Abstract
A fundamental property of choice functions is stability, which, loosely speaking, prescribes that choice sets are invariant under adding and removing unchosen alternatives. We provide several structural insights that improve our understanding of stable choice functions. In particular, (1) we show that every stable choice function is generated by a unique simple choice function, which never excludes more than one alternative, (2) we completely characterize which simple choice functions give rise to stable choice functions, and (3) we prove a strong relationship between stability and a new property of tournament solutions called local reversal symmetry. Based on these findings, we provide the first concrete tournamentâ€”consisting of 24 alternativesâ€”in which the tournament equilibrium set fails to be stable. Furthermore, we prove that there is no more discriminating stable tournament solution than the bipartisan set and that the bipartisan set is the unique most discriminating tournament solution which satisfies standard properties proposed in the literature.
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Notes
We refer to Monjardet (2008) for a more thorough discussion of the origins of this condition.
Together with TheoremÂ 3, TheoremÂ 2 also implies that, for any stable tournament solution S, \(\lceil {S}\rceil \) is a coarsest generator of S. When only considering generators that satisfy \({\widehat{\alpha }}_{_\subseteq }\), \(\lceil {S}\rceil \) is also the coarsest generator of S. In addition, since simple choice functions trivially satisfy \({\widehat{\alpha }}_{_\subseteq }\), the two theorems imply that \(\lceil {S}\rceil \) is the unique simple choice function generating S.
It can be checked that we obtain an equivalent condition even if we require that all outside alternatives have to be removed. When defining local \({\widehat{\alpha }}\) in this way, it can be interpreted as some form of transitivity of stability: Stable sets of minimally stable sets are also stable within the original feasible set (cf. Brandt 2011, Lem.Â 3).
It is open whether there are tournaments in which \( BP \) and \( TEQ \) are disjoint.
Brandtâ€™s proof relies on a lemma that essentially showed that the generators he considers always satisfy local \({\widehat{\alpha }}\).
The name of this axiom is inspired by a social choice criterion called reversal symmetry. Reversal symmetry prescribes that a uniquely chosen alternative has to be unchosen when the preferences of all voters are reversed (Saari and Barney 2003). A stronger axiom, called ballot reversal symmetry, which demands that the choice set is inverted when all preferences are reversed, was recently introduced by Duddy etÂ al. (2014).
To see that discriminative power is not captured by the axioms, observe that the trivial tournament solution satisfies stability, monotonicity, regularity, and compositionconsistency.
However, these analytic results stand in sharp contrast to empirical observations that Condorcet winners are likely to exist in realworld settings, which implies that tournament solutions are much more discriminative than results for the uniform distribution suggest (Brandt and Seedig 2016).
Note that \((T_7)^g\) is the smallest tournament in which \({ BA}\) and \({ UC}\) differ (Brandt etÂ al. 2015).
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Acknowledgements
This material is based on work supported by Deutsche Forschungsgemeinschaft under Grants BRÂ 2312/71 and BRÂ 2312/72, by a Feodor Lynen Research Fellowship of the Alexander von Humboldt Foundation, by ERC Starting Grant 639945, by a Stanford Graduate Fellowship, and by the MITGermany program. The authors thank Christian Geist for insightful computer experiments and Paul Harrenstein for helpful discussions and preparing Fig.Â 1.
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Appendices
Appendix 1: Examples for RemarkÂ 2
Brandt and Harrenstein (2011,Â p.Â 1729) mention that \({\widehat{\alpha }}\) and \({\widehat{\gamma }}\) are independent from each other in the context of general choice functions. Here, we prove that the same holds even in the context of tournament solutions.
Proposition 1
There exists a tournament solution that satisfies \({{\widehat{\alpha }}}\), but not \({{\widehat{\gamma }}}\).
Proof
Let S be a stable tournament solution. As mentioned in Sect.Â 3, \(\lceil {S}\rceil \) satisfies \({\widehat{\alpha }}\). However, it is easily seen that, unless S is trivial, \(\lceil {S}\rceil \) violates \({\widehat{\gamma }}\). Hence, the statement follows from the existence of nontrivial stable tournament solutions (such as \( BP \)). \(\square \)
Proposition 2
There exists a tournament solution that satisfies \({{\widehat{\gamma }}}\), but not \({{\widehat{\alpha }}}\).
