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

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  • $\begingroup$ The last should be interpreted as “every pair of intersecting open sets contains a non-empty open set”? $\endgroup$ – Henno Brandsma Apr 15 '18 at 13:33
  • $\begingroup$ @HennoBrandsma I think his/her mean is that $(X,\tau )$ is a generalized topological space with an extra condition. Here the meaning of an open set is different from open set in topology. $\endgroup$ – M.Ramana Apr 20 '18 at 17:41
  • $\begingroup$ A generalized topology on $X$ is a set $\tau \subset 2^X$ with two conditions: (1)- $\emptyset , X\in \tau$ and (2)- $\tau$ is closed with respect to arbitrary unions of its elements. $\endgroup$ – M.Ramana Apr 20 '18 at 17:44
  • $\begingroup$ In generalized topology, intersection of two open sets is not necessary open. The last is a weaker condition. $\endgroup$ – M.Ramana Apr 20 '18 at 17:48
  • $\begingroup$ @M.Ramana is my formulation of that extra condition correct? $\endgroup$ – Henno Brandsma Apr 20 '18 at 20:08
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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.

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  • $\begingroup$ You proved that if $O_1$ and $O_2$ are open, then $O_1 \cap O_2$ is also open. This implies that $\tau$ is a topology on $X$ which is a contradiction. $\endgroup$ – M.Ramana Apr 21 '18 at 10:29
  • $\begingroup$ @M.Ramana indeed I showed that if the closure condition holds, we have a topology. So your original question is answered negatively. $\endgroup$ – Henno Brandsma Apr 21 '18 at 10:31
  • $\begingroup$ @M.Ramana it’s only a contradiction if we have a generalised topology which is not a topology. In that case I showed that the closure condition also must fail. $\endgroup$ – Henno Brandsma Apr 21 '18 at 10:33
  • $\begingroup$ Yes, Your answer is completely true. I'm wrong. I'm sorry. $\endgroup$ – M.Ramana Apr 21 '18 at 10:39

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