Use of the last axiom of neighbourhoods' topology proving equivalence with open sets

Question

I'm proving that the standard definition by open sets of a topology (closed under arbitrary unions and finite intersections) is equivalent to the definition by neighbourhoods. I'll give the precise definition I'm using below.

This is where I'm at: assuming neighbourhood axioms and defining open sets as sets that are neighbourhoods of all of its points, open sets satisfy the usual axioms of a topology. However, I believe I've proven this without the need of the fourth axiom, where is the flaw in my reasoning?:

• $\emptyset$ satisfies vacuously the open condition.
• Given $\bigcup O_i $ a union of open sets and any $x\in \bigcup O_i$, then $x\in O_k\in N(x)$ for some $k$ , since $O_k\subseteq \bigcup O_i$ then by the superset condition $\bigcup O_i \in N(x)$, since $x$ was any point then the union is a neighbourhood of all its points and therefore an open set.
• Given two open sets $U,V$ and any $x\in V\cap U$ then since the sets are open $V,U\in N(x)$ and by the intersection condition $U\cap V\in N(x)$. Since $x$ whas arbitrary $U\cap V$ is open.
• Since $N(x)$ is a non empty family of sets then for each $x\in X$ there exists at least one $V_x \in N(x)$, then by the superset contidion $V_x\subseteq X$, we have that $X\in N(x)$, being $x$ arbitrary then $X$ is open.

I believe my mistake is in the last item, since it's here that I've seen other people use the fourth axiom, to prove that $V_x$ has an open subset, or to take arbitrary unions over the other neighbourhood given by the axiom. But I don't need for $V_x$ to have an open subset in order for $X$ to be open.

Where is the flaw on my proof of this direction of the equivalence? Am I assuming the axiom of choice on the last item? Do I still need to prove something else to prove this direction of the equivalence? Thanks!

Here the precise definition of the axioms of neighbourhoods that I'm using.

Given $X$ a set and a function that takes any $x\in X$ and gives a non-empty family of subsets of $X$ denoted by $N(x)$:

1. $x\in V$ for all $V\in N(x)$ (the point belongs to all its neighbourhoods).
2. If $V\in N(x)$ and $V\subseteq U$ then $U\in N(x)$ (supersets of neighbourhoods are neighbourhoods).
3. If $U,V\in N(x)$ then $U\cap V \in N(x)$ (intersection of neighbourhoods are neighbourhoods).
4. If $U\in N(x)$ then there is a $V\in N(x), V\subseteq U$ such that for all $y\in V$ it follows that $U\in N(y)$ (every neighbourhood $U$ has a neighbourhood $V\subseteq U$ such that $U$ is neghbourhood of all $V$'s points).

Answer 1

Your only mistake is in not understanding the purpose of the fourth axiom.

As you've demonstrated, if you have a function $N : X \to \mathscr P(X)$ which satisfies axioms 1, 2 and 3, then $N$ induces a topology $\tau$ on $X$.

Also, if you have a topology $\tau$ on $X$, then you can define $$N' : X \to \mathscr P(X): x \mapsto \{U \subseteq X : \exists O \in \tau, x \in O \subseteq U\}$$ and show that $N'$ satisfies all four of your axioms on neighborhoods.

So note the difference: You can start with a function $N$ that satisfies only three of the axioms, use it to define a topology on $X$, then use that topology to induce another function $N'$ that now satisfies all four axioms. So in general $N' \ne N$. But if you then use $N'$ to induces a topology on $X$ again, there is no change. It is the same topology as induced by $N$.

With 3 axioms, there are multiple neighborhood functions $N$ which will generate the same topology $\tau$ on $X$. But if you add axiom 4, there is only one. That uniqueness is the purpose of axiom 4.