From ProofWiki
Jump to navigation Jump to search


Let $\left({L,\wedge_L,\vee_L,\preceq_L}\right)$ be a lattice.

Let $S$ be a subset of $L$ and let $\wedge_S$, $\vee_S$, and $\preceq_S$ be the restrictions to $S$ of $\wedge_L$, $\vee_L$, and $\preceq_L$, respectively.

Then $\left({S,\wedge_S,\vee_S,\preceq_S}\right)$ is a sublattice of $\left({L,\wedge_L,\vee_L,\preceq_L}\right)$ iff $S$ is closed under $\wedge_S$ and $\vee_S$.

If in addition $L$ is a bounded lattice and its top and bottom elements are in $S$, then $S$ is called a $0,1$-sublattice of $L$.

If in addition $L$ is a complete lattice and for each subset $T$ of $S$, $\sup_L T,\inf_L T \in S$, then $S$ is called a complete sublattice of $L$.

Also see