Is an "open system" just a topological space?

Question

I think of an "open set" as being "roomy" or "spacious," in the sense that around every point, there is a little bit of room. This motivates the following definition.

Definition. An "open system" consists of an underlying set $X$ together with a collection of subsets of $X$ that are considered "open," such that for all $A \subseteq X$ it holds that if

• for all $a \in A$ there exists open $B \subseteq A$ such that $a \in B$

then

• $A$ is open.

Question. Is an "open system" just a topological space?

Answer 1

Let $X$ be a set. Then a collection $\mathscr{C}$ of subsets of $X$ is an "open system" if and only if $\mathscr{C}$ is closed under unions.

One direction is easy. Suppose then that $\mathscr{C}$ is closed under unions. Let $A$ be a subset of $X$, and assume that for each $a \in A$, there exists a $B_a \in \mathscr{C}$ such that $a \in B_a \subseteq A$. Then $A = \cup_{a \in A} B_a$, so $A \in \mathscr{C}$ since $\mathscr{C}$ is closed under arbitrary unions.

Now it should be easy for you to find "open systems" that are not topological spaces, ie. collections of subsets that are closed under unions but not under finite intersections.

Answer 2

In a topological space you require the open sets to be closed under unions (which trivially holds in your case), but also under finite intersection.

Your "open system", on the other hand, either can be any collection of subsets closed under unions ($A\subseteq A$ is open if $A$ is open, and $a\in A$ satisfies $a\in A$ as well); or if you require $B\subsetneqq A$, you disallow any isolated points because $\{x\}$ cannot be open, and by induction no finite set can be open (except the empty set, perhaps). In either case it doesn't seem to follow that "open systems" are closed under intersections.