On Thu, 6 Feb 2003, Alex Small wrote: > Forest Simmons said: > > It seems reasonable that if S is a ballot set with a definite winner X, > > and T is any other ballot set, then sufficiently many copies of S added > > to T should result in a ballot set supporting X. > > Let me see if I can understand what you meant about the openness of the > victory regions. Say that S and T each have the same number of voters in > them. We can represent adding copies of S to T by writing the resulting > electorate as > > E = q*S + T (q is a real, positive number) > > For sufficiently large q, T is "drowned out" and we assume that X is the > winner. Now, since we assume that election methods shouldn't care how > many voters have each preference, only what fraction of the voters have > each preference, we can always normalize our new electorate as > > E(normalized) = (q*S + T)/(1+q) = S*q/(1+q) + T/(1+q) > > If we assume that X is the winner for an electorate that is purely S, and > also for an electorate that is "mostly S" (finite q) then we see that > regions in which X wins must have non-zero volume, as measured in whatever > space we're using. Is that equivalent to what you say below?
That's the right idea. But instead of saying "must have non-zero volume," say "must be open." A set is open if each member of the set is shielded from non-members by other members of the set, so that sufficiently small perturbations cannot knock it out of the set. In other words, each member is surrounded by a ball of positive radius entirely contained within the set. In this context, positive radius implies positive N!-1 dimensional volume, but the reverse implication is not true. To see this in a simpler setting consider in three dimensional space that every ball of positive radius has a positive volume, and every open set contains balls of positive radius, so every open set has positive volume. On the other hand, a ball of positive radius is not itself open if it contains its boundary sphere, yet it still has positive volume. [The points on the boundary sphere are not shielded from the complement of the ball.] > > > For those who have had a little topology, this condition can be > > interpreted as openness of the victory regions in the space of all > > possible elections associated with the method. > ... > > I believe that this condition and the Pareto condition taken together > > might be sufficient to rule out the Strong FBC. > > Do tell more when you get the chance. I also have some ideas for how to > resuscitate my treatment of strong FBC for 3 candidates, and perhaps on > how to generalize to 4+ candidates. > Note that the Pareto condition and the Openness condition together imply that sufficiently many ballots with X in first place will swamp any other set of ballots and allow X to win. So these two conditions together yield a kind of "Overwhelming Majority Condition" which is weaker than the Majority Criterion, but (I believe) still strong enough to rule out the Strong FBC. To be continued ... Forest ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
