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.
$\blacksquare$