This is an exercise from Willard & Stephen, General Topology. Exercise $4E$ , adapted from theorem $4.5$.
Starting from a set $X$. In which we define at every $x$ a neighborhood system $\mathcal V _x$ such that:
- $\forall U \in \mathcal V _x , x\in U$
- $\forall U,V \in \mathcal V _x , U\cap V \in \mathcal V _ x$
- $U\in \mathcal V _x $ and $U\subseteq V \implies V\in \mathcal V _x$
- $\forall U \in \mathcal V _x, \exists V\in \mathcal V_x , \forall y \in V, U \in \mathcal V_y$
we proceed to show that the sets $O$ such that $\forall x\in O , O \in \mathcal V _x$ form a topology and moreover, that the neighborhood system of a point $x$ in this topology (which is defined in a topology as $\{ N \subseteq X \mid x\in N^{\circ} \}$) is the neighborhood system we started from ($\mathcal V _x$).
Showing that $\mathcal V _x$ induces a topology was easy (and I only used $(3)$ and $(4)$ in doing it).
Showing that $\mathcal V _x = \{ N \subseteq X \mid x\in N^{\circ} \}$ is yet less easy.
Here is how I've proceeded:
First, I've noticed that the alternative definition of a neighborhood in a topology could be more handy. So $\{ N \subseteq X \mid x\in N^{\circ} \}$ turned into $\{ N \subseteq X \mid \exists M \text{ open}, M\subseteq N, x\in M \}$.
So let $U\in \mathcal V _x$. Using $(4)$ we have $\exists V\in\mathcal V _x, \forall y \in V , U\in \mathcal V_ y$. Trivially we have $x\in V$ (using $(1)$). Using $(1)$ again we have that every element of $V$ is also an element of $U$, so that $V\subseteq U$.
Now remains to prove that $V$ is an open set. In the general case, I don't think that $(4)$ provides us the existence of an open set $V$. It can very well happen that $V$ isn't open. That leads me to think that, maybe, we should go for a "smaller" set within $V$ (i.e. construct a smaller set inside of $V$ that, this time, is open).
Thanks very much in advance.