Two-player partial results for axiomatic containment of Mertens-stable sets without genericity

Let $V$ be a neighborhood of $(0,z)$ in $X$. There is $0<\kappa<\rho$ such that the graph over $P_\kappa$ lies in $V$. The graph over \[ U_\kappa=\{\eta:0<\delta(\eta)<\kappa,\ \eta_i(a)>0\ \forall i,a\} \] is connected: its base is homeomorphic to $(0,\kappa)\times\operatorname{relint}\Sigma$. It lies in $V\setminus\partial X_1$. Strictly positive directions are dense in $P_\kappa$, so its closure contains the graph over $P_{\kappa/2}$. Consequently, the component of $V\setminus\partial X_1$ containing that graph has closure containing a neighborhood of $(0,z)$ in $X$. This verifies connexity separately from cohomological essentiality. ∎

Next, all witness equilibria over sufficiently small common masses lie in $W$. If not, there are $(\eta^k,z^k)\in X$ with $\eta^k\to0$ and $z^k\notin W$. Compactness supplies a subsequence converging to $(0,z)\in X$ with $z\notin W$. But the zero fiber is $S\subset W$. This is impossible. Combining this observation with surjectivity onto the interior of $P_\epsilon$ proves the assertion. Essentiality at one $\epsilon$ suffices; no unproved passage of essentiality to smaller witnesses is used. ∎

To check connexity separately, let $V$ be any neighborhood of $X_0$ in $X$. Compactness implies that $X_\kappa\subseteq V$ for some $0<\kappa<\rho$: otherwise a sequence outside $V$ with masses tending to zero has a limit in $X_0$, contradicting the neighborhood property. The set \[ U_\kappa=\{\eta:0<\delta(\eta)<\kappa,\ \eta_i(a)>0\text{ for all }i,a\} \] is connected, being homeomorphic to $(0,\kappa)\times\operatorname{relint}\Sigma$. The image of $K\times U_\kappa$ is connected and lies in $V\setminus\partial X_1$. Its closure contains $X_{\kappa/2}$, by density of strictly positive perturbations and continuity of $s$. A component of $V\setminus\partial X_1$ containing this image therefore has closure containing a neighborhood of $X_0$. This is the required connexity. Restricting $s$ to any singleton parameter $z$ proves the last assertion by the same argument. ∎

We prove that $x^\eta$ is a constrained best reply to $y^\eta$. The column-player argument uses $g$ and interchanges the roles of $x,y$.

Case 1: $0<y^*<1$. Choose the mass small enough that $y^\eta=y^*$. The Nash condition gives $f(y^*)\le0$ if $x^*=0$, $f(y^*)\ge0$ if $x^*=1$, and $f(y^*)=0$ if $0<x^*<1$. In the first two cases the clamped $x^\eta$ is the appropriate feasible endpoint. In the last case every feasible $x$ is optimal.

Case 2: $y^*\in\{0,1\}$ and $0<x^*<1$. Perfection supplies $y^k\in(0,1)$ tending to $y^*$, with $x^*$ a best reply for every $k$. Since $x^*$ uses both actions, $f(y^k)=0$ for every $k$. A nonzero affine function cannot vanish at such a sequence approaching a boundary point. Therefore $f$ is identically zero, and every feasible $x^\eta$ is a best reply.

Case 3: $x^*,y^*\in\{0,1\}$. If $x^*=1$, perfection gives $f(y^k)\ge0$ for an interior sequence approaching $y^*$. An affine function with this property is nonnegative on a whole one-sided neighborhood of $y^*$. To check this explicitly, if $f(y^*)>0$ use continuity; if $f(y^*)=0$, the sign just inside the interval is fixed by its slope; and if $f(y^*)<0$, the sequence is impossible. Thus $f(y^\eta)\ge0$ for all sufficiently small bounds, and $x^\eta=u_1$ is optimal. When $x^*=0$ the same reasoning uses the weak inequality $f\le0$, and $x^\eta=\ell_1$ is optimal.

