8\right\} &{} \text {if }{\Sigma }_{3}\left( s\right) =1 \\ \left\{ 8\right\} &{} \text {if }{\Sigma }_{3}\left( s\right) <1\text {,} \end{array} \right. \end{aligned}$$

conclude that $\Gamma$ is a real $\Sigma$-interactive game with strategic substitutes and decreasing externalities. Note that $E_{STN}^{\Gamma }=Q_{2}\subset {\bigcup }_{i=1}^{6}Q_{i}=E_{N}^{\Gamma }$; thus $E_{wCPN}^{\Gamma }\supseteq Q_{2}\subseteq wF_{N}^{\Gamma }$ by Theorem 1. Note that

• $\left( u_{i}\left( s\right) \right) _{i\in M}=\left( 0,0,0\right) \$ if$\ \ s\in Q_{1}$,

• $\left( u_{i}\left( s\right) \right) _{i\in M}=\left( -1,-1,8\right) \$if$\ \ s\in Q_{2}$,

• $\left( u_{i}\left( s\right) \right) _{i\in M}=\left( 1-s_{3},0,0\right)$ if $s\in Q_{3}$ and $s_{3}\in \left( 0,\frac{1}{2} \right)$

• $\left( u_{i}\left( s\right) \right) _{i\in M}=\left( 1-s_{3},-1,0\right)$ if $s\in Q_{3}$ and $s_{3}\in \left[ \frac{1}{2} ,1\right)$,

• $\left( u_{i}\left( s\right) \right) _{i\in M}=\left( 0,1-s_{3},0\right)$ if $s\in Q_{4}$ and $s_{3}\in \left( 0,\frac{1}{2} \right)$,

• $\left( u_{i}\left( s\right) \right) _{i\in M}=\left( -1,1-s_{3},0\right)$ if $s\in Q_{4}$ and $s_{3}\in \left[ \frac{1}{2} ,1\right)$,

• $\left( u_{i}\left( s\right) \right) _{i\in M}=\left( 0,1-s_{1},-1\right) \ \$if $\ s\in Q_{5}$ and $s_{1}\in \left( 0,1\right)$,

• $\left( u_{i}\left( s\right) \right) _{i\in M}=\left( 1-s_{2},0,-1\right) \ \$if $\ s\in Q_{6}$ and $s_{2}\in \left( 0,1\right]$,

and conclude that $wF_{N}^{\Gamma }=Q_{2}$. Note that

• $s^{*}\in Q_{2}\cup Q_{5}\cup Q_{6}$ and $s^{**}\in Q_{1}$ implies that $s^{*}$ does not weakly Pareto dominate $s^{**}$ in $\Gamma$,

• if $s\in Q_{1}$ then $s$ is w-self-enforcing for $\Gamma$ as, for all $i\in M$, $s_{-i}$ is not weakly Pareto dominated in $\Gamma |_{s_{i}}$ by any other Nash equilibrium for $\Gamma |_{s_{i}}$,

• every strategy $s\in Q_{3}$ is not w-self-enforcing for $\Gamma$ as $s_{-2}$ is weakly Pareto dominated in $\Gamma |_{s_{2}}$ by $\left( s_{1}, \frac{1}{2}s_{3}\right) \in E_{N}^{\Gamma |_{s_{2}}}$,

• and every strategy $s\in Q_{4}$ is not w-self-enforcing for $\Gamma$ as $s_{-1}$ is weakly Pareto dominated in $\Gamma |_{s_{1}}$ by $\left( s_{2},\frac{1}{2}s_{3}\right) \in E_{N}^{\Gamma |_{s_{1}}}$,

and conclude that $Q_{1}\subseteq E_{wCPN}^{\Gamma }$. Note that $\left( Q_{5}\cup Q_{6}\right) \cap E_{wCPN}^{\Gamma }=\emptyset$ (consider the deviating coalition $\left\{ 1,2\right\}$). Thus $Q_{2}=wF_{N}^{\Gamma }\subset E_{wCPN}^{\Gamma }=Q_{1}\cup Q_{2}\subset E_{N}^{\Gamma }$ and in particular $E_{wCPN}^{\Gamma }\backslash wF_{N}^{\Gamma }\ne \emptyset$. $\square$

