Equivalence of Definitions of Bounded Lattice
 | This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
Theorem
The following definitions of the concept of Bounded Lattice are equivalent:
Definition 1
Let $\struct {S, \preceq}$ be an ordered set.
Let $S$ admit all finite suprema and finite infima.
Let $\vee$ and $\wedge$ be the join and meet operations on $S$, respectively.
Then the ordered structure $\struct {S, \vee, \wedge, \preceq}$ is a bounded lattice.
Definition 2
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $\vee$ and $\wedge$ have identity elements $\bot$ and $\top$ respectively.
Then $\struct {S, \vee, \wedge, \preceq}$ is a bounded lattice.
Definition 3
Let $\struct {S, \wedge, \vee, \preceq}$ be a lattice.
Let $S$ be bounded in $\struct {S, \preceq}$.
Then $\struct {S, \wedge, \vee, \preceq}$ is a bounded lattice.
Proof
Definition 1 if and only if Definition 2
From Supremum of Empty Set is Smallest Element, we have that:
- $\bot := \sup \O$
satisfies $\bot \preceq a$ for all $a \in S$.
From Join Semilattice has Smallest Element iff has Identity:
- $\bot$ is an identity element for $\vee$, and conversely.
That $\inf \O$ is an identity element for $\wedge$, and conversely, follows by the Duality Principle.
Hence the result.
$\Box$
Definition 1 if and only if Definition 3
Let $x \in S$.
From Supremum of Empty Set is Smallest Element:
- $x = \sup \O$ if and only if $x$ is the smallest element of $\struct{S, \preceq}$
From Ordered Set has Lower Bound iff has Smallest Element:
- $x$ is the smallest element of $\struct{S, \preceq}$ if and only if $x$ is a lower bound of $S$ in $\struct{S, \preceq}$
That $\inf \O$ is an upper bound for $S$, and conversely, follows by the Duality Principle.
Hence the result.
$\blacksquare$