[EM] Thoughts about set methods

[Kristofer Munsterhjelm]

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