There are only two players and finitely many neighborhood requirements. Taking their minimum yields one positive $\rho$ valid for every perturbation in $P_\rho$, including directions with some zero coordinates. Both players are constrained best replies. Lemma Lemma (Continuous equilibrium section) proves singleton stability. ∎

Let $\mu$ be the smallest positive coordinate of $p_I^*$ and $q_J^*$. It is positive because the supports are finite and nonempty. The changes in supported probabilities in (eq:affinesection) are bounded by a constant times $\delta(\alpha,\beta)$. Choosing that mass sufficiently small makes every supported coordinate exceed $\mu/2$ and all lower bounds less than $\mu/2$. Thus feasibility holds. Quasi-strictness and continuity preserve every strict payoff gap for unused actions. A player's constrained best reply must assign the lower bound to each strictly inferior action and may distribute the remaining probability among the tied best actions. That is exactly what (eq:affinesection) does. This proves (iv). Its continuity and Lemma Lemma (Continuous equilibrium section) give (iv)$\Rightarrow$(i).

It remains to prove (ii)$\Rightarrow$(iii). Choose a fixed neighborhood $W$ of $z^*$ in which all coordinates on $I,J$ exceed $\mu/2$, and each unused action remains strictly inferior to each originally supported action. Such a neighborhood exists by quasi-strictness and continuity. For bounds with common mass below $\mu/2$, any constrained equilibrium in $W$ must have \[ p_{I^c}=\alpha_{I^c},\quad q_{J^c}=\beta_{J^c},\quad Cq=0,\quad Dp=0. \] The supported coordinates are strictly above their lower bounds, so they must tie at the maximal payoff; inferior coordinates must be at their bounds. Subtracting the target equations gives the necessary systems \begin{equation}\label{wsref-0-65-71-3a-6e-65-63-65-73-73-61-72-79-73-79-73-74-65-6d-73} M_A(q_J-q_J^*)=-N_A\beta_{J^c},\qquad M_B(p_I-p_I^*)=-N_B\alpha_{I^c}. \end{equation}

Suppose, for instance, $\operatorname{im} N_A\not\subseteq\operatorname{im} M_A$. Finite-dimensional linear algebra gives a row vector $w$ with $wM_A=0$ but $wN_A\ne0$. There is a vector $b$ with all coordinates strictly positive and $wN_A b\ne0$: a nonzero linear form cannot vanish on the open positive orthant. Scale $b$ so $\sum b<1$, extend it on $J$ by strictly positive coordinates of total mass $1-\sum b$, and call the resulting full probability vector $\bar\beta$. Choose any completely mixed $\bar\alpha\in\Delta_m$. For every $\delta>0$ set $\alpha=\delta\bar\alpha$ and $\beta=\delta\bar\beta$. These are strictly positive perturbations with common mass $\delta$, but multiplying the first equation of (eq:necessarysystems) by $w$ gives \[ 0=-\delta wN_A b\ne0. \] No constrained equilibrium lies in $W$, however small $\delta$ is. This contradicts (ii). The second inclusion is proved by the same argument. ∎

For the optional stronger claim of properness, replace the column profile by $q^\epsilon=(1-2\epsilon^2,\epsilon^2,\epsilon^2)$, with $0<\epsilon<1/2$, and retain the row profile above. The row actions tie. Each inferior column action has probability $\epsilon^2$, whereas the better action has probability $1-2\epsilon^2$, and \[ \epsilon^2\le\epsilon(1-2\epsilon^2). \] Thus these profiles are $\epsilon$-proper: whenever an action has strictly lower payoff than another, its probability is at most $\epsilon$ times the probability of the better action. Their limits are the claimed proper equilibria.

Now impose row lower bounds $\alpha=(\delta/2,\delta/2)$. Consider the two strictly positive column perturbations, ordered as $(L,M,R)$, \[ \beta^+=(\delta/4,\delta/2,\delta/4),\qquad \beta^-=(\delta/4,\delta/4,\delta/2). \] All players' lower bounds have common mass $\delta$. Since $L$ is strictly best, the column player uniquely puts $M,R$ at their lower bounds and all remaining mass on $L$. Under $\beta^+$ the row payoff difference is $\delta/4>0$, giving the unique constrained equilibrium \[ p^+(U)=1-\delta/2,\qquad q^+=(1-3\delta/4,\delta/2,\delta/4). \] Under $\beta^-$ it is $-\delta/4<0$, giving \[ p^-(U)=\delta/2,\qquad q^-=(1-3\delta/4,\delta/4,\delta/2). \] The two equilibria converge respectively to $z_1$ and $z_0$. Lemma Lemma (Necessary local robustness) and closedness imply that every stable set must contain both $z_0$ and $z_1$: a compact set missing one of them has a neighborhood avoided by its corresponding equilibrium ray for all small $\delta$. Stable sets are connected and are subsets of $E$, so they must contain the whole interval. No singleton can do so. Finally, existence of Mertens stable sets in finite games, as used in [gw], implies that $E$ itself is the unique stable set. Only this last existence conclusion uses that general existence theorem. ∎

