Upper Semilattice on Classical Set is Semilattice

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Theorem

Let $\left({S, \vee}\right)$ be an upper semilattice on a classical set $S$.


Then $\left({S, \vee}\right)$ is a semilattice.


Proof

To show that the algebraic structure $\left({S, \vee}\right)$ is a semilattice, the following need to be verified:

Closure
Associativity
Commutativity
Idempotence


In order:

Closure

By definition of an upper semilattice:

$\forall x, y \in S: \sup \left\{{x, y}\right\} \in S$

Since $x \vee y = \sup \left\{{x, y}\right\} \in S$ for all $x, y \in S$:

$\left({S, \vee}\right)$ is closed.

$\Box$


Associativity

Let $x, y, z \in S$.

By definition of $\vee$:

\(\displaystyle \left({x \vee y}\right) \vee z\) \(=\) \(\displaystyle \sup \left\{ {x, y}\right\} \vee z\)
\(\displaystyle \) \(=\) \(\displaystyle \sup \left\{ {\sup \left\{ {x, y}\right\}, z}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle \sup \left\{ {x, y, z}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle \sup \left\{ {x, \sup \left\{ {y, z}\right\} }\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle x \vee \sup \left\{ {y, z}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle x \vee \left({y \vee z}\right)\)

Hence $\left({S, \vee}\right)$ is associative.

$\Box$


Commutativity

Let $x, y \in S$.

By definition of $\vee$:

\(\displaystyle x \vee y\) \(=\) \(\displaystyle \sup \left\{ {x, y}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle \sup \left\{ {y, x}\right\}\) Definition of Supremum
\(\displaystyle \) \(=\) \(\displaystyle y \vee x\)

Hence $\left({S, \vee}\right)$ is commutative.

$\Box$


Idempotence

For all $x \in S$:

\(\displaystyle x \vee x\) \(=\) \(\displaystyle \sup \left\{ {x, x}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle x\)

Hence $\vee$ is idempotent.

$\Box$


Having explicitly verified all prerequisites, it follows that $\left({S, \vee}\right)$ is a semilattice.

$\blacksquare$


Sources