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.
Best Answer
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.