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Econ-20-A1 · revision r1 · on Econ-20

Axiomatic containment of Mertens-stable sets without genericity: Merged partial contribution and proof audit

Partial result · LLM-generated · Submitted by Anonymous submitter · gpt 6 astra pro, chatgpt gpt-6, codex (gpt-6)

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Claim: partial result

Claims a largest globally axiom-compatible solution rule via compact pruning and exact elimination reformulation; continuous-section sufficient criterion, strict/fully-mixed and nongeneric zero-sum polymatrix cases, duplicate/strong-embedding transfers, two payoff-limit counterexamples, and a Hopf-map obstruction; universal S1 remains unresolved.

Assumptions: Exact global axioms B, D*, I* and Mertens characteristic-zero stability per Govindan–Wilson v2; Finite normal-form games including additional players/actions; no closedness assumed for the family of selected sets; Uses literature all-game stability properties, stable-image and stable-lift results

Limitations: No universal containment proof or globally compatible counterexample; No novelty or formal/human/referee verification certification; Payoff-limit and Hopf-map obstructions do not disprove S1; Source identity metadata was corrected editorially; original files and normalization audit are archived.; Submitted research claim; correctness and novelty have not been independently established.

Manuscript

Contents
  1. Axiomatic containment of Mertens-stable sets without genericity: Merged partial contribution and proof audit
  2. Contribution metadata
  3. Write-up
  4. Executive statement.
  5. Exact target and external inputs
  6. External game-theoretic input.
  7. What is new relative to the preceding draft.
  8. Compact spaces of candidate solutions
  9. A largest axiom-compatible rule by compact pruning
  10. Comparison maps
  11. The recursion
  12. An exact reformulation of the unsolved implication
  13. What this does not prove.
  14. A constructive criterion for singleton stability
  15. A whole family with non-isolated quasi-perfect equilibria
  16. An explicit three-player instance without duplicate actions
  17. Two useful transfer results
  18. Lifting a section through exact duplicates
  19. Containment descends through strong embeddings
  20. Why a limit of generic-game conclusions is insufficient
  21. A polynomial topological obstruction to a second shortcut
  22. Exact outstanding assertion and proof audit
  23. Quantifiers checked.
  24. Scope of the counterexamples.
  25. Checks and reproducibility.
  26. Novelty and provenance.
  27. Recommended submission metadata.
  28. Models used
  29. Human contributors and roles
  30. References

Axiomatic containment of Mertens-stable sets without genericity:
Merged partial contribution and proof audit

OpenAI GPT-6 assisted research draft

18 October 2026  •  Merged contribution, version 3

Contribution metadata

Permanent problem identifier: WS-107793.
Claim type: Partial result, with full written proof arguments and an exact global reduction; not a complete solution.
Scope: Largest globally axiom-compatible solution rule by compact pruning; an exact reformulation of the missing step; a continuous-section criterion (including fully mixed and strict equilibria); a nongeneric zero-sum polymatrix family; duplicate-action and strong-embedding transfers; two payoff-limit counterexamples; and a polynomial Hopf-map obstruction to replacing rational-cohomological essentiality by a coincidence property.
Assumptions and limitations: Exact axioms $B,D^*,I^*$ and Mertens $0$-stability in Govindan--Wilson $2026$. No proof or globally compatible counterexample for the unrestricted implication; no novelty, publication, independent referee, or formal-verification claim.
Source relation: Consolidates the two user-supplied local drafts for WS-107793. The source author reports proof and boundary checks against the pinned primary-source definitions; these reports are author declarations, not independent verification.

Write-up

Executive statement.
The generic-game implication is due to Govindan--Wilson. The precise nongeneric implication is still unresolved. This manuscript gives several sufficient subclasses, a global reduction, and explicit obstructions to plausible but invalid extensions of the generic proof. The constructions are deliberately not described as a universal solution.

Exact target and external inputs

A finite normal-form game is $G=(I,(A_i)_{i\in I},(u_i)_{i\in I})$, with mixed-strategy space $\Sigma_G=\prod_i\Delta(A_i)$ and Nash set $\operatorname{NE}(G)$. All games have at least one player and nonempty finite action sets. We use canonically labelled finite player and action sets; this makes the collection of games with real payoff arrays a set, not a proper class. Relabellings and all permitted representations remain included in the axioms.

The source's definition of Mertens $0$-stability, including its perturbation space, boundary conventions and connexity requirement, is normative [gw, Definition 2.4]. Here $0$ denotes characteristic zero, not a local fixed-point index. The problem is the implication \begin{equation}\label{wsref-0-65-71-3a-74-61-72-67-65-74} \Phi\text{ satisfies }B,D^*,I^* \quad\Longrightarrow\quad \forall G\ \forall K\in\Phi(G)\ \exists S\in\operatorname{Stab}_0(G): S\subseteq K. \end{equation} The selected sets are nonempty, closed and connected subsets of $\operatorname{NE}(G)$. No additional closedness of the family $\Phi(G)$ is assumed.

We retain precisely the source's permitted comparisons. Axiom $B$ requires intersection with profiles equivalent to quasi-perfect equilibria for every perfect-recall extensive-form representation. Axiom $D^*$ compares the source's strongly irrelevant dominated-strategy extensions, which have the same Nash set under the natural inclusion. Axiom $I^*$ requires equality between a solution family and the family of images of solutions under every strong embedding. Arbitrary payoff perturbations and arbitrary dominated-strategy extensions are not added to these axioms [gw, Axioms 3.1, 4.7, 5.4; Definitions 4.4, 5.2].

