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

*

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

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

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

*$\eta(A\cap B)\subseteq A\cup B\cup \eta(A\cup B)$ for any $A,B\subseteq X$.

So putting
$$
\tau*=\{A\subseteq X: \eta(A)\subseteq A\}
$$
we let 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
$$
\eta(A)=\operatorname{bd} A
$$
for any $A\in\mathcal P(X)$
So previously we observe that
$$
\eta(A)=\eta(X\setminus A)
$$
since the third point implies that
$$
\eta(A)=\eta(\varnothing\cup A)\subseteq\eta(\varnothing)\cup \eta(X\setminus A)=\eta(X\setminus A)\\\text{and}\\\eta(X\setminus A)=\eta(\varnothing\cup X\setminus A)\subseteq\eta(\varnothing)\cup\eta(X\setminus(X\setminus A))=\eta(A)
$$
Well to show that the operator $\eta$ is the boundary operator for $\tau$ I thought up to defining the operator
$$
\theta:\mathcal{P}(X)\owns A\rightarrow \big(A\cup \eta(A)\big)\in\mathcal{P}(X)
$$
so that I tried to show that it is a Kuratowski operator because in this case the identity
$$
\operatorname{bd} A=
\operatorname{cl}{A}\cap\operatorname{cl}({X\setminus A})=
(A\cup\eta(A))\cap(X\setminus A\cup\eta(X\setminus A))=\\(A\cup\eta(A))\cap(X\setminus A\cup\eta(A))=
((A\cup\eta(A))\cap X\setminus A)\cup((A\cup\eta(A))\cap\eta(A))=\\
((A\cap X\setminus A)\cup(\eta(A)\cap X\setminus A))\cup((A\cap\eta(A))\cup(\eta(A)\cap\eta(A))=\\
(\varnothing\cup(\eta(A)\cap X\setminus A))\cup((A\cap\eta(A))\cup\eta(A))=(\eta(A)\cap X\setminus A)\cup \eta(A)=
\eta(A)
$$
would holds.
So to do this I must 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 the property 1-4 of $\theta$:

*

*So first of all we observe that
$$
\theta(\varnothing)=\varnothing\cup\eta(\varnothing)=\varnothing
$$


*Then we observe that
$$
A\subseteq A\cup\eta(A)=\theta(A)
$$


*Now we first observe that
$$
\theta(\theta(A))=
\theta(A\cup\eta(A))=\\
(A\cup\eta(A))\cup\eta(A\cup\eta(A))\subseteq (A\cup\eta(A))\cup(\eta(A)\cup\eta(X\setminus\eta(A)))=\\(A\cup\eta(A))\cup(\eta(A)\cup\eta(\eta(A)))=\\(A\cup\eta(A))\cup\eta(\eta(A))\subseteq
{A}\cup\eta(A)\cup\eta(A)=
A\cup\eta(A)=
\theta(A)$$
and then we observe that
$$
A\cup\theta(A)=A\cup(A\cup\eta(A))=A\cup\eta(A)=\theta(A)
$$ i
so that we conclude that
$$
\theta(A)\subseteq\theta(A)\cup\theta(\theta(A))=\theta(\theta(A))
$$


*So finally we observe that
$$
\theta(A\cup B)=(A\cup B)\cup\eta(A\cup B)\subseteq(A\cup B)\cup(\eta(A)\cup\eta(X\setminus B))=(A\cup B)\cup(\eta(A)\cup\eta(B))=(A\cup\eta(A))\cup(B\cup\eta(B))=\theta(A)\cup\theta(B)
$$
Unfortunately how you can observe I'm not be able to demonstrate the inclusion
$$
\theta(A)\cup\theta(B)\subseteq\theta(A\cup B)
$$
so that I ask if it is possible that the position
$$
\theta(A):=A\cup\eta(A)
$$
is uncorrect. Could someone help me, please?
 A: A closure operator is a map $\theta$ from a poset to itself satisfying


*

*$A\leq \theta(A)$ (increasing)

*$A\leq B\implies \theta(A)\leq \theta(B)$ (isotonoic)

*$\theta(\theta(A))=\theta(A)$ (idempotent).


If in addition $\theta(A\vee B)\leq \theta(A)\vee\theta(B)$, then the operator is said to be topological. (This will imply $\theta(A\vee B)=\theta(A)\vee\theta(B)$, as noted below). 
If in addition $\theta(\mathbf{0}) = \mathbf{0}$, then it is a topological closure operator (though moding out by $\downarrow\!\!(\theta(\mathbf{0}))$ will turn the operator into such an operator, so this is usually ignored). 
In the case at hand, the condition $\theta(A)\cup \theta(B)\subseteq \theta(A\cup B)$ is true of any closure operator, not just the topological ones, and so does not depend on the precisely definition of $\theta$, but only on its general properties. In fact, it follows from $B\subseteq C\implies \theta(B)\subseteq \theta(C)$ (isotonicity of the operator), property 2.
Since $A\subseteq A\cup B$, by isotonicity we have $\theta(A)\subseteq \theta(A\cup B)$. Since $B\subseteq A\cup B$, we likewise have $\theta(B)\subseteq \theta(A\cup B)$. 
Since $\theta(A)$ and $\theta(B)$ are both contained in $\theta(A\cup B)$, it follows that so is their union. So $\theta(A)\cup\theta(B)\subseteq \theta(A\cup B)$.
This property is part of the usual definition, though you do not seem to list it.
To prove isotonicity for your particular case, suppose $A\subseteq B$. Then certainly $A\subseteq B\subseteq \theta(B)$. You just need to show that $\eta(A)\subseteq \theta(B)$. You know that
$$\eta(A)=\eta(A\cap B)\subseteq A\cup B\cup\eta(A\cup B) = B\cup \eta(B) = \theta(B).$$
Thus, $A\subseteq \theta(B)$, $\eta(A)\subseteq \theta(B)$, hence $\theta(A) = A\cup\eta(A)\subseteq \theta(B)$, proving the isotonicity of $\theta$. 
