# Neighborhood System Axioms

Question: Let $$\mathcal{N}$$ be a neighborhood system. Prove the neighborhood system induced by the topology induced by $$\mathcal{N}$$ is $$\mathcal{N}$$ itself.

Definition: A neighborhood system $$\mathcal{N}:X\rightarrow \mathcal{P}(\mathcal{P}(X))$$ is a function that satisfy

1. $$\mathcal{N}(x)$$ is non-empty
2. $$\forall N \in \mathcal{N}(x), x \in N$$
3. $$N \in \mathcal{N}(x)$$ and $$N \subseteq M \implies M \in \mathcal{N}(x)$$
4. If $$N, M \in \mathcal{N}(x)$$ then $$N \cap M \in \mathcal{N}(x)$$.
5. For every $$N \in \mathcal{N}(x)$$, there exists $$M \in \mathcal{N}(x)$$ such that $$M \subseteq N$$ and for every $$y \in M$$, $$N \in \mathcal{N}(y)$$.

Definition: Neighborhoods system $$\mathcal{N}$$ induces a topology $$\mathfrak{O}$$ by defining open sets as the sets $$O$$ such that $$O$$ is a neighborhood around each point in $$O$$.

Definition: A topology $$\mathfrak{O}$$ induces a neighborhood system by defining neighborhoods around $$x$$ as the sets containing an open set containing $$x$$.

I have proved $$\forall x,\mathcal{M}(x)\subseteq \mathcal{N} (x)$$, where $$\mathcal{M}$$ is the neighborhood system double induced by $$\mathcal{N}$$. But I don't know how to prove the other way around. Thanks for any help. I have stucked here for a week already.

Suppose $$N \in \cal N(x)$$ and let $$U = \{y \in N \mid N \in \cal N(y)\}$$.
Lemma: $$U$$ is open in the topology induced by $$\cal N$$, that is $$U \in \cal N(y)$$ for every $$y \in U$$.
Proof: Suppose $$y \in U$$. Then $$N \in \cal N(y)$$, so (by property 5) there exits $$M \in \cal N(y)$$ such that $$M \subseteq N$$ and $$N \in \cal N(z)$$ for every $$z \in M$$. Now $$M \subseteq U$$, because for every $$z \in M$$ we have $$z \in N$$ and $$N \in \cal N(z)$$, so (by property 3) we have $$U \in \cal N(y)$$.
So $$U$$ in open in the topology induced by $$\cal N$$, and clearly $$x \in U$$ and $$U \subseteq N$$, so $$N$$ is a neighborhood of $$x$$ in the topology induced by $$\cal N$$.