Subset of Join Semilattice on Total Ordering is Closed

Theorem

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

Let $\struct {S, \vee, \preccurlyeq}$ be the join semilattice with respect to $\preccurlyeq$.

Let $T \subseteq S$ be an arbitrary subset of $S$.

Then $\struct {T, \vee, \preccurlyeq}$ is closed under $\vee$.

Proof

Let $x \in T$.

Then from Supremum of Singleton:

$\sup \set x = x$

That is:

$x \vee x = x$

and so:

$x \vee x \in T$

Let $x, y \in T$.

From Supremum of Doubleton in Totally Ordered Set:

$\sup \set {x, y} \in \set {x, y}$

That is:

$x \vee y \in \set {x, y}$

But:

$\set {x, y} \subseteq T$

Hence by definition of subset:

$x \vee y \in T$

That is:

$\struct {T, \vee, \preccurlyeq}$ is closed under $\vee$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.22 \ \text {(a)}$