Characterization of Topology Inspired by this question Neighborhood Topology and Open Set Topology: their Equivalence and Comparison as well as many other similar questions asked here on Math SE. However, I could not prove it given my setup:
Suppose that $\mathcal T\subseteq\mathcal P(X)$ and $\mathfrak N:X\rightarrow\mathcal P(\mathcal P(X))$. Then it holds that


*

*$\mathcal T$ is a topology over $X$ and

*for each $x\in X$ such that there exists a $O\in\mathcal T$ with $x\in O\subseteq N$ it holds that $N\in\mathfrak N$


if and only if


*

*for each $x\in X$ it holds that


*

*$\mathfrak N(x)$ is a filter over $X$,

*for each $N\in\mathfrak N(x)$ it holds that $x\in N$ and

*for each $N\in\mathfrak N(x)$ it holds that $\{z\in N:N\in\mathfrak N(z)\}\in\mathfrak N(x)$
and 


*$\mathcal T = \{O\subseteq X\mid\forall x\in X:O\in\mathfrak N(x)\}$.


$$\\$$
In other words:


*

*$\mathcal T$ is a topology over $X$ and

*$N$ is a $\mathcal T$-neighborhood for $x\in X$ (i.e. $N\in\mathfrak N(x)$) iff there exists a $O\in\mathcal T$ such that $x\in O\subseteq N$


if and only if


*

*for each $x\in X$ it holds that


*

*$\mathfrak N(x)$ is a filter over $X$,

*for each $N\in\mathfrak N(x)$ it holds that $x\in N$ and

*for each $N\in\mathfrak N(x)$ it holds that $\{z\in N:N\in\mathfrak N(z)\}\in\mathfrak N(x)$


*$O\subseteq X$ is $\mathfrak N$-open (i.e. $O\in\mathcal T$) iff for each $x\in O$ it holds that $O\in\mathfrak N(x)$


The other questions could help me figuring out the importance and meaning of the respective axioms. Proving the equivalence is hard for me though. I have proved the "$\Rightarrow$"-direction except for "for each $N\in\mathfrak N(x)$ it holds that $\{z\in N:N\in\mathfrak N(z)\}\in\mathfrak N(x)$". Any hints?
 A: Let $\mathbf{Top}_X$ be the set of topologies on a set $X$, and let 
$\mathbf{Neigh}_X$ be the set of all neighbourhood spaces on $X$, defined by the three axioms you described. I'll denote $\{z \in A: A \in \mathcal{N}(z)\}$ for a set $A$ as $A^\ast$, so that the third axiom can then be written as $$\forall x \in X: \forall N \in \mathcal{N}(x): N^\ast \in \mathcal{N}(x)\text{.}$$

There is a map $\mathbb{N}: \mathbf{Top}_X \to \mathbf{Neigh}_X$, defined by: 

$\mathbb{N}((X, \mathcal{T}))$ is the neighbourhood system on $X$ defined by:
  $$\forall x \in X: \mathfrak{N}(x) = \{A \subseteq X: \exists O \in \mathcal{T}: x \in O \subseteq A\}$$

Let's prove, (although you probably already know), that this is well-defined, i.e. the axioms for a neighbourhood space are fulfilled:


*

*$X \in \mathfrak{N}(x)$ (as we can take $O=X$ as $X \in \mathcal{T}$), so that $\mathfrak{N}(x) \neq \emptyset$. (part of being a filter)

*Clearly all members of $\mathfrak{N}(x)$ contain $x$, so $\emptyset \notin \mathfrak{N}(x)$ (also part of being a filter), and also axiom 2 is satisfied.

*If $A_1, A_2 \in \mathfrak{N}(x)$ then we have $O_1, O_2 \in \mathcal{T}$ such that $x \in O_1 \subseteq A_1, x \in O_2 \subseteq A_2$. But then $O_1 \cap O_2 \in \mathcal{T}$ (finite intersection axiom of topologies) and $x \in O_1 \cap O_2 \subseteq A_1 \cap A_2$, so by definition, $A_1 \cap A_2 \in \mathfrak{N}(x)$ (other filter axiom)

*If $A \in \mathfrak{N}(x)$ and $A \subseteq B$, then we have $O \in \mathcal{T}$ with $x \in O \subseteq A$ and so also $x \in O \subseteq B$, and so $B \in \mathfrak{N}(x)$ (final filter axiom).

*The third axiom: Let $N \in \mathfrak{N}(x)$. Then we have $x \in U \subseteq N$ for some $U \in \mathcal{T}$. Note that for each $z \in U$ we can pick $O=U$ in the definition of $\mathfrak{N}(z)$ to see that $N \in \mathfrak{N}(z)$. So $x \in U \subseteq \{z \in N: N \in \mathfrak{N}(z)\}$ so that $\{z \in N: N \in \mathfrak{N}(z)\} \in \mathfrak{N}(x)$, as required.



There is also a mapping in the reverse direction:
$\mathbb{T}: \mathbf{Neigh}_X \to \mathbf{Top}_X$ defined by 

$\mathbb{T}((X, \mathfrak{N}))$ is the topology on $X$ defined by: $$\mathcal{T} = \{O \subseteq X: \forall x \in O: O \in \mathfrak{N}(x)\}$$

Let's check this is a topology:


*

*$\emptyset \in \mathcal{T}$ as the $\forall x$-condition is voidly fulfilled.