Define $f_1(p)=p$ and \[ f_2(\bar q)=\bar q_{\rm old}+\sum_t\bar q_t\tau^t. \] This affine map is the identity on the old simplex. For any pair of original row strategies $p,p'$, payoff comparisons against $\bar q$ are exactly the original comparisons against $f_2(\bar q)$, by the first equality of (eq:penalized). For any pair of original column strategies, payoff comparisons against $p$ are unchanged. These statements hold on the entire enlarged strategy space, so in particular on a neighborhood of the original face. They verify every condition of Definition 4.4, not merely dominance. ∎

A pure column plan specifies a root action and an off-path continuation. Every plan choosing a particular $j\in J$ directly is a duplicate of $j$, regardless of its continuation. A plan choosing $\mathrm{Out}$ and then $k$ is identified with $k$. Collapsing these duplicates is an affine surjection. Payoffs in $H$ are exactly those of $\bar G$ after collapse. This is a strong embedding with no outsiders: the outsider face is a point and the preference-equivalence condition holds globally.

All actual action pairs are distinct terminal outcomes. Since the row player's information set is always reached, a mixed profile realization- equivalent to a behavioral profile has the same collapsed row and column marginals. Thus a $B$-witness in a lifted set projects into its specified image in $\bar G$; the unreached continuation is not incorrectly treated as an observed action. ∎

For each $(p,q)\in K$, choose $j\in\operatorname{BR}_2(p)\subseteq J$. The payoff from its associated continuation action satisfies \begin{align*} p^{\mathsf T}\bar B_{:,t_j} &= (1-\epsilon)v(p)+\epsilon p^{\mathsf T}B_{:,\ell}-c\\ &\ge v(p)-\epsilon L-c =v(p)-\gamma/2 \ge p^{\mathsf T}B_{:,k}+\gamma/2\quad(k\in J^c). \end{align*} Thus every original action in $J^c$ is strictly worse than some new action at the continuation information set. The inequality persists against row profiles sufficiently close to the projection of $K$.

Assume for contradiction that $K\in\Phi(G)$. Axiom $D^*$ makes $K$, padded by zero probabilities, a selected set of $\bar G$. Apply Lemma Lemma (Duplicate plans and a hidden continuation) and the existence-of-lifts direction of $I^*$ to obtain $K_H\in\Phi(H)$ with $f(K_H)=K$. Axiom $B$ supplies a profile in $K_H$ realization-equivalent to a quasi-perfect behavioral equilibrium $b$ of the hidden-continuation tree. Its collapsed image lies in $K$. Let $b^r\to b$ be the completely mixed behavioral sequence in the definition of quasi-perfection.

The continuation's limiting strategy puts total probability one on the new actions: any positive limiting probability on an old $k\in J^c$ would contradict its strict suboptimality against the approximating row strategies. Denote its probabilities along $b^r$ by $\zeta_k^r$, and let $\theta_r>0$ be the root probability of $\mathrm{Out}$. We have \[ \sum_{j\in J}\zeta_{t_j}^r\longrightarrow1, \qquad \zeta_k^r\longrightarrow0\quad(k\in J^c). \] Writing $a=w^{\mathsf T}A_{:,\ell}\ne0$, equation (eq:annihilate) gives the same value for every new action: \[ w^{\mathsf T}\bar A_{:,t_j}=\epsilon a. \] All root-direct actions have weighted row payoff zero. The weighted row payoff against the full induced column distribution $\bar q^r$ is therefore \begin{equation}\label{wsref-0-65-71-3a-63-6f-6e-74-72-61-64-69-63-74-69-6f-6e} w^{\mathsf T}\bar A\bar q^r =\theta_r\left( \epsilon a\sum_{j\in J}\zeta_{t_j}^r +\sum_{k\in J^c}\zeta_k^r w^{\mathsf T}A_{:,k}\right). \end{equation} The bracket converges to $\epsilon a\ne0$; since $\theta_r>0$, the expression is nonzero for all sufficiently large $r$.

