[EM] Name that Criterion
fsimmons at pcc.edu
Thu Feb 6 15:51:55 PST 2003
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
> > 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 ...
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