Upper and Lower Closures of Open Set in GO-Space are Open

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Theorem

Let $\struct {X, \preceq, \tau}$ be a Generalized Ordered Space/Definition 1.

Let $A$ be open in $X$.


Then the upper and lower closures of $A$ are open.


Proof

We will show that the upper closure $U$ of $A$ is open.

The lower closure can be proven open by the same method.

By the definition of upper closure:

$U = \set {u \in X: \exists a \in A: a \preceq u}$

But then:

\(\ds U\) \(=\) \(\ds \set {u \in X: \paren {u \in A} \lor \paren {\exists a \in A: a \prec u} }\)
\(\ds \) \(=\) \(\ds A \cup \bigcup \set {a^\succeq: a \in A}\)

where $a^\preceq$ denotes the upper closure of $a$.

By Open Ray is Open in GO-Space/Definition 1, each $a^\succeq$ is open.

Thus $U$ is a union of open sets.

Thus $U$ is open by the definition of a topology.

$\blacksquare$