Order Generating iff Every Element is Infimum

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.

Let $X$ be a subset of $S$.

Then
 * $X$ is order generating


 * $\forall x \in S: \exists Y \subseteq X: x = \inf Y$

Sufficient Condition
Let $X$ be order generating.

Let $x \in S$.

By definition of order generating:
 * $x = \map \inf {x^\succeq \cap X}$

Thus
 * $\exists Y \subseteq X: x = \inf Y$

Necessary Condition
Let:
 * $\forall x \in S: \exists Y \subseteq X: x = \inf Y$

Let $x \in S$.

By assumption:
 * $\exists Y \subseteq X: x = \inf Y$

Define $Z := x^\succeq \cap X$.

We will prove that
 * $\forall b \in S: b$ is lower bound for $Z \implies b \preceq x$

Let $b \in S$ such that
 * $b$ is lower bound for $Z$.

Let $c \in Y$.

By definition of infimum:
 * $x$ is lower bound for $Y$.

By definition of lower bound:
 * $x \preceq c$

By definition of upper closure of element:
 * $c \in x^\succeq$

By definition of subset:
 * $c \in X$

By definition of intersection:
 * $c \in Z$

Thus by definition of lower bound:
 * $b \preceq c$

By definition of lower bound:
 * $b$ is lower bound for $Y$

Thus by definition of infimum:
 * $b \preceq x$

We will prove that
 * $x$ is lower bound for $Z$

Let $a \in Z$

By definition of intersection:
 * $a \in x^\succeq$

Thus by definition of upper closure of element:
 * $x \preceq a$

Thus by definitions of complete lattice and infimum:
 * $x^\succeq \cap X$ admits an infimum and $x = \map \inf {x^\succeq \cap X}$