Opens in a convergence space By the book "contemporary mathematics", Beyond Topology (F.mynard , E.Pearl)
I am now studying convergence spaces on the book mentioned above. On this book (p.123) I find this definition:
A subset O of a convergence space is open if lim $\mathcal{F}$ $\cap$ O $\neq$ $\emptyset$ implies that O $\in lim \mathcal{F}$
By definition lim $\mathcal{F}$ is the set of points in relation with the filter $\mathcal{F}$ throught a relation $\xi$
How can O (that is a subset) be in that set? What is your definition for opens in a convergence space?
Moreover, after a few lines it talk about a "topologization" of a relation $\xi$. Searching online or on references didn't get me any results on what this may be. Suggestions?
Thanks!

Well, for a topological space $X$ a set $O \subset X$ is open iff for all $x \in O$, for any filter $\mathcal{F}$ on $X$ with $\mathcal{F} \rightarrow x$ we have $O \in \mathcal{F}$.
(Because $O$ is a neighbourhood of $X$ and a filter convergent to $x$ in a topological space means that all neighbourhoods of $x$ are in the filter.)
So this comes down to: for every filter $\mathcal{F}$ on $X$, $O \cap \lim\mathcal{F} \neq \emptyset$ implies $O \in \mathcal{F}$. (Just apply the previous to $x$ in this intersection.)
So it seems that there is a typo in your book, and the second statement above is what the authors meant as the definition of an open set in a convergence space.
Without more context I couldn't say what the "topologization" would be. Maybe this definition of open sets (though it is not a topology in general, but a so-called pre topology)?
