consequence operator determined by a class of subsets

Theorem 1.

Let $L$ be a set and let $K$ be a subset of $\mathrm{P}\mathit{}\mathrm{(}L\mathrm{)}$. The the mapping$C\mathrm{:}\mathrm{P}\mathit{}\mathrm{(}L\mathrm{)}\mathrm{\to }\mathrm{P}\mathit{}\mathrm{(}L\mathrm{)}$ defined as $C\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}\mathrm{\cap }\mathrm{\{}Y\mathrm{\in }K\mathrm{\mid }X\mathrm{\subseteq }Y\mathrm{\}}$ is a consequence operator.

Proof.

We need to check that $C$ satisfies the defining properties.

Property 1:Since every element of the set $\{Y\in K\mid X\subseteq Y\}$contains $X$, we have $X\subseteq C(X)$.

Property 2:For every element $Y$ of $K$ such that $X\subseteq Y$, it also is the case that$C(X)\subseteq Y$ because an intersection of a family of sets is a subset ofany member of the family. In other words (or rather, symbols),

$$\{Y\in K\mid X\subseteq Y\}\subseteq \{Y\in K\mid C(X)\subseteq Y\},$$

hence $C(C(X))\subseteq C(X)$. By the first property proven above, $C(X)\subseteq C(C(X))$ so $C(C(X))=C(X)$. Thus, $C\circ C=C$.

Property 3:Let $X$ and $Y$ be two subsets of $L$ such that $X\subseteq Y$. Then if,for some other subset $Z$ of $L$, we have $Y\subset Z$, it follows that$X\subset Z$. Hence,

$$\{Z\in K\mid Y\subseteq Z\}\subseteq \{Z\in K\mid X\subseteq Z\},$$

so $C(X)\subseteq C(Y)$.

∎