Equivalence of two different formulations of neighborhood systems in a set

Question

When defining a non-empty neighborhood system $\mathcal V _x$ of a set $X$, the formulation of some authors differ.

Some provide these $4$ criteria:

$\forall U \in \mathcal V _x , x\in U$
$\forall U,V \in \mathcal V _x , U\cap V \in \mathcal V _ x$
$U\in \mathcal V _x $ and $U\subseteq V \implies V\in \mathcal V _x$
$\forall U \in \mathcal V _x, \exists V\in \mathcal V_x , \forall y \in V, U \in \mathcal V_y$

Some others provide:

$\forall U \in \mathcal V _x , x\in U$
$\forall U,V \in \mathcal V _x , U\cap V \in \mathcal V _ x$
$U\in \mathcal V _x $ and $U\subseteq V \implies V\in \mathcal V _x$
$\forall U \in \mathcal V _x, \exists V\in \mathcal V_x , V\subseteq U, \forall y \in V, V \in \mathcal V_y$

In both formulations, the criteria $(1)$ to $(3)$ are common. I'd like to prove the equivalence of these two formulations.

The second formulation implies the first one because $\forall y \in V, V\in \mathcal V_ y$ and $V\subseteq U$ imply (by $(3)$) that $\forall y \in V, U \in \mathcal V _y$.

The first formulation implies the second one because if we simply take $V=U^{\circ}$ (note that the open sets can be defined by the sets such that they are neighborhood of each of their points), then $V\subseteq U$ and $V$ is a neighborhood of each of its points.

However, I am not convinced by this solution, because I don't have the guarantee that $U$ has an interior (i.e. that there exists open sets that are included in $U$).

Another flaw in this proof is that I never used the $4^{th}$ criterion from the first formulation or any other criterion from the first formulation. Don't we need to use at least one of the criteria of the first formulation in order show $1^{st}$ formulation implies the $2^{nd}$ one ?

Thanks in advance.

Are you sure about your second 4th point? What is the point of having a $U$? Should not the last bit be $U \in \mathcal{V}_y$? — julio_es_sui_glace, Jun 15 '23 at 11:13
That's what a Stackexchange user told me in the comment section of this discussion — niobium, Jun 15 '23 at 14:18
If you change this, they are almost the same and equivalent, it can be a typo... — julio_es_sui_glace, Jun 15 '23 at 15:32
No, I think Arturo Magidin is right (his definition makes my other SE post's proof easier). But, to be completely honest Wikipedia, Wolfram, and most of the ressources I've seen use your definition (i.e. my first formulation). Arturo Magidin cites the book Topology from Kelley, but I couldn't find the reference in the book. — niobium, Jun 15 '23 at 16:20

score 1 · Accepted Answer · answered Jun 16 '23 at 02:16

Your proof that the second definition implies the first is fine. The other way is also not bad.

Indeed, if we are defining the neighborhood system by the criteria sets you listed, then it might be better to avoid explicitly dealing with "open" sets (which aren't defined at the outset). Nonetheless, thinking about the "interior" is good.

Assume the first definition's set of criteria, and let $U \in \mathcal{V}_x$ be an arbitrary neighborhood of $x$. Consider the set $W = \{y \in X | U \in \mathcal{V}_y\}$; that is, the set of elements of $X$ for which $U$ is a neighborhood. By hypothesis, $x \in W$.

Take any element $y \in W$. By definition of $W$, we have that $U \in \mathcal{V}_y$. By criterion (1), we see that $y \in U$. This much shows that $W \subseteq U$.

Again, take arbitrary $y \in W$, meaning that $U \in \mathcal{V}_y$. By criterion (4), there exists $V \in \mathcal{V}_y$ such that for each $z \in V$, $U \in \mathcal{V}_z$. But this means for every $z \in V$, we have $z \in W$, or in other words, $V \subseteq W$. Since $V \in \mathcal{V}_y$, by criterion (3), we have that $W \in \mathcal{V}_y$. Since $y$ was arbitrary, $W$ is a neighborhood of all its elements (including $x$, as shown earlier), and we are finished.

Equivalence of two different formulations of neighborhood systems in a set

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