Difference between revisions of "Complete lattice"
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x \mapsto \{x\}^\nabla = \{ y \in X : y \le x \} | x \mapsto \{x\}^\nabla = \{ y \in X : y \le x \} | ||
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− | maps $P$ into the complete lattice $\mathcal{P}(P)$, and hence defines a completion $\mathcal{O}(P)$ of $P$. However this has this disadvantage that if $P$ is already a complete lattice, and hence a completion of itself, then the completion $\mathcal{O}(P)$ is larger than $P$ itself. The [[Completion, MacNeille (of a partially ordered set)|Dedkind–MacNeille completion]] | + | maps $P$ into the complete lattice $\mathcal{P}(P)$, and hence defines a completion $\mathcal{O}(P)$ of $P$. However this has this disadvantage that if $P$ is already a complete lattice, and hence a completion of itself, then the completion $\mathcal{O}(P)$ is larger than $P$ itself. The [[Completion, MacNeille (of a partially ordered set)|Dedkind–MacNeille completion]] is the least of all completions of a given partially ordered set. |
Complete lattices are formed by the set of all subalgebras in a universal algebra, by the set of all congruences in a universal algebra, and by the set of all closed subsets in a topological space (note that while the meet of a family of closed sets is their set-theoretic intersection, the join of a family of closed sets is the closure of their set-theoretic union). | Complete lattices are formed by the set of all subalgebras in a universal algebra, by the set of all congruences in a universal algebra, and by the set of all closed subsets in a topological space (note that while the meet of a family of closed sets is their set-theoretic intersection, the join of a family of closed sets is the closure of their set-theoretic union). |
Revision as of 15:01, 21 May 2016
2020 Mathematics Subject Classification: Primary: 06B23 [MSN][ZBL]
A partially ordered set in which any subset $A$ has a least upper bound and a greatest lower bound. These are usually called the join and the meet of $A$ and are denoted by $\wedge_{a \in A} a$ and and $\vee_{a \in A} a$ or simply by $\vee A$ and $\wedge A$ (respectively). If a partially ordered set has a largest element and each non-empty subset of it has a greatest lower bound, then it is a complete lattice. A lattice $L$ is complete if and only if any isotone mapping $\phi$ of the lattice into itself has a fixed point, i.e. an element $a \in L$ such that $a \phi = a$. If $\mathcal{P}(M)$ is the set of subsets of a set $M$ ordered by inclusion and $\phi$ is a closure operation on $\mathcal{P}(M)$, then the set of all $\phi$-closed subsets is a complete lattice.
Any partially ordered set $P$ can be isomorphically imbedded in a complete lattice, which in that case is called a completion of $P$. For example, the map $$ x \mapsto \{x\}^\nabla = \{ y \in X : y \le x \} $$ maps $P$ into the complete lattice $\mathcal{P}(P)$, and hence defines a completion $\mathcal{O}(P)$ of $P$. However this has this disadvantage that if $P$ is already a complete lattice, and hence a completion of itself, then the completion $\mathcal{O}(P)$ is larger than $P$ itself. The Dedkind–MacNeille completion is the least of all completions of a given partially ordered set.
Complete lattices are formed by the set of all subalgebras in a universal algebra, by the set of all congruences in a universal algebra, and by the set of all closed subsets in a topological space (note that while the meet of a family of closed sets is their set-theoretic intersection, the join of a family of closed sets is the closure of their set-theoretic union).
References
[1] | G. Birkhoff, "Lattice theory" , Colloq. Publ. , 25 , Amer. Math. Soc. (1973) |
[2] | L.A. Skornyakov, "Elements of lattice theory" , Hindushtan Publ. Comp. (1977) (Translated from Russian) |
[a1] | B. A. Davey, H. A. Priestley, Introduction to lattices and order, 2nd ed. Cambridge University Press (2002) ISBN 978-0-521-78451-1 |
Complete lattice. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Complete_lattice&oldid=38828