Lattices of sets and algebraic closure operator
Vestnik Tverskogo gosudarstvennogo universiteta. Seriâ Prikladnaâ matematika, no. 4 (2017), pp. 35-42
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It is well known that a lattice of closed sets is algebraic lattice if a closure operator is algebraic. The converse is not true. In this paper we give an example of an algebraic lattice the closure operator of which is not algebraic.
The exact criterion that the closure operator of an algebraic lattice is algebraic is found. It is proved that the closure operator of an algebraic lattice ${\mathcal T}$ is algebraic if and only if for any $X\in{\mathcal T}$ and for any
$x\in X$, there exists a compact element $K_x$ such that $x\in K_x$ and $K_x\subseteq X$.
Keywords:
algebraic lattice, algebraic closure operator, closure system.
@article{VTPMK_2017_4_a2,
author = {I. A. Gorbunov},
title = {Lattices of sets and algebraic closure operator},
journal = {Vestnik Tverskogo gosudarstvennogo universiteta. Seri\^a Prikladna\^a matematika},
pages = {35--42},
publisher = {mathdoc},
number = {4},
year = {2017},
language = {ru},
url = {http://geodesic.mathdoc.fr/item/VTPMK_2017_4_a2/}
}
I. A. Gorbunov. Lattices of sets and algebraic closure operator. Vestnik Tverskogo gosudarstvennogo universiteta. Seriâ Prikladnaâ matematika, no. 4 (2017), pp. 35-42. http://geodesic.mathdoc.fr/item/VTPMK_2017_4_a2/