Relational Structure with Topology of Subsets with Property (S) is Topological Space
Theorem
Let $T = \struct {S, \preceq, \tau}$ be a relational structure with topology
where
- $\struct {S, \preceq}$ is an up-complete ordered set
- $\tau$ is the set of all subsets of $S$ with property (S).
Then $\struct {S, \tau}$ is topological space.
Proof
We will prove that
- $S$ has property (S).
Let $D$ be a directed subset of $S$ such that
- $\sup D \in S$
By definition of non-empty set:
- $\exists y: y \in D$
Thus $y \in D$.
Thus by definition of subset:
- $\forall x \in D: y \preceq x \implies x \in S$
$\Box$
Then:
$(\text O 3): \quad S \in \tau$
We will prove that:
- $(\text O 1): \quad \forall F \subseteq \tau: \bigcup F \in \tau$
Let $F \subseteq \tau$.
To prove that $\bigcup F \in \tau$ it should be proved that
- $\bigcup F$ has property (S).
Let $D$ be a directed subset of $S$ such that
- $\sup D \in \bigcup F$
By definition of union:
- $\exists X \in F: \sup D \in X$
By definition of subset:
- $X \in \tau$
By assumption:
- $X$ has property (S).
By definition of property (S):
- $\exists y \in D: \forall x \in D: y \preceq x \implies x \in X$
Thus $y \in D$.
Let $x \in D$ such that
- $y \preceq x$
Then $x \in X$.
Thus by definition of union:
- $x \in \bigcup F$
$\Box$
We will prove that
- $(\text O 2): \quad \forall X, Y \in \tau: X \cap Y \in \tau$
Let $X, Y \in \tau$.
By assumption:
- $X$ and $Y$ have property (S).
To prove that $X \cap Y \in \tau$ it should be proved that:
- $X \cap Y$ has property (S).
Let $D$ be a directed subset of $S$ such that:
- $\sup D \in X \cap Y$
By definition of intersection:
- $\sup D \in X$ and $\sup D \in Y$
By definition of property (S):
- $\exists x \in D: \forall z \in D: x \preceq z \implies z \in X$
and
- $\exists y \in D: \forall z \in D: y \preceq z \implies z \in X$
By definition of directed:
- $\exists z \in D: x \preceq z \land y \preceq z$
Thus $z \in D$.
Let $u \in D$ such that
- $z \preceq u$
By definition of transitivity:
- $x \preceq u$ and $y \preceq u$
Then:
- $u \in X$ and $u \in Y$
Thus by definition of intersection:
- $u \in X \cap Y$
$\Box$
Hence by $(\text O 1)$, $(\text O 2)$, $(\text O 3)$, and definition:
- $\struct {S, \tau}$ is topological space.
$\blacksquare$
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 WAYBEL11:16