Editing Smith set (section)

===Properties of dominating sets===
'''''Theorem:''''' Dominating sets are ''nested''; that is, of any two dominating sets in an election, one is a subset of the other.

'''''Proof:''''' Suppose on the contrary that there exist two dominating sets, ''D'' and ''E'', neither of which is a subset of the other. Then there must exist candidates {{nowrap|''d'' ∈ ''D'',}} {{nowrap|''e'' ∈ ''E''}} such that {{nowrap|''d'' ∉ ''E''}} and {{nowrap|''e'' ∉ ''D''.}} But by hypothesis ''d'' defeats every candidate not in ''D'' (including ''e'') while ''e'' defeats every candidate not in ''E'' (including ''d''), a contradiction. ∎

'''''Corollary:''''' It follows that the Smith set is the smallest non-empty dominating set, and that it is well defined.

'''''Theorem:''''' If ''D'' is a dominating set, then there is some threshold θ<sub>''D''</sub> such that the elements of ''D'' are precisely the candidates whose [[Copeland's method|Copeland scores]] are at least θ<sub>''D''</sub>. (A candidate's Copeland score is the number of other candidates whom he or she defeats plus half the number of other candidates with whom he or she is tied.)

'''''Proof:''''' Choose ''d'' as an element of ''D'' with minimum Copeland score, and identify this score with θ<sub>''D''</sub>. Now suppose that some candidate {{nowrap|''e'' ∉ ''D''}} has a Copeland score not less than θ<sub>''D''</sub>. Then since ''d'' belongs to ''D'' and ''e'' doesn't, it follows that ''d'' defeats ''e''; and in order for ''e''{{'}}s Copeland score to be at least equal to ''d''{{'}}s, there must be some third candidate ''f'' against whom ''e'' gets a better score than does ''d''. If {{nowrap|''f'' ∈ ''D'',}} then we have an element of ''D'' who does not defeat ''e'', and if {{nowrap|''f'' ∉ ''D''}} then we have a candidate outside of ''D'' whom ''d'' does not defeat, leading to a contradiction either way. ∎