Proof of Claim 2

Consider the game $\Gamma$ with $M=\left\{ 1,2,3\right\}$, $S_{1}=S_{2}= \left[ 0,1\right]$, $S_{3}=\left[ -1,0\right]$, and

$$\begin{aligned} u_{1}\left( s\right)&= 99\left( s_{2}+s_{3}\right) ^{2}+400\left( s_{2}+s_{3}\right) +2s_{1}\left( s_{2}+s_{3}\right) -s_{1}^{2}\text {,} \\ u_{2}\left( s\right)&= 99\left( s_{1}+s_{3}\right) ^{2}+400\left( s_{1}+s_{3}\right) +2s_{2}\left( s_{1}+s_{3}\right) -s_{2}^{2}\text {,} \\ u_{3}\left( s\right)&= 99\left( s_{1}+s_{2}\right) ^{2}+400\left( s_{1}+s_{2}\right) -s_{3}\left( s_{1}+s_{2}\right) -s_{3}^{2} \text {.} \end{aligned}$$

In this game $b_{1}\left( s\right) =\left\{ \max \left\{ 0,s_{2}+s_{3}\right\} \right\}$, $b_{2}\left( s\right) =\left\{ \max \left\{ 0,s_{1}+s_{3}\right\} \right\}$ and $b_{3}\left( s\right) =\left\{ - \frac{1}{2}\left( s_{1}+s_{2}\right) \right\}$. Let $e:=\left( 0,0,0\right)$. It can be easily verified that $E_{N}^{\Gamma }=\left\{ e\right\} =sF_{N}^{\Gamma }$. All conditions of the Theorem in Yi (1999) hold (in particular the condition of “ (strong) strategic substitutes in equilibrium” holds vacuously); however the thesis of the Theorem in Yi (1999) does not hold: $e\notin E_{sCPN}^{\Gamma }=\emptyset$ since $\left( 1,1\right) \in E_{N}^{\Gamma |_{e_{-3}}}$ and $\left( 1,1\right)$ strongly Pareto dominates $e_{-3}$ in $\Gamma |_{e_{-3}}$ . Therefore the statement of Yi’s theorem is false.

We can provide also a second counterexample with four players (in fact it suffices to add a player): consider the game $\Gamma$ with $M=\left\{ 1,2,3,4\right\}$, $S_{1}=S_{2}=S_{4}=\left[ 0,1\right]$, $S_{3}=\left[ -1,0 \right]$, and

$$\begin{aligned} u_{1}\left( s\right)&= 99\left( s_{2}+s_{3}+s_{4}\right) ^{2}+400\left( s_{2}+s_{3}+s_{4}\right) +2s_{1}\left( s_{2}+s_{3}+s_{4}\right) -s_{1}^{2} \text {,} \\ u_{2}\left( s\right)&= 99\left( s_{1}+s_{3}+s_{4}\right) ^{2}+400\left( s_{1}+s_{3}+s_{4}\right) +2s_{2}\left( s_{1}+s_{3}+s_{4}\right) -s_{2}^{2} \text {,} \\ u_{3}\left( s\right)&= 99\left( s_{1}+s_{2}+s_{4}\right) ^{2}+400\left( s_{1}+s_{2}+s_{4}\right) -s_{3}\left( s_{1}+s_{2}+s_{4}\right) -s_{3}^{2}\text {,} \\ u_{4}\left( s\right)&= s_{1}+s_{2}+s_{3}-s_{4}^{2}\text {.} \end{aligned}$$

All hypotheses of Yi’s Theorem hold but $E_{N}^{\Gamma }=\left\{ \left( 0,0,0,0\right) \right\} =sF_{N}^{\Gamma }\ne \emptyset$ and $\emptyset =E_{sCPN}^{\Gamma }$ (consider again the deviating coalition $\left\{ 1,2\right\}$). $\square$

Proof of Claim 3

Consider the game $\Gamma$ with $M=\left\{ 1,2,3,4\right\}$ and, for all $i\in M$, $S_{i}=\left\{ s_{i}^{*},s_{i}^{**}\right\}$ and $u_{i}$ is specified by Table 1 (the $l$-th number in each entry is the $l$-th player’s payoff). Put, for all $i\in M$, $s_{i}^{*}=0$ and $s_{i}^{**}={\sum }_{l=1}^{i}10^{l-1}$. (Note that every four-player game with the just defined strategy sets satisfies condition (1) in the statement of Yi’s theorem). Apart from condition (3), all the conditions of Yi’s Theorem hold and we have $\left( s_{1}^{**},s_{2}^{**},s_{3}^{*},s_{4}^{*}\right) \in sF_{N}^{\Gamma }\backslash E_{sCPN}^{\Gamma }\ne \emptyset$ (this time consider the deviating coalition $\left\{ 1,2,3\right\}$).$\square$

Table 1 Counterexample to a remark in Yi (1999)