Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We first start with following definitions.

Definition 1. A family $\mathcal{P}$ of seminorms on a real vector space $X$ is called filtering if for any $p_1,p_2\in \mathcal{P}$ there exsist $q\in \mathcal{P}$ and $r_1,r_2>0$ such that the two inequalities $r_1p_1\le q$ and $r_2p_2\le q$ hold on $X$.

Definition 2. A family $\mathcal{P}$ of seminorms on a real vector space $X$ is called separating if for every $x\in X$ with $x\neq 0$, there exists $p\in \mathcal{P}$ such that $p(x)>0$.

My question is this: Suppose that $(X,\tau)$ is a Hausdorff locally convex space. How can we construct a family of seminorms on $X$ which is both separating and filtering?

My idea. If $(X,\tau)$ is a Hausdorff locally convex space, then as pointed out in Rudin's Functional Analysis book, we can construct a separating family $\mathcal{P}$ of seminorms on $X$. At this point, I don't have any idea if $\mathcal{P}$ is filtering.

Tips are very much appreciated.

share|cite|improve this question
The bottom line is that I want to verify if the topology of the Hausdorff locally space can be given by a filtering family of seminorms. I would be happy if this can be done.:) – juniven Feb 6 '13 at 4:43
up vote 1 down vote accepted

Since you only want a filtering family of seminorms you could simply take the family of all continuous seminorms $$ \mathcal{P} = \{p \colon X \to [0,\infty) \mid p \text{ is continuous}\} $$ this is clearly a filtering family and it is separating if and only if $X$ is Hausdorff.

More useful is the following construction: Let $\mathscr{U}$ a neighborhood basis of $0$ consisting of open, convex, balanced, circled and absorbing sets and for $U \in \mathscr{U}$ let $p_U(x) = \inf\{\lambda \geq 0 \mid x \in \lambda U\}$ be the Minkowski functional associated with $U$. Then $$ \mathcal{P} = \{p_U \mid U \in \mathscr{U}\} $$ is a family of continuous seminorms on $X$.

  • The family induces the given locally convex topology $\tau$ on $X$: Since all $p_U$ are continuous, the topology induced by $\mathcal{P}$ is weaker than $\tau$ and since $U = \{x \in X \mid p_U(x) \lt 1\}$, the topology induced by $\mathcal{P}$ is at least as strong as $\tau$.

  • If $X$ is Hausdorff, $\mathcal{P}$ is separating: if $x \neq 0$ there is $U \in \mathscr{U}$ such that $x \notin U$, so $p_U(x) \geq 1$.

  • The family $\mathcal{P}$ is filtering since for $U_1, U_2 \in \mathscr{U}$ there is $V \in \mathscr{U}$ such that $V \subseteq U_1 \cap U_2$ by the definition of a neighborhood basis, and therefore $p_V \geq p_{U_1}, p_{U_2}$.

share|cite|improve this answer
Are you still there? Thanks to your answer. Its nice. What guarantees that the family of all continuous seminorms is indeed filtering?:) – juniven Feb 6 '13 at 15:46
If $p_1$ and $p_2$ are continuous seminorms then so is $q = p_1 + p_2$ and $p_1,p_2 \leq q$ – Martin Feb 6 '13 at 15:47
That's it. In the definition of filtering, do we really need the existence of $r_1$ and $r_2$?:) – juniven Feb 6 '13 at 15:52
It's slightly more general. For example, on $\mathbb R^2$ consider the semi-norms $|(x_1,x_2)|_i = |x_i|$ and $|(x_1,x_2)|_3 = \frac12 \max\{|x_1|, |x_2|\}$. This is filtering according to the definition with $r_i$, but not without the $r_i$. I can't think of a natural example off the top of my head. – Martin Feb 6 '13 at 15:57
Now, I see.Thank you very much. Im now happy...:)God bless... – juniven Feb 6 '13 at 16:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.