Every Element is Lower implies Union is Lower

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Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $A$ be a set of subsets of $S$.

Let

$\forall X \in A: X$ is a lower section.


Then $\bigcup A$ is a lower section.


Proof

Let $x \in \bigcup A, y \in S$ such that:

$y \preceq x$

By definition of union:

$\exists X \in A: x \in X$

By assumption:

$X$ is a lower section.

By definition of lower section:

$y \in X$

Thus by definition of union:

$y \in \bigcup A$

$\blacksquare$


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