External game-theoretic input.
We use the established all-game properties of the Mertens stability rule: nonemptiness and satisfaction of $B,D^*,I^*$, as stated by Govindan--Wilson and proved in their axioms discussion and Appendices A--B. The separate generic necessity conclusion of their Theorem 6.1 is not assumed for nongeneric games. The stable-image and stable-lift properties under strong embeddings are in Appendix B. These literature theorems are inputs; we do not claim to reprove or independently referee their full proofs.
What is new relative to the preceding draft.
The original local attempt addressed strict quasi-perfect equilibria; the user-supplied manuscript adds explicit fully mixed and strict-equilibrium sections and a particularly economical nongeneric payoff-limit example. The compact pruning arguments address global axiom quantifiers, while Subsection sec:polymatrix supplies games with only non-strict equilibria. The full target (eq:target) remains explicitly separate.

Compact spaces of candidate solutions

Write \[ \mathscr C(G)=\{K\subseteq\operatorname{NE}(G):K\ne\varnothing,\ K\text{ compact and connected}\}, \] with the Hausdorff metric $d_H$. Compactness and closedness of individual sets are equivalent here because $\operatorname{NE}(G)$ is compact.
Lemma: For every finite game, $\mathscr C(G)$ is compact. If $f:\operatorname{NE}(H)\to\operatorname{NE}(G)$ is continuous, then $f_*:\mathscr C(H)\to\mathscr C(G)$, defined by $f_*(L)=f(L)$, is continuous.
Proof: The Nash set is closed in the compact product of simplexes: each best-response inequality is closed. For completeness, the hyperspace of nonempty compact subsets of any compact metric space $M$ is compact. Finite nets in $M$ yield finite nets in its hyperspace by taking all nonempty subsets of a finite net. It is complete: for a Hausdorff-Cauchy sequence $K_j$, the functions $z\mapsto\operatorname{dist}(z,K_j)$ converge uniformly to a continuous function $v$. The zero set $K=\{v=0\}$ is nonempty by compactness of $M$. Uniform convergence implies $\sup_{z\in K}\operatorname{dist}(z,K_j)\to0$. If the other Hausdorff distance did not tend to zero, points in $K_j$ bounded away from $K$ would have a limit in $M$; uniform convergence would put that limit in $K$, a contradiction. Thus $K_j\to K$.

Connected subsets form a closed subspace of this hyperspace. Indeed, if a Hausdorff limit $K$ of connected compact sets were disconnected, a separation of $K$ into two nonempty compact sets would give disjoint open neighborhoods with positive separation. All sufficiently close $K_j$ would lie in their union and meet both, contradicting connectedness. This proves compactness of $\mathscr C(G)$. Finally, $f$ is uniformly continuous on its compact domain. A modulus of uniform continuity bounds $d_H(f(L),f(L'))$ in terms of $d_H(L,L')$. Continuous images preserve nonemptiness, compactness and connectedness. ∎

For a fixed perfect-recall representation $\Gamma$ of $G$, let $Q_\Gamma\subseteq\operatorname{NE}(G)$ denote exactly the mixed profiles equivalent to quasi-perfect equilibria in the source's sense.

Lemma: $Q_\Gamma$ is compact. Consequently $\{K\in\mathscr C(G):K\cap Q_\Gamma\ne\varnothing\}$ is closed in $\mathscr C(G)$.
Proof: First show that quasi-perfect behavioral profiles form a closed set for a fixed finite $\Gamma$. Suppose $b^j\to b$ and every $b^j$ is quasi-perfect. There are finitely many information-set actions. After discarding finitely many terms, every action with positive probability in $b$ also has positive probability in $b^j$. From a defining completely mixed witness sequence for $b^j$, choose $c^j$ with $\|c^j-b^j\|<1/j$. Every action supported by $b^j$ is optimal against $c^j_{-i}$ at its information set, by [gw, Definition 2.1]. Therefore every action supported by $b$ has the same property. As $c^j\to b$, this is a quasi-perfect witness for $b$. The behavioral strategy space is compact, so its quasi-perfect subset is compact.

The mixed/behavioral equivalence relation used here is closed. In the source's realization or enabling representation it is expressed by equality of finitely many continuous realization coordinates: they are linear in mixed strategies and polynomial in behavioral strategies. Thus the set of pairs $(x,b)$ with $b$ quasi-perfect and $x$ equivalent to $b$ is compact. Its projection to mixed-strategy space is $Q_\Gamma$ and is compact. The usual Nash inequalities are preserved by this equivalence.

Finally, if $K_j\to K$ and $x_j\in K_j\cap Q_\Gamma$, take a convergent subsequence in compact $Q_\Gamma$. Its limit lies in $K$ by Hausdorff convergence. Hence $K\cap Q_\Gamma\ne\varnothing$. ∎

A largest axiom-compatible rule by compact pruning

Comparison maps

Introduce one arrow $e:H\to G$ for each of the following exactly permitted comparisons:
  1. A strong embedding of $G$ in $H$, with the map $f_e:\operatorname{NE}(H)\to\operatorname{NE}(G)$ supplied by Definition 5.2 of the source.
  2. An extension $H$ of $G$ satisfying Definition 4.4 of the source, with $f_e$ the natural deletion of the unused added coordinates on the Nash sets.
In (ii), the two Nash sets are naturally homeomorphic. In (i), Nash profiles indeed map to Nash profiles: the outsiders of a Nash profile lie in the specified face; surjectivity of each insider's strategy map implements any original deviation; and the source's preference preservation would turn a profitable original deviation into a profitable insider deviation in $H$. The other insiders' mapped strategies remain unchanged because their maps depend only on their own strategies and the outsiders' strategies.

