Consider a Boolean set algebra on a set $\Omega$. Let $\sigma$ be a set function on $\Omega$ such that for all $m\in \Omega$ $\sigma(m)\subset \Omega$. The operator $\square_\sigma$ is defined by $\square_\sigma P=\{m:\sigma(m)\subset P\}$ for all $P\subset \Omega$. It has the following properties:
- $\square_\sigma \Omega= \Omega$
- $\square_\sigma(P\cap Q)=\square_\sigma P \cap \square_\sigma Q$
- $\square_\sigma(P\to Q)\to (\square_\sigma P\to \square_\sigma Q)=\Omega$
If for all $m\in \Omega$ $m\in \sigma(m)$, then
- $\square_\sigma \emptyset = \emptyset$
- $\square_\sigma P \to P= \Omega$ $\quad (\square_\sigma P\subset P)$
- if $\square_\sigma P=\Omega$, then $P=\Omega$
and if for all $m\in \Omega$ and for all $k \in \sigma(m)$ $\sigma(k)\subset \sigma(m)$, then
- $\square_\sigma \square_\sigma P= \square_\sigma P$.
Is this definition of the modal operator $\square_\sigma$ general?
