# Equivalence of Definitions of Bounded Lattice

## Theorem

The following definitions of the concept of **Bounded Lattice** are equivalent:

### Definition 1

Let $\left({S, \preceq}\right)$ 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 $\left({S, \vee, \wedge, \preceq}\right)$ is a **bounded lattice**.

### Definition 2

Let $\left({S, \vee, \wedge, \preceq}\right)$ be a lattice.

Let $\vee$ and $\wedge$ have identity elements $\bot$ and $\top$ respectively.

Then $\left({S, \vee, \wedge, \preceq}\right)$ is a **bounded lattice**.

### Definition 3

Let $\left({S, \wedge, \vee, \preceq}\right)$ be a lattice.

Let $S$ be bounded in $\left({S,\preceq}\right)$.

Then $\left({S, \wedge, \vee, \preceq}\right)$ is a **bounded lattice**.

## Proof

From Supremum of Empty Set is Smallest Element, we have that:

- $\bot := \sup \varnothing$

satisfies $\bot \preceq a$ for all $a \in S$.

By Ordering in terms of Join, this is equivalent to:

- $\forall a \in S: a \vee \bot = a = \bot \vee a$

where the last equality follows as Join is Commutative.

Thus $\bot$ is an identity element for $\vee$, and conversely.

That $\inf \varnothing$ is an identity element for $\wedge$, and conversely, follows by the Duality Principle.

Hence the result.

$\blacksquare$