Both kinds of $f_e$ are continuous. Put $e_*(L)=f_e(L)$. For these arrows, the two invariance axioms become the single family of equalities \begin{equation}\label{wsref-0-65-71-3a-61-72-72-6f-77-73} \Phi(G)=e_*(\Phi(H)) :=\{e_*(L):L\in\Phi(H)\}\qquad(e:H\to G). \end{equation} For a dominated-strategy arrow this is just the source's equality under the natural face identification, not an added projection axiom.

The recursion

Let $\mathfrak E(G)$ be the set of all permitted perfect-recall representations of $G$. Define \begin{equation}\label{wsref-0-65-71-3a-41-30} \mathcal A_0(G)=\bigcap_{\Gamma\in\mathfrak E(G)} \{K\in\mathscr C(G):K\cap Q_\Gamma\ne\varnothing\}. \end{equation} Thus $\mathcal A_0$ imposes exactly $B$ on individual candidate sets. For $r\ge0$, define \begin{equation}\label{wsref-0-65-71-3a-72-65-63-75-72-73-69-6f-6e} \begin{split} \mathcal A_{r+1}(G)={}&\mathcal A_r(G)\ \cap\! \bigcap_{e:G\to F}e_*^{-1}(\mathcal A_r(F))\ \cap\! \bigcap_{e:H\to G}e_*(\mathcal A_r(H)),\\ \mathcal A_\infty(G)={}&\bigcap_{r=0}^{\infty}\mathcal A_r(G). \end{split} \end{equation} The first intersection of comparisons enforces image compatibility; the second enforces the existence of a compatible lift for each incoming comparison. Different arrows may require different lifts, exactly as in (eq:arrows).
Theorem (Largest compatible solution concept): For every $G$, all families $\mathcal A_r(G)$ and $\mathcal A_\infty(G)$ are nonempty compact subsets of $\mathscr C(G)$. The rule $\mathcal A_\infty$ satisfies $B,D^*,I^*$ globally. Moreover, every globally defined solution concept $\Phi$ satisfying these axioms obeys \begin{equation}\label{wsref-0-65-71-3a-6c-61-72-67-65-73-74} \Phi(G)\subseteq\mathcal A_\infty(G)\qquad\text{for every finite game }G. \end{equation} Thus $\mathcal A_\infty$ is the largest such rule, ordered by inclusion of solution families game by game.
Proof: Compactness. Lemma Lemma makes $\mathcal A_0(G)$ closed in compact $\mathscr C(G)$. Suppose all $\mathcal A_r(F)$ are compact. Each inverse image in (eq:recursion) is closed by Lemma Lemma. Each image $e_*(\mathcal A_r(H))$ is compact, hence closed in the metric space $\mathscr C(G)$. Arbitrary intersections of these closed sets are closed. This proves compactness by induction. The families decrease with $r$.

Nonemptiness. We prove the stronger assertion \begin{equation}\label{wsref-0-65-71-3a-73-74-61-62-6c-65-70-72-6f-74-65-63-74-65-64} \operatorname{Stab}_0(G)\subseteq\mathcal A_r(G)\qquad\text{for every }G,r. \end{equation} At $r=0$, it follows from stability satisfying $B$. Suppose it holds at stage $r$. For a stable set $S$ in $G$, every outgoing arrow has stable image in its target, hence an image in that target's $\mathcal A_r$. Every incoming arrow admits a stable lift of $S$, hence a lift in its source's $\mathcal A_r$. These statements are exactly the established $D^*,I^*$ properties of $\operatorname{Stab}_0$. Thus $S\in\mathcal A_{r+1}(G)$. Existence of stable sets proves nonemptiness, including at the intersection $\mathcal A_\infty(G)$.

The axioms at the limit. Axiom $B$ holds because $\mathcal A_\infty(G)\subseteq\mathcal A_0(G)$. Fix an arrow $e:H\to G$. For $L\in\mathcal A_\infty(H)$, membership in $\mathcal A_{r+1}(H)$ implies $e_*(L)\in\mathcal A_r(G)$ for every $r$. Hence \[ e_*(\mathcal A_\infty(H))\subseteq\mathcal A_\infty(G). \] For the converse fix $K\in\mathcal A_\infty(G)$. For every $r$, the fiber \[ F_r=\{L\in\mathcal A_r(H):e_*(L)=K\} \] is nonempty: use $K\in\mathcal A_{r+1}(G)$ and the incoming-arrow condition. It is compact by continuity of $e_*$, and $F_{r+1}\subseteq F_r$. Compactness gives a point in $\bigcap_rF_r$. Such a point is a set $L\in\mathcal A_\infty(H)$ with $e_*(L)=K$. This proves the reverse inclusion and therefore (eq:arrows). It is the step that justifies stopping at the countable intersection; independently choosing a different lift at each stage would not suffice without compactness.

Maximality. Let $\Phi$ satisfy the three axioms. Axiom $B$ gives $\Phi(G)\subseteq\mathcal A_0(G)$ for every $G$. Suppose inclusion holds at stage $r$. For $K\in\Phi(G)$, its image under every outgoing arrow belongs to the corresponding $\Phi(F)$ and hence $\mathcal A_r(F)$. Under every incoming arrow it has a lift in $\Phi(H)$ and hence $\mathcal A_r(H)$. Thus $K\in\mathcal A_{r+1}(G)$. Induction and intersection prove (eq:largest). No compactness assumption on the original family $\Phi(G)$ has been used. ∎

Remark: The existence of a largest rule also follows by taking the union of all compatible rules. The additional content of Theorem Theorem (Largest compatible solution concept) is the explicit compact recursion, the compactness of its families, and the fact that countably many stages suffice. The recursion is set-theoretic: its individual stages quantify over all permitted finite games and comparisons.

