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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:

  1. $\forall U \in \mathcal V _x , x\in U$
  2. $\forall U,V \in \mathcal V _x , U\cap V \in \mathcal V _ x$
  3. $U\in \mathcal V _x $ and $U\subseteq V \implies V\in \mathcal V _x$
  4. $\forall U \in \mathcal V _x, \exists V\in \mathcal V_x , \forall y \in V, U \in \mathcal V_y$

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.

niobium
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    Property $4$ produces a set $V$ such that $V\in\mathcal{V}_x$, and such that for all $y\in V$, $V\in\mathcal{V}_y$. The topology induced is defined to consist of sets $O$ such that for all $x\in O$, $O\in\mathcal{V}_x$. But $V$ precisely satisfies that property: for all $y\in V$, you know $V\in\mathcal{V}_y$. Therefore, $V$ is necessarily open in the topology you are defining. – Arturo Magidin Jun 14 '23 at 20:02
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    P.S. You need to assume the systems $\mathcal{V}_x$ are nonempty; the empty collections satisfies all properties 1 through 4, but do not give you a topology. – Arturo Magidin Jun 14 '23 at 20:04
  • I think $(4)$ is: $\forall U \in \mathcal V _x, \exists V\in \mathcal V_x , \forall y \in V, U \in \mathcal V_y$. It is not $V\in \mathcal V _y$ as you say, but $U\in \mathcal V _ y$. – niobium Jun 14 '23 at 20:09
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    You should double-check that. Kelley's "General Topology" has "If $U\in\mathcal{V}_x$, then there is a member $V$ of $\mathcal{V}_x$ such that $V\subset U$ and $V\in\mathcal{V}_y$ for every $y\in V$ (that is, $V$ is a neighborhood of each of its points)", as I said. That is the standard. (Of course, that would imply what you write when combined with 3...) – Arturo Magidin Jun 14 '23 at 20:24

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