Union of Upper Sections is Upper

Theorem

Let $\struct {S, \preceq}$ be a preordered set.

Let $A$ be a set of subsets of $S$ such that:

$\forall X \in A: X$ is an upper section.

Then:

$\bigcup A$ is also an upper section.

Proof

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

$x \preceq y$

By definition of union:

$\exists Y \in A: x \in Y$

By assumption:

$Y$ is an upper section.

By definition of upper section:

$y \in Y$

Thus by definition of union:

$y \in \bigcup A$

Hence

$\bigcup A$ is an upper section.

$\blacksquare$

Sources

• 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices

• Mizar article WAYBEL_0:28