An exact reformulation of the unsolved implication

Define the upward containment family \[ \mathcal U_0(G)=\{K\in\mathscr C(G):\exists S\in\operatorname{Stab}_0(G),\ S\subseteq K\}. \] This definition does not say that $K$ itself is stable.
Corollary: The following are equivalent:
  1. The unrestricted implication (eq:target) holds.
  2. $\mathcal A_\infty(G)\subseteq\mathcal U_0(G)$ for every finite game $G$.
  3. For every $G$ and every $K\in\mathscr C(G)\setminus\mathcal U_0(G)$, there is a finite integer $r\ge0$ such that $K\notin\mathcal A_r(G)$.
If an explicit $G,K$ violating (b) is found, $\Phi=\mathcal A_\infty$ supplies the required globally defined counterexample solution concept.
Proof: Apply (a) to $\Phi=\mathcal A_\infty$ to obtain (b). Conversely, maximality in Theorem Theorem (Largest compatible solution concept) makes (b) imply (a) for every compatible $\Phi$. Finally, (b) and (c) are equivalent by the definition $\mathcal A_\infty(G)=\bigcap_{r\ge0}\mathcal A_r(G)$. ∎
What this does not prove.
The protected inclusion (eq:stableprotected) has the opposite direction from the desired containment assertion. We have not proved that every survivor contains a stable set. Nor is finite-stage elimination an effective finite certificate: one stage already quantifies over an unrestricted set of representations and comparisons. There is no algorithm, uniform depth bound, or test of all incoming embeddings in this contribution.

For games covered by the source's generic theorem, Corollary Corollary does imply finite-stage elimination of each candidate containing no stable set. Removing the genericity assumption from that consequence is exactly the remaining issue, not an additional established result.

A constructive criterion for singleton stability

Fix $G$, write $m_i=|A_i|$, and use the common-mass perturbation polytope \[ P_\varepsilon=\{\delta\tau:0\le\delta\le\varepsilon,\ \tau\in\Sigma_G\}. \] For $\eta\in P_1$, its common mass is $\delta(\eta)=\sum_{a\in A_i}\eta_i(a)$, independent of $i$. The perturbed feasible set is $\Sigma_i(\eta)=\{x_i\in\Delta(A_i):x_i(a)\ge\eta_i(a)\}$. Let $E_G$ be the graph of constrained Nash equilibria over $P_1$ and $\pi:E_G\to P_1$ its projection.

The polytope $P_\varepsilon$ has dimension \[ d=1+\sum_i(m_i-1) \] in its linear span and is homeomorphic, with its relative boundary, to $(D^d,S^{d-1})$. For $X\subseteq E_G$ put $X_\varepsilon=X\cap\pi^{-1}(P_\varepsilon)$ and $\partial X_\varepsilon=X\cap\pi^{-1}(\partial P_\varepsilon)$. In particular $\partial X_1$ refers to the boundary of $P_1$, not an arbitrarily chosen boundary of $X$.

Proposition (Continuous-section criterion): Let $x\in\operatorname{NE}(G)$. Suppose that for some $0<\rho<1$ there is a continuous map $s:P_\rho\to\Sigma_G$ with \[ s(0)=x,\qquad s(\eta)\in\operatorname{NE}(G(\eta))\quad(\eta\in P_\rho). \] Then $\{x\}\in\operatorname{Stab}_0(G)$.
Proof: Let $X=\{(\eta,s(\eta)):\eta\in P_\rho\}$. This is a compact subset of $E_G$ with $X_0=\{(0,x)\}$. For every $0<\varepsilon\le\rho$, projection is a homeomorphism of pairs \[ (X_\varepsilon,\partial X_\varepsilon) \longrightarrow(P_\varepsilon,\partial P_\varepsilon). \] Consequently its induced map on degree-$d$ relative Čech cohomology with coefficients in $\mathbb Q$ is an isomorphism between one-dimensional vector spaces, so it is nonzero.

We verify the source's connexity condition separately. Any neighborhood $V$ of $(0,x)$ in $X$ contains the graph over $P_\kappa$ for some $0<\kappa<\rho$: continuity at zero gives this after shrinking $\kappa$. The graph over \[ B_\kappa=\{\eta:0<\delta(\eta)<\kappa,\ \eta_i(a)>0\text{ for all }i,a\} \] is connected and lies in $V\setminus\partial X_1$. Connectedness follows, for example, by parameterizing $B_\kappa$ by $(0,\kappa)\times\operatorname{relint}\Sigma_G$. The closure of this graph contains the graph over $P_{\kappa/2}$, since strictly positive directions are dense. Hence the connected component of $V\setminus\partial X_1$ containing this graph has closure containing a neighborhood of $(0,x)$ in $X$. The same conclusion holds whether closure is taken relative to $V$ or to $X$, since that smaller graph lies in $V$. This is precisely connexity. Together with essentiality it proves the assertion. ∎

