How to show that $\theta(A):=A\cup\mathscr{der}(A)$ is a kuratowski operator? Let be $\eta:\mathcal{P}(X)\rightarrow\mathcal{P}(X)$ a function such that:

*

*$\eta(\varnothing)=\varnothing$;

*$\eta(\eta(A))\subseteq A\cup\eta(A)$, for any $A\subseteq X$;

*$\eta(A\cup B)=\eta(A)\cup\eta(B)$, for any $A,B\subseteq X$;

*$x\notin \eta(\{x\})$, for any $x\in X$.

So putting
$$
\tau*=\{A\subseteq X: \phi(A)\subseteq A\}
$$
I am trying to prove that the collection
$$
\tau=\{V\subseteq X:X\setminus V\in\tau*\}
$$
is a topology on $X$ equal to $\tau^*$ and such that
$$
\eta(A)=\operatorname{der}A
$$
for any $A\in\mathcal P(X)$.
Well to prove the statement we define
$$
\theta:\mathcal{P}(X)\owns A\rightarrow\big(A\cup\eta(A)\big)\in\mathcal{P}(X)
$$
so that we let to prove that $\theta$ is a Kuratowski operator. So to do this we have to demonstrate that

*

*$\theta(\varnothing)=\varnothing$;

*$A\subseteq\theta(A)$ for any ${A}\subseteq{X}$;

*$\theta(\theta(A))=\theta(A)$ for any ${A}\subseteq{X}$;

*$\theta(A\cup{B})=\theta(A)\cup\theta(B)$ for any $A,B\subseteq{X}$.

Well let's start to prove this:

*

*first of all we observe that
$$
\theta(\varnothing)=\varnothing\cup\eta(\varnothing)=\varnothing
$$
so that $1$ holds;

*then we observe that
$$
A\subseteq A\cup\eta(A)=\theta(A)
$$
so that $2$ holds;

*moreover we observe that
$$
\theta(\theta(A))=
\theta(A\cup\eta(A))=
(A\cup\eta(A))\cup\eta(A\cup\eta(A))=\\(A\cup\eta(A))\cup(\eta(A)\cup\eta(\eta(A)))=
A\cup\eta(A)\cup\eta(\eta(A))=
A\cup\eta(A)=
\theta(A)
$$
so that $3$ holds;

*finally we observe that
$$
\theta(A\cup B)=(A\cup B)\cup\eta(A\cup B)=A\cup\eta(A)\cup B\cup\eta(B)=\theta(A)\cup\theta(B)
$$
so that $4$ holds.

Well we proved that $\theta$ is a kuratowski operator so we have
$$
\theta(A)=\operatorname{cl}A
$$
that is
$$
A\cup\eta(A)=A\cup\mathscr{der}(A)
$$
for any $A\subseteq X$: could I argue from this that the identity
$$
\eta(A)=\mathscr{der}(A)
$$
holds? if not how to prove it?
Could someone help me, please?
 A: No, you can't argue so, because condition 4. was not used, and without that your reasoning would also work for the closure operator $\eta:=\theta$  itself, concluding that closure is always the same as the derived set.
Nevertheless, $A\cup B=A\cup C$ implies $B\setminus A\subseteq C$ and $C\setminus A\subseteq B$, which will be used below. 
If $x\in A\setminus \mathrm{der}(A)$, then there is an open $U\ni x$ such that $U\cap A=\{x\}$, then $\eta(A\setminus U) \subseteq \theta(A\setminus U) \subseteq \theta(X\setminus U) =X\setminus U$, which in particular doesn't contain $x$, so by conditions 3. and 4. for $\eta$, we get
$$x\notin \eta(A\cap U)\cup\eta(A\setminus U) =\eta(A)\,.$$
If $x\notin{\rm der}(A)$ then the case $x\in A$ was just concluded, and in the case $x\notin A$, we have $x\notin \bar A=A\cup \eta(A)$, thus again $x\notin \eta(A)$. 
These imply $\eta(A)\subseteq {\rm der}(A)$. 
For the other direction, we have ${\rm der}(A)\setminus A=\bar A\setminus A\subseteq \eta(A)$ for any set $A$. 
So, let $a\in{\rm der}(A)\cap A$, but then, for $B:=A\setminus\{a\}$, we still have $a\in{\rm der}(B)$, so we can apply the above observation to obtain $a\in\eta(B)$, and conclude by $\eta(A) =\eta(B) \cup\eta(\{a\})$. 
