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?
Best Answer
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".