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

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Theorem

Let $\left({X, \preceq, \tau}\right)$ 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:

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

But then:

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

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.

$\blacksquare$