prove that the Hashimoto topology is a topology Let $\mathcal{I}$ be an ideal of subsets of the real line.
The collection of sets of the form $O\setminus I$, where $O$ is open (in natural topology) and $I\in\mathcal{I}$, is a topology.
I'm having difficulties in proving this.
I have done the first two parts (empty set and finite intersections). For the unions, I tried to use the fact that open set in $R$ is a countable sum of disjoint open intervals, but it doesn't led me to anything.
 A: There's a good reason you're having trouble!
As written, that's not a topology: for example, let $\mathcal{I}$ be the ideal generated by all finite unions of open intervals of the form $(x, x+{1\over 2})$ for $x\in\mathbb{Z}$. Then the set $\bigcup_{x\in\mathbb{Z}} [x+{1\over 2}, x+1)$ is a union of sets of the form $O\setminus I$ ($O$ open, $I$ in the ideal $\mathcal{I})$ but is obviously not of that form itself.
The collection of sets of the form $O\setminus I$ does generate a topology, though. And assumptions on the ideal $\mathcal{I}$ may fix things: is $\mathcal{I}$ supposed to be any special kind of ideal?

EDIT: It seems there are supposed to be some assumptions about the ideal $\mathcal{I}$; see e.g. the first page of https://www.sav.sk/journals/uploads/1030154902terep.pdf, which assumes that $\mathcal{I}$ is proper, contains every singleton, and satisfies the additional property $$(1)\quad A\cap A^\mathcal{I}=\emptyset\iff A\in\mathcal{I},$$ where $A^\mathcal{I}$ is the set of $x$ such that every open neighborhood of $x$, intersected with $A$, is "large" (= not in $\mathcal{I}$).
