I'm new on the locally convex spaces. I know that if $X$ is a vector space and $S$ an irreducible set of seminorms defined in $X$, $(X,S)$ is a locally convex vector space. The first question is, how the convergence is defined in $(X,S)$ and how is a Cauchy sequence defined? And second, yet is not clear for me which are the elements that are in the topology induced by the seminorms in $S$? If you could recommend me a book about this I'll be grateful.

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    $\begingroup$ seminorms induce a uniformity (and thus a notion of Cauchy sequences and filters/nets) and a uniformity induces a topology. Both are standard constructions. $\endgroup$ Commented Feb 28, 2019 at 5:14

2 Answers 2


First you need to define the topology $\mathscr{T}$ on $X$. For each seminorm $\rho\in S$, $\epsilon>0$, and point $x\in X$, define the open ball $$ B(x,\rho,\epsilon)=\{y\in X\ |\ \rho(y-x)<\epsilon\}\ . $$ Take the collection of all such balls. That is a subbasis for the topology $\mathscr{T}$. Namely, by taking finite intersections and then arbitrary unions of these balls you get all the open subsets of $X$.

Convergence is defined in the usual way. A sequence $(x_n)$ in $X$ converges to some $x\in X$, iff $$ \forall V\ {\rm open\ neighborhood\ of\ } x, \exists N, \forall n\ge N,\ x_n\in V $$ Similarly unsurprising is the definition of $(x_n)$ being Cauchy: $$ \forall V\ {\rm open\ neighborhood\ of\ } 0, \exists N, \forall m,n\ge N,\ x_m-x_n\in V $$

That being said, a few remarks are in order. In this general setting, sequences are not the best thing to work with, especially if $S$ is uncountable. You should use nets instead, with similar definitions for being Cauchy and convergence. A good introductory reference is the book by Osborne on "Locally Convex Spaces". To go a bit further, "Introduction to Functional Analysis" by Meise and Vogt is also good.

  • $\begingroup$ Thank you for your answer, and the continuity of $f \rightarrow \mathbb{R}$ in $x_0$ as follows right? $ \forall \epsilon > 0 \hspace{.1cm} \exists q \in S: \hspace{.1cm} q(x-x_0) \leq 1 \Rightarrow |f(x) - f(x_0)| \leq \epsilon $ $\endgroup$ Commented Mar 3, 2019 at 2:13
  • $\begingroup$ You mean $f:X\rightarrow\mathbb{R}$, right? Your definition of continuity is correct except for the quantifier $\exists q\in S$. It should be $\exists q\in \bar{S}$ where $\bar{S}$ is the set of all seminorms on $X$ that continuous with respect to the topology $\mathscr{T}$ defined by $S$. BTW, you used the word "irreducible" to describe $S$. What do you mean by that? $\endgroup$ Commented Mar 3, 2019 at 16:51
  • $\begingroup$ A set S of seminorm is irreducible if it verifies 3 properties: 1.- $\forall t \geq 0$ and $q \in S: tq \in S$. 2.- $If q \in S $ and $p$ is a seminorm in $X$ such that $p \leq q$: $q \in S$. 3.- $sup(q,p) \in S \hspace{.1cm} \forall q,p \in S$. For 1, can be proved that every seminorm in $X$ is continuos and moreover is uniformely continuos. $\endgroup$ Commented Mar 3, 2019 at 18:18
  • $\begingroup$ OK then with this definition of "irreducible" then $\bar{S}=S$ and we agree. Where did you see this terminology? It sounds weird to me because irreducible usually means minimal, whereas your $S$ is maximal as far as having the property of defining the topology. $\endgroup$ Commented Mar 3, 2019 at 19:36
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    $\begingroup$ The term "set irreducible" of seminorms is in the book "Locally Convex Spaces and Linear Partial Differential Equations" by Francois Treves. $\endgroup$ Commented Mar 5, 2019 at 4:00

A basis for the topology on $(X,S)$ consists of the sets of the form $$\{x \in X : \|x-x_0\|_1 < \varepsilon, \ldots, \|x-x_0\|_n < \varepsilon\}$$ for some $x_0 \in X$, $\varepsilon > 0$ and a finite collection of seminorms $\|\cdot\|_1, \ldots, \|\cdot\|_n \in S$.

Using this you can show that a net $(x_\lambda)_{\lambda\in \Lambda}$ converges to $x \in X$ if and only if $\|x- x_\lambda\| \to 0$ for every seminorm $\|\cdot\| \in S$.

Similarly, a net $(x_\lambda)_{\lambda\in \Lambda}$ is Cauchy if and only if for every $\varepsilon > 0$ and seminorm $\|\cdot\| \in S$ there exists $\lambda_0 \in \Lambda$ such that $\lambda, \mu \ge \lambda_0 \implies \|x_\lambda-x_\mu\| < \varepsilon$.


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