Relational Structure admits Lower Topology

Theorem
Let $R = \left({S, \preceq}\right)$ be a relational structure.

Then there exists a relational structure with lower topology $T = \left({S, \preceq, \tau}\right)$ such that $T$ is a topological space.

Proof
Define $B := \left\{ {\complement_S\left({x^\succeq}\right): x \in S}\right\}$

where $x^\succeq$ denotes the upper closure of $x$.

By definition of generated topology:
 * $\tau\left({B}\right)$ is a topology on $S$

where $B$ is a sub-basis of $\tau \left({B}\right)$.

Thus by definition of lower topology:
 * $T := \left({S, \preceq, \tau\left({B}\right)}\right)$ has a lower topology.

Thus by definition:
 * $T$ is a topological space.