A locally convex space (LCS) can be defined as a topological vector space (i.e. scalar product and sum are continuous) whose topology is generated by translation of a family of balanced and absorbent convex sets. Now let's imagine that I want to trim this definition as much as possible considering that we are in a topological vector space. For example it would be enough to say that a base of open neighborhoods of $0$ is given by a family $\{U_\alpha\}_\alpha$ of balaced absorbent convex sets since then a base of neighborhoods for $x$ would be $\{x+U_\alpha\}_\alpha$.

A set $U$ in $X$ is absorbent if given $x\in X$ there is $r\geq 0$ such that $x\in rU$. My question is

Isn't every neighborhood of $0$ in a topological vector space absorbent?.

This is because given $x\in X$ the function from $\mathbb{R}$ or $\mathbb{C}$ that does $\lambda\mapsto \lambda x$ is continuous and so for $\lambda>0$ sufficiently small we have $$ \lambda x \in U \Rightarrow x\in \frac{1}{\lambda} U. $$

Hence it is enough to say a that a LCS is a top. vector space where a base of open neighborhoods of $0$ is given by a family of balanced convex sets, and being absorbent would be but a necessary condition for the neighborhoods (I always thought before that the absorbent condition was made so that the seminorms could be defined).

Extra question: Is every family closed under intersection of absorbent sets containing zero sufficient to define by translation a base for a topology in a vector space where sum and scalar product are continuous? Or in other words, is any of the other conditions (being balaced, convex or any other) also necessary given that a LCS is a topological vector space?

EDIT: Okay by looking at possible basis of open neighborhood of $0$ in $\mathbb{R}$ it becomes clear that they need not be formed by convex or balanced sets, but my question still remains as if there is any other condition that is necessary if we want them to generate a topological vector space.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.