Topological "closure" of a binary relation Let $f$ be a binary relation on a set $U$.
Topology $T f = \{ E \in \mathscr{P} U \mid
f [E] \subseteq E \}$ (here $f[E]$ is the image of a set $E$ by binary relation $f$).
Conjecture Closure operator $\operatorname{cl}$ of $T f$ is equal to $E \mapsto ( \operatorname{id}_U \cup f \cup
  f^2 \cup f^3 \cup \ldots ) [E]$.

$E \mapsto ( \operatorname{id}_U \cup f \cup f^2 \cup f^3 \cup \ldots ) [E]$ maps open sets to itself. So, it can be closure only if all open sets are closed.
For a counterexample for the conjecture take $f = \{ (0, 0), (1, 1), (0, 1) \}$. Open sets are $\{ \}$, $\{ 1 \}$, $\{ 0, 1 \}$. $\{ 1 \}$ is open but not closed.
