# Strongest topology that makes vector space locally convex

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?

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 $$\mathbb{F}\times\{v\}$$.