Proof
Let S be a stable tournament solution. Define the tournament solution \(S'\) such that for each tournament \(T=(A,\succ )\),
It can be shown that \(S'\) satisfies \({{\widehat{\gamma }}}\), but may violate \({{\widehat{\alpha }}}\). For the latter, let \(S= TC \) and consider a transitive tournament \((\{a,b,c\},\succ )\) such that \(a \succ b\), \(b \succ c\), and \(a \succ c\). By definition, \(S'(\{a,b,c\})=\{a\}\), but \(S'(\{a,b\})=\{a,b\}\). \(\square \)
Appendix 2: Examples for RemarkÂ 3
\(\textit{BA}\) satisfies local \({\widehat{\alpha }}\), but \({\widehat{\textit{BA}}}={ ME}\) violates \({\widehat{\alpha }}\) (Brandt etÂ al. 2016b).
Similarly, there exists a tournament solution S for which \({\widehat{S}}\) is welldefined, but \({\widehat{S}}\) is not stable. For a stable tournament solution S, we have by definition that \(S={\widehat{S}}\) and hence that \({\widehat{S}}\) is also stable. The following proposition shows that \({\widehat{\alpha }}\) does not carry over from S to \({\widehat{S}}\) even if S is simple and \({\widehat{S}}\) is welldefined.
Proposition 3
There exists a simple tournament solution S satisfying \({\widehat{\alpha }}\) such that \({\widehat{S}}\) is welldefined but \({\widehat{S}}\) does not satisfy \({\widehat{\alpha }}\).
Proof
Let S be the tournament solution that always chooses all alternatives, with two exceptions:

If the tournament is of order 2, then S chooses only the Condorcet winner.

If the tournament is the tournament \(T_4\) given in Fig.Â 6, then S chooses alternatives a,Â b,Â and c.
Clearly, S is simple and satisfies \({\widehat{\alpha }}\). Since \({\widehat{S}}\) chooses alternatives a,Â b,Â and c from \(T_4\), but chooses only the Condorcet winner from the transitive tournament of order 3, it does not satisfy \({\widehat{\alpha }}\).
It remains to show that \({\widehat{S}}\) is welldefined. One can check that every tournament contains an Sstable set. Suppose for contradiction that some tournament T contains two distinct minimal Sstable sets, which we denote by B and C. Then B and C are also Sstable in \(B\cup C\). If B is a singleton, then B is the Condorcet winner in \(B\cup C\), which means C cannot be Sstable, a contradiction. Hence both B and C are transitive tournaments of order 3, and \(4\le B\cup C\le 6\). One can check all the possibilities of \(B\cup C\) to conclude that this case is also impossible. \(\square \)
For the tournament solution S defined in the proof of PropositionÂ 3, we have that \(\widehat{{\widehat{S}}}\) is not welldefined. Even though no tournament contains two distinct minimal \({\widehat{S}}\)stable sets, \(T_4\) does not contain any \({\widehat{S}}\)stable set. This example also shows that for a tournament solution \(S'\), \(\widehat{S'}\) may fail to be welldefined not because it allows two distinct minimal \(S'\)stable sets in a tournament but because some tournament contains no \(S'\)stable set.
Appendix 3: Example for RemarkÂ 6
We show that there exists a tournament solution different from \( BP \) that satisfies \( LRS \), monotonicity, regularity, and Condorcet consistency.
To this end, we define a new tournament solution called \( POS \) which chooses all alternatives with positive relative degree. More precisely, an alternative is chosen by \( POS \) if it dominates strictly more than half of the remaining alternatives, is not chosen if it dominates strictly less than half of the remaining alternatives, and goes to a â€śtiebreakâ€ť to determine whether it is chosen if it dominates exactly half of the remaining alternatives.
For tournaments of even size, \( POS \) chooses exactly the alternatives that dominate at least (or equivalently, more than) half of the remaining alternatives. Hence we do not need a tiebreak for tournaments of even size. The tiebreaking rule for tournaments of odd size \(2n+1\) is as follows: For any (unlabeled) tournament T of order 2n and any partition of it into two sets B and C of size n, consider two tournaments \(T_1\) and \(T_2\) of order \(2n+1\). The tournament \(T_1\) contains T and another alternative a that dominates B but is dominated by C, while the tournament \(T_2\) contains T and another alternative a that dominates C but is dominated by B. If \(T_1\) or \(T_2\) is regular, \( POS \) chooses a in that tournament and not in the other one. Otherwise, \( POS \) arbitrarily chooses a in exactly one of \(T_1\) and \(T_2\).
Proposition 4
\( POS \) satisfies \( LRS \), monotonicity, and regularity.
Proof
We need to show that the tiebreaking rule in the definition of \( POS \) is welldefined. First, we show that if we perform a local reversal on alternative a, we do not get an isomorphic tournament with alternative a mapped to itself. Indeed, if a were mapped to itself, it would mean that no tournament solution satisfies \( LRS \), which we know is not true since \( BP \) satisfies \( LRS \). Secondly, the tournament obtained by performing a local reversal on an alternative in a regular tournament is not regular. Hence we do not obtain a conflict within the tiebreaking rule.
