Here is an exercise from Barvinok's "A Course in Convexity" (ex. III.3.3.3, p.119):

Prove that the strongest topology that makes a vector space $V$ a locally convex topological vector space is the topology where $U \subseteq V$ is open if and only if it is a union of convex algebraically open sets.

Isn't the discrete topology (all sets are open) also turning $V$ into a locally convex TVS? Indeed, every singleton set $\{x\}$ is convex and open, the operations are continuous, and every singleton set is also closed.

Am I missing something or is there a problem with the exercise? If the statement is wrong, then any clues as to what should be the correct statement?


1 Answer 1


No! The scalar multiplication $\mathbb{F}\times V\to V$ ($\mathbb{F}=\mathbb{R},\mathbb{C}$) ceases to be continuous if you put the discrete topology on a nontrivial $V$. Indeed, fixing $v\neq 0$, the inverse image of the open set $\{v\}$ intersecting the open $\mathbb{F}\times\{v\}\subset\mathbb{F}\times V$ is a singleton $\{(1,v)\}$, which is not open in $\mathbb{F}\times\{v\}$.

This should also hint at how to recover the "algebraically open".

  • $\begingroup$ Very useful! Thanks! $\endgroup$
    – Fernando
    Jun 20, 2019 at 14:23

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