consequence operator is determined by its fixed points

Theorem 1

Suppose that $C_{\mathrm{1}}$ and $C_{\mathrm{2}}$ are consequence operators on a set $L$ and that, for every $X\mathrm{\subseteq }L$, it happens that $C_{\mathrm{1}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}X$ if and only if $C_{\mathrm{2}}\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}X$. Then $C_{\mathrm{1}}\mathrm{=}{C}_{\mathrm{2}}$.

Theorem 2

Suppose that $C$ is a consequence operators on a set $L$. Define $K\mathrm{=}\mathrm{\{}X\mathrm{\subseteq }L\mathrm{\mid }C\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}X\mathrm{\}}$. Then, for every $X\mathrm{\in }L$, there exists a $Y\mathrm{\in }K$ such that $X\mathrm{\subseteq }Y$ and, for every $Z\mathrm{\in }K$ such that $X\mathrm{\subseteq }Z$, one has $Y\mathrm{\subseteq }Z$.

Theorem 3

Given a set $L$, suppose that $K$ is a subset of $L$ such that, for every $X\mathrm{\in }L$, there exists a $Y\mathrm{\in }K$ such that $X\mathrm{\subseteq }Y$ and, for every $Z\mathrm{\in }K$ such that $X\mathrm{\subseteq }Z$, one has $Y\mathrm{\subseteq }Z$. Then there exists a consequence operator $C\mathrm{:}\mathrm{P}\mathit{}\mathrm{(}L\mathrm{)}\mathrm{\to }\mathrm{P}\mathit{}\mathrm{(}L\mathrm{)}$ such that $C\mathit{}\mathrm{(}X\mathrm{)}\mathrm{=}X$ if and only if $X\mathrm{\in }K$.

Title	consequence operator is determined by its fixed points
Canonical name	ConsequenceOperatorIsDeterminedByItsFixedPoints
Date of creation	2013-03-22 16:29:48
Last modified on	2013-03-22 16:29:48
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	6
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 03G10
Classification	msc 03B22
Classification	msc 03G25