# Equivalence of Definitions of Order Complete Set

## Contents

## Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.

The following definitions of the concept of **Order Complete Set** are equivalent:

### Definition 1

$\left({S, \preceq}\right)$ is **order complete** if and only if:

- Each non-empty subset $H \subseteq S$ which has an upper bound admits a supremum.

### Definition 2

$\left({S, \preceq}\right)$ is **order complete** if and only if:

- Each non-empty subset $H \subseteq S$ which has a lower bound admits an infimum.

## Proof

### Definition 1 implies Definition 2

Let $\left({S, \preceq}\right)$ be an order complete set by definition 1.

Let $H \subseteq S$ have a lower bound.

Let $K$ be the set of all lower bounds of $H$.

Then $K \ne \varnothing$.

By definition of lower bound:

- $\forall x \in K: \forall y \in H: x \le y$

and so all elements of $H$ are upper bounds of $K$.

Thus by hypothesis $K$ admits a supremum.

Let $k = \sup \left({K}\right)$.

By definition of supremum, $k$ precedes every upper bound of $K$.

In particular, $k$ precedes every element of $H$.

Thus $k$ is a lower bound of $H$.

But we have that $k$ is an upper bound of $K$.

That is, $k$ succeeds every lower bound of $H$.

That is, $k$ is an infimum of $H$.

Thus $\left({S, \preceq}\right)$ is an order complete set by definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $\left({S, \preceq}\right)$ be an order complete set by definition 2.

Let $H \subseteq S$ have an upper bound.

Let $K$ be the set of all upper bounds of $H$.

Then $K \ne \varnothing$.

By definition of upper bound:

- $\forall x \in K: \forall y \in H: y \le x$

and so all elements of $H$ are lower bounds of $K$.

Thus by hypothesis $K$ admits an infimum.

Let $k = \inf \left({K}\right)$.

By definition of infimum, $k$ succeeds every lower bound of $K$.

In particular, $k$ succeeds every element of $H$.

Thus $k$ is an upper bound of $H$.

But we have that $k$ is a lower bound of $K$.

That is, $k$ precedes every upper bound of $H$.

That is, $k$ is a supremum of $H$.

Thus $\left({S, \preceq}\right)$ is an order complete set by definition 1.

$\blacksquare$

## Sources

- 1955: John L. Kelley:
*General Topology*... (previous) ... (next): Chapter $0$: Orderings: Theorem $9$