How to show from the definition of topological space that the topology $\tau$ is the family of open sets of a topology with a filter base?

Question

I am starting with the given definitions

1. A filter $ \mathcal{F}$ on $X$ is a family of subsets of $X$ such that

   • $\varnothing \not\in \mathcal{F}.$
   • $\mathcal{F}$ is algebraically closed under finite intersections
   • $ \mathcal{F} $ is an upper family.

2. A neigbhourhood 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; The filter $ \mathcal{N}(x)$ called the neighbourhood filter of the point $x$ in $X$.

3. A set $T \subset X$ is open if, and only if, for all $x \in T$, there exists $N \in \mathcal N(x)$ such that $N \subset A$

4. A filter base $ \mathcal{B} $ is a family of non-empty subsets of $X$ such that if $A,B \in \mathcal{B}$ there exist $C \in \mathcal{B} $ then there exists $C \in \mathcal{B}$ such that $C \subset A \cap B$.

5. A non-empty collection $ \mathcal{B} $ of subsets of $X$ is a base for a specific filter $ \mathcal{F} $ on $X$ if and only if

   • $ \mathcal{B} \subset \mathcal{F}$
   • If $ A \in \mathcal{F}$, there exists $ B \in \mathcal{B}$ such that $ B \subset A$.

6. A topological space is a neighbourhood space $( X, \mathcal{N})$ in which, for all $x \in X$ and for all $ N \in \mathcal{N} $, there exists $N_1 \in \mathcal{N}(x)$ such that, for all $y \in N_1 , N\in \mathcal{N}(y)$.

Theorem 1 : A neighbourhood space $( X, \mathcal N)$ is a topological space if and only if each filter $ \mathcal N(x)$ has a filter base consisting of open sets.

The proof of this theorem starting with the definition of topological space can be found here

Now, one can also speak of this theorem,

Theorem 2 : Let $\tau$ be any family of subsets of a set $X$ that satisfies the three conditions:

1. $\varnothing$ and $X$ belong to $\tau$
2. an arbitrary union of elements of $\tau$ belongs to $\tau$.
3. any finite intersection of elements of $\tau$ belongs to $\tau$

Then $\tau$ is the family of open sets of a topology on $X$ with a neighbourhood base $\mathcal B(x) :=\{O \in \tau | x \in O\}$ for all $x$ in $X$.

Here the filter base $\mathcal B(x)$ generates the neighbourhood filter $\mathcal N(x)$ at $x \in X$.

I want to prove theorem 2 starting either from definition-6 or theorem-1.

I know certain things as -

• I only know whether a subset of $X$ is open or not only from the definition-3.
• An arbitrary union and finite intersection of open sets is an open set.
• Closer under finite intersection is a property of filters. (Here the Neighbourhood filters $\mathcal N(x)$ which are generated by the filter base $\mathcal B(x)$)
• $\forall x \in X, $ $X \in \mathcal B(x)$, and $\varnothing \not\in \mathcal B(x)$ (by definition)

I want to prove that the sets $O \in \tau $ are open as per the definition-3 of open sets

How, to go from here as the proof is concerned?

Answer 1

It’s immediate from the definition: if $U\in\tau$ and $x\in U$, then $U\in\mathscr{B}(x)\subseteq\mathscr{N}(x)$, and certainly $U\subseteq U$, so $U$ is open in the sense of (3). Of course to use Theorem 1 to prove Theorem 2 you still have to show that each $\mathscr{B}(x)$ really is a filter base, but that’s very straightforward.