Totally Ordered Set is Lattice

Theorem

Every totally ordered set is a lattice.


Proof

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

Then we have:

$\forall x, y \in S: x \preceq y \lor y \preceq x$
$\forall x, y \in S: x \preceq y \implies \sup \set {x, y} = y \land \inf \set {x, y} = x$
$\forall x, y \in S: y \preceq x \implies \sup \set {x, y} = x \land \inf \set {x, y} = y$


Thus the conditions for $\struct {S, \preceq}$ to be a lattice are fulfilled.

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
  • 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): partial order
  • 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): partial order
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