Relational Structure with Topology of Subsets with Property (S) is Topological Space

From ProofWiki
Jump to navigation Jump to search

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