Strict Lower Closure of Element is Proper Lower Section
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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 $A$.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 1$ A few preliminaries