In our functional analysis course, we defined the locally convex topology induced by a family of seminorms as follows:

Let $X$ be a vector space, $I \neq \emptyset$ an index set and $\{p_{\alpha}\}_{\alpha \in I}$ a family of seminorms. Let $U_{\alpha, r}(x)$ be the (open) ball with radius $r > 0$ around $x \in X$ w.r.t. the seminorm $p_\alpha$. Define

$$ \mathcal N_x := \Bigl\{ \bigcap_{j=1}^n U_{\alpha_j, r_j}(x) : n \in \mathbb N, \,\,\alpha_j \in I,\,\, r_j > 0,\,\, 1 \leq j \leq n \Bigr\}. $$

Then, a set $Y \subseteq X$ is open in the locally convex topology induced by $\{ p_\alpha\}_{\alpha \in I}$ iff $\forall x \in Y \, \exists V_x \in \mathcal N_x: V_x \subseteq Y$.

What it says to me: We make $\mathcal N_x$ to a general neighbourhood base by force. A set $Y \subseteq X$ is open iff $\mathcal N_x$ is a neighbourhood base for all $x \in Y$. And $\mathcal N_x$ itself is somehow a collection of all intersections of finitly many open balls w.r.t. any individual seminorms $p_{\alpha_j}$ with any radius $r_j$ around a fixed point $x$.

There are two big questions in my head:

  1. $\mathcal N_x$ seems a bit weird... especially the "combination" of open balls w.r.t. different seminorms in the intersection. Why does one do this? Does the locally convex topology somehow gets some "nice" properties by this construction?

  2. The whole locally convex topology seems like a strange construction. How can I understand the locally convex topology intuitively? Why is this a central topology, what makes this topology so important?

Thanks in advance.

  • $\begingroup$ You're constructing the weakest (i.e. least amount of open sets) topology such that the $p_\alpha$ are all continuous. Try yourself to build such a topology and you'll see why taking finite intersections is natural. The local convexity comes from the fact that the $p_\alpha$ are seminorms, for instance it works the same to give local convexity of the weak topology on a Banach space (where you're constructing the weakest topology making all (continuous in the norm) linear functionals continuous). $\endgroup$
    – Jose27
    Sep 4 at 3:06
  • 2
    $\begingroup$ @Jose27 That is not true. Consider for example $\mathbb{R} , | \cdot | $ (absolute value). Then the initial topology $\tau$ is simply given by $ \tau = \{ | \cdot |^{-1}(U) : U \subset \mathbb{R} \ \text{open} \} $. Let $U \in \tau$ then $x \in U \Rightarrow -x \in U$, since $|x| = |-x|$. But this means that for example $(1/2,3/2)$ is not open in the intial topology, but it obviously is in the norm topology. $\endgroup$
    – jd27
    Sep 4 at 6:58
  • $\begingroup$ @jd27: You're absolutely right, one needs to account for translates of the seminorms as well (see this question). $\endgroup$
    – Jose27
    Sep 4 at 16:51

1 Answer 1


To your second question: The defined topology is the smallest topological vector space (TVS) topology so that all semi-norms are continuous.

To understand why the neighborhood bases are defined this way (and why such a definition is natural) a review of the topology of TVS's is in order:

We need the following proposition (for a proof see here):

Proposition: Let $X$ be a set and $\mathcal{N} : X \to \mathcal{P}(\mathcal{P}(X))$ have the following properties for all $x \in X$:

  • $\mathcal{N}(x) $ is a filter
  • $ x \in N$ for all $N \in \mathcal{N}(x)$
  • The set $\{ z \in N : N \in \mathcal{N}(z) \} $ is in $\mathcal{N}(x)$ for all $N \in \mathcal{N}(x)$

Then the set $$\tau = \{ U \subset X: U \in \mathcal{N}(x) \, \forall x \in U \}$$ defines a topology on $X$ and the below defined neighborhood system agrees with $\mathcal{N}$. (So the bizzare looking third condition actually says that the interior of every neighborhood $N$ of a point $x$ is a neighborhoood itself.)

