# 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.

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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.

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Thank you for answering my question. I don't understand the last part of your explanation. Let $O$ be a set satisfying $f+U \subset O$ for some $f\in C(D)$ and $U\in {\mathfrak B}$. Why can we say that it is always a union of translates of set from the fundamental system of neighborhoods ? –  Aki Oct 3 '11 at 15:01
Note the quantifier! A set is open $O$ if for every $f\in O$ there is $U\in\mathcal B$ such that $f\in f+U\subseteq O$. Now it should be easy to show that a set is open iff it is the union of translates of memebers of $\mathcal B$. –  Stefan Geschke Oct 3 '11 at 17:56
Now I understand how to prove the necessity. To prove the sufficiency, I suppose that we need to show that any translate of set from the fundamental system of neighborhoods is open. But, for that, it seems to me that any set from the fundamental system of neighborhoods need to be a subgroup. –  Aki Oct 4 '11 at 7:42
You certainly don't need that the sets in the fundamental system are subgroups. In case of the $V(K,\varepsilon)$ you can use fairly straight forward estimation to show that they are indeed unions of translates of other sets of the form $V(K',\varepsilon')$. –  Stefan Geschke Oct 4 '11 at 16:45
I believe if you want to axiomatize what a fundamental system of neighborhoods has to satisfy in a topologicial group (abelian, for simplicity), you would have an axiom saying something like for all $U$ and $V$ in your system there is $W$ in the system such that $W\subseteq U-V$. –  Stefan Geschke Oct 4 '11 at 16:50

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 $x$.

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Thank you for your help. I'm not sure if $V(K,\epsilon)$'s satisfy the condition (NB3) in your note. Take $V(K',\epsilon')$ as V where $K'$ is a compact set such that $K\subset K' \subset D$ and $0< \epsilon' \leq \epsilon$. For $f\in V$, the set $f+V(K'',\epsilon'')$ should be identical to V for some compact set $K'' \subset D$ and $\epsilon''>0$. It seems impossible to me. –  Aki Oct 3 '11 at 14:40