A Question on the Topology Determined by a Neighborhood System In the lecture note of my professor, he claimed the following theorem without any proof:
Theorem:
Let $X$ be a non-empty set. Suppose there exists a mapping $\tau : X \rightarrow P(P(X))$ defined by $\tau(x)=\mathcal{N}_{x}$, where $\mathcal{N}_{x}$ satisfies the following properties:


*

*For any $N \in \mathcal{N}_{x}, x \in N$.

*$\mathcal{N}_{x}$ is closed under finite intersection.

*If $N\in \mathcal{N}_{x}$ and $N \subset M \subset X$, then $M \in \mathcal{N}_{x}$.

*Fix $A \subset X$. Consider the set $A^I = \{ y \in A : A \in \mathcal{N}_{y} \}$.
If $A \in \mathcal{N}_{z}$ for some $z \in A$, then $A^I \in \mathcal{N}_{z}$.
Then there exists a unique topology $\mathcal{T}$ on $X$ such that $\mathcal{N}_{x}$ contains all the neighborhood of $x$ with respect to this topology. Moreover, $A^I=\mathring{A}$ for any $A \subset X$.
I attempted a proof but I cannot proceed due to the following reasons.
The following are two guesses on $\mathcal{T}$:


*

*$\displaystyle \left\{{\bigcup G: G \subseteq \bigcup_{x \mathop \in X} \mathcal N_x}\right\}$

*$\mathcal{T}$=the topology generated by the subbase $\bigcup_{x \in X} \mathcal{N}_x$.
To prove that the first set is a topology, it suffices to prove that $\bigcup_{x \in X} \mathcal{N}_x$ is a base. One of the way is to show that $X$ is a union of sets in $\bigcup_{x \in X} \mathcal{N}_x$ and any finite intersections of sets in $\bigcup_{x \in X} \mathcal{N}_x$ is a union of the sets in $\bigcup_{x \in X} \mathcal{N}_x$. However, I cannot figure out how to do the second part.
On the other hand, for the second set, although it is a topology, I am confused on how to show that $\mathcal{N}_x$ contains all the neighborhoods of $x$.
Any suggestions on how to proceed on the above sets? Or are there any other good hints on how the form of $\mathcal{T}$ should be?
Thank for assistance!
 A: I believe property 4 should be changed from "for all" to "for some $z \in A$". If such modification is done, here is a part of my proof.
Define $\mathcal{T}=\{ A \subset X : A \in \mathcal{N}_x$ for any $x \in A \}$. We first prove that it is indeed a topology. Note $\emptyset \in \mathcal{T}$ vacuously. It is obvious that $X \in \mathcal{T}$. Let $U,V \in \mathcal{T}$. Then $U \in \mathcal{N}_x$ for any $x \in U$ and $V \in \mathcal{N}_y$ for any $y \in V$. Consider $U \cap V$. Pick any $z \in U \cap V$. Then $z \in U \implies U \in \mathcal{N}_z$. Similarly, $V \in \mathcal{N}_z$. Hence, $U \cap V \in \mathcal{N}_z$ for any $z\in U \cap V$, which implies that $U \cap V \in \mathcal{T}$. Next, let $\{ U_{\alpha} \}_{\alpha \in \Phi}$ be a family of sets in $\mathcal{T}$. Consider $\bigcup_{\alpha \in \Phi}  U_{\alpha}$. Pick any $t \in \bigcup_{\alpha \in \Phi}  U_{\alpha}$. Then there exist some $\alpha_0$ such that $t \in U_{\alpha_0}$. Thus, $U_{\alpha_0} \in \mathcal{N}_t$. Property 3 implies that $\bigcup_{\alpha \in \Phi}  U_{\alpha} \in \mathcal{N}_t$, which implies that $\bigcup_{\alpha \in \Phi}  U_{\alpha} \in \mathcal{T}$. Thus $\mathcal{T}$ is a topology.
Let $x \in X$. Let $N$ be a neighborhood of $x$. Then there exist some open set $A$ such that $x \in A \subset N$. Note $A \in \mathcal{N}_x$ and $A \subset N$. Therefore, property 3 implies that $N \in \mathcal{N}_x$. Therefore, $\mathcal{N}_x$ contains every neighborhood of $x$ with respect to $\mathcal{T}$.
Pick any $ y \in A^I$. Then $A \in \mathcal{N}_y$. By property 4, $A^I \in \mathcal{N}_y$. Therefore, $A^I$ is an open set by the definition of the topology $\mathcal{T}$. Since $\mathring{A}$ is the largest open set contained in $A$, we must have $A^I \subset \mathring{A}$. Pick any $y \in \mathring{A}$. Then $A$ is a neighborhood of $y$. Since $\mathcal{N}_y$ contains every neighborhood of $y$, $A \in \mathcal{N}_y$. Hence, $y \in A^I$, which implies  $A^I \supset \mathring{A}$. Thus, we have $A^I=\mathring{A}$.
However, I am struggling in proving the uniqueness of such topology:
"Finally, we want to prove that such topology is unique. Suppose $\mathcal{T}'$ is another topology satisfying the above requirements. Let $N$ be an open set in $\mathcal{T}$. Then $N \in \mathcal{N}_x$ for any $x \in N$....."
I have no idea how to proceed. Anyone can give me some hints? Thank you!