Corollary (Equilibrium segments): Suppose $x,y\in\operatorname{NE}(G)$, $y$ is completely mixed, and the whole segment $[x,y]$ lies in $\operatorname{NE}(G)$. Then $\{x\}$ is stable. Consequently, if $\operatorname{NE}(G)$ is star-shaped about a completely mixed equilibrium, every singleton equilibrium is stable and (eq:target) holds for $G$ for any rule selecting nonempty subsets of $\operatorname{NE}(G)$.
Proof: Set $\mu=\min_{i,a}y_i(a)>0$ and choose $0<\rho<\min\{\mu,1\}$. For $\eta\in P_\rho$, define \[ s(\eta)=\left(1-\frac{\delta(\eta)}{\mu}\right)x+ \frac{\delta(\eta)}{\mu}y. \] It is an original Nash equilibrium by the segment assumption. Every coordinate is at least $\delta(\eta)y_i(a)/\mu\ge\delta(\eta)\ge\eta_i(a)$, so it is feasible for the perturbed game. Restricting deviations cannot invalidate an original equilibrium that remains feasible. The continuous-section criterion applies. Every nonempty selected set contains a point $x$, hence its stable singleton. ∎
Corollary (Strict and completely mixed equilibria): Every strict equilibrium and every completely mixed equilibrium has a stable singleton, under the stipulated strategy-perturbation definition.
Proof: For a completely mixed equilibrium take the constant section for sufficiently small perturbations, or put $x=y$ in Corollary Corollary (Equilibrium segments). For a strict equilibrium $a^*$, assign exactly $\eta_i(a)$ to every $a\ne a_i^*$ and all remaining probability to $a_i^*$. For sufficiently small $\rho$, continuity and the finitely many strict payoff gaps ensure that $a_i^*$ is the unique best reply at every such profile. Thus every player allocates all discretionary probability optimally. This is a continuous constrained-equilibrium section through $a^*$. ∎
Remark: These assertions concern strategy lower bounds in a fixed game, not arbitrary perturbations of its payoff array. Section sec:payofflimit shows why that distinction matters. The section criterion is sufficient, not a claimed characterization of all stable singleton equilibria.
Corollary (Backward induction and the earlier subclass): Suppose, for one permitted representation $\Gamma$ of $G$, every point of $Q_\Gamma$ admits a section as in Proposition Proposition (Continuous-section criterion). Then every solution concept satisfying $B$ obeys stable-set containment at $G$. In particular, if every quasi-perfect equilibrium in the simultaneous-move representation is strict, every selected connected set is a stable singleton.
Proof: Axiom $B$ puts some $x\in Q_\Gamma$ in any selected $K$; the section gives $\{x\}\in\operatorname{Stab}_0(G)$. In the strict case, strict payoff gaps persist in a neighborhood of $x$. Every equilibrium in that neighborhood must put probability one on the same uniquely optimal actions, so $x$ is isolated in $\operatorname{NE}(G)$. Its singleton is relatively open and closed in the Nash set. Connectedness of $K$ containing $x$ forces $K=\{x\}$. ∎

This retains the earlier draft's example: \[ \begin{array}{c|ccc}&L&M&R\\\hline U&(2,2)&(0,0)&(0,0)\\ D&(0,0)&(0,0)&(0,0) \end{array}. \] Writing $p=\Pr(U)$ and $q=\Pr(L)$, the payoff advantages of $U$ over $D$ and of $L$ over either alternative are $2q$ and $2p$. If $p>0$, equilibrium requires $q=1$ and then $p=1$. If $p=0$, action $D$ can be optimal only when $q=0$. Hence the Nash set is $(U,L)$ together with the segment $\{(D,tM+(1-t)R):0\le t\le1\}$. Against every completely mixed opponent, $D$ is strictly inferior to $U$, so no point on the segment is quasi-perfect. The strict point $(U,L)$ is quasi-perfect by continuity of its payoff gaps. Corollary Corollary (Backward induction and the earlier subclass) therefore selects only its singleton.

A whole family with non-isolated quasi-perfect equilibria

Let $n\ge3$ and let $A\in\mathbb R^{n\times n}$ satisfy \[ A^{\mathsf T}=-A,\qquad A\mathbf 1=0. \] Player $i$ chooses $a_i\in\{0,1\}$ and receives \begin{equation}\label{wsref-0-65-71-3a-73-6b-65-77-67-61-6d-65} u_i(a)=a_i\sum_{j=1}^n A_{ij}a_j. \end{equation} This is a finite polymatrix game: each payoff is a sum of pairwise terms. It is zero-sum because $a^{\mathsf T}Aa=0$ for every pure profile. Write $x_i=\Pr(a_i=1)$, so the mixed-strategy space is $[0,1]^n$.
Theorem: For every game (eq:skewgame), \begin{equation}\label{wsref-0-65-71-3a-6b-65-72-6e-65-6c-4e-45} \operatorname{NE}(G)=\ker A\cap[0,1]^n. \end{equation} Every equilibrium is quasi-perfect in the canonical simultaneous-move representation and is non-strict. Every equilibrium singleton is Mertens $0$-stable. Hence stable-set containment holds for every nonempty selected subset, without needing $B,D^*,I^*$.
Proof: The payoff advantage of action $1$ over $0$ is $d_i=(Ax)_i$. At a Nash profile, comparison with the feasible mixture $1/2$ gives $d_i(x_i-1/2)\ge0$. Yet \[ \sum_i d_i(x_i-1/2) =x^{\mathsf T}Ax-\tfrac12\mathbf1^{\mathsf T}Ax=0, \] using skew-symmetry and $A\mathbf1=0$. Thus every term is zero. If $d_i>0$, optimality requires $x_i=1$, making the term $d_i/2>0$. If $d_i<0$, it requires $x_i=0$, making the term $-d_i/2>0$. Both contradict the zero sum. Hence $Ax=0$. Conversely, $Ax=0$ makes both actions best replies for every player, proving (eq:kernelNE) and non-strictness.

The equilibrium set is convex and contains the completely mixed point $c=\frac12\mathbf1$. For any $x\in\operatorname{NE}(G)$, the profiles $x^k=(1-1/k)x+(1/k)c$ are completely mixed equilibria converging to $x$. Against every $x^k_{-i}$ both actions are optimal. This proves quasi-perfection using its defining completely mixed witnesses.

