How to show that $\theta(A):=A\cup\mathscr{der}(A)$ is a kuratowski operator?

Question

Let be $\eta:\mathcal{P}(X)\rightarrow\mathcal{P}(X)$ a function such that:

1. $\eta(\varnothing)=\varnothing$;
2. $\eta(\eta(A))\subseteq A\cup\eta(A)$, for any $A\subseteq X$;
3. $\eta(A\cup B)=\eta(A)\cup\eta(B)$, for any $A,B\subseteq X$;
4. $x\notin \eta(\{x\})$, 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

1. $\theta(\varnothing)=\varnothing$;
2. $A\subseteq\theta(A)$ for any ${A}\subseteq{X}$;
3. $\theta(\theta(A))=\theta(A)$ for any ${A}\subseteq{X}$;
4. $\theta(A\cup{B})=\theta(A)\cup\theta(B)$ for any $A,B\subseteq{X}$.

Well let's start to prove this:

1. first of all we observe that $$ \theta(\varnothing)=\varnothing\cup\eta(\varnothing)=\varnothing $$ so that $1$ holds;
2. then we observe that $$ A\subseteq A\cup\eta(A)=\theta(A) $$ so that $2$ holds;
3. 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;
4. 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?

Answer 1

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 $\eta(A) =\eta(B) \cup\eta(\{a\})$.