# Lattice is Complete iff it Admits All Suprema

Jump to navigation
Jump to search

## Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Then $\struct {S, \preceq}$ is a complete lattice if and only if

- $\forall X \subseteq S: X$ admits a supremum.

## Proof

### Sufficient Condition

Let $\struct {S, \preceq}$ be a complete lattice.

Thus by definition of complete lattice:

- $\forall X \subseteq S: X$ admits a supremum.

$\Box$

### Necessary Condition

Let $\struct {S, \preceq}$ be such that

- $\forall X \subseteq S: X$ admits a supremum.

We will prove that

Let $X \subseteq S$.

Thus by assumption:

- $X$ admits a supremum.

Define:

- $Y := \leftset {s \in S: s}$ is a lower bound for $\rightset X$

By assumption:

- $Y$ admits a supremum

We will prove that

- $\sup Y$ is a lower bound for $X$

Let $x \in X$.

By definition of lower bound:

- $x$ is an upper bound for $Y$

By definition of supremum:

- $\sup Y \preceq x$

Thus by definition:

- $\sup Y$ is a lower bound for $X$

We will prove that

- $\forall s \in S: s$ is a lower bound for $X \implies s \preceq \sup Y$

Let $s \in S$ such that:

- $s$ is a lower bound for $X$.

By definition of $Y$:

- $s \in Y$

By definition of supremum:

- $\sup Y$ is an upper bound for $Y$.

Thus by definition of upper bound:

- $s \preceq \sup Y$

Thus by definition

- $X$ admits an infimum

By definition

- $\struct {S, \preceq}$ is a lattice.

Thus the result follows by definition of complete lattice.

$\blacksquare$

## Sources

- 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott:
*A Compendium of Continuous Lattices*

- Mizar article YELLOW_2:24