For clarity, singleton stability also has a direct affine certificate. Given $0<\rho<1/2$ and $\eta\in P_\rho$, set \begin{equation}\label{wsref-0-65-71-3a-61-66-66-69-6e-65-73-65-63-74-69-6f-6e} s_x(\eta)=(1-2\delta(\eta))x+\delta(\eta)\mathbf1. \end{equation} It belongs to $\ker A$ and every coordinate is in $[\delta(\eta),1-\delta(\eta)]$. Both action probabilities therefore meet all lower bounds, each of which is at most the common mass $\delta(\eta)$. It is an original Nash equilibrium and remains a constrained one. Proposition Proposition (Continuous-section criterion) proves stability of $\{x\}$. ∎

An explicit three-player instance without duplicate actions

Take \[ A=\begin{pmatrix}0&1&-1\\-1&0&1\\1&-1&0\end{pmatrix}, \qquad u_i(a)=a_i(a_{i+1}-a_{i+2}) \] with cyclic indices. Its equilibrium set is exactly \[ D=\{(t,t,t):0\le t\le1\}. \] No point of $D$ is isolated, and none is strict. Both actions of each player are genuinely distinct: their payoff difference is $1$ at one opponents' pure profile and $-1$ at another. Thus this continuum is not merely created by duplicating payoff-equivalent actions.

For every $t\in[0,1]$, formula (eq:affinesection) becomes \[ s_t(\eta)=\left(t+\delta(\eta)(1-2t)\right)\mathbf1. \] It explicitly handles both boundary equilibria $t=0,1$ as well as all interior equilibria. Block-diagonal matrices containing $k$ copies of this $3\times3$ matrix yield $3k$-player examples with a $k$-dimensional equilibrium cube and no strict equilibria. The same proof and section apply.

Two useful transfer results

Lifting a section through exact duplicates

Proposition: Suppose $H$ is obtained by exact duplication of pure actions of $G$. If $x\in\operatorname{NE}(G)$ has a continuous lower-bound equilibrium section, then every mixed-profile lift $\widetilde x$ of $x$ under the duplication map has such a section in $H$. In particular, every singleton $\{\widetilde x\}$ is stable.
Proof: Let $q_i:B_i\twoheadrightarrow A_i$ identify duplicates, so $u_i^H(b)=u_i^G(q(b))$. Given lower bounds $\widetilde\eta$, aggregate them as $\eta_i(a)=\sum_{q_i(b)=a}\widetilde\eta_i(b)$. Common perturbation mass is preserved. Write $s(\eta)$ for a section through $x$ in $G$. For each group $q_i^{-1}(a)$ choose weights $\lambda_i(b)\ge0$ summing to $1$. When $x_i(a)>0$, choose $\lambda_i(b)=\widetilde x_i(b)/x_i(a)$; when $x_i(a)=0$, all corresponding $\widetilde x_i(b)$ are zero and arbitrary weights suffice. Define \[ \widetilde s_i(b)=\widetilde\eta_i(b)+\lambda_i(b) \left(s_i(a;\eta)-\eta_i(a)\right),\qquad a=q_i(b). \] These coordinates meet their lower bounds, sum to one, and aggregate to $s_i(\eta)$. They are continuous and equal $\widetilde x$ at zero. Any feasible deviation in $H$ aggregates to a feasible deviation in $G$; its payoff is exactly the corresponding original payoff. Since $s(\eta)$ is a constrained equilibrium, no such deviation is profitable. The section criterion completes the proof, including zero-probability boundary groups. ∎

Containment descends through strong embeddings

Proposition: Let $\Phi$ satisfy $I^*$ and let $H$ strongly embed $G$ by $f$. If every $L\in\Phi(H)$ contains a stable subset, then every $K\in\Phi(G)$ contains a stable subset. For $D^*$ extensions, containment transfers in both directions under the natural identification.
Proof: Given $K\in\Phi(G)$, choose $L\in\Phi(H)$ with $f(L)=K$ by $I^*$. By hypothesis, some $S_H\in\operatorname{Stab}_0(H)$ is contained in $L$. The established stable-image property gives $f(S_H)\in\operatorname{Stab}_0(G)$, and $f(S_H)\subseteq K$. For a $D^*$ extension the Nash sets, selected families, and stable families are identified, so the conclusion is immediate in either direction. ∎

The reverse strong-embedding inference is not supplied by this proof. A stable $S\subseteq f(L)$ has some stable lift under the literature's stable-lift property; that property does not place this lift inside the particular selected set $L$. Confusing these two quantifiers would silently assume the missing containment.

Why a limit of generic-game conclusions is insufficient

Proposition (One-sided tie payoff-limit obstruction): There exist two-player games $H_\varepsilon\to H_0$ and stable singleton sets $S_\varepsilon\to S_0$ in Hausdorff distance such that $S_0$ is not stable. The limiting game has a nontrivial Nash-equilibrium continuum.
Proof: Let \[ H_\varepsilon=\begin{array}{c|cc} &L&R\\\hline T&(1,1)&(1,0)\\ B&(1+\varepsilon,1)&(0,0) \end{array},\qquad \varepsilon\ge0. \] For every $\varepsilon>0$, $(B,L)$ is strict: $B$ strictly beats $T$ against $L$, and $L$ strictly beats $R$ against $B$. Hence $S_\varepsilon=\{(B,L)\}$ is $0$-stable by the strict-equilibrium section, and $S_\varepsilon\to S_0=\{(B,L)\}$.

