Example of a preclosure that is not a closure For a set $X$ we can define an operator $cl:\mathscr{P}(X)\rightarrow\mathscr{P}(X)$ satisfying for all $A,B\subseteq X$.
$$cl(\emptyset)=\emptyset\tag{1}$$
$$A\subseteq cl(A)\tag{2}$$
$$cl(cl(A))=cl(A)\tag{3}$$
$$cl(A\cup B)=cl(A)\cup cl(B)\tag{4}$$
Kuratowski's theorem says that if such operator is found we can define a topology $\tau$ on X where $cl(A)$ is the $\tau$-closure of $A$.
My goal was to find operators say $\varphi:\mathscr{P}(X)\rightarrow \mathscr{P}(X)$ s.t. for each (1)-(4) it satisfies the other (3) so for example $\varphi_1:A\mapsto X$ for arbitrary set $X$ satisfies 2,3,4 but not 1, for $\varphi_2:A\mapsto\emptyset$ for arbitrary set $X$ satisfies 1,3,4 but not 2. For 3rd assume $X=\mathbb{N}$ and consider the map $\varphi_4:A\mapsto A\cup\{\min{A}\cdot 2\}$. Where for example for sets $A=\{2,3\},B=\{3,4\}$ this fails the condition (4). What could be an example of $\varphi_3$? Something that satisfies 1,2,4 but not 3? We can assume any set $X$.
 A: A topological example: let $X$ be a space and define $\phi(A) = \{x \in X: \exists (a_n) \in A^\mathbb{N}: a_n \to x\}$, the sequential closure operation.
For metric spaces this would just be the normal closure, but for the space $X$ the Arens' space, as defined here, we have that $\phi(\mathbb{N}\times \mathbb{N}) = (\mathbb{N} \times \mathbb{N}) \cup \mathbb{N}$ and $\phi(\phi(\mathbb{N} \times \mathbb{N})) = X \neq \phi(\mathbb{N} \times \mathbb{N})$, showing that $\phi$ does not obey 3.
Sequential closure is one of the motivating examples for studying the Cech closure spaces as a generalisation of topological spaces. There are also analysis examples in measure theory IIRC (convergence a.e. or convergence in measure or some such notion)
A: The answer is already in the comments so to post it here explicitly, set $X=\{1,2,3\}$, define $\varphi_3:\mathscr{P}(X)\rightarrow \mathscr{P}(X)$ as following: $\varphi_3:A\mapsto A\cup\{n-1\mid n\in A\setminus\{1\}\}$. That means:
$$\emptyset\mapsto\emptyset,\{1,2,3\}\mapsto \{1,2,3\}$$
$$\{1\}\mapsto \{1\}, \{2\}\mapsto\{1,2\},\{3\}\mapsto\{2,3\}$$
$$\{1,2\}\mapsto \{1,2\}, \{1,3\}\mapsto \{1,2,3\}, \{2,3\}\mapsto\{1,2,3\}$$