On the other hand, the target row strategy uses every action in $I$. Quasi-perfection at its sole information set makes each such action a best reply to the opponent's behavioral strategy $b_2^r$ for every $r$. Their payoffs are equal. Since $w$ is supported on $I$ and its coordinates sum to zero, their weighted payoff must be zero. This contradicts (eq:contradiction). No globally compatible rule can select $K$. ∎

Choose matrices $R_A,R_B$ solving $M_AR_A=N_A$, $M_BR_B=N_B$. For $z=(p^*,q^*)\in K$ and row/column lower bounds $\alpha,\beta$, put \begin{equation}\label{wsref-0-65-71-3a-75-6e-69-66-6f-72-6d-73-65-63-74-69-6f-6e} \begin{aligned} q_{J^c}&=\beta_{J^c},&q_J&=q_J^*-R_A\beta_{J^c},\\ p_{I^c}&=\alpha_{I^c},&p_I&=p_I^*-R_B\alpha_{I^c}. \end{aligned} \end{equation} The probability-sum rows of the identities make $p,q$ distributions; the other rows give $Cq=0$, $Dp=0$. Thus supported actions stay tied.

All coordinates in $p_I^*,q_J^*$ have a common positive lower bound as $z$ ranges over compact $K$. Likewise the unused-action payoff gaps have a common positive lower bound whenever such actions exist. The changes in (eq:uniformsection) are bounded uniformly by a constant times the perturbation mass. Hence, for a common sufficiently small $\rho>0$, all supported coordinates exceed their required lower bounds and all unused actions remain strictly inferior, uniformly over $K$. Inferior actions receive exactly their lower bounds; all remaining probability is assigned to tied best replies. Therefore (eq:uniformsection) is a jointly continuous family of constrained equilibria. Lemma Lemma (A continuous family of sections) proves both conclusions about stability. ∎

At the continuation information set, $T^+$ gives the column player $-3/4$, strictly exceeding the payoff $-1$ of either alternative, against every row strategy. A quasi-perfect equilibrium must therefore choose $T^+$ there. At the root it chooses $L$, which gives $0$. In a completely mixed behavioral sequence witnessing quasi-perfection, write $\theta_r>0$ for the probability of $\mathrm{Out}$ and $\zeta^r$ for the continuation distribution. Then \[ u_1(U,b_2^r)-u_1(D,b_2^r) =\theta_r(\zeta_M^r-\zeta_R^r+\tfrac12\zeta_{T^+}^r)>0 \] for all sufficiently large $r$, since $\zeta_{T^+}^r\to1$. Every quasi-perfect outcome of this tree is therefore $(U,L)$. Lifting an arbitrary $K\in\Phi(G)$ through $D^*$ and the duplicate embedding, then applying $B$, forces $z_1\in K$.

Now instead add the penalized mixture $\tau^-=(L+R)/2$ with the same penalty. Its row payoffs are $-1/2,0$ and its column payoff is again $-3/4$. The identical continuation argument makes the row payoff difference strictly negative along every quasi-perfect witness, so it forces $z_0=(D,L)$ into the same original selected set $K$. A nonempty connected subset of the segment $E$ containing its two endpoints is $E$. Thus every selected set is $E$; nonemptiness of the solution family gives $\Phi(G)=\{E\}$.

Apply this proved selection statement to the known all-game solution concept $\Phi=\operatorname{Stab}_0$. Its existence and satisfaction of the axioms are independent literature results cited in Section 1. It follows that $\operatorname{Stab}_0(G)=\{E\}$. This application does not invoke the unproved nongeneric containment converse. ∎

Now let $d_0=0$, so the Nash set is $E$. If all $d_j=0$, every target $t\in[0,1]$ has the section \[ p_U=\alpha_U+(1-\delta)t,\quad p_D=\alpha_D+(1-\delta)(1-t),\quad q_j=\beta_j\ (j\ne0),\quad q_0=1-\sum_{j\ne0}\beta_j. \] Both row actions always tie, and column $0$ remains strictly best. Every singleton is stable, so every nonempty selected set contains one.

