I'm currently reading Mendelson's Introduction to topology and have came across this theorem:
Theorem 3.8:
Let a neighborhood in a topological space be defined by Definition 2.2 and an open set in a neighborhood space be defined by Definition 3.5. Then the neighborhoods of a topological space $( X,\mathfrak T)$ give rise to a neighborhood space $ (X,\mathfrak N) = \mathfrak a(X,\mathfrak T)$ and the open sets of a neighborhood space $ (Y,\mathfrak N')$ give rise to a topological space $ (Y,\mathfrak T')=\mathfrak a' (Y,\mathfrak N')$. Furthermore, for each topological space $(X,\mathfrak T)$, $$(X,\mathfrak T)=\mathfrak a'(\mathfrak a(X,\mathfrak T))$$ and for each neighborhood space $(X,\mathfrak N)$, $$ (X,\mathfrak N)=\mathfrak a(\mathfrak a'(X,\mathfrak N)),$$ thus establishing a one-one correspondence between the collection of all topological spaces and the collection of all neighbourhood spaces.
Definition 2.2: Given a topological space $(X,\mathfrak T)$, a subset $\mathrm N$ of $X$ is called a neighborhood of a point $a \in X$ if $N$ contains an open set that contains $a$.
Definition 3.5: In a neighborhood space, a subset $O$ is said to be open if it's a neighborhood of each of it's points.
So, my question is:
What are $\mathfrak a$ and $\mathfrak a'$?
The mappings $\mathfrak{a}$ and $\mathfrak{a}'$ are implicitly defined in the statement of the theorem.
$\mathfrak{a}$ maps a topological space $(X,\mathfrak{T})$ to the neighbourhood space $(X,\mathfrak{N})$, where $\mathfrak{N}$ is the family of neighbourhoods of all $x\in X$ in the topology $\mathfrak{T}$, and $\mathfrak{a}'$ is the map in the other direction, that constructs a topological space $(X,\mathfrak{T})$ from a neighbourhood space $(X,\mathfrak{N})$ by letting $\mathfrak{T}$ be the family of sets that are neighbourhoods of all their elements.
The theorem asserts that
Put in another way, neighbourhood spaces and topological spaces are equivalent, and you can choose whichever viewpoint happens to be more convenient.