Inquiry regarding neighborhood bases I just want to see if I have the right angle here.
We have a set $S$, and for each  $x\in S$ there's a collection of subsets in $S$, $N(x)$ which satisfies:


*

*$N \in N(x) \rightarrow x \in N$

*$N,M \in N(x) \rightarrow \exists P \in N(x): P \subset N\cap M$

*$x \in S \rightarrow N(x) \neq \phi$
then there's a unique topology on S such that $N(x)$ is a basis neighbourhood basis at x, for $x \in S$.
Am I right in thinking that topology is generated by the following open sets:
$$ \cup_{x \in S \ N \in N(x)} N$$
In which case it's unique by this very definition, right?
Thanks, I am quite rusty with topology, I took my first course in it 3 years ago.
 A: Since $$\bigcup\limits_{x\in S,N\in N(x)}N$$ is simply the single set $S$, I doubt it’s what you actually had in mind. 
Edit: On the (foolish) assumption that the nbhd bases were intended to be open nbhd bases, I originally wrote:

Let $$\mathscr{N} = \bigcup\limits_{x\in S}N(x),$$ the collection of all sets appearing in any of the families $N(x)$. Then $\mathscr{N}$ is a base for a topology $\mathscr{T}$ members of $-$ i.e., the open sets in the space $-$ are the unions of subcollections of $\mathscr{N}$.

Now that the wording of the problem has been confirmed, I realize that this assumption was unjustified, and what I wrote above isn’t necessarily true. Here’s a correct argument.
Let $\mathscr{T} = \{U \subseteq S:\forall x\in U\exists N\in N(x)[N \subseteq U]\}$. The first step is to prove that $\mathscr{T}$ is a topology on $S$.


*

*$\varnothing \in \mathscr{T}$ vacuously, and (1) and (3) imply that $S \in \mathscr{T}$.  

*Suppose that $U,V \in \mathscr{T}$; we must show that $U\cap V \in \mathscr{T}$. To that end let $x \in U\cap V$. Then $x \in U$, so there is an $N_U \in N(x)$ such that $x \in N_U \subseteq U$, and $x \in V$, so there is an $N_V \in N(x)$ such that $x \in N_V \subseteq V$. By (2) there is $N \in N(x)$ such that $x \in N \subseteq N_U \cap N_V \subseteq U \cap V$. Since $x$ was an arbitrary element of $U \cap V$, it follows that $U \cap V \in \mathscr{T}$.

*Now let $\mathscr{H}$ be any subfamily of $\mathscr{T}$; we must show that $\bigcup\mathscr{H} \in \mathscr{T}$. Suppose that $x \in \bigcup\mathscr{H}$; then $x \in H$ for some $H \in \mathscr{H}$. $H \in \mathscr{T}$, so there is some $N \in N(x)$ such that $x \in N \subseteq H \subseteq \bigcup \mathscr{H}$, and it follows as before that $\bigcup\mathscr{H} \in \mathscr{T}$.


It remains to show that $\mathscr{T}$ is unique. Suppose that $\mathscr{U}$ is a topology such that for each $x \in S$, $N(x)$ is a nbhd base at $x$ with respect to $\mathscr{U}$. Then for any $U \in \mathscr{U}$ and any $x \in U$ there must be some $N \in N(x)$ such that $x \in N \subseteq U$, which of course means that $U \in \mathscr{T}$. This shows that $\mathscr{U} \subseteq \mathscr{T}$. 
Suppose now that $U \in \mathscr{T}$, so that for each $x \in U$ there is an $N_x \in N(x)$ such that $x \in N_x \subseteq U$. $N_x$ is a $\mathscr{U}$-nbhd of $x$, so there must be some $V_x \in \mathscr{U}$ such that $x \in V_x \subseteq N_x$. But then for each $x \in U$ we have $x \in V_x \subseteq U$, so $U = \bigcup\limits_{x \in U} V_x$. $U$ is therefore a union of members of the topology $\mathscr{U}$, so $U \in \mathscr{U}$, and hence $\mathscr{T} \subseteq \mathscr{U}$.
Putting the pieces together, we have $\mathscr{U} = \mathscr{T}$, and it follows immediately that $\mathscr{T}$ is unique.
A: (Edited in response to a comment by jdc)
In your list of axioms the equivalent of the "topological triangle inequality" is missing. It says the following: For any neighborhood $U$ of the point $x$ there exists a neighborhood $V$ of $x$ such that $U$ is also a neighborhood of all points $y\in V$. If one expresses this in terms of your neighborhood bases $N(x)$ one arrives at the axiom
$4$. For any $N\in N(x)$ there is an $M\in N(x)$ such that for all $y\in M$ there is an $N'\in N(y)$ with $N'\subset N$.
The answer by Brian M. Scott above produces a topology ${\mathscr T}$ related to the family $\bigl(N(x)\bigr)_{x\in S}$ alright, but the given $N\in N(x)$ need not be neighborhoods of $x$ with respect to this topology. He then proves that if they are indeed neighborhoods then everything is in order.
Consider the following example: In $S:={\mathbb R}^2$ let $N(z)$ consist of all crosses $+$ with center $z$. The family ${\mathscr N}:=\bigl(N(z)\bigr)_{z\in S}$ violates axiom $4$ above. Now open sets in the topology ${\mathscr T}$ defined from ${\mathscr N}$ extend in two dimensions at all of their points. Therefore no $N\in N(0)$ can contain an open neighborhood of $0$.
