How can you construct a topology from a fundamental system of neighborhoods? How can you construct a topology from a fundamental system of neighborhoods ?
In "Elementary Theory of Analytic Functions of One or Several Complex Variables" by Henri Cartan, it seems that a topology is uniquely determined in C(D), the vector space of continuous complex-valued functions in the open set D, by a fundamental system of neighborhoods.
The fundamental system of neighborhoods of o is defined as follows:
For any pair $(K,\epsilon)$ consisting of a compact subset $K \subset D$ and a number $\epsilon > 0$, we consider the subset $V(K,\epsilon)$ of C(D) defined by 
$$f \in V(K,\epsilon) \Leftrightarrow |f(x)|\leq \epsilon, \; x \in K. $$
The neighborhoods of a point f are defined by translating the neighborhoods of o by f.
Then,  Proposition 3.I. follows
Proposition 3.I.
C(D) has indeed a topology (invariant under translation) in which the sets $V(K,\epsilon)$ form a fundamental system of neighborhoods of o. This topology is unique and can be defined by a distance which is invariant under translation.
Proof. 
The uniqueness of the topology is obvious, because we know a fundamental system of neighborhoods of o, and ...
I know that a topology can be constructed by specifying all neighborhoods of each point x (for example Bourbaki "Elements of Mathematics: General Topology I.1.2 Proposition 2"), but I cannot understand how a topology is defined from a fundamental system of neighborhoods.
 A: I would assume that what Cartan calls a "fundamental system of neighborhoods" is what I would call a neighborhood base at $0$ (for a topological group, abelian, in our case).
This would be a collection $\mathcal B$ of open neighborhoods of $0$ such that for each neighborhood $U$ of zero, there is $V\in\mathcal B$ such that $V\subseteq U$.
Now a set $O$ is open iff for each $f\in O$ there is $U\in\mathcal B$ such that $f+U\subseteq O$.
In other words, a set is open iff it is a union of translates of set from the fundamental system of neighborhoods.
A: In contemporary times the term neighborhood base (or "basis") is more common than fundamental system of neighborhoods, I believe.  But by any name the way to get from a fundamental system $\mathcal{B}_x$ of neighborhoods of a point $x$ to the set of all neighborhoods of $x$ is simply to define a neighborhood of $x$ to be a subset $V$ of $X$ such that there exists some $U_x \in \mathcal{B}_x$ with $x \in U_x \subset V$.
See $\S 0.1$ of these notes for a little more information on this, including the axioms that a family of subsets $\mathcal{B}_x$ must satisfy in order to be a neighborhood base at $x$.
A: There are, in fact, two questions in your post.

The first question is: given, for every $x \in X$, a family $\mathcal F_x = (U_{x, i_x})_{i \in I_x}$ of subsets of $X$ such that:


*

*$\mathcal F_x$ is lower directed (i.e. for every $U_{x, i_x}$ and $U_{x, j_x}$ there exists some $U_{x, k_x}$ such that $U_{x, k_x} \subseteq U_{x, i_x} \cap U_{x, j_x}$),

*$\mathcal F_x$ does not contain $\emptyset$,

*every $U_{x, i_x} \in \mathcal F$ contains $x$,
how do we construct a topology that has $(U_{i_x,x})_{i \in I_x}$ for fundamental system of open neighbourhoods around $x$?
One approach is to:


*

*first define convergent nets: a net $(x_a)_{a \in A}$ converges to $x \in X$ if and only if for every $U_{x, i_x} \in \mathcal F_x$ there exists $a_0 \in A$ such that for all $a \ge a_0$ we have $x_a \in U_{x, i_x}$;

*next define continuous functions at a point: $f : X \to \mathbb R$ is continuous at $x \in X$ if and only if for every net $(x_a)_{a \in A}$ that converges to $x$ we have that the net $(f(x_a))_{a \in A}$ converges to $f(x)$;

*next, define continuous functions on $X$: $f : X \to \mathbb R$ is continuous if and only if it is continuous at every $x \in X$; let $C(X, \mathbb R)$ be the set of all these functions;

*finally, let $\mathcal T_X$ be the initial topology on $X$ generated by the family $C(X, \mathbb R)$, i.e. the coarsest topology on $X$ making all the functions in $C(X, \mathbb R)$ continuous; a base for $\mathcal T _X$ is formed by the subsets of the form $f^{-1}(V)$ where $f \in C(X, \mathbb R)$ and $V \subseteq \mathbb R$ is open.
Notice that $\mathcal T_X$ constructed above is unique: any other topology having the same fundamental systems of open neighbourhoods around each point would necessarily have the same convergent nets, therefore the same continuous functions, therefore would coincide with $\mathcal T_X$.

The second question is: can we give a more convenient description of the topology constructed by Cartan on $C(D)$? It turns out that if we translate the sets $V(K,\epsilon)$ by $f \in C(D)$ we get a fundamental system of open neighbourhoods $V(K,\epsilon,f)$ around $f$, and these form a subbase for the topology known as "the compact-open topology" on $C(D)$, which in this case coincides with the topology of compact convergence, i.e. a net $(f_i)_{i \in I} \subset C(D)$ converges to $f \in C(D)$ if and only if $\sup _{x \in K} |f_i (x) - f(x)| \to 0$ for every compact $K \subseteq D$. It may be shown, next, that $D \subseteq \mathbb C$ is $\sigma$-compact, i.e. there exists a family $(K_n)_{n \in \mathbb N}$ of compact subsets of $D$ such that $D = \bigcup _{n \in \mathbb N} K_n$. Consequently, the said topology on $C(D)$ can be given the the family of seminorms $p_n (f) = \sup _{x \in K_n} |f(x)|$, and one may next define the distance $d(f,g) = \sum _{n \in \mathbb N} \frac 1 {2^n} \frac {p_n (f-g)} {1 + p_n (f-g)}$ which is easily seen to give the same topology. This distance is translation-invariant and topologically complete, making $C(D)$ a Fréchet space.
