Strict Lower Closure of Element is Proper Lower Section

Theorem
Let $A$ be a class under an ordering $\preccurlyeq$.

Let $a \in A$.

Let $a^\prec$ denote the strict lower closure of $a$ in $A$.

Then $a^\prec$ is a proper lower section of $A$.

Proof
By definition of strict lower closure:


 * $a^\prec := \set {b \in S: b \prec a}$

By definition of $\prec$:
 * $a \not \prec a$

and so:
 * $a \notin a^\prec$

and so:
 * $a \in A \setminus a^\prec$

Let $x \in a^\prec$.

Then by definition of $a^\prec$:
 * $x \prec a$

and so:
 * $\exists a \in A \setminus a^\prec: x \prec a$

and so $x$ is in a proper lower section of $A$.

As $x$ is arbitrary, it follows that every element of $a^\prec$ is in a proper lower section of $A$.

Hence $a^\prec$ is a proper lower section of $A$.