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? > 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. Alex ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
