Voting system is a concept in political science. Here I give the mathematical definition of a voting system.

A (binary) voting system is a tuple $(P,V,q)$, where $P$ is any set, called the set of proposals, and $V$ is a finite set of preference relations on $P$, called the set of voters, and $q$ is an integer between (inclusive) $0$ and $\left|V\right|$, called the quota.

For each voter $v\in V$ and two proposals $x,y\in P$, we denote “$v$ prefers $x$ to $y$” by

\[x\succeq_vy.\]

A proposal $x\in P$ is a defeat of $y\in P$ if

\[\left|\left\{v\in V\,\middle|\,x\succeq_vy\right\}\right|\geq q,\]

denoted as $x\succsim_{V,q}y$ (despite this notation, $\succsim_{V,q}$ is not necessarily a preference relation on $P$ because it is not transitive generally, which is actually a well-known example of irrationality).

The core $\mathcal C(P,V,q)$ of the voting system is the set of such element $x\in P$: $x$ does not have any defeat other than $x$ itself (non-trivial defeat).


Pareto sets are common concepts in economics. To clarify, I also give the mathematical definition of them here.

Let $P$ be a set and $Q$ be a family of preference relations on $P$. Then, $x\in P$ is called a (weak) $Q$-Pareto improvement of $y\in P$ if $\forall v\in V:x\succeq_vy$, denoted as $x\succsim_Qy$ (despite the notation, $\succsim_Q$ is not necessarily a preference relation on $P$).

The Pareto set $\mathcal P(P,Q)$ is the set of all such element $x\in P$: $x$ does not have any $Q$-Pareto improvement other than $x$ itself (non-trivial $Q$-Pareto improvement).


Here is the main result. For a voting system $(P,V,q)$,

\[\mathcal C(P,V,q)=\bigcap_{Q\subseteq V,\left|Q\right|=q}\mathcal P(P,Q).\]

Proof. To prove this, we need to show that $x\in P$ does not have any non-trivial Pareto improvement for any $q$ voters iff $x$ does not have any non-trivial defeat.

To prove the forward direction, suppose that $x\in P$ does not have any non-trivial Pareto improvement for any $q$ voters. Let $y\in P$ such that $y\ne x$, and the goal is to prove that $y$ is not a defeat of $x$.

Let

\[Y:=\left\{v\in V\,\middle|\,y\succeq_vx\right\}.\]

Then, $y$ is a $Y$-Pareto improvement of $x$, so we have $\left|Y\right|<q$ (because otherwise there is a subset of $Y$ with $q$ voters for which $y$ is a Pareto improvement of $x$). Therefore, $y$ is not a defeat of $x$.

To prove the backward direction, suppose that $x\in P$ has a non-trivial $Q$-Pareto improvement, where $Q\subseteq V$ and $\left|Q\right|=q$. Denote the improvement as $y$. Let

\[Y:=\left\{v\in V\,\middle|\,y\succeq_vx\right\}.\]

because $y$ is a $Q$-Pareto improvement of $x$, we have $Q\subseteq Y$. Therefore, $\left|Y\right|\geq\left|Q\right|=q$. Therefore, $y$ is a defeat of $x$. $\square$


Specially, we have

\[\mathcal C\!\left(P,V,\left|V\right|\right)=\mathcal P(P,V).\]

Here is an example. Suppose we have 5 voters, and the set of proposals is $\mathbb R^2$. Each voter has an ideal point and prefers points nearer to the ideal point. The 5 ideal points form a convex pentagon. Then we can find the core easily by the conclusion above:

The core of the example