# Variation on neighbourhood base

Suppose $\{\mathscr B(x) \mid x \in X\}$ is a collection of filters (or filter bases) on a set X, with each $x \in \cap\mathscr B(x)$. Then $$\mathscr T = \{U \subseteq X \mid (\forall x \in U)(\exists B \in \mathscr B(x)) B \subseteq U \}$$ defines a topology $\mathscr T$ on X. Note that $\mathscr B(x)$ is not necessarily a neighbourhood base at x, as its members need not be neighbourhoods of x.

Is there an accepted name for such collections, or any interesting known results?

(If we call a space quasi-first-countable if it has such a collection with each $\mathscr B(x)$ having a countable base then.$$\text{First-countable} \implies \text{Quasi-first-countable} \implies \text{Sequential}$$ with none of the implications being reversible. Most of that has been shown in answers to recent questions about d-metrisable spaces. (E.g d-metrizable spaces are sequential. ) Quasi-first-countable is equivalent to d-metrisable.)

• Sorry David, are you saying it's not necessarily a neighbourhood base because its members aren't necessarily open, or for a different reason? Terminology varies as to whether a neighbourhood must be open. – Mike Nov 16 '14 at 23:24
• You may want to look at en.wikipedia.org/wiki/Pretopological_space – user2345215 Nov 16 '14 at 23:26
• @Mike. The members of $\mathscr B(x)$ needn't contain an open set which has x as a member. In fact they can have no open (non-empty) subsets at all. For example, take X to be $\Bbb R$ and let $\mathscr B(x)$ be the filter base formed by taking open neighbourhoods of x in the usual topology and removing all irrationals < x and all rationals > x. (An example of a space that is quasi-first-countable but not first-countable.) – David Hartley Nov 17 '14 at 0:19
• @user2345215. Thanks.That looks very close but is slightly different. The collections I'm suggesting are indeed neighbourhood systems for pretopologies as defined in your link, but the associated topology is not the same. For instance, with the example in my previous comment the net consisting of all irrationals < x does not converge to x in the associated pretopology but does in the topology I suggest. Perhaps they will be the same whenever the pretopology is a topology, but it's late so I'll leave that for tomorrow. – David Hartley Nov 17 '14 at 0:38
• @DavidHartley Good example. – Mike Nov 17 '14 at 1:02

tl;dr; I would call such collection a (local) network system.

Let $X$ be a topological space, $x ∈ X$. We call a nonempty collection $\mathscr{B}(x) ⊆ \mathscr{P}(X)$ such that $(∀B ∈ \mathscr{B}(x))\ x ∈ B$.

• (local) network at $x$ if for every neighborhood $U$ of $x$ there is some $B ∈ \mathscr{B}(x)$ such that $x ∈ B ⊆ U$;
• (local) (neighborhood} base at $x$ if in addition every $B ∈ \mathscr{B}(x)$ is a neighborhood of $x$;
• (local) open (neighborhood) base at $x$ if in addition every $B ∈ \mathscr{B}(x)$ is open.

Then the corresponding mapping $\mathscr{B} := \{x \mapsto \mathscr{B}(x): x ∈ X\}$ is called (local) network system, (local) neighborhood system, or (local) open neighborhood system, respectivelly. One can observe that the topology is uniquely determined by such system. As you write, $\mathscr T_\mathscr{B} = \{U \subseteq X \mid (\forall x \in U)(\exists B \in \mathscr B(x))\ B \subseteq U \}$.

On the other hand we may start with a set $X$ and a such mapping $\mathscr{B}$ and ask, when $\mathscr{T}_\mathscr{B}$ is a topology and under what conditions $\mathscr{B}$ is its network system or even (open) neighborhood system.

When we have a set $X$ and a mapping $\mathscr{B}$, by the topology generated by the network / neighborhood / open neighborhood system, we mean just $\mathscr{T}_\mathscr{B}$ and the claim that $\mathscr{B}$ is a corresponding system of induced topology.

So I would say, that $X$ is first-countable if it is generated by some neighborhood system $\mathscr{B}$ such that $\mathscr{B}(x)$ is countable for each $x$. And similarily $X$ is quasi-first-countable if it is generated by some network system $\mathscr{B}$ such that $\mathscr{B}(x)$ is countable for each $x$.

• I take it that's your suggestion, not an already established convention? I have been calling them quasi-neighbourhood systems but wanted to know if there is an existing standard. – David Hartley Nov 17 '14 at 17:08
• Well, it's an extension of established convention. There is a notion of network similar to the notion of base, but the sets are not required to be open. So (local) network at a point is similar to (local) (neighborhood) base at a point. And Suggested network system is analoguous to neighborhood system. – user87690 Nov 17 '14 at 17:31
• (Accidently posted the comment before I'd finished) Quasi-neighbourhood systems differ from the other two you mention in that they are less determined by the topology: every neighbourhood base is a base for the filter of all neighbourhoods which is unique to the topology, but a space can have different quasi-neighbour bases which do not determine the same filter (at the same point). I suspect it is possible to have two systems st. $\mathscr B(x) = \mathscr C(x)$ at some point x but the neighbourhoods of x in the corresponding topologies are different. – David Hartley Nov 17 '14 at 17:34
• That's possible. I think it is not in contradiction with anything in my answer. – user87690 Nov 17 '14 at 17:42
• Yes, there's no contradiction. Sorry. I was just a little uneasy at defining a (local) network at a point x in isolation, it being of little use without the rest of the system. But of course you did from then on use only the whole system. Can you give a link to any discussion of this notion of network? My searches just turn up discussions of the topology of computer networks. – David Hartley Nov 17 '14 at 23:01