It follows directly from the definition that \( POS \) satisfies \( LRS \), monotonicity, and regularity. \(\square \)
The tournament \(T_4\) given in Fig.Â 6 shows that \( POS \) violates compositionconsistency and \({\widehat{\alpha }}\) (and hence stability).
Interestingly, \( BP \) (and all of its coarsenings) always intersect with \( POS \) while there exists a tournament for which \( BA \) (and all of its refinements such as \( TEQ \) and \( ME \)) do not overlap with \( POS \). This follows from results on the Copeland value by Laffond etÂ al. (1993b, 1994).
Appendix 4: Example for RemarkÂ 7
We construct a tournament solution that satisfies monotonicity and stability, but violates regularity and compositionconsistency.
Every tournament solution has to be regular on tournaments of order 5 or less because of nontrivial automorphisms. Consider the tournament \(T_7\) shown in Fig.Â 7. \(T_7\) admits a unique nontrivial automorphism that maps each of the six alternatives in the two 3cycles to the next alternative in its 3cycle and maps alternative g to itself.^{Footnote 14}
Now, define the simple tournament solution \(S_7\), which always returns all alternatives unless the tournament is \(T_7\) or it can be modified from \(T_7\) by weakening alternative g. In the latter case, \(S_7\) returns all alternatives except g.
We check that this definition is sound. First, we know that in \(T_7\), there is no automorphism that maps alternative g to another alternative. When we weaken g, it is the unique alternative with the smallest outdegree and hence cannot be mapped by an automorphism to another alternative. Now, the alternatives a,Â b,Â c form an orbit, and \(S_7\) excludes g whenever it is dominated by d,Â e,Â f (and has any dominance relationship to a,Â b,Â c). This yields four nonisomorphic tournaments for which \(S_7\) excludes g.
Proposition 5
\(\widehat{S_7}\) satisfies stability and monotonicity.
Proof
First, observe that \(S_7\) trivially satisfies local \({\widehat{\alpha }}\) because \(S_7\) only excludes an alternative in tournaments of order 7. By virtue of TheoremÂ 3, it therefore suffices to show that \(\widehat{S_7}\) is welldefined.
One can check that every tournament contains an \(S_7\)stable set. Let \(T_6\) denote the tournament obtained by removing alternative g from \(T_7\). Suppose for contradiction that there exists a tournament T that contains two distinct minimal \(S_7\)stable sets, which we denote by B and C. Then B and C are also \(S_7\)stable in \(B\cup C\). Moreover, \(T_{B}\) must correspond to the tournament \(T_6\), and each alternative in \(C\backslash B\) either has the same dominance relation to B as the alternative g does to \(T_6\) or has a dominance relation that is a weakening of g. The same statement holds for C. We consider the following cases.
Case 1: \(10\le B\cup C\le 11\). The tournament \(T_{B}\) has one of its alternatives corresponding to alternatives d, e, and f in Fig.Â 7 outside of \(B\cap C\), and this alternative must dominate all of the alternatives in C. Similarly, there exists an alternative in \(C\backslash B\) that dominates all of the alternatives in B. But this implies that some two alternatives dominate each other, a contradiction.
Case 2: \(7\le B\cup C\le 9\). At least one of the two tournaments \(T_{B}\) and \(T_{C}\) must have all of its alternatives corresponding to alternatives d, e, and f in Fig.Â 7 in the intersection \(B\cap C\), for otherwise we obtain a contradiction in the same way as in Case 1. Assume without loss of generality that \(T_{B}\) has its alternatives corresponding to d, e, and f in the intersection. Hence three alternatives in \(B\cap C\) that form a cycle dominate the same alternative in C. But this does not occur in \(T_6\), a contradiction.
It follows from TheoremÂ 4 that \(\widehat{S_7}\) satisfies monotonicity. \(\square \)
Clearly, \(\widehat{S_7}\) is not regular since it excludes an alternative from the regular tournament \(T_7\). Moreover, it is not a coarsening of \( BP \) since \( BP \) selects all of the alternatives in \(T_7\). Hence we have that stable and monotonic tournament solutions are not necessarily coarsenings of \( BP \).
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Brandt, F., Brill, M., Seedig, H.G. et al. On the structure of stable tournament solutions. Econ Theory 65, 483â€“507 (2018). https://doi.org/10.1007/s001990161024x
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DOI: https://doi.org/10.1007/s001990161024x