If all $d_j\ge0$ and at least one is positive, $U$ is the unique best reply against every completely mixed column strategy. Axiom $B$, applied to the simultaneous-move representation, therefore places $(U,0)$ in each selected set. The formula $p_U=1-\alpha_D$, together with the column formula above, is a continuous equilibrium section because $\sum_jd_jq_j\ge0$ for every nonnegative lower-bound vector. Its zero fiber is $(U,0)$, proving the requisite stability. The case with all $d_j\le0$ and at least one negative is symmetric.

It remains to treat (eq:oppositesigns). Put \[ \gamma=\min_{i,j\ne0}(B_{i0}-B_{ij})>0,\qquad M=\max_{i,j\ne0}(B_{i0}-B_{ij})\ge\gamma, \qquad \epsilon=\frac{\gamma}{4M},\quad c=\frac\gamma4. \] Choose $h$ with $d_h>0$, and add the penalized column mixture $\tau=(1-\epsilon)e_0+\epsilon e_h$. For every row action $i$ and original nonzero column $j$, \begin{align*} \bar B_{it}-B_{ij} &= B_{i0}-B_{ij}-\epsilon(B_{i0}-B_{ih})-c\\ &\ge\gamma-\epsilon M-c=\gamma/2>0. \end{align*} Column $0$ strictly beats $t$, and $t$ strictly beats every original nonzero column. Its row payoff difference is $\epsilon d_h>0$. The root choice $0/\mathrm{Out}$ followed by all nonzero original columns and $t$ is therefore exactly the forcing construction of Theorem Theorem (The axioms force the entire segment). It forces $(U,0)$ into every selected $K$. Repeating with a column $h$ satisfying $d_h<0$ forces $(D,0)$ into the same $K$. Connectedness gives $K=E$. Finally apply the proved conclusion to the known nonempty stable-set solution concept to get $\operatorname{Stab}_0(G)=\{E\}$.

The cases $d_0\ne0$, all differences zero, one weak sign, and both strict signs exhaust all possibilities. The proof permits any finite $n$; $n=1$ falls within the first or all-zero case. ∎

For Definition 4.4 use the affine dictionary \[ f_1(p)=p,\qquad f_2(\bar q)=\bar q_{\rm old}+\sum_{j=1}^k\bar q_{t_j}\tau^j. \] It is the identity on the original face. For original row strategies $p,p'$, the difference in their enlarged-game payoffs against $\bar q$ is exactly $(p-p')^TAf_2(\bar q)$, because the new row-payoff columns are $A\tau^j$. For any two original column strategies, their payoff comparison is unchanged and evaluated against $f_1(p)=p$. The required preference equivalences therefore hold everywhere, not merely in a neighborhood. Together with domination and equality of Nash sets, these facts establish the exact $D^*$ comparison. ∎

At the continuation node, every target action with positive probability is optimal against the row component of $b^\nu$. Passing to the limit and deleting the common term $p^TB_{:,0}-c$ from (eq:new) gives (eq:colbr).

For completeness the row-player argument must account for the fact that Out is unreached at $b$. Let $\theta_\nu>0$ be its probability in $b^\nu$, and let $\zeta^\nu$ be the conditional continuation distribution. Its mass on original nonzero columns tends to zero, while its probe probabilities tend to $y$. For $i\in\operatorname{supp}(p)$ and $h\in I$, quasi-perfection requires the payoff difference of row $i$ over row $h$ to be nonnegative against $b_2^\nu$. Both rows have the same payoff against root action $0$. Dividing the inequality by $\theta_\nu>0$ and taking the limit yields \[ (C_Iy)_i\ge(C_Iy)_h. \] Every row used by $p$ is therefore a best reply to $y$ among $I$. Comparing with $r\in\Delta(I)$ gives (eq:rowbr). No comparison between two vanishing error rates has been assumed: the continuation probabilities are conditioned on the single positive factor $\theta_\nu$, which cancels exactly. ∎

Apply that equality to the source's stability solution concept itself. The independent source inputs specified in Section 1 say that it exists and satisfies the three axioms on all games. Therefore it selects exactly $E$ in this game. This identifies the genuine Mertens-stable set, not merely a topologically robust substitute. The nongeneric converse (T) has not been assumed at any stage. ∎