# 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 of
any 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 $C(X)\subseteq C(Y)$.

∎

Title | consequence operator determined by a class of subsets |
---|---|

Canonical name | ConsequenceOperatorDeterminedByAClassOfSubsets |

Date of creation | 2013-03-22 16:29:45 |

Last modified on | 2013-03-22 16:29:45 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 10 |

Author | rspuzio (6075) |

Entry type | Theorem^{} |

Classification | msc 03G25 |

Classification | msc 03G10 |

Classification | msc 03B22 |