In $H_0$, player 2 strictly prefers $L$ independently of the opponent; if player 2 chooses $R$ with probability $r$, player 1's payoff advantage of $T$ over $B$ is exactly $r$. Thus \[ \operatorname{NE}(H_0)=\{(pT+(1-p)B,L):0\le p\le1\}. \] For a perturbation vector $\eta$ with every lower bound strictly positive and $0<\delta(\eta)<1$, the constrained game has a unique Nash profile \[ x_1(T)=1-\eta_1(B),\qquad x_2(L)=1-\eta_2(R). \] As $\eta\to0$ through such interior perturbations, these profiles converge to $(T,L)$, not to $(B,L)$. To verify failure of the actual connexity requirement, suppose $S_0$ had a stability witness $X$ with $X_0=\{(0,(B,L))\}$. Choose $V$ a small relative neighborhood of $X_0$ in $X$ so that no profile in $V$ places more than $1/4$ on $T$ and $\delta(\eta)<1/4$. There are no graph points in $V$ over $\operatorname{relint}P_1$, because every such equilibrium places $1-\eta_1(B)>3/4$ on $T$. Hence $V\setminus\partial X_1=\varnothing$, violating connexity. Therefore $S_0$ is not stable. A continuous section through $(T,L)$ exists by the displayed formula, so this failure is specific to the other endpoint and does not imply that $H_0$ lacks stable sets. ∎

The preceding example is incorporated from the user-supplied WS-107793-attempt.tex. It refutes only the proposed passage to payoff limits; it does not give a counterexample to the globally quantified axiomatic problem. In particular, backward induction in the simultaneous-move representation excludes the endpoint $(B,L)$.

Proposition: There are games $G_k\to G_0$ and singleton stable equilibria $K_k\to K_0$ in Hausdorff distance such that $K_0$ contains no stable subset.
Proof: For $\theta\ge0$ consider \[ G_\theta=\begin{array}{c|cc} &L&R\\ \hline U&(1,1)&(0,0)\\ D&(0,0)&(\theta,\theta) \end{array}. \] For every $\theta>0$, $(D,R)$ is strict, so $K_\theta=\{(D,R)\}$ is stable by Corollary Corollary (Strict and completely mixed equilibria). As $\theta\downarrow0$, the payoff arrays converge and these singleton sets have the same limit $K_0$.

We show directly that $K_0$ is not stable in $G_0$, using the stipulated connexity condition. Suppose it had a witness $X\subseteq E_{G_0}$ with $X_0=\{(0,(D,R))\}$. Let $p=\Pr(U)$ and $q=\Pr(L)$, and consider the relative neighborhood \[ V=X\cap\{\delta(\eta)<1/4,\ p<1/4,\ q<1/4\} \] of $X_0$. If all coordinates of $\eta$ are strictly positive, then $q\ge\eta_2(L)>0$ makes $U$ strictly preferable to $D$, and $p\ge\eta_1(U)>0$ makes $L$ strictly preferable to $R$. Every constrained equilibrium must therefore satisfy \[ p=1-\eta_1(D)>3/4,\qquad q=1-\eta_2(R)>3/4 \] when $\delta(\eta)<1/4$. No such profile belongs to $V$. Consequently every parameter occurring in $V$ lies on the relative boundary of $P_1$: it has a zero coordinate, including the zero-mass case. Thus $V\setminus\partial X_1$ is empty, contradicting connexity. Hence $K_0$ is not stable. Its only nonempty subset is itself. ∎

The positive-$\theta$ games are already nondegenerate bimatrix games. If a particular stronger genericity convention is desired, take generic payoff arrays arbitrarily close to each $G_{1/k}$ within the open region where $(D,R)$ remains strict. They still converge to $G_0$ and give the same counterexample. This invalidates the unqualified step "approximate by generic games and pass stability to the limit." It does not refute (eq:target): no globally compatible rule selecting $K_0$ is constructed. Indeed the simultaneous-move backward-induction requirement already excludes this particular $K_0$.

A polynomial topological obstruction to a second shortcut

The source identifies a gap between coincidence properties and rational cohomological essentiality in the nongeneric case [gw, Section 10]. The following classical Hopf-map construction makes the distinction explicit. It is not asserted to be a Nash perturbation graph.

Let $D^m$ be the closed Euclidean unit ball. Regard $D^4$ as the unit ball in $\mathbb C^2$ and define the real polynomial map \begin{equation}\label{wsref-0-65-71-3a-68-6f-70-66} F(z_1,z_2)=\left(2\operatorname{Re}(z_1\overline z_2),\, 2\operatorname{Im}(z_1\overline z_2),\,|z_1|^2-|z_2|^2\right). \end{equation} Its boundary restriction is a Hopf map, up to the harmless choice of orientation. The non-null-homotopy of this map is a standard input; see [hatcher, Example 4.51, p. 380; Section 4.B, p. 427].

Proposition: The map $F:D^4\to D^3$ has a coincidence with every continuous map $g:D^4\to D^3$. Nevertheless, \[ F^*:\check H^3(D^3,S^2;\mathbb Q)\longrightarrow \check H^3(D^4,S^3;\mathbb Q) \] is zero. The restriction of $F$ to any compact subpair $(Y,Y\cap S^3)$ also induces the zero map in degree three.
Proof: Direct calculation gives $\|F(z)\|^2=(|z_1|^2+|z_2|^2)^2$; hence $F^{-1}(S^2)=S^3$. Suppose a continuous $g:D^4\to D^3$ has no coincidence with $F$. Then \[ R(z)=\frac{F(z)-g(z)}{\|F(z)-g(z)\|} \] is a continuous map $D^4\to S^2$. On the boundary $S^3$, consider \[ h_t(z)=\frac{F(z)-t\,g(z)}{\|F(z)-t\,g(z)\|},\qquad 0\le t\le1. \] For $t<1$, the denominator is nonzero because $\|F(z)\|=1$ and $\|t g(z)\|\le t<1$; for $t=1$ nonvanishing follows from the no-coincidence hypothesis. The map $h_t$ is a homotopy from $F|_{S^3}$ to $R|_{S^3}$. But $R|_{S^3}$ extends over $D^4$ and is therefore null-homotopic, which contradicts the non-null-homotopy of the Hopf map. Thus a coincidence exists for every $g$.

