Proof that a neighbourhood space is a topological space under certain condition

Question

I'm trying to prove, as a book excercise, the following statement:

A neighbourhood space $(X, \mathcal{N})$ is a topological space iff each neighbourhood filter $\mathcal{N}(x)$ has a filter base consisting of open sets.

I'm using the following definitions and theorems:

1. A neighbourhood structure $\mathcal{N}$ on a set $X$ is an assignment to each $x \in X$ of a filter $\mathcal{N}(x)$ on $X$ all of whose elements contain the point $x$. The pair $(X,\mathcal{N})$ is called a neighbourhood space and the filter $\mathcal{N}(x)$ is called the neighbourhood filter of the point $x \in X$.
2. A topological space is a neighbourhood space $(X,\mathcal{N})$ in which, for all $x \in X$ and for all $N \in \mathcal{N}(x)$, there exists $N^\star \in \mathcal{N}(x)$ such that, for all $y \in N^\star$, $N \in \mathcal{N}(y)$.
3. A non-empty collection $\mathcal{B}$ of subsets of $X$ is a base for a specific filter $\mathcal{F}$ on $X$ iff (i) $\mathcal{B} \subseteq \mathcal{F}$ and (ii) if $A \in \mathcal{F}$, there exists $B \in \mathcal{B}$ such that $B \subseteq A$.

I was able to prove that a neighbourhood space satisfying the condition is a topological space: as all neighbourhood filters have an open base, there exists $B$ open and $N$ neighbourhood of $x \in X$ such that $B \subseteq N$. As $B$ is open, it contains a neigbourhood for each of its points, therefore $N$ is a neigbourhood of all points in $B$ because filters are upward closed, and also $B$ is a neighbourhood of $x$, so the definition of a topological space is met.

However, I'm having trouble with the other implication. Let $x \in X$ and $N \in \mathcal{N}(x)$. I know that $\exists N^\star \in \mathcal{N}(x)$ such that $\forall y \in N^\star : N \in \mathcal{N}(y)$, and want to see that there exists an open neighbourhood $B \subseteq N$; that would mean there exists a collection of open neighbourhoods $\mathcal{B}$ finer than $\mathcal{N}(x)$, and that would mean $\mathcal{B}$ is a filter base of open sets. However I can't find a way of arriving at that statement.

Am I on the right track, or is there something I'm overlooking/making some mistake?

Answer 1

Suppose $(X,\mathcal{N})$ is a topological space, so it satisfies the second axiom in your question. This is quite an important axiom as it ties the different neighbourhood filters (for different points) together. Otherwise we could just arbitarily assign filters for each point and we'd get no real spatial "structure".

Suppose $x \in X$ and $N \in \mathcal{N}(x)$. Define $$N^\circ = \{y \in X: N \in \mathcal{N}(y)\}\text{.}$$

This is (intuitively, and later more formally) the interior of $N$ in this topology, as it will turn out. In words, all $y$ that $N$ is a neighbourhood of.

1. $x \in N^\circ$, which is trivial, as this is how $N$ was chosen in the first place.

2. $N^\circ \subseteq N$, for pick $y \in N^\circ$, then $N \in \mathcal{N}(y)$ which means, by part of axiom 1, that $y \in N$.

3. Suppose that $y \in N^\circ$, so $N \in \mathcal{N}(y)$. Then apply axiom 2. to this $y$ and $N$ to get $N^\ast \in \mathcal{N}(y)$ such that for all $z \in N^\ast$ we have $N \in \mathcal{N}(z)$. Note that this means that all these $z$ are in $N^\circ$ by definition, and so $N^\ast \subseteq N^\circ$ (we won't have equality in general, but we don't need it). But as $N^\ast \in \mathcal{N}(y)$, as $\mathcal{N}(y)$ is a filter, so upwards closed, $N^\circ \in \mathcal{N}(y)$ as well. So note what we have shown: we start with any $y \in N^\circ$ and show $N^\circ$ is a neighbourhood of $y$. So $N^\circ$ is open by the usual definition of openness in neighbourhood spaces.

So for every $N \in \mathcal{N}(x)$ we have found an open subset $N^\circ$ of $N$ that is open and a neighbourhood of $x$ as well (as $x \in N^\circ$ and $N^\circ$ is open!). So the open neighbourhoods of $x$ form an open filter base at $x$. We can just define $\mathcal{B}(x) = \{N^\circ: N \in \mathcal{N}(x)\}$ to make it explicit.