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.


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.

  • $\begingroup$ 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 ? $\endgroup$ – Aki Oct 3 '11 at 15:01
  • 1
    $\begingroup$ 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 $\mathcal B$. $\endgroup$ – Stefan Geschke Oct 3 '11 at 17:56
  • $\begingroup$ 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. $\endgroup$ – Aki Oct 4 '11 at 7:42
  • $\begingroup$ 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 $V(K',\varepsilon')$. $\endgroup$ – Stefan Geschke Oct 4 '11 at 16:45
  • $\begingroup$ 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 $W\subseteq U-V$. $\endgroup$ – 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 $x$.

  • $\begingroup$ 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. $\endgroup$ – Aki Oct 3 '11 at 14:40

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.


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