A Kuratowski closure operator induces the unique topology There is a conclusion :A Kuratowski closure operator $f:P(X)\to P(X)$ has the property that there is a (unique) topology $\tau$ on $X$ such that $f$ is the closure operator for that topology.
We can define the topology $\tau =\{X\setminus f(A): A\subseteq X\}$. How the show the topology satisfying the above property is unique?
 A: The definition of the topology $\tau$ is fine, now you have to show first that the closure $\operatorname{cl}_\tau$ of taking the closure in $\tau$ actually equals $f$, so that for all subsets of $X$ we have $\operatorname{cl}_\tau(A)= f(A)$.
So fix $A$.
First note that a set is closed in $\tau$ iff it is of the form $f(B)$ for some $B$, and $\operatorname{cl}_\tau(A)$ is defined as $$\operatorname{cl}_\tau(A) = \bigcap \{C: C \text{ closed in } \tau, A \subseteq C\}$$
the smallest $\tau$-closed containing $A$.
Now $f(A)$ is closed by definition of $\tau$ and one axiom of the closure tellsus that $A \subseteq f(A)$, so $f(A)$ is one of the $\tau$-closed sets in the intersection that defines $\operatorname{cl}_\tau(A)$ so from that it follows that 
$$\operatorname{cl}_\tau(A) \subseteq f(A)$$
as the intersection is a subset of each of the sets it's an intersection of, by definition.
On the other hand if $C$ is any closed set containing $A$, we know $A \subseteq C$ and $C=f(B)$ for some $B$, as said. But then monotonicity of $f$ (e.g. follows from union axiom), and idempotence of $f$ tells us that $f(A) \subseteq f(C)= f(f(B))=f(B)=C$, and as $C$ was arbitrary, $f(A)$ is a subset of all sets in the definining intersection of $\operatorname{cl}_\tau(A)$ so
$$f(A) \subseteq \operatorname{cl}_\tau(A)$$ and we have the equality, we wanted.
If $\tau'$ is a topology on $X$ so that $f(A)=\operatorname{cl}_{\tau'}(A)$ for all $A$, then we have that $\operatorname{cl}_\tau(A)=\operatorname{cl}_{\tau'}(A)$ for all $A$ for the $\tau$ we defined. And so $\tau'$ has the same closed sets as $\tau$ (a set is closed iff it equals its closure) and so the same open sets too (as a set is open iff its complement is closed). So $\tau=\tau'$ and we have unicity. 
The unicity argument alone doesn't need us to define $\tau$, it follows quite easily by the above argument that topologies that induce the same closure operation for all sets, are the same topologies.
In formulae, if the topologies $\tau,\tau'$ both "work" for the given $f$:
$$O \in \tau \iff O^\complement \text{ closed in }\tau \iff O^\complement=\operatorname{cl}_\tau(O^\complement)=f(O^\complement)=\operatorname{cl}_{\tau'}(O^\complement) \iff \\
O^\complement \text{ closed in }\tau' \iff O \in \tau'$$
