[EM] Thoughts about set methods

[Forest Simmons]

It looks right ... and a valuable study that should be part of any
systematic exposition of Election Methods  ...taking up where Woodall left
off.

On Sat, Nov 4, 2023, 2:47 PM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> Here are some thoughts about methods made by composing sets and base
> methods. Could you check if I'm right?
>
> Define a "set" as a method that accepts an election and returns a set of
> candidates. E.g. the Smith set is a set by this definition.
>
> And define the method A,B where A is a set and B is a method, as picking
> the member of A that's highest ranking according to B's social order (or
> members, in case of a tie).
>
> For a set, let the monotone inclusion criterion (or inclusion
> monotonicity) be defined like this:
>         1. For any election where A is in the set, raising A can't insert
> anyone else into the set;
>         2. lowering A can't boot anyone from the set before A is booted
> out;
>         3. and raising A can never boot A from the set if he's already in
> it,
> nor can lowering A insert him into it if he's not.
>
> (The Smith set, for instance, passes this criterion.)
>
> First claim: Suppose X and Y inclusion-monotone sets, and Z is their
> intersection (i.e. a candidate is in Z iff he's in both X and Y). Then Z
> is inclusion-monotone.
>         1 and 2. If candidate C is in Z, then C is in both X and Y by
> definition. Since both X and Y pass monotone inclusion, nobody can be
> inserted into either of them by raising C. Thus nobody can be inserted
> into Z either. The reasoning is analogous for the second condition.
>         3. C can't be booted from X by raising nor from Y since they're
> both
> inclusion-monotone. Therefore C will remain in both if raised. (The
> reasoning is analogous for the "lowering A" criterion.)
>
> Second: Let A be a set and B be a method. If A is inclusion-monotone and
> B passes mono-raise, then A,B passes mono-raise.
>
> Proof idea: Consider an election e where candidate C is the winner
> according to A,B. Then raising C could lead someone else to win if:
>         1a. B's social ordering ranks candidate C2 above C, and raising C
> inserts C2 into the A set, or
>         2a. B's social ordering ranks candidate C right above C2, and
> raising C
> removes C from the A set, or
>         3a. B is nonmonotone and raising C leads C2, who is also a member
> of A,
> to be ranked ahead of C according to method B.
>
> Lowering C could lead someone C to win if:
>         1b. B's social ordering ranks C over C2, and lowering C inserts
> him
> into the A set, or
>         2b. B's social ordering ranks C2 right above C, and lowering C
> removes
> C2 from the set, or
>         3b. B is nonmonotone (etc.)
>
> In both cases, the first two possibilities are eliminated by the
> inclusion monotonicity of A, and the third by B passing monotonicity. In
> particular:
>
> - Inclusion monotonicity condition 1 makes 1a impossible,
> - Inclusion monotonicity condition 3 makes 2a impossible,
> - B passing mono-raise makes 3a impossible,
> - Inclusion monotonicity condition 3 makes 1b impossible,
> - Inclusion monotonicity condition 2 makes 2b impossible,
> - B passing mono-raise makes 3b impossible.
>
> Hence raising C can't make him lose according to method A,B; nor can
> lowering C make him win.
>
> Does that seem right?
>
> -km
> ----
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> info
>
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