Meet Irreducible iff Finite Infimum equals Element

Theorem
Let $L = \left({S, \wedge, \preceq}\right)$ be a meet semilattice.

Let $x \in S$.

Then
 * $x$ is meet irreducible


 * for every non-empty finite subset $A$ of $S$: $x = \inf A \implies x \in A$

Sufficient Condition
Let $x$ be meet irreducible.

We will prove the result by induction on cardinality of $A$.

Base case

 * for every non-empty subset $A$ of $S$: $\left\vert{A}\right\vert = 1 \land x = \inf A \implies x \in A$

Let $A$ be a non-empty subset of $S$ such that
 * $\left\vert{A}\right\vert = 1 \land x = \inf A$

Then
 * $A = \left\{ {a}\right\}$

By Infimum of Singleton:
 * $x = a$

Thus by definition of singleton:
 * $x \in A$

Induction Hypothesis

 * $n \ge 1$ and for every non-empty subset $A$ of $S$: $\left\vert{A}\right\vert = n \land x = \inf A \implies x \in A$

Induction Step

 * $n \ge 1$ and for every non-empty subset $A$ of $S$: $\left\vert{A}\right\vert = n+1 \land x = \inf A \implies x \in A$

Let $A$ be a non-empty subset of $S$ such that
 * $\left\vert{A}\right\vert = n+1 \land x = \inf A$

Then
 * $A = \left\{ {a_1, \dots, a_n, a_{n+1} }\right\}$

By definition of infimum:
 * $\inf A = \left({a_1 \wedge \dots \wedge a_n}\right) \wedge a_{n+1}$

By definition of meet irreducible:
 * $x = a_1 \wedge \dots \wedge a_n$ or $x = a_{n+1}$

By Induction Hypothesis:
 * $x \in \left\{ {a_1, \dots, a_n}\right\}$ or $x = a_{n+1}$

Thus by definition of unordered tuple:
 * $x \in A$

Necessary Condition
Assume that
 * for every non-empty finite subset $A$ of $S$: $x = \inf A \implies x \in A$

Let $y, z \in S$ such that
 * $x = y \wedge z$

By definition of meet:
 * $x = \inf \left\{ {y, z}\right\}$

By assumption:
 * $x \in \left\{ {y, z}\right\}$

Thus by definition of unordered tuple:
 * $x = y$ or $x = z$