For the ball pairs, relative Čech cohomology agrees with ordinary relative cohomology, giving \[ \check H^3(D^3,S^2;\mathbb Q)\cong\mathbb Q,\qquad \check H^3(D^4,S^3;\mathbb Q)=0. \] Therefore $F^*=0$. For a compact $Y\subseteq D^4$, restriction of the map of pairs to $(Y,Y\cap S^3)$ factors through $(D^4,S^3)$; functoriality shows that the induced degree-three map remains zero. ∎

The graph $\{(F(z),z):z\in D^4\}\subseteq D^3\times D^4$ is compact and semialgebraic, and projection to its first coordinate is $F$ under a homeomorphism. Thus merely insisting that a map be a polynomially described coordinate projection does not cure the obstruction. The example rules out a general topological inference from coincidence to rational essentiality. To turn it into a counterexample to WS-107793, one would still need to realize suitable structure in an actual Nash perturbation graph and prove compatibility with all comparisons in $B,D^*,I^*$. Neither requirement is established here.

Exact outstanding assertion and proof audit

The general theorem would follow from \begin{equation}\label{wsref-0-65-71-3a-67-61-70} K\in\mathcal A_\infty(G)\quad\Longrightarrow\quad \text{a stability witness has nonempty zero fiber contained in }\{0\}\times K. \end{equation} Such a witness must satisfy both the source's connexity condition and nonzero rational relative cohomology. The compactness argument constructs compatible selected sets; it does not construct this perturbation witness. The continuous-section argument supplies witnesses in the stated subclasses, not for every survivor. No line of the manuscript proves (eq:gap).
Quantifiers checked.
The construction uses a global domain and both directions of every comparison. Lifts at the limit come from nested compact fibers, not from an unlicensed exchange of existence and limits. Boundary lower bounds, zero-probability duplicate groups, and the pure endpoints of the equilibrium continuum are included. The equilibrium examples do not use genericity or isolation. Changing games in Proposition Proposition is expressly distinguished from fixed-game hyperspace limits in Theorem Theorem (Largest compatible solution concept).
Scope of the counterexamples.
Proposition Proposition refutes payoff-limit closure of singleton stability. Proposition Proposition refutes an unrestricted topological implication. Neither is a counterexample to the posted game-theoretic statement.
Checks and reproducibility.
The accompanying short Python program checks the displayed polynomial identities symbolically and the explicit section inequalities using exact rational test data. Those checks supplement the proofs; they do not establish the universal axioms, the Hopf-map theorem, or the missing assertion (eq:gap). The manuscript is not a Lean or other proof-assistant formalization, and no independent referee verification is claimed.
Novelty and provenance.
The source's definitions, known stability properties and generic theorem remain credited to their authors. Hyperspace compactness, section arguments, and the Hopf obstruction use standard mathematics. The compact pruning formulation and explicit examples were developed in this attempt; no claim that they are absent from the literature is made. The prior manuscripts reported assistance from GPT-6 Astra Pro; the present consolidation used ChatGPT GPT-6, without an available build identifier or independently verified deployment settings. The user requested and directed the work; no human verification is attributed to them. This version supersedes the earlier local draft as a research write-up, not as a change to the website's problem status.
Recommended submission metadata.
Problem: WS-107793. Claim type: partial result. Scope: largest-compatible-rule construction, exact elimination reformulation, continuous-section criterion, a nongeneric polymatrix family, and two obstructions to proposed proof routes. Status of WS-107793: unresolved. No public issue, repository edit, or solved-status update is made by preparing this manuscript.

Models used

This consolidation used ChatGPT GPT-6 to compare the two supplied LaTeX attempts, scrutinize the arguments, and prepare this merged draft. Exact deployment snapshot and reasoning settings were not independently available. The supplied manuscript declares "OpenAI GPT 6 Astra Pro"; that is preserved as the supplied manuscript's attribution, not asserted as this run's verified model setting. No Lean or other formal proof assistant was used.

Human contributors and roles

The user supplied the WS-labelled attempt and the earlier local draft, requested combination, and specified the target. No human mathematician's proof endorsement or referee review is claimed. Govindan and Wilson retain sole credit for the cited prior theory, and Hatcher for the cited topological background. The added combination is an LLM-assisted research draft with unverified novelty.

References

  • [gw] Srihari Govindan and Robert Wilson. Axiomatic Equilibrium Selection in Generic Extensive-Form Games. arXiv:2504.16908v2, 26 August 2026. The PDF title reads Axiomatic Equilibrium Selection: The Case of Generic Extensive Form Games. Definitions 2.1, 2.4, 2.5, 4.4, 5.2; Axioms 3.1, 4.7, 5.4; Theorem 6.1; Section 10; Appendices A and B. https://arxiv.org/html/2504.16908v2. https://arxiv.org/pdf/2504.16908v2.
  • [hatcher] Allen Hatcher. Algebraic Topology. Cambridge University Press, 2002; author-hosted electronic edition. Example 4.51, p. 380; Section 4.B, p. 427; the relative cohomology of balls. https://pi.math.cornell.edu/~hatcher/AT/AT.pdf.
  • [attachment] User-supplied Local witnesses for nongeneric stability and two obstructions to extending the generic theorem, 18 October 2026, file WS-107793-attempt.tex (supplied working draft; not a peer-reviewed source).
  • [problem] We-Solve. Axiomatic containment of Mertens-stable sets without genericity. Permanent public problem identifier WS-107793. Consulted predecessor statement revision stm_32cd4f81d2514415920d3ba7e50d23af. Permanent problem lookup. Published problem record. Prompt blob SHA: f34db89188d9a2b326623f8ccb664f0956da7f4b.