Conversely if $\tau$ is a topology on $X$ then the neighborhood system $\mathcal{N}: X \to \mathcal{P}(\mathcal{P}(X))$ defined by $$\mathcal{N}(x)= \{ N \subset X : \exists U \in \tau \ \text{with} \ x \in U\subset N \}$$ satisfies the above properties and the topology induced by it is the same as the original topology.

For a TVS $X$ it is true that $\mathcal{N}(x) = x + \mathcal{N}(0)$ (here $\mathcal{N}$ is the neighborhood system of the TVS) for all $x \in X$, because $+$ is a homeomorphism. Therefore the topology is fully determined by the neighborhood filter of $0$.

Define a filter base (of the neighborhood filter at $0$) to be a subset $\mathcal{B} \subset \mathcal{N}(0)$ so that $$\mathcal{N}(0) = \{ N \subset X : \exists B \in \mathcal{B} \ \text{with} \ B \subset N \}.$$

A filter base (of the neighborhood filter at $0$) must satisfies the following:

  • $\mathcal{B} \neq \varnothing$
  • $A \in \mathcal{B} \Rightarrow 0 \in A$
  • $A, B \in \mathcal{B} \Rightarrow \exists C \in \mathcal{B} : C \subset A \cap B$

and conversely any $\mathcal{B} \subset\mathcal{P}(X)$ with the above properties is a filter base for the filter of sets containing $0$ defined by $$\mathcal{N}(0) = \{ N \subset X : \exists B \in \mathcal{B} \ \text{with} \ B \subset N \}.$$

The next theorem tells us that for a TVS we only need to specify a filter base with certain properties:

Theorem: Let $X$ be a vector space. Let $\mathcal{N}(0)$ be a filter of sets containing $0$ that has a filter base $\mathcal{B}$ (as defined above) with the following properties:

  • $B \in \mathcal{B} \Rightarrow \exists A\in \mathcal{B} : A +A \subset B $
  • every subset of $ \mathcal{B}$ is balanced and absorbing

Then the topology induced (as in the above proposition) by the system $x \mapsto \mathcal{N}(x)= x+ \mathcal{N}(0)$ is a TVS topology and $\mathcal{N}(0)$ is the neighborhood filter of $0$. Conversely given a TVS topology on $X$ the neighborhood filter satisfies the above (and induces the same topology).

For a statement of this theorem with source see here.

Furthermore a semi-norm is continuous with respect to a TVS topology if and only if it is continuous at $0$ (follows from the reverse triangle inequality).

Therefore it is sufficient and necessary that $B_\alpha(r) = p_\alpha^{-1}([0,r)) \in \mathcal{N}(0)$ for any $\alpha \in I ,r >0$ for the continuity of the semi-norms.

Lets consider the case of a single semi-norm $p$ first. Then it is not hard to see that $$ \mathcal{B}= \{ B(r) : r > 0 \}$$ is a filter base that satisfies all the assumptions of the theorem. The induced topology is also the smallest TVS topology for which the semi-norm is continuous. Since for a second TVS topology in which the semi-norm is continuous there is for every $B \in \mathcal{B}$ an open (in the second topology) set $U$ with $0 \in U \subset B$ and so if a set is a neighborhood of all of its points in the first topology then it also is in the second.

Now if there are 2 semi-norms $p_1, p_2$, then maybe we want to define $$ \mathcal{B}= \{ B_{\alpha}(r) : r > 0, \alpha = 1,2 \}.$$ But this is not a filter base, because given $B_1 (r_1) $ and $B_2(r_2)$ there is in general no ball $B \in \mathcal{B}$ with $B \subset B_1 (r_1) \cap B_2(r_2)$.

Now the obvious fix is to simply include these intersections: $$ \mathcal{B}= \{ \cap_{j=1}^n B_{\alpha_j}(r_j) : n \in \{ 1,2 \}, r_j > 0, \alpha_j \in \{1,2 \} \}.$$ It is not hard to check either that $\mathcal{B}$ satisfies all the conditions in the theorem and that this yields the minimal topology (since the finite intersection of open sets containing $0$ is an open set containing $0$). This should answer your first question.

The same idea applies to any number of semi-norms and we arrive at the filter base described in your post.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .