# Defining a topology in terms of filter neighbourhoods

I am a little confused about the following:

if $$X$$ is a set for which there is a filter $$\mathcal{F}(x)$$ of sets containing $$x\in X$$ assigned to every point $$x\in X$$, and these filters are such that for all $$U \in \mathcal{F}(x)$$ there is some set $$V \in \mathcal{F}(x)$$ such that for all $$y \in V$$, $$U \in \mathcal{F}(y)$$, if we say a set $$O$$ is open if it is empty or such that for all $$o \in O,O \in \mathcal{F}(o)$$, then why is it true that the closure of a set $$A$$ is the set of all points $$x$$ for which all members of $$\mathcal{F}(x)$$ have nonempty intersection with $$A$$?

I can prove this if I know that every member of $$\mathcal{F}(x)$$ contains an open set containing $$x$$, because then it just comes down to proving the following statement: 'The closure of a set $$A$$ in a topological space is the set of all points $$x \in X$$ such that for all open sets $$U$$ containing $$x$$, $$U \cap A \neq \emptyset$$' which I am familiar with and know how to prove.

The issue is although it seems like intuitively, all members of the filter of 'neighbourhoods' of a point should all themselves contain open sets containing that point, I cannot seem to easily deduce it from the definition of open set in my first paragraph.

This definition is taken from 'Topological Vector Spaces,Distributions and Kernels' by Francois Treves.

In this post I give a full proof of equivalence between neighbourhood spaces and corresponding topological spaces. Given a neighbourhood space, we can define open set as you describe (the empty set is not a special case BTW, because of void truth) and we get that the neighbourhood system we start with is exactly the one corresponding to that topology.

In particular the axiom: "for all $$U \in \mathcal{F}(x)$$ there is some set $$V \in \mathcal{F}(x)$$ such that for all $$y \in V$$, $$U \in \mathcal{F}(y)$$" (quoting your post) really "says" that each neighbourhood of $$x$$ contains an open neighbourhood of $$x$$.

The above equivalence proves also that the closure of a set is indeed

$$\operatorname{cl}(A) = \{x \in X\mid \forall N \in \mathcal{N}_x: N \cap A \neq \emptyset\}$$

because that's indeed true for neighbourhoods in all topological spaces. We cold have taken that as the definition of closed sets in this correspondence too (though it's slightly more convoluted, unless you're allowed to use the Kuratowski closure axioms as the foundations of topology, and check directly that cl as defined above obeys these).

• Thanks for the reply. I don't see anywhere in that post this exact condition, rather I see the condition that if $U \in \mathcal{F}(x)$, $\{z \in U : U \in \mathcal{F}(z) \} \in \mathcal{F}(x)$, are you saying these are equivalent? Commented Mar 26, 2021 at 9:48
• @porridgemathematics yes, they are, if you think about it a bit. It's easy to see that your formulation implies that one. Commented Mar 26, 2021 at 9:55
• Ok so my condition implies this one because this set contains $V$, which is in $\mathcal{F}(x)$ and filters are closed under supersets, apologies but even so, I don't see where the desired conclusion is implied, is it easy to show that for example,$\{z \in U : U \in \mathcal{F}(z) \}$ is open as per the definition I gave? Commented Mar 26, 2021 at 9:59
• @porridgemathematics Yes I call that set $U^\ast$ and I show at the end it is open (for $N^\ast$, different letters, same meaning). Commented Mar 26, 2021 at 10:09