*$X \in \mathcal{T}$ as by the filter property of $\mathfrak{N}(x)$ we always have $X \in \mathfrak{N}(x)$ for any $x$ ($X$ being a superset of all sets here).

*If $O_1, O_2 \in \mathcal{T}$, let $x \in O_1 \cap O_2$. As $x \in O_1$ we know that $O_1 \in \mathfrak{N}(x)$ and we likewise know that $O_2 \in \mathfrak{N}(x)$ and as $\mathfrak{N}(x)$ is a filter, $O_1 \cap O_2 \in \mathfrak{N}(x)$ and as $x \in O_1 \cap O_2$ was arbitary, $O_1 \cap O_2 \in \mathcal{T}$, making the set closed under finite intersections.

*If $O_i \in \mathcal{T}$, where $i \in I$ (for some index set), let $x \in \bigcup_i O_i$. This means there is some $j \in I$ such that $x \in O_j$ and this again means that $O_j \in \mathfrak{N}(x)$. As $O_j \subseteq \bigcup_i O_i$, by being a filter we have that $\bigcup_i O_i \in \mathfrak{N}(x)$, and as this holds for all $x$ in it, $\bigcup_i O_i \in \mathcal{T}$.


So we indeed have a topology (where we have not yet used the third axiom, nor the second one).

The equivalence of both viewpoints can now be stated as 


*

*$\mathbb{N} \circ \mathbb{T}$ is the identity map on $\mathbf{Neigh}_X$.

*$\mathbb{T} \circ \mathbb{N}$ is the identity map on $\mathbf{Top}_X$.


In words: if we start with a neighbourhood space and define its corresponding topology then that topology's induced neighbourhood space is exactly the one you started with. And mutatis mutandis for topological spaces.
Proof that $\mathbb{T}(\mathbb{N}((X,\mathcal{T})))  =(X,\mathcal{T})$
So we have to show two inclusions (we have two topologies on $X$ that we have to show to be equal). First, let $O \in \mathcal{T}$. Let $x \in O$. Then by definition, $O \in \mathcal{N}(x)$ for the neighbourhood space $\mathbb{N}((X,\mathcal{T}))$. And as this holds for all $x \in O$, $O$ is open in the topology $\mathbb{T}(\mathbb{N}((X, \mathcal{T})))$. That shows one inclusion.
Now, let $O$ be open in $\mathbb{T}(\mathbb{N}((X, \mathcal{T})))$. So for every $x \in O$, $O \in \mathcal{N}(x)$ in the neighbourhood space $\mathbb{N}((X, \mathcal{T}))$, which in turn means there is some $O_x \in \mathcal{T}$ such that $x \in O_x \subseteq O$. But then $$O \subseteq \bigcup_{x \in O} O_x \subseteq O$$ as $x \in O_x$ for each $x \in O$ and all $O_x$ are subsets of $O$. This shows that $O$ is a union of members of $\mathcal{T}$ so $O \in \mathcal{T}$, showing the other inclusion.
Proof that $\mathbb{N}(\mathbb{T}((X,\mathcal{N})))  =(X, \mathcal{N})$
This again comes down to proving two inclusions: for each $x\in X$ $\mathcal{N}(x)$ is the same set in the original sense and in the left hand sense. So let $x \in X$. If $N$ is a neighbourhood of $x$ in the neighbourhood space $\mathbb{N}(\mathbb{T}((X,\mathcal{N})))$, then there is an open $O$, in the sense of $\mathbb{T}((X,\mathcal{N}))$,  such that $x \in O \subseteq N$. That $O$ is open in $\mathbb{T}((X,\mathcal{N}))$ implies that $O \in \mathcal{N}(x)$ ($O$ is a neighbourhood of all its points, including $x$) and as $\mathcal{N}(x)$ is a filter, $N \in \mathcal{N}(x)$. This shows the left to right inclusion we want.
On the other hand, if $N \in \mathcal{N}(x)$, then consider $N^\ast \in \mathcal{N}(x)$. I claim that $N^\ast$ is open in $\mathbb{T}((X,\mathcal{N}))$:
To see this: let $z \in N^\ast$. By definition, $N \in \mathcal{N}(z)$. 
Apply axiom 4 to get an $N' \in \mathcal{N}(z)$ such that $\forall y \in N': N \in \mathcal{N}(y)$, so $N' \subseteq N^\ast$ and so $N^\ast \in \mathcal{N}(z)$ by the enlargement property. Hence $\forall z \in N^\ast: N^\ast \in \mathcal{N}(z)$ which means that $N^\ast$ is open, as claimed. 
Also, note that $N^\ast \subseteq N$ ($z \in N^\ast$ means $N \in \mathcal{N}(z)$ which by axiom 2 of a neighbourhood space implies $z \in N$). So we have $N^\ast \in \mathbb{T}((X,\mathcal{N}))$ with $x \in N^\ast \subseteq N$. This means by definition that $N$ is a neighbourhood of $x$ in $\mathbb{N}(\mathbb{T}((X,\mathcal{N})))$, showing the right to left inclusion.

Most of the ideas are in the post you linked to in the question, as well. In my first general topology course we did all these equivalences at the beginning  (closure spaces, interior axioms, net convergence etc. can all be used to give alternative axiomisations for topological spaces, and it's good to come to terms with all these to enhance one's understanding of the basics).
