$\overline{A\cup B} \subset \bar{A}\cup \bar{B}$ with a weaker condition in topology

Question

We know that in a topological space $(X,\tau )$, we have $\overline{A\cup B} \subset \bar{A}\cup \bar{B}$ for every subset $A$ and $B$ of $X$.

Is the above statement true when instead of intersection condition in definition of topology we have the following condition?

$U\cap V\neq \emptyset\; $ implies $\; int (U\cap V)\neq \emptyset$

Answer 1

Suppose that your condition on the closure operation holds for all $A,B$ (I assume you're working in the context of so-called generalised topological spaces, where only the union axiom of a topology holds, but not necessarily the finite intersection one), then $X$ is in fact a (usual) topology:

For if $O_1, O_2$ are open, $X\setminus O_1$ and $X\setminus O_2$ are closed, so they equal their own closure. Then (by de Morgan): $$\overline{X\setminus (O_1 \cap O_2)} = \overline{(X\setminus O_1) \cup (X\setminus O_2)} \subseteq \overline{X\setminus O_1} \cup \overline{X\setminus O_2} \\ = (X\setminus O_1) \cup (X\setminus O_2) = X\setminus (O_1 \cap O_2) \subseteq \overline{X \setminus (O_1 \cap O_2)}$$

(where the last one is from the always true $A \subseteq \overline{A}$) which shows that $X\setminus(O_1 \cap O_2)$ is closed and so $O_1 \cap O_2$ is open.

All this is classical: generalised topological spaces are just a manifestation of Čech-closure spaces, that have a closure operation that obeys the following axioms:

1. $\overline{\emptyset} = \emptyset$.
2. $\forall A \subseteq X: A \subseteq \overline{A}$
3. $\forall A,B \subseteq X: A \subseteq B \to \overline{A} \subseteq \overline{B}$.
4. $\forall A \subseteq X: \overline{\overline{A}} = \overline{A}$

If $X$ is a generalised topological space and the complements of "open" subsets are called closed, and we define $\overline{A} = \bigcap \{C: C \text{ closed and } A \subseteq C\}$ then this is a Čech-closure space. If $X$ is such a space, the collection $\tau=\{O \subseteq X: \overline{X\setminus O} = X\setminus O\}$ is a generalised topology. These constructions are each other's inverse.

It is known that $X$ is a topological closure space iff it obeys the extra "topology closure axiom" $\forall A,B \subseteq X$: $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$ (which in fact implies, by using 3. that $\overline{A \cup B} = \overline{A} \cup \overline{B}$ always.) The $\tau$ above is then exactly a topology, and closures defined from a topology always obey this axiom.